1.
LRFD Steel Design
AASHTO LRFD Bridge Design
Specifications
Example Problems
Created July 2007
2.
This material is copyrighted by
The University of Cincinnati
and
Dr. James A Swanson.
It may not be reproduced, distributed, sold,
or stored by any means, electrical or
mechanical, without the expressed written
consent of The University of Cincinnati and
Dr. James A Swanson.
July 31, 2007
3.
LRFD Steel Design
AASHTO LRFD Bridge Design Specification
Example Problems
Case Study: 2Span Continuous Bridge.......................................................................................1
Case Study: 1Span SimplySupported Bridge .........................................................................63
Case Study: 1Span Truss Bridge...............................................................................................87
AdHoc Tension Member Examples
Tension Member Example #1 ..........................................................................................105
Tension Member Example #2 ..........................................................................................106
Tension Member Example #3 ..........................................................................................108
Tension Member Example #4 ..........................................................................................110
AdHoc Compression Member Examples
Compression Member Example #1 .................................................................................111
Compression Member Example #2 .................................................................................112
Compression Member Example #3 .................................................................................114
Compression Member Example #4 .................................................................................116
Compression Member Example #5 .................................................................................119
Compression Member Example #6 .................................................................................121
Compression Member Example #7 .................................................................................123
AdHoc Flexural Member Examples
Flexure Example #1 ..........................................................................................................127
Flexure Example #2 ..........................................................................................................129
Flexure Example #3 ..........................................................................................................131
Flexure Example #4 ..........................................................................................................134
Flexure Example #5a ........................................................................................................137
Flexure Example #5b ........................................................................................................141
Flexure Example #6a ........................................................................................................147
Flexure Example #6b ........................................................................................................152
AdHoc Shear Strength Examples
Shear Strength Example #1 .............................................................................................159
Shear Strength Example #2 .............................................................................................161
4.
AdHoc Web Strength and Stiffener Examples
Web Strength Example #1 ...............................................................................................165
Web Strength Example #2 ...............................................................................................168
AdHoc Connection and Splice Examples
Connection Example #1....................................................................................................175
Connection Example #2....................................................................................................179
Connection Example #3....................................................................................................181
Connection Example #4....................................................................................................182
Connection Example #5....................................................................................................185
Connection Example #6a..................................................................................................187
Connection Example #6b .................................................................................................189
Connection Example #7....................................................................................................190
5.
James A Swanson
Associate Professor
University of Cincinnati
Dept of Civil & Env. Engineering
765 Baldwin Hall
Cincinnati, OH 452210071
Ph: (513) 5563774
Fx: (513) 5562599
James.Swanson@uc.edu
6.
1. PROBLEM STATEMENT AND ASSUMPTIONS:
A twospan continuous composite Igirder bridge has two equal spans of 165’ and a 42’ deck width. The
steel girders have Fy = 50ksi and all concrete has a 28day compressive strength of
f’c = 4.5ksi. The concrete slab is 91/2” thick. A typical 2¾” haunch was used in the section properties.
Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as
a composite dead load.
HL93 loading was used per AASHTO (2004), including dynamic load allowance.
42'  0" Out to Out of Deck
39'  0" Roadway Width
9½” (typ)
23/4" Haunch (typ)
3'0"
3 spaces @ 12'  0" = 36'  0"
3'0"
References:
Barth, K.E., Hartnagel, B.A., White, D.W., and Barker, M.G., 2004, “Recommended Procedures for
Simplified Inelastic Design of Steel IGirder Bridges,” ASCE Journal of Bridge Engineering, May/June
Vol. 9, No. 3
“Four LRFD Design Examples of Steel Highway Bridges,” Vol. II, Chapter 1A Highway Structures
Design Handbook, Published by American Iron and Steel Institute in cooperation with HDR Engineering,
Inc. Available at http://www.aisc.org/
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 1 of 62
 1 
7.
Negative Bending Section (Section 2)
Positive Bending Section (Section 1)
2. LOAD CALCULATIONS:
DC dead loads (structural components) include:
• Steel girder self weight (DC1)
• Concrete deck self weight (DC1)
• Haunch self weight (DC1)
• Barrier walls (DC2)
DW dead loads (structural attachments) include:
• Wearing surface (DW)
2.1: Dead Load Calculations
Steel Girder SelfWeight (DC1): (Add 15% for Miscellaneous Steel)
(a) Section 1 (Positive Bending)
A = (15”)(3/4”) + (69”)(9/16”) + (21”)(1”) = 71.06 in2
⎛
2 ⎜ 490
Wsec tion1 = 71.06 in ⎜
⎜
⎝
⎞
pcf ⎟
(
12 in
ft
)
2
⎟ (1.15 ) = 278.1 ft
⎟
⎠
Lb
per girder
(b) Section 2 (Negative Bending)
A = (21”)(1”) + (69”)(9/16”) + (21”)(21/2”) = 112.3 in2
⎛
2⎜
Wsec tion 2 = 112.3 in ⎜
⎜
⎝
⎞
490 pcf ⎟
(
12 in
ft
)
2
⎟ (1.15 ) = 439.5
⎟
⎠
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
Lb
ft
per girder
AASHTOLRFD 2007
Page 2 of 62
 2 
8.
Deck SelfWeight (DC1):
Wdeck
⎛
⎜ 150 pcf
= (9.5")(144") ⎜
2
⎜ 12 in
ft
⎝
(
)
⎞
⎟
Lb
⎟ = 1, 425 ft per girder
⎟
⎠
Haunch SelfWeight (DC1):
⎛ 21"(66') + 15"(264') ⎞
⎟ = 16.2"
66'+ 264'
⎝
⎠
Average width of flange: ⎜
Average width of haunch:
Whaunch
( 1 2 ) ⎡(16.2"+ (2)(9") ) + 16.2"⎤ = 25.2"
⎣
⎦
⎛
⎞
⎜ ( 2")( 25.2") ⎟ (150 pcf ) = 52.5 Lb per girder
=⎜
ft
2
⎟
⎜ 12 in
⎟
ft
⎝
⎠
(
)
Barrier Walls (DC2):
⎛ (2 each) ( 640 plf ) ⎞
⎟ = 320.0 Lb ft per girder
⎜
⎟
4 girders
⎝
⎠
Wbarriers = ⎜
Wearing Surface (DW):
W fws =
(39')(60 psf )
4 girders
= 585 Lb per girder
ft
The moment effect due to dead loads was found using an FE model composed of four frame elements.
This data was input into Excel to be combined with data from moving live load analyses performed in
SAP 2000. DC1 dead loads were applied to the noncomposite section (bare steel). All live loads were
applied to the shortterm composite section (1n = 8). DW (barriers) and DC2 (wearing surface) dead
loads were applied to the longterm composite section (3n = 24).
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 3 of 62
 3 
9.
Unfactored Dead Load Moment Diagrams from SAP
4,000
DC1
3,000
2,000
DW
1,000
Moment (kipft)
0
DC2
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
0
30
60
90
120
150
180
210
240
270
300
330
Station (ft)
Unfactored Dead Load Shear Diagrams from SAP
200
DC1
150
100
DW
Shear (kip)
50
DC2
0
50
100
150
200
0
30
60
90
120
150
180
210
240
270
300
330
Station (ft)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 4 of 62
 4 
10.
The following Dead Load results were obtained from the FE analysis:
•
The maximum positive liveload moments occur at stations 58.7’ and 271.3’
•
The maximum negative liveload moments occur over the center support at station 165.0’
DC1  Steel:
DC1  Deck:
DC1  Haunch:
DC1  Total:
DC2:
DW
Max (+) Moment
Stations 58.7’ and 271.3’
475kft
2,415kft
89kft
2,979kft
553kft
1,011kft
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
Max () Moment
Station 165.0’
1,189kft
5,708kft
210kft
7,107kft
1,251kft
2,286kft
AASHTOLRFD 2007
Page 5 of 62
 5 
11.
2.2: Live Load Calculations
The following design vehicular live load cases described in AASHTOLRFD are considered:
1) The effect of a design tandem combined with the effect of the lane loading. The design
tandem consists of two 25kip axles spaced 4.0’ apart. The lane loading consists of a 0.64klf
uniform load on all spans of the bridge. (HL93M in SAP)
2) The effect of one design truck with variable axle spacing combined with the effect of the
0.64klf lane loading. (HL93K in SAP)
3) For negative moment between points of contraflexure only: 90% of the effect of a trucktrain
combined with 90% of the effect of the lane loading. The truck train consists of two design
trucks (shown below) spaced a minimum of 50’ between the lead axle of one truck and the rear
axle of the other truck. The distance between the two 32kip axles should be taken as 14’ for each
truck. The points of contraflexure were taken as the field splices at 132’ and 198’ from the left
end of the bridge. (HL93S in SAP)
4) The effect of one design truck with fixed axle spacing used for fatigue loading.
All live load calculations were performed in SAP 2000 using a beam line analysis. The nominal
moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the
truck and tandem loads and an impact factor of 1.15 was applied to the fatigue loads within SAP.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 6 of 62
 6 
12.
Unfactored Moving Load Moment Envelopes from SAP
6,000
Single Truck
4,000
Tandem
Moment (kipft)
2,000
Fatigue
0
Fatigue
Tandem
2,000
Contraflexure Point
Contraflexure Point
4,000
Single Truck
Two Trucks
6,000
0
30
60
90
120
150
180
210
240
270
300
330
270
300
330
Station (ft)
Unfactored Moving Load Shear Envelopes from SAP
200
Single Truck
150
Tandem
100
Fatigue
Shear (kip)
50
0
50
100
150
200
0
30
60
90
120
150
180
210
240
Station (ft)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 7 of 62
 7 
13.
The following Live Load results were obtained from the SAP analysis:
•
The maximum positive liveload moments occur at stations 73.3’ and 256.7’
•
The maximum negative liveload moments occur over the center support at station 165.0’
HL93M
HL93K
HL93S
Fatigue
Max (+) Moment
Stations 73.3’ and 256’
3,725kft
4,396kft
N/A
2,327kft
Max () Moment
Station 165’
3,737kft
4,261kft
5,317kft
1,095kft
Before proceeding, these liveload moments will be confirmed with an influence line analysis.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 8 of 62
 8 
14.
2.2.1: Verify the Maximum Positive LiveLoad Moment at Station 73.3’:
25kip
25kip
Tandem:
32kip 32kip
8kip
Single Truck:
0.640kip/ft
Lane:
Moment (kft / kip)
40
30
20
10
0
10
20
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
330
Station (ft)
Tandem:
Single Truck:
Lane Load:
( 25 ) ( 33.00 ) + ( 25 ) ( 31.11 ) = 1, 603
( 8 ) ( 26.13 ) + ( 32 ) ( 33.00 ) + ( 32 ) ( 26.33 ) = 2,108
kip
kip
kft
kip
kip
kip
kft
kft
kip
( 0.640 ) ( 2, 491 ) = 1,594
ft
kip
kft
kip
kip
kft
kft
kip
kft
2
kft
kip
kft
kip
(IM)(Tandem) + Lane:
(1.33 ) (1, 603
(IM)(Single Truck) + Lane:
(1.33 ) ( 2,108
kft
kft
) + 1,594
kft
) + 1,594
kft
= 3,726
kft
= 4,397
kft
GOVERNS
The case of two trucks is not considered here because it is only used when computing negative moments.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 9 of 62
 9 
16.
Based on the influence line analysis, we can say that the moments obtained from SAP appear to be
reasonable and will be used for design.
Before these Service moments can be factored and combined, we must compute the distribution factors.
Since the distribution factors are a function of Kg, the longitudinal stiffness parameter, we must first
compute the sections properties of the girders.
2.3: Braking Force
The Breaking Force, BR, is taken as the maximum of:
A) 25% of the Design Truck
BRSingle Lane = ( 0.25 ) ( 8kip + 32kip + 32kip ) = 18.00kip
B) 25% of the Design Tandem
BRSingle Lane = ( 0.25 ) ( 25kip + 25kip ) = 12.50kip
C) 5% of the Design Truck with the Lane Load.
(
)
BRSingle Lane = ( 0.05 ) ⎡( 8kip + 32kip + 32kip ) + ( 2 )(165') 0.640 kip ⎤ = 14.16kip
ft ⎦
⎣
D) 5% of the Design Tandem with the Lane Load.
(
)
BRSingle Lane = ( 0.05 ) ⎡( 25kip + 25kip ) + ( 2 )(165' ) 0.640 kip ⎤ = 13.06kip
ft ⎦
⎣
Case (A) Governs:
BRNet = ( BRSingle Lane ) ( # Lanes )( MPF )
= (18.00kip ) ( 3)( 0.85 ) = 45.90kip
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
This load has not been factored…
AASHTOLRFD 2007
Page 11 of 62
 11 
17.
2.4: Centrifugal Force
A centrifugal force results when a vehicle turns on a structure. Although a centrifugal force doesn’t apply
to this bridge since it is straight, the centrifugal load that would result from a hypothetical horizontal
curve will be computed to illustrate the procedure.
The centrifugal force is computed as the product of the axle loads and the factor, C.
C= f
v2
gR
(3.6.31)
where:
ft
( sec )
v
 Highway design speed
f
 4/3 for all load combinations except for Fatigue, in which case it is 1.0
g
 The acceleration of gravity
R
 The radius of curvature for the traffic lane (ft).
( )
ft
sec 2
Suppose that we have a radius of R = 600’ and a design speed of v = 65mph = 95.33ft/sec.
ft 2
⎤
⎛ 4 ⎞ ⎡ ( 95.33 sec )
⎥ = 0.6272
C = ⎜ ⎟⎢
ft
⎝ 3 ⎠ ⎢ 32.2 sec2 ( 600 ') ⎥
⎣
⎦
(
)
CE = ( Axle Loads )( C )( # Lanes )( MPF )
= ( 72kip ) ( 0.6272 )( 3)( 0.85 ) = 115.2kip
This force has not been factored…
The centrifugal force acts horizontally in the direction pointing away from the center of curvature and at a
height of 6’ above the deck. Design the cross frames at the supports to carry this horizontal force into the
bearings and design the bearings to resist the horizontal force and the resulting overturning moment.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 12 of 62
 12 
18.
2.5: Wind Loads
For the calculation of wind loads, assume that the bridge is located in the “open country” at an elevation
of 40’ above the ground.
Take Z = 40’
V o = 8.20mph
Z o = 0.23ft
Open Country
Horizontal Wind Load on Structure: (WS)
Design Pressure:
2
⎛V ⎞
VDZ 2
PD = PB ⎜ D Z ⎟ = PB
2
10, 000mph
⎝ VB ⎠
(3.8.1.2.11)
 Base Pressure  For beams, PB = 50psf when VB = 100mph.
 Base Wind Velocity, typically taken as 100mph.
 Wind Velocity at an elevation of Z = 30’ (mph)
 Design Wind Velocity (mph)
PB
VB
V30
VDZ
(Table 3.8.1.2.11)
Design Wind Velocity:
⎛V ⎞ ⎛ Z ⎞
VDZ = 2.5Vo ⎜ 30 ⎟ ln ⎜ ⎟
⎝ VB ⎠ ⎝ Z o ⎠
= ( 2.5 ) ( 8.20
PD = ( 50
psf
)
(105.8 )
mph
(3.8.1.11)
ft
⎛ 100 ⎞ ⎛ 40 ⎞
) ⎜ 100 ⎟ Ln ⎜ 0.23ft ⎟ = 105.8mph
⎝
⎠ ⎝
⎠
mph 2
(10, 000 )
mph 2
PD
= 55.92psf
The height of exposure, hexp, for the finished bridge is computed as
hexp
hexp = 71.5"+ 11.75"+ 42" = 125.3" = 10.44 '
The wind load per unit length of the bridge, W, is then computed as:
W = ( 55.92psf ) (10.44 ' ) = 583.7 lbs
ft
= ( 583.7 lbs ) ( 2 )(165') = 192.6kip
ft
Total Wind Load:
WS H ,Total
For End Abutments:
WS H , Abt = ( 583.7 lbs ) ( 1 ) (165' ) = 48.16kip
ft
2
For Center Pier:
WS H , Pier
= ( 583.7 lbs ) ( 2 ) ( 1 ) (165' ) = 96.31kip
ft
2
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 13 of 62
 13 
19.
Vertical Wind Load on Structure: (WS)
When no traffic is on the bridge, a vertical uplift (a line load) with a magnitude equal to 20psf times the
overall width of the structure, w, acts at the windward quarter point of the deck.
P = ( 20psf ) ( w ) = ( 20psf ) ( 42 ' ) = 840 lbs
V
ft
Total Uplift:
(840 lbs ) ( 2 )(165') = 277.2kip
ft
For End Abutments:
(840 lbs ) ( 1 ) (165') = 69.30kip
ft
2
For Center Pier: ( 840 lbs ) ( 2 ) ( 1 ) (165') = 138.6kip
ft
2
Wind Load on Live Load: (WL)
The wind acting on live load is applied as a line load of 100 lbs/ft acting at a distance of 6’ above the
deck, as is shown below. This is applied along with the horizontal wind load on the structure but in the
absence of the vertical wind load on the structure.
WL
PD
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 14 of 62
 14 
20.
3. SECTION PROPERTIES AND CALCULATIONS:
3.1: Effective Flange Width, beff:
For an interior beam, beff is the lesser of:
⎧ Leff 132'
=
= 33' = 396"
⎪•
4
⎪ 4
bf
⎪
15"
= (12)(8.5") +
= 109.5"
⎨• 12ts +
2
2
⎪
⎪• S = (12')(12 in ft ) = 144"
⎪
⎩
For an exterior beam, beff is the lesser of:
⎧ Leff 132'
=
= 33' = 198.0"
⎪•
4
⎪ 4
bf
⎪
15"
= (12)(8.5") +
= 109.5"
⎨• 12ts +
2
2
⎪
⎪ S
⎛ 12'
⎞
+ 3' ⎟ (12 in ) = 108.0"
⎪• + d e = ⎜
ft
⎝ 2
⎠
⎩ 2
Note that Leff was taken as 132.0’ in the above calculations since for the case of effective width in
continuous bridges, the span length is taken as the distance from the support to the point of dead load
contra flexure.
For computing the section properties shown on the two pages that follow, reinforcing steel in the deck
was ignored for shortterm and longterm composite calculations but was included for the cracked
section. The properties for the cracked Section #1 are not used in this example, thus the amount of rebar
included is moot. For the properties of cracked Section #2, As = 13.02 in2 located 4.5” from the top of the
slab was taken from an underlying example problem first presented by Barth (2004).
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 15 of 62
 15 
21.
3.2: Section 1 Flexural Properties
Bare Steel
Ix
Ay
791.72
0.53
1,377.84 15,398.86
10.50
1.75
Ad2
b
15.00
69.00
21.00
A
11.25
38.81
21.00
Y=
ITotal =
53,157
30.68
71.06
IX
17,729
16,301
19,127
2,180.06
y
70.38
35.50
0.50
17,728
902
19,125
SBS1,top =
1,327
SBS1,bot =
Top Flange
Web
Bot Flange
t
0.7500
0.5625
1.0000
1,733
d
39.70
4.82
30.18
ShortTerm Composite (n = 8)
b
109.50
15.00
15.0000
69.0000
21.0000
n:
Slab
Haunch
Top Flange
Web
Bot Flange
t
8.5000
0.0000
0.7500
0.5625
1.0000
A
116.34
0.00
11.25
38.81
21.00
187.41
y
75.00
70.75
70.38
35.50
0.50
Ix
Ay
8,725.78
700.49
0.00
0.00
791.72
0.53
1,377.84 15,398.86
10.50
1.75
10,905.84
Ad2
IX
8.00
Y=
d
16.81
12.56
12.18
22.69
57.69
33,562
0
1,670
35,387
69,901
140,521
SST1,top =
SST1,bot =
58.19
32,862
0
1,669
19,988
69,900
ITotal =
11,191
2,415
LongTerm Composite (n = 24)
b
109.50
15.00
15.0000
69.0000
21.0000
n:
Slab
Haunch
Top Flange
Web
Bot Flange
t
8.5000
0.0000
0.7500
0.5625
1.0000
A
38.78
0.00
11.25
38.81
21.00
109.84
y
75.00
70.75
70.38
35.50
0.50
Ix
Ay
2,908.59
233.50
0.00
0.00
791.72
0.53
1,377.84 15,398.86
10.50
1.75
5,088.66
Ad2
31,885
0
6,506
4,549
44,101
ITotal =
IX
32,119
0
6,507
19,948
44,103
102,676
SLT1,top =
SLT1,bot =
4,204
2,216
24.00
Y=
d
28.67
24.42
24.05
10.83
45.83
46.33
Cracked Section
Rebar
Top Flange
Web
Bot Flange
t
4.5000
0.7500
0.5625
1.0000
b
15.0000
69.0000
21.0000
A
13.02
11.25
38.81
21.00
84.08
y
75.25
70.38
35.50
0.50
Y=
Ix
Ay
979.76
791.72
0.53
1,377.84 15,398.86
10.50
1.75
3,159.82
37.58
Ad2
73,727
55,717
48,913
5
ITotal =
IX
73,727
55,718
64,312
7
193,764
SCR1,top =
SCR1,bot =
5,842
5,156
d
75.25
70.38
35.50
0.50
These section properties do NOT include the haunch or sacrificial wearing surface.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 16 of 62
 16 
22.
3.3: Section 2 Flexural Properties
Bare Steel
Ix
Ay
1,512.00
1.75
1,436.06 15,398.86
65.63
27.34
Ad2
b
21.00
69.00
21.00
A
21.00
38.81
52.50
Y=
ITotal =
96,642
26.83
112.31
IX
42,843
19,411
34,388
3,013.69
y
72.00
37.00
1.25
42,841
4,012
34,361
SBS2,top =
2,116
SBS2,bot =
Top Flange
Web
Bot Flange
t
1.0000
0.5625
2.5000
3,602
d
45.17
10.17
25.58
Short Term Composite (n = 8)
b
109.50
21.00
21.0000
69.0000
21.0000
n:
Slab
Haunch
Top Flange
Web
Bot Flange
t
8.5000
0.0000
1.0000
0.5625
2.5000
A
116.34
0.00
21.00
38.81
52.50
228.66
y
76.75
72.50
72.00
37.00
1.25
Ix
Ay
8,929.38
700.49
0.00
0.00
1,512.00
1.75
1,436.06 15,398.86
65.63
27.34
11,943.07
Ad2
IX
8.00
Y=
d
24.52
20.27
19.77
15.23
50.98
70,641
0
8,208
24,403
136,481
239,734
SST2,top =
SST2,bot =
52.23
69,941
0
8,207
9,005
136,454
ITotal =
11,828
4,590
LongTerm Composite (n = 24)
b
109.50
15.00
21.0000
69.0000
21.0000
n:
Slab
Haunch
Top Flange
Web
Bot Flange
t
8.5000
0.0000
1.0000
0.5625
2.5000
A
38.78
0.00
21.00
38.81
52.50
151.09
y
76.75
72.50
72.00
37.00
1.25
Ix
Ay
2,976.46
233.50
0.00
0.00
1,512.00
1.75
1,436.06 15,398.86
65.63
27.34
5,990.15
Ad2
53,393
0
21,983
272
77,395
ITotal =
IX
53,626
0
21,985
15,670
77,423
168,704
SLT2,top =
SLT2,bot =
5,135
4,255
24.00
Y=
d
37.10
32.85
32.35
2.65
38.40
39.65
Cracked Section
Rebar
Top Flange
Web
Bot Flange
t
4.5000
1.0000
0.5625
2.5000
b
21.0000
69.0000
21.0000
A
13.02
21.00
38.81
52.50
125.33
y
77.00
72.00
37.00
1.25
Y=
Ix
Ay
1,002.54
1,512.00
1.75
1,436.06 15,398.86
65.63
27.34
4,016.23
32.04
Ad2
26,313
33,525
953
49,786
ITotal =
IX
26,313
33,527
16,352
49,813
126,006
SCR2,top =
SCR2,bot =
3,115
3,932
d
44.96
39.96
4.96
30.79
These section properties do NOT include the haunch or sacrificial wearing surface.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 17 of 62
 17 
23.
4. DISTRIBUTION FACTOR FOR MOMENT
4.1: Positive Moment Region (Section 1):
Interior Girder –
One Lane Loaded:
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg
⎟ ⎜ ⎟ ⎜
⎝ 14 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3
⎝
DFM 1, Int + = 0.06 + ⎜
⎞
⎟
⎟
⎠
0.1
2
K g = n( I + Aeg )
4
2
2
K g = 8(53,157 in + (71.06 in )(46.82") )
K g = 1, 672, 000 in
4
0.4
0.3
4
⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞
⎟
⎟ ⎜
⎟ ⎜
⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠
0.1
DFM 1, Int + = 0.06 + ⎜
DFM 1, Int + = 0.5021
In these calculations, the terms eg and Kg include the haunch and sacrificial wearing surface since
doing so increases the resulting factor. Note that ts in the denominator of the final term excludes
the sacrificial wearing surface since excluding it increases the resulting factor.
Two or More Lanes Loaded:
⎞
⎟
⎟
⎠
0.1
DFM 2, Int +
0.6
0.2
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg
= 0.075 + ⎜
⎟ ⎜ ⎟ ⎜
⎝ 9.5 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3
⎝
DFM 2, Int +
0.6
0.2
4
⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞
= 0.075 + ⎜
⎟
⎟ ⎜
⎟ ⎜
⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ 12(165 ')(8.5")3 ⎠
0.1
DFM 2, Int + = 0.7781
Exterior Girder –
One Lane Loaded:
The lever rule is applied by assuming that a hinge forms
over the first interior girder as a truck load is applied near
the parapet. The resulting reaction in the exterior girder is
the distribution factor.
DFM 1, Ext + =
8.5
12
= 0.7083
Multiple Presence: DFM1,Ext+ = (1.2) (0.7083) = 0.8500
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 18 of 62
 18 
24.
Two or More Lanes Loaded:
DFM2,Ext+ = e DFM2,Int+
e = 0.77 +
= 0.77 +
de
9.1
1.5
9.1
= 0.9348
DFM2,Ext+ = (0.9348) (0.7781) = 0.7274
4.2: Negative Moment Region (Section 2):
The span length used for negative moment near the pier is the average of the lengths of the adjacent
spans. In this case, it is the average of 165.0’ and 165.0’ = 165.0’.
Interior Girder –
One Lane Loaded:
DFM 1, Int −
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg
= 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜
⎝ 14 ⎠ ⎝ L ⎠ ⎜ 12 Lts3
⎝
⎞
⎟
⎟
⎠
0.1
2
K g = n( I + Aeg )
K g = 8(96, 642 in 4 + (112.3 in 2 )(52.17 ") 2 )
K g = 3, 218, 000 in 4
0.4
0.3
4
⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞
⎟
⎜
⎟ ⎜
⎟
⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠
0.1
DFM 1, Int − = 0.06 + ⎜
DFM 1, Int − = 0.5321
Two or More Lanes Loaded:
⎞
⎟
⎟
⎠
0.1
DFM 2, Int −
0.6
0.2
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg
= 0.075 + ⎜
⎟ ⎜ ⎟ ⎜
⎝ 9.5 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3
⎝
DFM 2, Int −
0.6
0.2
4
⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞
= 0.075 + ⎜
⎟
⎟ ⎜
⎟ ⎜
⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠
0.1
DFM 2, Int − = 0.8257
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 19 of 62
 19 
25.
Exterior Girder –
One Lane Loaded:
Same as for the positive moment section: DFM1,Ext = 0.8500
Two or More Lanes Loaded:
DFM2,Ext = e DFM2,Intd
e = 0.77 + e
9.1
= 0.77 +
1.5
9.1
= 0.9348
DFM2,Ext = (0.9348) (0.8257) = 0.7719
4.3: Minimum Exterior Girder Distribution Factor:
NL
DF
Ext , Min
=
NL
Nb
+
X Ext ∑ e
Nb
∑x
2
One Lane Loaded:
DF
M 1, Ext , Min
=
1
4
+
(18.0 ')(14.5 ')
(2) ⎡(18 ') 2 + (6 ') 2 ⎤
⎣
⎦
= 0.6125
Multiple Presence:
DFM1,Ext,Min = (1.2) (0.6125) = 0.7350
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 20 of 62
 20 
26.
Two Lanes Loaded:
14.5'
2.5'
2'
3'
3'
2'
3'
3'
P1
P2
DF
M 2 , Ext , Min
=
2
4
+
(18.0 ')(14.5 '+ 2.5 ')
(2) ⎡(18 ') 2 + (6 ') 2 ⎤
⎣
⎦
= 0.9250
Multiple Presence:
DFM2,Ext,Min = (1.0) (0.9250) = 0.9250
Lane 1 (12')
3'
Lane 2 (12')
12'
6'
Three Lanes Loaded:
The case of three lanes loaded is not considered for the minimum exterior distribution factor since
the third truck will be placed to the right of the center of gravity of the girders, which will
stabilize the rigid body rotation effect resulting in a lower factor.
4.4: Moment Distribution Factor Summary
Strength and Service Moment Distribution:
1 Lane Loaded:
2 Lanes Loaded:
Positive Moment
Interior
Exterior
0.5021
0.8500 ≥ 0.7350
0.7781
0.7274 ≥ 0.9250
Negative Moment
Interior
Exterior
0.5321
0.8500 ≥ 0.7350
0.8257
0.7719 ≥ 0.9250
For Simplicity, take the Moment Distribution Factor as 0.9250 everywhere for the Strength and Service
load combinations.
Fatigue Moment Distribution:
For Fatigue, the distribution factor is based on the onelaneloaded situations with a multiple presence
factor of 1.00. Since the multiple presence factor for 1lane loaded is 1.2, these factors can be obtained
by divided the first row of the table above by 1.2.
1 Lane Loaded:
Positive Moment
Interior
Exterior
0.4184
0.7083 ≥ 0.6125
Negative Moment
Interior
Exterior
0.4434
0.7083 ≥ 0.6125
For Simplicity, take the Moment Distribution Factor as 0.7083 everywhere for the Fatigue load
combination
Multiplying the live load moments by this distribution factor of 0.9250 yields the table of “nominal”
girder moments shown on the following page.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 21 of 62
 21 
28.
5. DISTRIBUTION FACTOR FOR SHEAR
The distribution factors for shear are independent of the section properties and span length. Thus, the
only one set of calculations are need  they apply to both the section 1 and section 2
5.1: Interior Girder –
One Lane Loaded:
S
25.0
12 '
= 0.36 +
= 0.8400
25.0
DFV 1,Int = 0.36 +
Two or More Lanes Loaded:
DFV 2 ,Int
S ⎛ S ⎞
= 0.2 + − ⎜ ⎟
12 ⎝ 35 ⎠
2
2
= 0.2 +
12 ' ⎛ 12 ' ⎞
−⎜
⎟ = 1.082
12 ⎝ 35 ⎠
5.2: Exterior Girder –
One Lane Loaded:
Lever Rule, which is the same as for moment: DFV1,Ext = 0.8500
Two or More Lanes Loaded:
DFV2,Ext = e DFV2,Int
de
10
1.5'
= 0.60 +
= 0.7500
10
e = 0.60 +
DFV2,Ext = (0.7500) (1.082) = 0.8115
5.3: Minimum Exterior Girder Distribution Factor 
The minimum exterior girder distribution factor applies to shear as well as moment.
DFV1,Ext,Min = 0.7350
DFV2,Ext,Min = 0.9250
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 23 of 62
 23 
29.
5.4: Shear Distribution Factor Summary
Strength and Service Shear Distribution:
1 Lane Loaded:
2 Lanes Loaded:
Shear Distribution
Interior
Exterior
0.8400
0.8500 ≥ 0.7350
1.082
0.6300 ≥ 0.9250
For Simplicity, take the Shear Distribution Factor as 1.082 everywhere for Strength and Service load
combinations.
Fatigue Shear Distribution:
For Fatigue, the distribution factor is based on the onelaneloaded situations with a multiple presence
factor of 1.00. Since the multiple presence factor for 1lane loaded is 1.2, these factors can be obtained
by divided the first row of the table above by 1.2.
1 Lane Loaded:
Shear Distribution
Interior
Exterior
0.7000
0.7083 ≥ 0.6125
For Simplicity, take the Shear Distribution Factor as 0.7083 everywhere for the Fatigue load combination.
Multiplying the live load shears by these distribution factors yields the table of “nominal” girder
shears shown on the following page.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 24 of 62
 24 
31.
6. FACTORED SHEAR AND MOMENT ENVELOPES
The following load combinations were considered in this example:
Strength I:
Strength IV:
1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW
1.50DC1 + 1.50DC2 + 1.50DW
Service II:
1.3(LL + IM) + 1.0DC1 + 1.0DC2 + 1.0DW
Fatigue:
0.75(LL + IM)
(IM = 15% for Fatigue; IM = 33% otherwise)
Strength II is not considered since this deals with special permit loads. Strength III and V are not
considered as they include wind effects, which will be handled separately as needed. Strength IV is
considered but is not expected to govern since it addresses situations with high dead load that come into
play for longer spans. Extreme Event load combinations are not included as they are also beyond the
scope of this example. Service I again applies to wind loads and is not considered (except for deflection)
and Service III and Service IV correspond to tension in prestressed concrete elements and are therefore
not included in this example.
In addition to the factors shown above, a load modifier, η, was applied as is shown below.
Q = ∑ηiγ i Qi
η is taken as the product of ηD, ηR, and ηI, and is taken as not less than 0.95. For this example,
ηD and ηI are taken as 1.00 while ηR is taken as 1.05 since the bridge has 4 girders with a
spacing greater than or equal to 12’.
Using these load combinations, the shear and moment envelopes shown on the following pages
were developed.
Note that for the calculation of the Fatigue moments and shears that η is taken as 1.00 and the
distribution factor is based on the onelaneloaded situations with a multiple presence factor of
1.00 (AASHTO Sections 6.6.1.2.2, Page 629 and 3.6.1.4.3b, Page 325).
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 26 of 62
 26 
32.
Strength Limit Moment Envelopes
20,000
15,000
Strength I
10,000
Strength IV
Moment (kipft)
5,000
0
5,000
Strength IV
10,000
15,000
Strength I
20,000
25,000
0
30
60
90
120
150
180
210
240
270
300
330
270
300
330
Station (ft)
Strength Limit Shear Force Envelope
800
Strength I
600
400
Strength IV
Shear (kip)
200
0
200
400
600
800
0
30
60
90
120
150
180
210
240
Station (ft)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 27 of 62
 27 
33.
Service II Moment Envelope
12,500
10,000
7,500
5,000
Moment (kipft)
2,500
0
2,500
5,000
7,500
10,000
12,500
15,000
17,500
20,000
0
30
60
90
120
150
180
210
240
270
300
330
270
300
330
Station (ft)
Service II Shear Envelope
600
400
Shear (kip)
200
0
200
400
600
0
30
60
90
120
150
180
210
240
Station (ft)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 28 of 62
 28 
37.
7. FATIGUE CHECKS
7.1: Check transverse stiffener to flange weld at Station 73.3:
Traffic information:
ADTT given as 2400. Three lanes are available to trucks. (ADTT)SL = (0.80) (2,400) = 1,920
N = (ADTT)SL (365) (75) n = (1,920) (365) (75) (1) = 52.56M Cycles
Check Top Flange Weld:
Fatigue need only be checked when the compressive stress due to unfactored permanent loads is
less than twice the maximum tensile stress due to factored fatigue loads.
?
Check f comp , DL ≤ 2 f Fat
Distance from bottom of section to the detail under investigation
y = tf,bottom + D = 1.00” + 69.00” = 70”
(Pg 16)
(Pg 24)
( 2, 779 ) (12 ) ( 70"− 30.68") = 24.67
kft
M DC1 = 2, 779kft
f DC1 =
in
ft
ksi
53,157 in 4
(Pg 16)
(Pg 24)
(Pg 16)
( 515.8 ) (12 ) ( 70"− 46.33") = 1.427
kft
M DC 2 = 515.8kft
f DC 2 =
in
ft
102, 676 in
ksi
4
(Pg 16)
f comp , DL = 24.67 ksi + 1.427 ksi = 26.09ksi
(Pg 16)
(Pg 30)
( 258.6 ) (12 ) ( 70"− 58.19") = 0.261
kft
M Fat , Neg = 258.6kft
f Fat =
in
ft
140,521 in
ksi
4
(Pg 16)
?
Check f comp , DL ≤ 2 f Fat
26.09ksi ≤ ( 2 ) ( 0.261ksi ) = 0.521ksi , No.
?
Fatigue need not be checked on the top flange at Station 73.3.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 32 of 62
 32 
38.
Check Bottom Flange Weld:
The permanent loads at Station 73.3 cause tension in the bottom flange, thus by inspection
fatigue needs to be checked.
γ ( Δf ) ≤ ( ΔF )n
1
( ΔF ) n
⎛ A ⎞ 3 ( ΔF )TH
=⎜ ⎟ ≥
2
⎝N⎠
γ is a load factor of 0.75, which is already included in the fatigue moments.
γ ( Δf ) =
(1, 236 ) (12 ) ( 58.19"− 1.00") = 6.036
kft
in
ft
140,521 in
ksi
4
The detail under consideration is a Category C’ detail.
A = 44.0 x 108 ksi3 and (ΔF)TH = 12.0 ksi
( ΔF )TH
2
=
12.0ksi
= 6.00ksi
2
The stress in the detail is almost less than the
infinite life threshold
1
1
8
3
⎛ A ⎞ 3 ⎛ 44 × 10 ksi ⎞ 3
=⎜
= 4.375ksi
⎜ ⎟
6 ⎟
⎝ N ⎠ ⎝ 52.56 × 10 ⎠
1
( ΔF )TH
⎛ A ⎞3
= 6.00ksi , the infinite life governs.
Since ⎜ ⎟ = 4.375ksi is less than
2
⎝N⎠
( ΔF )n = 6.00ksi
Since γ ( Δf ) = 6.036ksi > ( F )n = 6.00ksi , the detail is not satisfactory.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 33 of 62
 33 
39.
Calculate the design life of the part under consideration:
Since γ ( Δf ) is greater than
( ΔF )TH
2
, solve for N in the following equation.
1
⎛ A ⎞3
γ ( Δf ) ≤ ⎜ ⎟
⎝N⎠
N≤
A
γ ( Δf )
3
=
44 × 108 ksi3
( 6.036 )
ksi 3
= 20.01×106 cycles
20.01×106 cycles
= 10, 420 days
1,920
10, 420 days
= 28.55 years
365
( = 28y, 6m, 19d, 2h, 38min...)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
☺
AASHTOLRFD 2007
Page 34 of 62
 34 
40.
8. CHECK CROSS_SECTION PROPORTION LIMITS
Web Proportions
•
69"
D
≤ 150
tw
9 "
16
= 122.7 ≤ 150
O.K.
Flange Proportions
•
bf
≤ 12
2t f
15"
= 10.00 ≤12
(2)( 3 4 ")
O.K.
•
bf
≤ 12
2t f
21"
= 10.50 ≤12
(2)(1")
O.K.
•
bf
≤ 12
2t f
21"
= 4.200 ≤12
(2)(2 12 ")
O.K.
Check ODOT Criteria for Flange Width
?
⎛D
⎞
b f ≥ ⎜ + 2.5 ⎟ ≥ 12"
⎝6
⎠
• b f ,min =
→
⎛ 69"
⎞
+ 2.5 ⎟ = 14"
⎜
⎝ 6
⎠
D 69"
=
=11.50"
6
6
• t f ,min = 1.1tw =(1.1)( 916 ")
O.K.
O.K.
5
O.K.
8"
• 0.1 ≤
I yc
≤ 10
I yt
0.1 ≤
( 3 4 ")(15")3
= 0.2733 ≤ 10
(1")(21")3
O.K.
• 0.1 ≤
I yc
≤ 10
I yt
0.1 ≤
(2.5")(21")3
= 2.500 ≤ 10
(1")(21")3
O.K.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 35 of 62
 35 
41.
9. CHECK SERVICE LIMIT STATE
9.1: Check Absolute Deflection of the Bridge: Section 6.10.4.1
Section 1
The cross section of Section 1 that is used for computing deflections is shown above. The entire deck
width is used (as opposed to just the effective width that was used earlier) and the haunch and sacrificial
wearing surface have been neglected. AASHTO permits the use of the stiffness of parapets and
structurally continuous railing but ODOT does not.
The transformed width of the bridge deck is w ' =
( 42 ') (12 in )
ft
8
= 63.00"
Using the bottom of the steel as a datum, the location of the CG of the deck can be found as:
yc = 1"+ 69"+ 3 4 "+
8.5"
= 75.00"
2
The CG of this composite cross section is found as:
Y
( 63")(8.5")i( 75.00") + ( 4 ) ( 71.06 in 2 )i( 30.68")
=
= 59.63"
( 63")( 8.5") + ( 4 ) ( 71.06 in 2 )
Now the moment of inertia of the section can be found as:
Concrete
→
Steel
→
( 63")( 8.5")
12
3
+ ( 63")( 8.5") [ 75.00"− 59.63"] =
2
129, 700 in 4
( 4 ) ( 53,160 in 4 ) + ( 4 ) ( 71.06 in 2 ) [30.68"− 59.63"]
2
=
450,900 in 4
I1,total = 580, 600 in 4
I1 =
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
580, 600 in 4
in 4
= 145,100 Girder
4 Girders
AASHTOLRFD 2007
Page 36 of 62
 36 
42.
Section 2
The cross section of Section 2 that is used for computing deflections is shown above.
The transformed width of the bridge deck is w ' =
( 42 ') (12 in ) = 63.00"
ft
8
Using the bottom of the steel as a datum, the location of the CG of the deck can be found as:
yc = 2 1 2 "+ 69"+ 1"+
8.5"
= 76.75"
2
The CG of this composite cross section is found as:
Y
( 63")(8.5")i( 76.75") + ( 4 ) (112.3 in 2 )i( 26.83")
=
= 53.98"
( 63")( 8.5") + ( 4 ) (112.3 in 2 )
Now the moment of inertia of the section can be found as:
Concrete
→
Steel
→
( 63")( 8.5")
12
3
+ ( 63")( 8.5") [ 76.75"− 53.98"] =
2
280,900 in 4
( 4 ) ( 96, 640 in 4 ) + ( 4 ) (112.3 in 2 ) [ 26.83"− 53.98"]
2
=
717, 700 in 4
I total = 998, 600 in 4
I2 =
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
998, 600 in 4
in 4
= 249, 700 Girder
4 Girders
AASHTOLRFD 2007
Page 37 of 62
 37 
43.
The following model, which represents the stiffness of a single girder, was used to compute absolute liveload deflections assuming the entire width of the deck to be effective in both compression and tension.
The live load component of the Service I load combination is applied. Based on AASHTO Section
3.6.1.3.2, the loading includes (1) the design truck alone and (2) the lane load with 25% of the design
truck. The design truck and design lane load were applied separately in the model and will be combined
below. The design truck included 33% impact.
I = 145,100 in4
I = 249,700 in4
I = 145,100 in4
From the analysis:
Deflection due to the Design Truck with Impact: ΔTruck = 2.442”
Deflection due to the Design Lane Load:
ΔLane = 0.8442”
These deflections are taken at Stations 79.2’ and 250.8’. The model was broken into segments roughly
25’ long in the positive moment region and 7’ long in the negative moment region. A higher level of
discretization may result in slightly different deflections but it is felt that this level of accuracy was
acceptable for deflection calculations.
Since the above results are from a singlegirder model subjected to one lane’s worth of loading,
distribution factors must be applied to obtain actual bridge deflections. Since it is the absolute deflection
that is being investigated, all lanes are loaded (multiple presence factor apply) and it is assumed that all
girders deflect equally. Given these assumptions, the distribution factor for deflection is simply the
number of lanes times the multiple presence factor divided by the number of girders.
Looking at the two loading criteria described above:
⎛ ( 0.85 )( 3) ⎞
Δ1 = ⎜
⎜ ( 4 ) ⎟ ( 2.442") = 1.558"
⎟
⎝
⎠
← Governs
⎛ ( 0.85 )( 3) ⎞
Δ2 = ⎜
⎦
⎜ ( 4 ) ⎟ ⎡( 0.8442") + ( 0.25 )( 2.442") ⎤ = 0.9274"
⎟⎣
⎝
⎠
The limiting deflection for this bridge is:
Δ Limit =
(165') (12 in )
L
ft
=
= 2.475"
800
800
← OK
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 38 of 62
 38 
44.
9.2: Check the Maximum SpantoDepth Ratio: Section 6.10.4.1
From Table 2.5.2.6.31, (1) the overall depth of a composite Ibeam in a continuous span must not be less
than 0.032L and (2) the depth of the steel in a composite Ibeam in a continuous span must not be less
than 0.027L.
(1)
( 2)
→
0.032 L = ( 0.032 )(165') (12 in ) = 63.36"
ft
OK
→
0.027 L = ( 0.027 )(165') (12 in ) = 53.46"
ft
OK
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 39 of 62
 39 
45.
9.3: Permanent Deformations  Section 1
At the Service Limit State, the following shall be satisfied for composite sections
Top Flange:
f f ≤ 0.95Rh Fyf
Bottom Flange
ff +
fl
≤ 0.95Rh Fyf
2
Per §6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of stresses
caused by loads applied separately to the bare steel, shortterm composite section, and longterm
composite section.
⎛ M DC 1 ⎞
⎛ M DC 2 ⎞
⎛ M DW
⎟ + 1.00 ⎜ S ⎟ + 1.00 ⎜ S
⎝ LT ⎠
⎝ LT
⎝ S BS ⎠
fc = 1.00 ⎜
⎛ M LL + IM ⎞
⎞
⎟ + 1.30 ⎜ S ⎟
⎠
⎝ ST ⎠
Top Flange, Positive Moment
It is not immediately evident to me whether the factored stress at 58.7’ or 73.3’ will govern.
⎛ (2, 979
f f ,58.7 = 1.00 ⎜
⎝
kft
)(12 in ) ⎞
ft
1,327 in
3
⎛ (549.7
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,204 in
3
⎛ (1, 008
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,204 in
3
⎛ (3, 999
⎟ + 1.30 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
11,191 in
3
⎟
⎠
f f ,58.7 = 36.96ksi
⎛ (2, 779
f f , 73.3 = 1.00 ⎜
⎝
kft
)(12 in ) ⎞
ft
1,327 in
3
⎛ (515.8
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,204 in
3
⎛ (946.1
kft
⎟ + 1.00 ⎜
⎠
⎝
)(12 in ) ⎞
ft
4,204 in
3
⎛ (4, 067
⎟ + 1.30 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
11,191 in
3
⎟
⎠
f f ,73.3 = 34.97 ksi
The stress at 58.7’ governs. ff = 36.96ksi.
f f ≤ 0.95Rh Fyf
→
36.96ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi
O.K.
Note: The bending moments in the above calculations come from page 22 while the moments of
inertia are found on page 16.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 40 of 62
 40 
46.
Bottom Flange, Positive Moment
⎛ (2, 979
f f ,58.7 = 1.00 ⎜
⎝
kft
)(12 in ) ⎞
ft
1,733 in
⎛ (549.7
⎟ + 1.00 ⎜
⎠
3
⎝
kft
)(12 in ) ⎞
ft
2,216 in
3
⎛ (1, 008
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
2,216 in
3
⎛ (3, 999
⎟ + 1.30 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
2,415 in
3
⎟
⎠
f f ,58.7 = 54.90ksi
⎛ (2, 779 )(12 in ) ⎞
⎛ (515.8 )(12 in ) ⎞
⎛ (946.1 )(12 in ) ⎞
⎛ (4, 067 )(12 in ) ⎞
ft
ft
ft
ft
= 1.00 ⎜
+ 1.00 ⎜
+ 1.00 ⎜
+ 1.30 ⎜
⎟
⎟
⎟
⎟
⎝ 1,733 in
⎠
⎝ 2,216 in
⎠
⎝ 2,216 in
⎠
⎝ 2,415 in
⎠
kft
f f , 73.3
kft
3
kft
3
kft
3
3
f f ,73.3 = 53.43ksi
The stress at 58.7’ governs. ff = 54.90ksi.
The load factor for wind under Service II is 0.00, ∴ fl = 0ksi
ff +
fl
≤ 0.95Rh Fyf
2
→
54.90ksi +
0ksi
≤ (0.95)(1.00)(50ksi ) = 47.50ksi
2
No Good.
Note: The bending moments in the above calculations come from page 22 while the moments of
inertia are found on page 17.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 41 of 62
 41 
47.
9.4: Permanent Deformations  Section 2
Top Flange, Negative Moment
⎛ (7,109 )(12 in ) ⎞
⎛ (1, 250 )(12 in ) ⎞
⎛ (2, 292 )(12 in ) ⎞
⎛ (4, 918 )(12 in ) ⎞
ft
ft
ft
ft
= 1.00 ⎜
+ 1.00 ⎜
+ 1.00 ⎜
+ 1.30 ⎜
⎟
⎟
⎟
⎟
⎝ 2,116 in
⎠
⎝ 5,135 in
⎠
⎝ 5,135 in
⎠
⎝ 11,828 in ⎠
kft
f f ,165
kft
kft
3
kft
3
3
3
f f ,165 = 55.08ksi
f f ≤ 0.95 Rh Fyf
?
→
55.08ksi ≤(0.95)(1.00)(50ksi ) = 47.50ksi
No Good.
Bottom Flange, Negative Moment
⎛ (7,108
f f = 1.00 ⎜
⎝
kft
)(12 in ) ⎞
ft
3,602 in
3
⎛ (1, 250
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,255 in
3
⎛ (2, 292
⎟ + 1.00 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,255 in
3
⎛ (4, 918
⎟ + 1.30 ⎜
⎠
⎝
kft
)(12 in ) ⎞
ft
4,590 in
3
⎟
⎠
f f = 50.39 ksi
The load factor for wind under Service II is 0.00, ∴ fl = 0ksi
ff +
fl
≤ 0.95 Rh Fyf
2
→
50.39 ksi +
0ksi
≤ (0.95)(1.00)(50ksi ) = 47.50ksi
2
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
No Good.
AASHTOLRFD 2007
Page 42 of 62
 42 
48.
9.5: Bend Buckling Checks
At the Service Limit State, all sections except composite sections in positive flexure shall satisfy:
f c ≤ Fcrw
where:
Fcrw =
0.9 Ek
⎛D⎞
⎜ ⎟
⎝ tw ⎠
k=
and
2
9
( Dc / D )
2
Section 1
Not Applicable
Section 2
⎛ (7,108 )(12
⎝ 3,602 in
kft
fc = 1.00 ⎜
)⎞
in
⎛ (1, 250 )(12
⎝ 4,255 in
kft
⎟ + 1.00 ⎜
ft
⎠
3
in
ft
)⎞
⎠
3
⎛ (2, 292 )(12
⎝ 4,255 in
kft
⎟ + 1.00 ⎜
in
ft
)⎞
⎠
3
⎛ (4, 918 )(12
⎝ 4,590 in
kft
⎟ + 1.30 ⎜
ft
)⎞
ft
)⎞
in
3
⎟
⎠
f c = 50.39 ksi
⎛ (7,108 )(12
⎝ 2,116 in
kft
ft = 1.00 ⎜
in
ft
)⎞
⎛ (1, 250 )(12
⎝ 5,135 in
kft
⎟ + 1.00 ⎜
⎠
3
3
in
ft
)⎞
⎛ (2, 292 )(12
⎝ 5,135 in
⎟ + 1.00 ⎜
⎠
kft
3
in
ft
)⎞
⎛ (4, 918 )(12
⎝ 11,828 in
kft
⎟ + 1.30 ⎜
⎠
3
in
⎟
⎠
ft = 55.08ksi
⎛ − fc ⎞
0
Dc = ⎜
⎜ f + f ⎟ d − tcf ≥
⎟
t ⎠
⎝ c
⎛
⎞
50.39ksi
72.5") − 2.5" ≥ 0
=⎜
ksi
ksi ⎟ (
⎝ 50.39 + 55.08 ⎠
= 32.14′′
k=
9
( Dc / D )
Fcrw =
2
=
9
⎛ 32.14" ⎞
⎜
⎟
⎝ 69" ⎠
2
= 41.49
(0.90)(29, 000ksi )(41.49)
⎛ 69" ⎞
⎜ 9 "⎟
⎝ 16 ⎠
2
= 71.96 ksi
This is larger than fc…O.K.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 43 of 62
 43 
49.
10. CHECK STRENGTH LIMIT STATE
10.1: Section 1 Positive Flexure
Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)
Check
2 Dcp
E
≤ 3.76
tw
Fyc
Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored).
Pt = Fyt bt tt = (50ksi ) ( 21")(1") = 1,050kip
Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip
Pc = Fyc bc tc = (50ksi )(15")( 3 4 ") = 562.5kip
Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip
3,554kip < 3,560kip, the PNA lies in the slab.
Since Pt + Pw +Pc < Ps
⎡ P + Pw + Pt ⎤
⎡ 3,554kip ⎤
Y = ( ts ) ⎢ c
= ( 8.5") ⎢
⎥
kip ⎥
Ps
⎣ 3,560 ⎦
⎣
⎦
Y = 8.486 " ↓ from top of slab ∴ D p = Y = 8.486 "
Since none of the web is in compression, Dcp = 0 and the web is compact.
For Composite Sections in Positive Flexure, (§6.10.7.1, Pg. 6.101 – 6.102)
Mu +
1
f S ≤ φf Mn
3 l xt
Mu = 13,568kft from Page 30; take fl = 0
Dt = 1” + 69” + 3/4” + 8.5” = 79.25”
0.1Dt = 7.925”
(The haunch is not included in Dt, as per ODOT Exceptions)
Dp ⎞
⎛
Since Dp =8.486 > 0.1Dt = 7.925, M n = M p ⎜ 1.07 − 0.7
⎟
Dt ⎠
⎝
⎛
M p = (3,554kip ) ⎜ 79.25"−
⎝
= 157,500
kin
8.486"
⎞
− 30.68" ⎟
2
⎠
= 13,130kft
⎡
⎛ 8.486" ⎞ ⎤
kft
M n = 13,130kft ⎢1.07 − ( 0.7 ) ⎜
⎟ ⎥ = 13,060
⎝ 79.25" ⎠ ⎦
⎣
(
)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 44 of 62
 44 
50.
?
M u + 13 f l S xt ≤ φ f M n
?
(13,568kft ) + (0) ≤(1.00)(13,060kft )
No Good.
Note that the check of M n ≤ 1.3Rh M y has not been made in the above calculations. This
section would satisfy the Article B6.2 so this check doesn’t need to be made.
Check the ductility requirement to prevent crushing of the slab:
?
D p ≤ 0.42 Dt
?
→ 8.486" ≤ ( 0.42 )( 79.25") = 33.29"
O.K.
The Section is NOT Adequate for Positive Flexure at Stations 58.7’ and 271.3’
The Girder failed the checks for service limits and has failed the first of several checks at the
strength limit state. At this point I will investigate the strength of a section with 70ksi steel in the top
and bottom flanges.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 45 of 62
 45 
52.
11. RECHECK SERVICE LIMIT STATE WITH 70KSI FLANGES
11.1: Permanent Deformations  Section 1
At the Service Limit State, the following shall be satisfied for composite sections
Top Flange:
f f ≤ 0.95Rh Fyf
Bottom Flange
ff +
fl
≤ 0.95Rh Fyf
2
Top Flange, Positive Moment
From before: f f ,58.7 = 36.96ksi
f f ≤ 0.95Rh Fyf
?
36.96ksi ≤(0.95)(0.9626)(70ksi ) = 64.01ksi
→
O.K.
Bottom Flange, Positive Moment
f f ,58.7 = 54.90ksi
ff +
The load factor for wind under Service II is 0.00, ∴ fl = 0ksi
fl
≤ 0.95Rh Fyf
2
→
54.90ksi +
0ksi ?
≤(0.95)(0.9626)(70ksi ) = 64.01ksi
2
O.K.
11.2: Permanent Deformations  Section 2
Top Flange, Negative Moment
From before: f f ,165 = 55.08ksi
f f ≤ 0.95Rh Fyf
?
→
55.08ksi ≤(0.95)(0.9833)(70ksi ) = 65.39ksi
O.K.
Bottom Flange, Negative Moment
f f = 50.39 ksi The load factor for wind under Service II is 0.00, ∴ fl = 0ksi
ff +
fl
≤ 0.95 Rh Fyf
2
→
50.39ksi +
0ksi ?
≤(0.95)(0.9833)(70ksi ) = 65.39ksi
2
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
O.K.
AASHTOLRFD 2007
Page 47 of 62
 47 
53.
11.3: Bend Buckling Checks
At the Service Limit State, all sections except composite sections in positive flexure shall satisfy:
f c ≤ Fcrw
where:
Fcrw =
0.9 Ek
⎛D⎞
⎜ ⎟
⎝ tw ⎠
k=
and
2
9
( Dc / D )
2
Not Applicable
Section 1 
Section 2
⎛ (7,108 )(12
⎝ 3,602 in
kft
fc = 1.00 ⎜
)⎞
in
⎛ (1, 250 )(12
⎝ 4,255 in
kft
⎟ + 1.00 ⎜
ft
⎠
3
in
ft
)⎞
⎠
3
⎛ (2, 292 )(12
⎝ 4,255 in
kft
⎟ + 1.00 ⎜
in
ft
)⎞
⎠
3
⎛ (4, 918 )(12
⎝ 4,590 in
kft
⎟ + 1.30 ⎜
ft
)⎞
ft
)⎞
in
3
⎟
⎠
f c = 50.39ksi
⎛ (7,108 )(12
⎝ 2,116 in
kft
ft = 1.00 ⎜
in
ft
)⎞
⎛ (1, 250 )(12
⎝ 5,135 in
kft
⎟ + 1.00 ⎜
⎠
3
3
in
ft
)⎞
⎛ (2, 292 )(12
⎝ 5,135 in
⎟ + 1.00 ⎜
⎠
kft
3
in
ft
)⎞
⎛ (4, 918 )(12
⎝ 11,828 in
kft
⎟ + 1.30 ⎜
⎠
3
in
⎟
⎠
ft = 55.08ksi
⎛ − fc ⎞
0
Dc = ⎜
⎜ f + f ⎟ d − tcf ≥
⎟
t ⎠
⎝ c
⎛
⎞
50.39ksi
72.5") − 2.5" ≥ 0
=⎜
ksi
ksi ⎟ (
⎝ 50.39 + 55.08 ⎠
= 32.14′′
k=
9
( Dc / D )
Fcrw =
2
=
9
⎛ 32.14" ⎞
⎜
⎟
⎝ 69" ⎠
2
= 41.49
(0.90)(29, 000ksi )(41.49)
⎛ 69" ⎞
⎜ 9 "⎟
⎝ 16 ⎠
2
= 71.96 ksi
This is larger than fc…O.K.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 48 of 62
 48 
54.
12. RECHECK STRENGTH LIMIT STATE WITH 70KSI FLANGES
12.1: Section 1  Positive Flexure
Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)
Check
2 Dcp
E
≤ 3.76
tw
Fyc
Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored).
Pt = Fyt bt tt = (70ksi ) ( 21")(1") = 1, 470kip
Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip
Pc = Fyc bc tc = (70ksi )(15")( 3 4 ") = 787.5kip
Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip
4,199kip > 3,560kip, the PNA is NOT in the slab.
Since Pt + Pw +Pc > Ps
?
Check Case I Pt + Pw ≥ Pc + Ps
?
1, 470kip + 1,941kip ≥ 787.5kip + 3,560kip
NO
?
Check Case II Pt + Pw + Pc ≥ Ps
?
1, 470kip + 1,941kip + 787.5kip ≥ 3,560kip
YES  PNA in Top Flange
⎞
⎛ t ⎞ ⎛ P + P − Ps
Y = ⎜ c ⎟⎜ w t
+ 1⎟
Pc
⎝ 2 ⎠⎝
⎠
kip
kip
kip
⎞
⎛ 0.750" ⎞ ⎛ 1,941 + 1, 470 − 3,560
=⎜
+ 1⎟ = 0.3040" (from the top of steel)
⎟⎜
kip
787.5
⎝ 2 ⎠⎝
⎠
Dp = 8.5” + 0.3040” = 8.804”
Since none of the web is in compression, Dcp = 0 and the web is compact.
For Composite Sections in Positive Flexure, (§6.10.7.1, Pg. 6.101 – 6.102)
Mu +
1
f S ≤ φf Mn
3 l xt
Mu = 13,568kft from Page 30; take fl = 0
Dt = 1” + 69” + 3/4” + 8.5” = 79.25”
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
0.1Dt = 7.925”
AASHTOLRFD 2007
Page 49 of 62
 49 
55.
(The haunch is not included in Dt, as per ODOT Exceptions)
Dp ⎞
⎛
Since Dp = 8.804” > 0.1Dt = 7.925”, M n = M p ⎜1.07 − 0.7
⎟
Dt ⎠
⎝
Determine Mp:
The distances from the component forces to the PNA are calculated.
8.5"
+ 0.3040" = 4.554"
2
69"
− ( 0.75"− 0.3040") = 34.05"
dw =
2
1"
dt = 70.75"− − 0.3040" = 69.95"
2
ds =
The plastic moment is computed.
⎛P ⎞
2
M p = ⎜ c ⎟ ⎡Y 2 + ( tc − Y ) ⎤ + [ Ps d s + Pw d w + Pdt ]
t
⎦
⎝ 2tc ⎠ ⎣
⎛ 787.5kip ⎞ ⎡
2
2
=⎜
⎟ ⎣( 0.3040") + ( 0.750"− 0.3040") ⎤ + ...
⎦
⎝ (2)(0.750") ⎠
... + ⎡( 3,560kip ) ( 4.554") + (1,941kip ) ( 34.05") + (1, 470kip ) ( 69.95") ⎤
⎣
⎦
= ( 525 kip in ) ⎡0.2913 in 2 ⎤ + ⎡185,100kin ⎤
⎣
⎦ ⎣
⎦
= 185,300kin = 15, 440kft
⎡
⎛ 8.804" ⎞ ⎤
kft
M n = 15, 440kft ⎢1.07 − ( 0.7 ) ⎜
⎟ ⎥ = 15,320
⎝ 79.25" ⎠ ⎦
⎣
(
)
?
M u + 13 f l S xt ≤ φ f M n
?
(13,568kft ) + (0) ≤(1.00)(15,320 kft )
O.K.
Note that the check of M n ≤ 1.3Rh M y has not been made in the above calculations. This
section would satisfy the Article B6.2 so this check doesn’t need to be made.
Check the ductility requirement to prevent crushing of the slab:
?
D p ≤ 0.42 Dt
?
→ 8.804" ≤ ( 0.42 )( 79.25") = 33.29"
O.K.
The Section is Adequate for Positive Flexure at Stations 58.7’ and 271.3’ with 70ksi Flanges
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 50 of 62
 50 
56.
12.2: Section 2  Negative Flexure
Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)
Check
2 Dc
E
≤ 5.70
tw
Fyc
Dc is the depth of the web in compression for the cracked section.
Dc = 32.04” – 21/2” = 29.54”
2 Dc (2)(29.54")
E
29,000ksi
=
= 105.0 < 5.70
= 5.70
= 137.3
tw
(916 ")
Fyc
50ksi
The web is nonslender. Since the web is nonslender we have the option of using the provisions
in Appendix A to determine the moment capacity. I will first determine the capacity using the
provisions in §6.10.8, which will provide a somewhat conservative determination of the flexural
resistance.
For Composite Sections in Negative Flexure, (§6.10.8.1, Pg. 6.105 – 6.114)
The Compression Flange must satisfy:
fbu +
1
f ≤ φ f Fnc
3 l
Per §6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of
stresses caused by loads applied separately to the bare steel, shortterm composite section, and
longterm composite section. In §6.10.1.1.1c, though, it states that for the Strength Limit, the
shortterm and longterm composite sections shall consist of the bare steel and the longitudinal
rebar. In other words, for determining negative moment stresses over the pier, we can use the
factored moment above with the properties for the cracked section.
⎛ M DC1 ⎞ ⎛ 1.25M DC 2 + 1.50 M DW + 1.75M LL − ⎞
⎟+⎜
⎟
SCR
⎝ S BS ⎠ ⎝
⎠
fbu = 1.25 ⎜
kft
kft
kft
⎛ (7,109 kft )(12 in ) ⎞ ⎛ ⎡ (1.25)(1, 250 ) + (1.50)(2, 292 ) + (1.75)(4, 918 ) ⎤ (12 in ) ⎞
⎣
⎦ ft
ft
fbu = 1.25 ⎜
⎟
⎟+⎜
3
3
3,932 in
⎝ 3,602 in
⎠ ⎝
⎠
fbu = 71.13ksi
Since fbu is greater than Fyc, it is obvious that a strength computed based on the provisions
in §6.10.8 will not be adequate.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 51 of 62
 51 
57.
As it stands here, this girder is clearly not adequate over the pier. The compression flange is
overstressed as per the provisions in §6.10.8. There are still other options to explore, though, before
increasing the plate dimensions.
1. Since the web is nonslender for Section 2 in Negative Flexure, we have the option of using
the provisions in Appendix A6 to determine moment capacity. This would provide an
upper bound strength of Mp instead of My as was determined in §6.10.8.
2. The provisions in Appendix B6 allow for redistribution of negative moment from the region
near the pier to the positive moment region near midspan for sections that satisfy stringent
compactness and stability criteria. If this section qualifies, as much as ~2,000kft may be able
to be redistributed from the pier to midspan, which could enable the plastic moment
strength from Appendix A6 to be adequate. (This solution may even work with the flange
strength at 50ksi, but I doubt it…)
Despite the fact that the girder appears to have failed our flexural capacity checks, let’s look at the
shear capacity.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 52 of 62
 52 
58.
12.3 Vertical Shear Capacity
At the strength Limit, the following must be satisfied
Vu ≤ φVn
For an unstiffened web,
Vn = Vcr = CV p
Check,
D
Ek
,
≤ 1.12
tw
Fyw
D 69"
=
= 122.7
tw 916 "
Since there are no transverse stiffeners, k = 5
(29, 000ksi )(5)
1.12
= 60.31
(50ksi )
Since
C=
(29, 000ksi )(5)
1.40
= 75.39
(50ksi )
D
Ek
, elastic shear buckling of the web controls.
> 1.40
tw
Fyw
1.57 ⎛ kE ⎞
1.57 ⎛ (5)(29, 000ksi ) ⎞
⎜
⎟=
⎟ = 0.3026
2
2 ⎜
(50ksi )
⎛ D ⎞ ⎜ Fyw ⎟ ⎛ 69" ⎞ ⎝
⎠
⎝
⎠
⎜ ⎟
⎜ 9 "⎟
⎝ 16 ⎠
⎝ tw ⎠
V p = 0.58 Fyw Dt w = (0.58)(50 ksi )(69")( 9 16 ") = 1,126 kip
Vn = CV p = (0.3026)(1,126 kip ) = 340.6 kip
(
)
φVn = (1.00 ) 340.6 kip = 340.6 kip
No Good.
This strength is adequate from 16’ – 100’ and 230’  314’.
This strength is not adequate near the end supports or near the pier, however.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 53 of 62
 53 
59.
Try adding transverse stiffeners spaced at do = 8’ = 96”
k =5+
5
⎛ do ⎞
⎜D⎟
⎝ ⎠
2
=5+
5
⎛ 96" ⎞
⎜ 69" ⎟
⎝
⎠
= 7.583
2
D
(29, 000ksi )(7.583)
(29, 000ksi )(7.583)
= 122.7 , 1.12
= 74.28 , 1.40
= 92.85
(50ksi )
(50ksi )
tw
Since
D
Ek
> 1.40
, elastic shear buckling of the web controls.
tw
Fyw
1.57 ⎛ Ek ⎞
1.57 ⎛ (29, 000ksi )(7.583) ⎞
C=
⎟=
⎟ = 0.4589
2 ⎜
2 ⎜
⎜
⎟
(50ksi )
⎛ D ⎞ ⎝ Fyw ⎠ ⎛ 69" ⎞ ⎝
⎠
⎜ ⎟
⎜ 9 "⎟
⎝ 16 ⎠
⎝ tw ⎠
φVn = φCV p = (1.00)(0.4589)(1,126 kip ) = 516.5kip
O.K.
This capacity is fine but we may be able to do better if we account for tension field action.
Try adding transverse stiffeners spaced at do = 12’ = 144”
k =5+
5
⎛ do ⎞
⎜D⎟
⎝ ⎠
2
=5+
5
⎛ 144" ⎞
⎜ 69" ⎟
⎝
⎠
2
= 6.148
D
(29, 000ksi )(6.148)
(29, 000ksi )(6.148)
= 122.7 , 1.12
= 66.88 , 1.40
= 83.60
(50ksi )
(50ksi )
tw
Since
C=
D
Ek
, elastic shear buckling of the web controls.
> 1.40
tw
Fyw
1.57 ⎛ Ek ⎞
1.57 ⎛ (29, 000ksi )(6.148) ⎞
=
⎟
⎟ = 0.3721
2 ⎜
2 ⎜
⎜
⎟
(50ksi )
⎛ D ⎞ ⎝ Fyw ⎠ ⎛ 69" ⎞ ⎝
⎠
⎜ ⎟
⎜ 9 "⎟
⎝ 16 ⎠
⎝ tw ⎠
Without TFA: Vn = CV p = (0.3721)(1,126 kip ) = 418.9 kip
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 54 of 62
 54 
60.
With TFA:
Since
2 Dtw
(2)(69")( 916 ")
=
= 1.056 ≤ 2.5 ,
( b fct fc + b ft t ft ) ( (21")(2 12 ") + (21")(1") )
⎡
⎤
⎡
⎤
⎢
⎥
⎢
⎥
0.87(1 − C ) ⎥
(0.87)(1 − 0.3721) ⎥
⎢
⎢
= (1,126kip ) ⎢0.3721 +
Vn = V p ⎢C +
⎥
2 ⎥
2
⎛ do ⎞ ⎥
⎛ 144" ⎞ ⎥
⎢
⎢
1+ ⎜
1+ ⎜ ⎟
⎟
⎢
⎥
⎢
⎝ 69" ⎠ ⎥
⎝D⎠ ⎦
⎣
⎦
⎣
Vn = (1,126kip )(0.6082) = 684.8kip
(
)
φVn = (1.00 ) 684.8kip = 684.8kip
O.K.
This TFA strength is adequate near the pier but TFA is not permitted in the end panels.
The following stiffener configuration should provide adequate shear strength.
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 55 of 62
 55 
61.
Strength Limit Shear Capacity
800
Tension Field Action
600
Strength I
400
Strength IV
Shear (kip)
200
0
200
400
600
800
0
30
60
90
120
150
180
210
240
270
300
330
Station (ft)
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 56 of 62
 56 
62.
12.4: Horizontal Shear Strength
Per ODOT Standard practice, shear studs will be used to transfer horizontal shear between the concrete
deck and top flange of the steel girder. ODOT prefers the use of 7/8”diameter studs.
Ideally, the studs should extend to the midthickness of the deck. Using this criterion, the height of the
studs can be determined.
be
ts
+ thaunch − t flange
2
9.5"
=
+ 2.75"− 0.75" = 6.75"
2
h=
ts
thaunch
tc
bc
Use 7/8” x 61/2” shear studs
D
tw
h
AASHTO requires that the ratio of /d be greater
than or equal to 4.0.
tt
bt
?
h
≥ 4.0
d
6 12 "
= 7.429 ≥ 4.0
7 "
8
OK
AASHTO requires a centertocenter transverse
spacing of 4d and a clear edge distance of 1”.
With 7/8” diameter studs, there is room enough transversely to use up to 4
studs in each row. With this in mind, I will investigate the option of
either 3 or 4 studs per row.
Fatigue Limit State:
The longitudinal pitch of the shear studs based on the Fatigue Limit is determined as
p≤
nZ r
Vsr
Vsr =
Vf Q
I
(6.10.10.1.21 & 3)
where:
n
Zr
Vsr
Vf
Q
I

Number of studs per row
Fatigue resistance of a single stud
Horizontal fatigue shear range per unit length
Vertical shear force under fatigue load combination
1st moment of inertia of the transformed slab about the shortterm NA
2nd moment of inertia of the shortterm composite section
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 57 of 62
 57 
63.
Zr = α d 2 ≥
5.5d 2
2
(6.10.10.21)
α = 34.5 − 4.28Log( N )
(6.10.10.22)
α = 34.5 − 4.28Log(55.84 ×106 ) = 1.343ksi
2
⎛ 5.5 ⎞ 7 2
Z r = (1.343ksi ) ( 7 8 ") ≥ ⎜
⎟ ( 8 ")
⎝ 2 ⎠
= 1.028kip ≥ 2.105kip
→
Z r = 2.105kip
Q = Atc d c
⎡ (109.5")( 9.5") ⎤ ⎛
9.5"
⎞
− 58.19" ⎟ = 2,511 in 3
QSection 1 = ⎢
⎥ ⎜ 1"+ 69"+ 2.75"+
8
2
⎠
⎣
⎦⎝
⎡ (109.5")( 9.5") ⎤ ⎛
9.5"
⎞
− 52.23" ⎟ = 3, 481 in 3
QSection 2 = ⎢
⎥ ⎜ 2.5"+ 69"+ 2.75"+
8
2
⎠
⎣
⎦⎝
I Section 1 = 140,500 in 4
I Section 2 = 239, 700 in 4
Since the fatigue shear varies along the length of the bridge, the longitudinal distribution of shear studs
based on the Fatigue Limit also varies. These results are presented in a tabular format on a subsequent
page. To illustrate the computations, I have chosen the shear at the abutment as an example.
(
)
At the abutment, V f = 38.13kip − −3.53kip = 41.66kip
( 41.66 )( 2,511 in ) = 0.7445
=
(140,500 in )
kip
Vsr
4
p≤
( 0.7445 )
kip
inch
= 8.482
kip
inch
For 4 studs in each row:
For 3 studs in each row:
( 3) ( 2.105ksi )
3
p≤
in
row
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
( 4 ) ( 2.105ksi )
( 0.7453 )
kip
inch
in
= 11.31 row
AASHTOLRFD 2007
Page 58 of 62
 58 
64.
Strength Limit:
Qr = φsc Qn
Qn = 0.5 Asc
φsc = 0.85
f c' Ec ≤ Asc Fu
(6.10.10.4.31)
2
⎛π ⎞
Asc = ⎜ ⎟ ( 7 8 ") = 0.6013 in 2
⎝4⎠
f c' = 4.5ksi
Since n = 8, Ec =
Es 29, 000ksi
=
= 3, 625ksi
8
n
Fu = 60ksi
Qn = ( 0.5 ) ( 0.6013 in 2 )
( 4.5 )( 3, 625 ) ≤ ( 0.6013 in )( 60 )
ksi
ksi
2
ksi
kip
kip
= 38.40 stud ≤ 36.08 stud
kip
kip
φscQn = ( 0.85 ) ( 36.08 stud ) = 30.67 stud
n+ =
Pp
n− =
Qr
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
Pp + Pn
Qr
AASHTOLRFD 2007
Page 59 of 62
 59 
65.
Positive moment  Section 1: Station 0.0’  73.3’
Pp = Min ( PConcrete , Psteel )
PConcrete = 0.85 f c'bets
= ( 0.85 ) ( 4.5ksi ) (109.5")( 9.5") = 3,979kip
PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc
= ( 70ksi ) ⎡(15")( 0.75") + ( 21")(1") ⎤ + ( 50ksi ) ( 69")( 0.5625") = 4,198kip
⎣
⎦
Pp = 3,979kip
n+ =
Pp
Qr
=
3,979kip
= 129.7studs
kip
30.67 stud
3 Studs per Row:
129.7studs
= 44rows
studs
3 row
→
p=
( 73.3'− 0 ') (12 in )
ft
= 20.46 inch
row
( 44 − 1)
→
Say 20"
→
p=
( 73.3'− 0 ') (12 in )
ft
= 27.49 inch
row
( 33 − 1)
→
Say 24"
4 Studs per Row:
129.7studs
= 33rows
studs
4 row
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 60 of 62
 60 
66.
Negative Moment  Section 2: Station 73.3’  165.0’
Pn = Min ( Psteel , PCrack )
PCrack = 0.45 f c'bets
= ( 0.45 ) ( 4.5ksi ) (109.5")( 9.5") = 2,107 kip
PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc
= ( 70ksi ) ⎡( 21")( 2.5") + ( 21")(1") ⎤ + ( 50ksi ) ( 69")( 0.5625") = 7, 086kip
⎣
⎦
Pn = 2,107 kip
n− =
Pp + Pn
Qr
=
3,979kip + 2,107 kip
= 198.4studs
kip
30.67 stud
3 Studs per Row:
198.4studs
= 67 rows
studs
3 row
→
p=
(165.0 '− 73.3') (12 in )
ft
= 16.67 inch
row
( 67 − 1)
→ Say 16"
→
p=
(165.0 '− 73.3') (12 in )
ft
= 22.48 inch
row
( 50 − 1)
→ Say 20"
4 Studs per Row:
198.4studs
= 50rows
studs
4 row
2 Span Continuous Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Page 61 of 62
 61 
68.
ONESPAN INELASTIC IGIRDER BRIDGE DESIGN EXAMPLE
1. PROBLEM STATEMENT AND ASSUMPTIONS:
A single span composite Igirder bridge has span length of 166.3’ and a 64’ deck width. The steel girders
have
Fy =
50ksi
and
all
concrete
has
a
28day
compressive
strength
of
ksi
f’c = 4.5 . The concrete slab is 9.5” thick. A typical 4” haunch was used in the section properties.
Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as
a composite dead load.
HL93 loading was used per AASHTO (2004), including dynamic load allowance.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 1 of 21
 63 
69.
172'  4" Total Girder Length
G
2
G
3
G
4
G
5
G
6
166'  4" cc Bearings
Cross Frames Spaced @ 22'  0" cc
G
1
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 2 of 21
 64 
70.
Positive Bending Section (Section 2)
Positive Bending Section (Section 1)
Positive Bending Section (Section 3)
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 3 of 23
 65 
71.
2. LOAD CALCULATIONS:
DC dead loads (structural components) include:
• Steel girder self weight (DC1)
• Concrete deck self weight (DC1)
• Haunch self weight (DC1)
• Barrier (DC2)
DW dead loads (structural attachments) include:
• Wearing surface (DW), Including FWS
2a. Dead Load Calculations
Steel Girder SelfWeight (DC1):
(a) Section 1
A = (14”)(1.125”) + (68”)(0.6875”) + (22”)(1.5”) = 95.5 in2
⎛
⎛ 490pcf ⎞ ⎞
⎟ ⎟ (1.15 ) = 373.7 lbs per girder
Wsection1 = ⎜ 95.5 in 2 ⎜
ft
⎜ (12 in )2 ⎟ ⎟
⎜
ft
⎝
⎠⎠
⎝
(b) Section 2
A = (14”)(2”) + (68”)(0.5625”) + (22”)(2”) = 110.25 in2
⎛
⎛ 490pcf ⎞ ⎞
⎟ ⎟ (1.15 ) = 431.4 lbs per girder
Wsection1 = ⎜110.3 in 2 ⎜
ft
⎜ (12 in )2 ⎟ ⎟
⎜
ft
⎝
⎠⎠
⎝
(c) Section 3
A = (14”)(2”) + (68”)(0.5625”) + (22”)(2.375”) = 118.5 in2
⎛
⎛ 490pcf ⎞ ⎞
⎟ ⎟ (1.15 ) = 463.7 lbs per girder
Wsection1 = ⎜118.5 in 2 ⎜
ft
⎜ (12 in )2 ⎟ ⎟
⎜
ft
⎝
⎠⎠
⎝
(d) Average Girder Self Weight
Wave =
( 2 )( 40.17 ') ( 373.7 lbs ) + ( 2 )(18.0 ') ( 431.4 lbs ) + ( 50.0 ') ( 463.7 lbs )
ft
ft
ft
166.3'
= 413.3 lbs
ft
Deck SelfWeight (DC1):
⎛ ( 9.5'')( 64.0 ') ⎞ ⎛ 150pcf ⎞
WDeck = ⎜
⎟ = 1,267 lbs per girder
⎟⎜
ft
6 Girders ⎠ ⎜ (12 in ) ⎟
⎝
ft ⎠
⎝
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 4 of 23
 66 
72.
Haunch SelfWeight (DC1):
Average width of haunch: 14’’
⎛ 150pcf ⎞
⎟ = 94.33 lbs per girder
Whaunch = (14 )( 4 ) + 2 ( ( 1 ) ( 9 '')( 4 '') ) ⎜
2
ft
⎜ (12 in )2 ⎟
ft
⎝
⎠
(
)
Barrier Walls (DC2):
Wbarriers
⎛ ( 2 each ) ( 640plf ) ⎞
⎟ = 213.3 lbs per girder
=⎜
ft
⎜
⎟
6 girders
⎝
⎠
Wearing Surface (DW):
Wwearing_surface =
( 61.0') ( 60psf )
6 Girders
= 610.0 lbs per girder
ft
The moment effect due to dead loads was found using an FE model composed of six frame elements to
model the bridge (a node was placed at midspan). This data was input into Excel to be combined with
data from moving live load analyses performed in SAP 2000. DC1 dead loads were applied to the noncomposite section (bare steel). All live loads were applied to the shortterm composite section (1n = 8).
DW (barriers) and DC2 (wearing surface) dead loads were applied to the longterm composite section
(3n = 24).
The maximum moments at midspan are easily computed since the bridge is statically determinate.
M DC1,Steel
2
⎛ wL2 ⎞ ⎡ ( 413.3 lbs ) (166.3') ⎤
ft
⎥ = 1, 429kft
=⎜
⎟=⎢
8 ⎠ ⎢
8
⎥
⎝
⎣
⎦
M DC1, Deck
2
⎛ wL2 ⎞ ⎡ (1, 267 lbs ) (166.3' ) ⎤
ft
⎥ = 4,379kft
=⎜
⎟=⎢
8
⎥
⎝ 8 ⎠ ⎢
⎣
⎦
M DC 2, Barriers
M DW
2
⎛ wL2 ⎞ ⎡ ( 213.3 lbs ) (166.3' ) ⎤
ft
⎥ = 737.4kft
=⎜
⎟=⎢
8
⎥
⎝ 8 ⎠ ⎢
⎣
⎦
2
⎛ wL2 ⎞ ⎡ ( 610.0 lbs ) (166.3' ) ⎤
ft
⎥ = 2,109kft
=⎜
⎟=⎢
8 ⎠ ⎢
8
⎥
⎝
⎣
⎦
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 5 of 23
 67 
73.
The maximum shear forces at the ends of the girder are also easily computed.
lbs
⎛ wL ⎞ ⎡ ( 413.3 ft ) (166.3' ) ⎤
kip
VDC1,Steel = ⎜
=⎢
⎥ = 34.37
⎟
2 ⎠ ⎢
2
⎝
⎥
⎣
⎦
VDC1, Deck
lbs
⎛ wL ⎞ ⎡ (1, 267 ft ) (166.3' ) ⎤
kipt
=⎜
⎥ = 105.4
⎟=⎢
2
⎝ 2 ⎠ ⎢
⎥
⎣
⎦
lbs
⎛ wL ⎞ ⎡ ( 213.3 ft ) (166.3' ) ⎤
kip
VDC 2, Barriers = ⎜
=⎢
⎥ = 17.74
⎟
2
⎝ 2 ⎠ ⎢
⎥
⎣
⎦
lbs
⎛ wL ⎞ ⎡ ( 610.0 ft ) (166.3' ) ⎤
kip
VDW = ⎜
=⎢
⎥ = 50.72
⎟
2 ⎠ ⎢
2
⎝
⎥
⎣
⎦
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 6 of 23
 68 
74.
2b. Live Load Calculations
The following design vehicular live load cases described in AASHTOLRFD are considered:
1) The effect of a design tandem combined with the effect of the lane loading. The design
tandem consists of two 25kip axles spaced 4.0’ apart. The lane loading consists of a 0.64klf
uniform load on all spans of the bridge. (HL93M in SAP)
2) The effect of one design truck with variable axle spacing combined with the effect of the
0.64klf lane loading. (HL93K in SAP)
3) For negative moment between points of contraflexure only: 90% of the effect of a trucktrain
combined with 90% of the effect of the lane loading. The truck train consists of two design
trucks (shown below) spaced a minimum of 50’ between the lead axle of one truck and the rear
axle of the other truck. The distance between the two 32kip axles should be taken as 14’ for each
truck. The points of contraflexure were taken as the field splices at 132’ and 198’ from the left
end of the bridge. (HL93S in SAP)
All live load calculations were performed in SAP 2000 using a beam line analysis. The nominal
moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the
truck and tandem loads within SAP.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 7 of 23
 69 
75.
Unfactored HL93 Moment Envelopes from SAP
6,000
Single Truck
4,000
Tandem
Moment (kipft)
2,000
0
2,000
4,000
6,000
0
30
60
90
120
150
Station (ft)
The following results were obtained from the SAP analysis:
•
The maximum positive liveload moments occur at stations 83.15’
HL93M
HL93K
HL93S
Station 40.16’ Section 1
3,614kft
4,322kft
N/A
Station 58.15’ Section 2
4,481kft
5,238kft
N/A
Station 83.15’ Section 3
4,911kft
5,821kft
N/A
Before proceeding, these liveload moments will be confirmed with an influence line analysis.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 8 of 23
 70 
76.
2c. Verify the Maximum Positive LiveLoad Moment at Station 83.15’:
kip
kip
25
25
Tandem:
kip
32
8
kip
32
kip
:
Single Truck
0.640kip/ft
Moment (kft / kip)
Lane
45
40
35
30
25
20
15
10
5
0
0
15
30
45
60
75
90
105
120
135
150
165
Station (ft)
Tandem:
Single Truck:
Lane Load:
( 25 ) ( 41.58 ) + ( 25 ) ( 39.58 ) = 2, 029
(8 ) ( 34.57 ) + ( 32 ) ( 41.58 ) + ( 32 ) ( 34.57 ) = 2, 713
kip
kip
kft
kip
kip
kip
kft
kip
kft
kip
( 0.640 )(3, 457 ) = 2, 212
kft
kip
kft
kip
kft
kft
kip
kip
kft
kip
kft
kft
(IM)(Tandem) + Lane:
(1.33 ) ( 2, 029 kft ) + 2, 212kft = 4, 911kft
kip
(IM)(Single Truck) + Lane:
(1.33 ) ( 2, 713 kft ) + 2, 212 kft = 5,821kft
kip
GOVERNS
The case of two trucks is not considered here because it is only used when computing negative moments.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 9 of 23
 71 
77.
Based on the influence line analysis, we can say that the moments obtained from SAP appear to be
reasonable and will be used for design.
Before these Service moments can be factored and combined, we must compute the distribution factors.
Since the distribution factors are a function of Kg, the longitudinal stiffness parameter, we must first
compute the sections properties of the girders.
3. SECTION PROPERTIES AND CALCULATIONS:
3a. Effective Flange Width, bs:
For an interior beam, bs is the lesser of:
bf
⎧
14"
= (12 )( 8.5") +
= 109 ''
⎪•12ts +
2
2
⎪
⎪
⎨•S = (11.33')(12 in ft ) = 135.96 ''
⎪ L
⎪• eff = 166.3' = 41.58' = 498.9 ''
⎪ 4
4
⎩
Therefore, bs = 109”
For computing the section properties shown on the two pages that follow, reinforcing steel in the deck
was ignored for shortterm and longterm composite calculations but was included for the cracked
section.
Note: At this point one should also check the effective of the outside girders as well. For this
example, however, I will proceed sing the effective width for the interior girders.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 10 of 23
 72 
78.
3b. Section 1 Flexural Properties
Single Span Bridge Example  Section 1
Bare Steel
t
b
A
y
Ay
d
Ix
Ad2
IX
Top Flange
1.1250
14.00
15.75
70.06
1,103.48
1.66
40.87
26,308
26,310
Web
Bot Flange
0.6875
1.5000
68.00
22.00
46.75
33.00
35.50
0.75
1,659.63 18,014.33
24.75
6.19
6.31
28.44
1,860
26,696
19,874
26,702
2,787.86
ITotal =
72,886
29.19
SBS1,top =
1,759
SBS1,bot =
2,497
95.50
Y=
ShortTerm Composite (N=8)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
115.81
74.88
8,671.46
697.29
20.65
49,365
50,062
Haunch
Top Flange
0.0000
1.1250
14.0000
14.0000
0.00
15.75
70.63
70.06
0.00
1,103.48
0.00
1.66
16.40
15.83
0
3,948
0
3,950
Web
Bot Flange
0.6875
1.5000
68.0000
22.0000
46.75
33.00
35.50
0.75
1,659.63 18,014.33
24.75
6.19
18.73
53.48
16,399
94,381
34,414
94,387
11,459.32
ITotal =
182,813
54.23
SST1,top =
11,150
SST1,bot =
3,371
211.31
n=
8.00
Y=
LongTerm Composite (N=24)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
38.60
74.88
2,890.49
232.43
32.53
40,856
41,089
Haunch
0.0000
14.00
0.00
70.63
0.00
0.00
28.28
0
0
Top Flange
1.1250
14.0000
15.75
70.06
1,103.48
1.66
27.72
12,102
12,104
Web
Bot Flange
0.6875
1.5000
68.0000
22.0000
46.75
33.00
35.50
0.75
1,659.63 18,014.33
24.75
6.19
6.84
41.59
2,189
57,089
20,203
57,095
5,678.35
ITotal =
130,491
42.34
SLT1,top =
4,614
SLT1,bot =
3,082
134.10
n=
24.00
Y=
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 11 of 23
 73 
79.
3c. Section 2 Flexural Properties
Single Span Bridge Example  Section 2
Bare Steel
t
b
A
y
Ay
d
Ix
Ad2
IX
Top Flange
2.0000
14.00
28.00
71.00
1,988.00
9.33
40.08
44,978
44,987
Web
Bot Flange
0.5625
2.0000
68.00
22.00
38.25
44.00
36.00
1.00
1,377.00 14,739.00
44.00
14.67
5.08
29.92
987
39,391
15,726
39,405
3,409.00
ITotal =
100,119
30.92
SBS1,top =
2,437
SBS1,bot =
3,238
110.25
Y=
ShortTerm Composite (N=8)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
115.81
76.25
8,830.70
697.29
22.11
56,600
57,297
Haunch
Top Flange
0.0000
2.0000
14.0000
14.0000
0.00
28.00
72.00
71.00
0.00
1,988.00
0.00
9.33
17.86
16.86
0
7,956
0
7,966
Web
Bot Flange
0.5625
2.0000
68.0000
22.0000
38.25
44.00
36.00
1.00
1,377.00 14,739.00
44.00
14.67
18.14
53.14
12,591
124,264
27,330
124,279
12,239.70
ITotal =
216,871
54.14
SST1,top =
12,145
SST1,bot =
4,006
226.06
n=
8.00
Y=
LongTerm Composite (N=24)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
38.60
76.25
2,943.57
232.43
33.57
43,514
43,746
Haunch
0.0000
14.00
0.00
72.00
0.00
0.00
29.32
0
0
Top Flange
2.0000
14.0000
28.00
71.00
1,988.00
9.33
28.32
22,462
22,472
Web
Bot Flange
0.5625
2.0000
68.0000
22.0000
38.25
44.00
36.00
1.00
1,377.00 14,739.00
44.00
14.67
6.68
41.68
1,705
76,425
16,444
76,439
6,352.57
ITotal =
159,101
42.68
SLT1,top =
5,426
SLT1,bot =
3,728
148.85
n=
24.00
Y=
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 12 of 23
 74 
80.
3d. Section 3 Flexural Properties
Single Span Bridge Example  Section 3
Bare Steel
t
b
A
y
Ay
d
Ix
Ad2
IX
Top Flange
2.0000
14.00
28.00
71.38
1,998.50
9.33
42.25
49,970
49,980
Web
Bot Flange
0.5625
2.3750
68.00
22.00
38.25
52.25
36.38
1.19
1,391.34 14,739.00
62.05
24.56
7.25
27.94
2,008
40,796
16,747
40,820
3,451.89
ITotal =
107,546
29.13
SBS1,top =
2,487
SBS1,bot =
3,692
118.50
Y=
ShortTerm Composite (N=8)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
115.81
76.63
8,874.13
697.29
24.02
66,819
67,516
Haunch
Top Flange
0.0000
2.0000
14.0000
14.0000
0.00
28.00
72.38
71.38
0.00
1,998.50
0.00
9.33
19.77
18.77
0
9,865
0
9,874
Web
Bot Flange
0.5625
2.3750
68.0000
22.0000
38.25
52.25
36.38
1.19
1,391.34 14,739.00
62.05
24.56
16.23
51.42
10,076
138,137
24,815
138,161
12,326.02
ITotal =
240,366
52.61
SST1,top =
12,158
SST1,bot =
4,569
234.31
n=
8.00
Y=
LongTerm Composite (N=24)
t
b
A
y
Ay
d
Ix
Ad2
IX
Slab
8.5000
109.00
38.60
76.63
2,958.04
232.43
35.82
49,544
49,777
Haunch
0.0000
14.00
0.00
72.38
0.00
0.00
31.57
0
0
Top Flange
2.0000
14.0000
28.00
71.38
1,998.50
9.33
30.57
26,174
26,184
Web
Bot Flange
0.5625
2.3750
68.0000
22.0000
38.25
52.25
36.38
1.19
1,391.34 14,739.00
62.05
24.56
4.43
39.61
749
81,990
15,488
82,015
6,409.93
ITotal =
173,463
40.80
SLT1,top =
5,494
SLT1,bot =
4,251
157.10
n=
24.00
Y=
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 13 of 23
 75 
81.
4. DISTRIBUTION FACTOR FOR MOMENT
4a. Section 1:
Interior Girder  One Lane Loaded:
DFM1,Int,Sec1
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
= 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠
0.1
2
K g = n ( I + Aeg )
(
K g = (8) 72,890 in 4 + ( 95.5 in 2 ) ( 49.06")
2
)
K g = 2, 422, 000 in 4
0.4
DFM1,Int,Sec1
⎛ 11.33' ⎞ ⎛ 11.33' ⎞
= 0.06 + ⎜
⎟ ⎜
⎟
⎝ 14 ⎠ ⎝ 166.3' ⎠
0.3
⎛ 2, 422, 000 in 4 ⎞
⎜
⎟
⎜ (12 )(166.3' )( 8.5")3 ⎟
⎝
⎠
0.1
DFM1,Int,Sec1 = 0.4994
Interior Girder  Two or More Lanes Loaded:
DFM2,Int,Sec1
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
= 0.075 + ⎜
⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠
DFM2,Int,Sec1
⎛ 11.33' ⎞ ⎛ 11.33' ⎞
= 0.075 + ⎜
⎟ ⎜
⎟
⎝ 9.5 ⎠ ⎝ 166.3' ⎠
0.6
0.2
0.1
⎛ 2, 422, 000 in 4 ⎞
⎜
⎟
⎜ (12 )(166.3')( 8.5")3 ⎟
⎝
⎠
0.1
DFM2,Int,Sec1 = 0.7703
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 14 of 23
 76 
82.
Exterior Girder – One Lane Loaded:
The lever rule is applied by assuming that a
hinge forms over the first interior girder as a
truck load is applied near the parapet. The
resulting reaction in the exterior girder is the
distribution factor.
DFM 1,Ext,Sec1 =
8.5'
= 0.7500
11.33'
Multiple Presence:
DFM1,Ext,Sec1 = (1.2) (0.7500) = 0.9000
Exterior Girder  Two or More Lanes Loaded:
DFM2,Ext,Sec1 = e DFM2,Int,Sec1
de
9.1
2.167 '
e = 0.77 +
= 1.008
9.1
e = 0.77 +
DFM2,Ext+ = (1.008) (0.7703) = 0.7765
4b. Section 2:
Interior Girder – One Lane Loaded:
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
DFM1,Int,Sec2 = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠
0.1
2
K g = n ( I + Aeg )
(
K g = (8) 100,100 in 4 + (110.3 in 2 ) ( 47.83")
2
)
K g = 2,819, 000 in 4
0.4
DFM1,Int,Sec2
⎛ 11.33' ⎞ ⎛ 11.33' ⎞
= 0.06 + ⎜
⎟ ⎜
⎟
⎝ 14 ⎠ ⎝ 166.3' ⎠
0.3
⎛ 2,819, 000 in 4 ⎞
⎜
⎟
⎜ (12 )(166.3' ) ( 8.5 ")3 ⎟
⎝
⎠
0.1
DFM1,Int,Sec2 = 0.5061
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 15 of 23
 77 
83.
Interior Girder – Two or More Lanes Loaded:
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
DFM2,Int,Sec2 = 0.075 + ⎜
⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠
DFM2,Int,Sec2
DFM2,Int,Sec2
0.1
0.6
0.2
4
⎞
⎛ 11.33' ⎞ ⎛ 11.33' ⎞ ⎛ 2,819, 000 in
= 0.075 + ⎜
⎜
⎟
⎟ ⎜
⎟ ⎜
3
⎝ 9.5 ⎠ ⎝ 166.3' ⎠ ⎝ (12 )(166.3')( 8.5") ⎟
⎠
= 0.7809
0.1
Exterior Girder  One Lane Loaded:
Same as for the positive moment section: DFM1,Ext,Sec2 = 0.9000
Exterior Girder  Two or More Lanes Loaded:
DFM2,Ext,Sec2 = e DFM2,Int,Sec2
e = 1.008 (same as before)
DFM2,Ext,Sec2 =(1.008) (0.7809) = 0.7871
4c. Section 3:
Interior Girder – One Lane Loaded:
DFM 1, Int ,Sec 3
(
⎛S ⎞
= 0.06 + ⎜ ⎟
⎝ 14 ⎠
2
K g = n I + Aeg
(
0.4
⎛S⎞
⎜ ⎟
⎝L⎠
0.3
)
(
⎛ Kg ⎞
⎜
3 ⎟
⎝ 12 LtS ⎠
)
0.1
K g = ( 8 ) 107,500 in 4 + 118.5 in 2 ( 50.00")
2
)
K g = 3, 230, 000 in 4
DFM 1, Int , Sec 3
⎛ 11.33' ⎞
= 0.06 + ⎜
⎟
⎝ 14 ⎠
0.4
⎛ 11.33' ⎞
⎜
⎟
⎝ 166.3' ⎠
0.3 ⎛
⎞
⎜
⎟
3 ⎟
⎜ (12 )(166.3')( 8.5")
⎝
⎠
3, 230, 000 in 4
0.1
DFM 1, Int ,Sec 3 = 0.5122
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 16 of 23
 78 
84.
Interior Girder  Two or More Lanes Loaded:
0.4
DFM 2, Int ,Sec 3
⎛ S ⎞
= 0.075 + ⎜
⎟
⎝ 9.5 ⎠
⎛S⎞
⎜ ⎟
⎝L⎠
DFM 2, Int ,Sec 3
⎛ 11.33' ⎞
= 0.075 + ⎜
⎟
⎝ 9.5 ⎠
0.6
0.3
⎛ Kg ⎞
⎜
3 ⎟
⎝ 12 LtS ⎠
⎛ 11.33' ⎞
⎜
⎟
⎝ 166.3' ⎠
0.2
0.1
⎛ 3, 230, 000 in 4 ⎞
⎜
⎟
⎜ (12 )(166.3')( 8.5")3 ⎟
⎝
⎠
0.1
DFM 2, Int ,Sec 3 = 0.7906
Exterior Girder – One Lane Loaded:
Same as for the positive moment section: DFM1,Ext,Sec3 = 0.9000
Exterior Girder  Two or More Lanes Loaded:
DFM2,Ext,Sec3 = e DFM2,Int,Sec3
e = 1.008 (same as before)
DFM2,Ext,Sec3 =(1.008) (0.7906) = 0.7969
NL
DF
Ext , Min
4d. Minimum Exterior Girder Distribution Factor:
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
=
NL
Nb
+
X Ext ∑ e
Nb
∑x
2
AASHTOLRFD 2007
Created July2007: Page 17 of 23
 79 
87.
5. FACTORED MOMENT ENVELOPES
The following load combinations were considered in this example:
Strength I:
Strength IV:
1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW
1.50DC1 + 1.50DC2 + 1.50DW
Service II:
1.3(LL + IM) + 1.0DC1 + 1.0DC2 + 1.0DW
Fatigue:
0.75(LL + IM)
(IM for Fatigue = 15%)
Strength II is not considered since this deals with special permit loads. Strength III and V are not
considered as they include wind effects, which will be handled separately as needed. Strength IV is
considered but is not expected to govern since it addresses situations with high dead load that come into
play for longer spans. Extreme Event load combinations are not included as they are also beyond the
scope of this example. Service I again applies to wind loads and is not considered and Service III and
Service IV correspond to tension in prestressed concrete elements and are therefore not included in this
example.
In addition to the factors shown above, a load modifier, η, was applied as is shown below.
Q = ∑ηiγ i Qi
η is taken as the product of ηD, ηR, and ηI, and is taken as not less than 0.95. For this example,
ηD, ηR, and ηI are taken as 1.00.
Using these load combinations, the shear and moment envelopes shown on the following pages
were developed.
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 20 of 23
 82 
90.
Strength Limit Moment Envelopes
25,000
Max (@ 83.14') = 21289kft
Strength I
Moment (kipft)
20,000
15,000
Strength IV
10,000
5,000
0
0
30
60
90
120
150
180
Station (ft)
Service II Moment Envelope
17,500
Max (@ 83.14') = 16,010
kft
15,000
Moment (kipft)
12,500
10,000
7,500
5,000
2,500
0
0
30
60
90
120
150
180
Station (ft)
SingleSpan Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July2007: Page 23 of 23
 85 
92.
SINGLESPAN TRUSS BRIDGE DESIGN EXAMPLE
1. PROBLEM STATEMENT AND ASSUMPTIONS:
Consider the truss bridge shown in Figure 1 below. The truss is simply supported with a span
length of 112’–0” and a width (cc of the trusses) of 19’–6”. The truss is made up of 7 panels
that are each 16’0” in length. Floor beams span between the truss panel points perpendicular to
traffic and support stringers that span 16’0” in the direction of traffic. Finally, the
noncomposite W10 x 88 stringers support a 6” thick reinforced concrete deck. The simply
supported stringers (6 across in each panel) are spaced at 3’  6” laterally.
1) Determine maximum and minimum axial forces in members 12, 14, 911, 910, and 1013
due to an HL93 Loading.
2) Determine the maximum moment in the stringer members due to the HL93 Loading
The entire truss superstructure is made up of W14 x 109 members except for the bottom chord,
which is made up of MC 12 x 35 members.
You may assume that the trucks drive down the center of the bridge (they really do, by the way)
and as a result, the truck loads are approximately equally distributed between the trusses. To be
on the safe side, however, assume that each truss carries 75% of the single lane. Model the truss
as a determinate structure with pinned joints even though the actual truss has very few joints that
are truly pinned. You may use a computer program for your truss analysis if you wish. I would
suggest that you use SAP2000, Visual Analysis, or another similar FE package to model the
truss.
Disregard the lower limit of L = 20’ on the span length for computing distribution factors for the
stringer members. Think about what is appropriate for the multiple presence factor.
Figure 1  Tyler Road Bridge, Delaware County, OH
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 1 of 17
 87 
93.
Figure 2  Truss Layout
6" Thick Reinforced Concrete Deck
6, W10 x 88 Stringers @ 3'6" cc
18'  0" Clear Roadway
19'  6" cc Trusses
Figure 3  Truss Cross Section
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 2 of 17
 88 
94.
Compute the Maximum and Minimum Forces in Critical Members of the Truss:
The following Influence Lines were obtained from SAP 2000:
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 3 of 17
 89 
95.
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 4 of 17
 90 
96.
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 5 of 17
 91 
100.
Consider Member 910 of the Truss:
Member 910 of the truss is a zero force member. It may see some bending moment due to its rigid
connection to the floor beam but it will not experience a net axial force due to live load.
P910 = 0.000kip
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 9 of 17
 95 
101.
Consider Member 1013 of the Truss:
Tandem:
⎡ (25kip ) + (25kip ) ⎤ (1.972 kip kip ) = 98.60kip
⎣
⎦
⎡
⎛ 48'− 12' ⎞
kip
kip ⎤
kip
kip
Truck: ⎢ (8kip ) ⎜
⎟ + (32 ) + (32 ) ⎥ (1.972 kip ) = 138.0
⎝ 48' ⎠
⎣
⎦
Lane:
( 0.640 kip ft )(1.972 kip kip ) ⎡( 12 ) (96') + (16') ⎤ = 80.77kip
⎣
⎦
Combining the HL93 Components with impact applied appropriately:
(IM)(Tandem) + Lane:
(1.33) ( 98.60kip ) + (80.77kip ) = 211.9kip
(IM)(Truck) + Lane:
(1.33) (138.0kip ) + (80.77kip ) = 264.3kip
GOVERNS
Apply the Truss Distribution Factor:
Assume that each truss carries 75% of the HL93 load effect
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
P1013 = 198.2kip
AASHTOLRFD 2007
Created July 2007: Page 10 of 17
 96 
104.
Member Force Summary:
Member
12
14
911
910
1013
1011
Max Tension
0.000kip
98.12kip
0.000kip
0.000kip
198.2kip
36.56kip
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
Max Compression
123.2kip
0.000kip
201.5kip
0.000kip
0.000kip
36.53kip
AASHTOLRFD 2007
Created July 2007: Page 13 of 17
 99 
105.
Compute the Moment Distribution Factor for the Stringers in the Floor System:
Interior Girder –
One Lane Loaded:
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
= 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 Lt s ⎠
0.4
DFM 1, Int
0.3
0.1
2
K g = n( I + Aeg )
K g = (8)(534 in 4 + (25.9 in 2 )(8.40") 2 )
K g =18,890 in 4
4
⎛ 3.5' ⎞ ⎛ 3.5' ⎞ ⎛ 18,890 in ⎞
= 0.06 + ⎜
⎟ ⎜
⎟ ⎜
3 ⎟
⎝ 14 ⎠ ⎝ 16 ' ⎠ ⎝ 12(16 ')(6.0") ⎠
0.4
DFM 1, Int
0.3
0.1
DFM 1, Int = 0.3965
Two or More Lanes Loaded:
The bridge is designed for a single traffic lane.
Exterior Girder –
One Lane Loaded:
The lever rule is applied by assuming that a
hinge forms over the first interior stringer as
a truck load is applied near the guard rail.
The resulting reaction in the exterior stringer
is the distribution factor.
R=
( P / 2 ) (1.75') = 0.2500 P
(3.50 ')
DFM 1, Ext = 0.2500
The Multiple Presence Factor would
generally be applied but in this case, there is
only a single design lane so it is not used.
Two or More Lanes Loaded:
The bridge is designed for a single traffic lane.
Minimum Exterior Girder Distribution Factor:
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 14 of 17
 100 
106.
NL
DF
Ext , Min
=
NL
Nb
+
X Ext ∑ e
Nb
∑x
2
One Lane Loaded:
4'0"
2'0"
3'0"
3'0"
P/2
P/2
1'9"
5'3"
8'9"
DF
M 1, Ext , Min
=
1
6
+
(4.00 ')(8.75')
(2) ⎡ (8.75') 2 + (5.25') 2 + (1.75') 2 ⎤
⎣
⎦
= 0.3299
The Multiple Presence Factor would generally be applied but in this case, there is only a
single design lane so it is not used.
Moment Distribution Factor Summary:
Interior Stringer:
Exterior Stringer (Lever Rule):
Exterior Stringer (Minimum):
DFM1,Int = 0.3965
DFM1, Ext = 0.2500
DFM1, Ext = 0.3299
For simplicity, take the moment distribution factor as 0.3965 for all stringers.
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 15 of 17
 101 
107.
Compute the Maximum Bending Moment in the Stringers of the Floor System:
25kip
25kip
Tandem:
32kip
Truck:
0.640kip/ft
Lane:
4.00kft/kip
IL Moment
@ CL Stringer
4 spaces @ 4'0" = 16'0"
Tandem:
⎡ kip
kip ⎛ 8'− 4' ⎞ ⎤
kft
kft
⎢ (25 ) + (25 ) ⎜ 8' ⎟ ⎥ ( 4.00 kip ) = 150.0
⎝
⎠⎦
⎣
Truck: (32kip ) ( 4.00 kft kip ) = 128.0kft
Lane:
( 0.640 kip ft ) ( 12 ) ( 4.00 kft kip ) (16') = 20.48kft
In this case, since the axle spacing is substantial relative to the beam length, we should consider the more
general approach for computing maximum moment. For two equal point loads, P, separated by a
distance, a, the maximum bending moment in a simply supported span is:
when a < 0.5858L,
2
P ⎛
a⎞
M Max =
L− ⎟
⎜
2L ⎝
2⎠
when a ≥ 0.5858L
M Max =
PL
4
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
AASHTOLRFD 2007
Created July 2007: Page 16 of 17
 102 
108.
2
Tandem:
M Max
(25kip ) ⎛
4' ⎞
kft
=
(an increase of 2.067%)
⎜ 16'− ⎟ = 153.1
(2)(16') ⎝
2⎠
Truck: M Max =
(32kip )(16')
= 128.0kft
(4)
(no change)
Combining the HL93 Components with impact applied appropriately:
(IM)(Tandem) + Lane:
(1.33) (153.1kft ) + ( 20.48kft ) = 224.1kft
(IM)(Truck) + Lane:
(1.33) (128.0kft ) + ( 20.48kft ) = 190.7 kft
GOVERNS
Apply the Stringer Distribution Factor:
Each stringer carries 0.3965 lanes of the HL93 loading
SingleSpan Truss Bridge Example
ODOT LRFD Short Course  Steel
MStringer = 88.86kft
AASHTOLRFD 2007
Created July 2007: Page 17 of 17
 103 
110.
AASHTO Tension Member Example #1:
Problem:
A tension member is made up from a bar of M27050 material that is 6” wide and 1” thick. It is
bolted at its ends by six, 7/8” diameter bolts arranged in two staggered rows as is shown below.
If the governing factored load, Pu, is 200kip, determine whether or not the member is adequate.
The member is 4’0” long.
Solution:
1.5"
1.5"
1"
3"
1.5"
Check Minimum Slenderness Ratio:
3"
I min
bt 3
t2
=
=
= 0.2887"
12bt
12
A
3"
rmin =
3"
3"
(4 '− 0")(12 in )
L
ft
=
= 166.2
0.2887"
rmin
3"
Since 140 < L / rmin <200, the slenderness is
ok so long as the member is not subjected to
stress reversals.
Compute the Design Strength:
Gross Section Yielding:
Pn = Fy Ag = (50ksi)(6”)(1”) = 300.0kip
φPn = (0.95)(300.0kip) = 285.0kip
Net Section Fracture:
U is 1.00 here because the section is
composed of a single element that is
connected.
Therefore the load is
“transmitted directly to each of the
elements within the cross section.”
Pn = Fu Ae = Fu U An
⎡
⎛ (1.5") 2 ⎞ ⎤
An = ⎢6"− (2) ( 7 8 "+ 1 8 ") + ⎜
⎟ ⎥ (1")
⎝ 4(3.0") ⎠ ⎦
⎣
= 4.188 in 2
Pn = (65ksi)(1.00)(4.188 in2) = 272.2kip
NSF Governs  φPn = 218kip
φPn = (0.80)(272.2kip) = 217.8kip
Since Pu < φPn (200kip < 218kip)
the member is adequate.
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #1
AASHTOLRFD 2007
Created July 2007: Page 1 of 1
 105 
111.
AASHTO Tension Member Example #2:
Problem:
A C12x30 is used as a tension member (L = 8’6”) as is shown in the sketch below. The channel
is made of M27036 material and is attached to the gusset plate with 7/8” diameter bolts.
Calculate the design tensile capacity, φPn, of the member considering the failure modes of gross
section yielding and net section fracture.
Solution:
Section AA
Check Minimum Slenderness Ratio:
rmin = 0.762” (from the AISC Manual)
(8.5')(12 in )
L
ft
=
= 133.9
0.762"
rmin
3"
6"
3"
Since L/rmin < 140, the slenderness is ok.
Compute the Design Strength:
Gross Section Yielding:
Pn = Fy Ag = (36ksi)(8.81 in2) = 317.2kip
A
A
φPn = (0.95)(317.2kip) = 301.3kip
C12 x 30
Net Section Fracture:
Pu
Pn = Fu Ae = Fu U An
An = ( 8.81 in 2 ) − (2) ( 7 8 "+ 18 ")( 0.510") = 7.790 in 2
U = 0.85 since there are ≥ 3 fasteners in the direction of stress
Pn = (58ksi)(0.85)(7.790 in2) = 384.0kip
φPn = (0.80)(384.0kip) = 307.2kip
Gross Section Yielding Governs  φPn = 301kip
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #2
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 106 
112.
Side Note:
Note that if the AISC shear lag provisions were used that Case 2 from AISC Table D3.1
would apply:
U = 1−
x
0.674"
= 1−
= 0.9251
9.00"
L
…. for net section fracture, φPn = 334.3kip
In this case, however, the design strength is unaffected since gross yielding governs.
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #2
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 107 
113.
AASHTO Tension Member Example #3:
Problem:
Determine the design strength of the W10x60 member of M27050 steel. As is shown, the
member is connected to two gusset plates – one on each flange. The end connection has two lines
of 3/4” diameter bolts in each flange  five in each line.
A
Gusset Plates
W10 x 60
Section AA
A
5 spaces @ 3”
Solution:
Check Minimum Slenderness Ratio:
rmin = 2.57” (from the AISC Manual)
L
rmin
=
L
≤ 140
2.57"
this is satisfied so long as L ≤ 359.8” = 29’113/4”
Compute the Design Strength:
Gross Section Yielding:
Pn = Fy Ag = (50ksi)(17.6 in2) = 880.0kip
φPn = (0.95)(880.0kip) = 836.0kip
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #3
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 108 
114.
Net Section Fracture:
Pn = Fu Ae = Fu U An
An = (17.6 in 2 ) − (4) ( 3 4 "+ 1 8 ")( 0.680") = 15.22 in 2
?
Check b f ≥ 2 3 d
?
(10.1") ≥ ( 2 3 )(10.2")
…
OK
U = 0.90 since bf > 2/3d and there are ≥ 3 fasteners in the direction of stress.
Pn = (65ksi)(0.90)(15.22 in2) = 890.4kip
φPn = (0.80)(890.4kip) = 712.3kip
Net Section Fracture Governs  φPn =712kip
Side Note:
Note that if the AISC shear lag provisions were used that Case 7a from AISC Table D3.1
would apply:
x
x
?
Check
bf ≥ 2 3 d
?
(10.1") ≥ ( 2 3 )(10.2")
OK
∴ U = 0.90
Alternatively, Table D3.1 Case 2 can be applied:
U = 1−
x
0.884"
= 1−
= 0.9263
L
12.0"
The value of U = 0.9263 can be used.
Pn = (65ksi)(0.9263)(15.22 in2) = 916.4kip
The connection eccentricity x is
taken as the distance from the
faying surface to the CG of a
WT5x30.
φPn = (0.80)(916.4kip) = 733.1kip
Since Net Section Fracture governs the capacity of this member, the overall design
strength of the member would be increased to 733kip.
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #3
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 109 
115.
AASHTO Tension Member Example #4:
Problem:
An L6x4x1/2, M27036, is welded to a gusset plate. The long leg of the angle is attached using
two, 8” long fillet welds. Compute the strength of the angle in tension.
Solution:
Check Minimum Slenderness Ratio:
rmin = rz = 0.864”
(from the AISC Manual)
L
L
=
≤ 240
rmin 0.864"
this is satisfied so long as L ≤ 207.4” = 17’33/8”
Compute the Design Strength:
Gross Section Yielding:
Pn = Fy Ag = (36ksi)(4.75 in2) = 171.0kip
φPn = (0.95)(171.0kip) = 162.5kip
Net Section Fracture:
Pn = Fu Ae = Fu U An
Lacking other guidance, AISC Table D3.1 Case 2 will be applied:
U = 1−
x
0.981"
= 1−
= 0.8774
L
8.0"
φPn = (0.80)(241.7kip) = 193.4kip
Pn = (58ksi) (0.8774)(4.75 in2) = 241.7kip
Gross Section Yielding Governs  φPn =163kip
ODOTLRFD Short Course  Steel
AASHTO Tension Member Example #4
AASHTOLRFD 2007
Created July 2007: Page 1 of 1
 110 
116.
AASHTO Compression Member Example #1:
Problem:
Compute the design compressive strength of a W14x74 made of M27050 steel. The column has
a length of 20 ft and can be treated as pinnedpinned.
Solution:
Check Local Buckling:
Flange:
Web:
bf
2t f
?
≤k
E
Fy
h ?
E
≤k
tw
Fy
?
29, 000
= 13.5 OK
50
?
29, 000
= 35.9
50
λf = 6.41 (Tabulated)
6.41 ≤ 0.56
λw = 25.4 (Tabulated)
25.4 ≤ 1.49
OK
Compute Flexural Buckling Capacity:
Slenderness Ratios:
(1.00)(20 ')(12 in )
⎛ KL ⎞
ft
=
= 39.74 < 120
⎜
⎟
6.04"
⎝ r ⎠x
OK
(1.00)(20 ')(12 in )
⎛ KL ⎞
ft
=
= 96.77 < 120
⎜
⎟
r ⎠y
2.48"
⎝
OK
Since the effective slenderness ratio is larger for the y axis than the x axis, yaxis buckling will
govern.
2
2
ksi
⎞
⎛ KL ⎞ Fy ⎛ 96.77 ⎞ ⎛ 50
λ=⎜
=⎜
= 1.636
⎟
⎟ ⎜
ksi ⎟
⎝ r π ⎠ y E ⎝ π ⎠ ⎝ 29, 000 ⎠
(6.9.4.13)
Since λ ≤ 2.25, Inelastic Buckling Governs
(
)
Pn = 0.66λ Fy As = 0.66(1.636) ( 50ksi )( 21.8 in 2 ) = 549.6kip
(6.9.4.11)
φPn = (0.90)(549.6kip)
φPn = 495kip
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #1
AASHTOLRFD 2007
Created July 2007: Page 1 of 1
 111 
117.
AASHTO Compression Member Example #2
Problem:
Compute the axial compressive design strength based on flexural buckling (no torsional or
flexuraltorsional buckling). Assume that the crosssectional elements are connected such that
the builtup shape is fully effective. All plates are 4” thick.
Solution:
Compute Section Properties:
I
A
r=
Ix = ∑
bh3
+ Ad 2
12
2
⎡ ( 4") ( 30" ( 2 × 4") )3 ⎤
⎡ ( 36")( 4")3
⎛ 30" 4" ⎞ ⎤
⎥ = 56,150 in 4
= 2⎢
+ ( 36" × 4") ⎜
− ⎟ ⎥ + 2⎢
12
2⎠ ⎥
12
⎢
⎥
⎝ 2
⎢
⎣
⎦
⎣
⎦
hb3
+ Ad 2
12
2
⎡ ( 30" ( 2 × 4") ) ( 4")3
⎡ ( 4")( 36")3 ⎤
⎛ 36" 4" ⎞ ⎤
= 2⎢
+ ( 30" ( 2 × 4") ) × 4" ⎜
− ⎟ ⎥
⎥ + 2⎢
12
12
2⎠ ⎥
⎢
⎝ 2
⎢
⎥
⎣
⎦
⎣
⎦
4
= 76,390 in
Iy = ∑
(
(
)
)
As = 2 ( 36" × 4") + 2 ( 30" ( 2 × 4") ) × 4" = 464.0 in 2
Since I x = 56,154.67 in 4 < I y = 76,394.67 in 4 , xaxis buckling controls
rx =
Ix
56,150 in 4
=
= 11.0 in
As
464.0 in 2
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #2
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 112 
118.
Check Local Buckling (Section 6.9.4.2):
b 36" − 2 ( 4")
=
= 7.00
t
4"
b?
E
≤k
t
Fy
(6.9.4.21)
7.00 ≤1.40
29, 000ksi
= 33.72
50ksi
OK
Calculate the Nominal Compressive Strength (Section E3 page 16.133):
Slenderness Ratios:
KL
r
where:
K = 0.8
(Section 4.6.2.5)
in = 480"
Lx = Ly = 40 ft × 12 ft
KLx ( 0.8 )( 480 in )
=
= 34.91
rx
(11.0 in )
2
2
⎛ KL ⎞ Fy ⎛ 34.91 ⎞
50ksi
λ=⎜
=⎜
= 0.2129
⎟
⎟
ksi
⎝ rs π ⎠ E ⎝ π ⎠ 29, 000
(6.9.4.13)
Since λ ≤ 2.25 , Inelastic Flexural Buckling Governs
(
Pn = 0.66λ Fy As = 0.660.2129 50ksi
(
)( 464.0 in ) = 21, 240
2
kip
(6.9.4.11)
)
φc Pn = ( 0.90 ) 21, 240kip = 19,110kip
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #2
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 113 
119.
AASHTO Compression Member Example #3:
Problem:
Determine the effective length factor, K, for column AB in the frame shown below. Column AB
is a W10x88 made of A992 steel. W16x36 beams frame into joint A and W16x77 beams frame
into joint B. The frame is unbraced and all connections are rigid. Consider only buckling in the
plane of the page about the sections’ strong axes.
A
W16 x 36
L=24'
B
W10 x 88
L=14'
B
8 @ 14'
A
W16 x 77
L=24'
4 @ 24'
Solution:
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #3 v2
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 114 
120.
Determine the Effective Length Factor:
GB
K
GA
⎛ (2)(534 in 4 ) ⎞
⎛I⎞
⎜
⎟
∑⎜ L ⎟
(14 ')
⎝ ⎠C
⎝
⎠ = 3.065
GA =
=
4
⎛I⎞
⎛
⎞
∑ ⎜ L ⎟ ⎛ 2 ⎞ ⎜ (2)(448')in ) ⎟
⎜ ⎟
⎝ ⎠G ⎝ 3 ⎠ ⎝
(24
⎠
⎛ (2)(534 in 4 ) ⎞
⎛I⎞
⎜
⎟
∑⎜ L ⎟
(14 ')
⎝ ⎠C
⎝
⎠ = 1.237
GB =
=
⎛I⎞
⎛ (2)(1,110 in 4 ) ⎞
∑ ⎜ L ⎟ ⎛ 2 ⎞ ⎜ (24 ') ⎟
⎜ ⎟
⎝ ⎠G ⎝ 3 ⎠ ⎝
⎠
For unbraced frames:
K=
K=
1.6G AG B + 4.0(G A + G B ) + 7.5
G A + G B + 7.5
(1.6)(3.065)(1.237) + (4.0)(3.065 + 1.237) + 7.5
= 1.615
(3.065 + 1.237 + 7.5)
The factor of 2/3 appears in the denominator to reflect the fact
that the far ends of the girders are “fixed” connections.
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #3 v2
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 115 
121.
AASHTO Compression Member Example #4:
Problem:
y
10"
Check to see if a builtup section will work to resist a factored load
of Pu = 209kip. The column is to be fabricated from two C10x15.3
as is shown in the figure to the right. The steel is M27036 and the
effective length is 20’ with respect to all axes.
If the column is adequate, determine the thickness of the battens.
The battens are 8” long and 6” deep and are also made of M27036
steel.
x
9"
Solution:
Check Local Buckling:
2.60"
b bf
=
=
= 5.96
t t f 0.436"
a
Flange:
b?
E
29, 000ksi
≤ 0.56
= 0.56
= 15.89 OK
t
Fy
36ksi
Web:
b d − 2t f 10"− (2)(0.436")
=
=
= 38.03
t
tw
0.240"
b?
E
29, 000ksi
≤ 1.49
= 1.49
= 42.29 OK
t
Fy
36ksi
Compute Section Properties:
As = (2) (4.48 in2) = 8.96 in2
IX = (2) (Ix) = (2)(67.3in4) = 134.6 in4
2
⎡
⎞ ⎤
4
2 ⎛ 9"
IY = (2) ⎢ 2.27 in + (4.48 in ) ⎜ − 0.634" ⎟ ⎥ = 138.5 in 4
⎝ 2
⎠ ⎥
⎢
⎣
⎦
rX =
IX
134.6 in 4
=
= 3.88"
8.96 in 2
A
rY =
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #4
IY
138.5 in 4
=
= 3.93"
8.96 in 2
A
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 116 
122.
Slenderness Ratios:
⎛ (20 ')(12 in ) ⎞
⎛ KL ⎞
ft
=⎜
⎜
⎟
⎟ = 61.86
⎝ r ⎠ X ⎝ 3.88" ⎠
⎛ KL ⎞ ⎛ (20 ')(12 in ) ⎞
ft
⎜
⎟ =⎜
⎟ = 61.07
⎝ r ⎠Y ⎝ 3.93" ⎠
It appears as though X axis buckling will govern but since the battens will be subjected to
shear if the section buckles about its Y axis, this slenderness ratio must be modified.
Batten Spacing:
⎛ 3 ⎞ ⎛ KL ⎞
a ≤ ri ⎜ ⎟ ⎜
⎟
⎝ 4 ⎠ ⎝ r ⎠ max
⎛ KL ⎞
⎛ KL ⎞
⎜
⎟ =⎜
⎟
⎝ r ⎠ max ⎝ r ⎠ X
ri = ry = 0.711” (for one channel)
a ≤ (0.711") ( 0.75 )( 61.86 )
a ≤ 32.98"
use 9 battens @ a = 30”
Modified Slenderness Ratio – Yaxis Buckling:
The modified slenderness ratio is calculated as,
2
α2 ⎛ a ⎞
⎛ KL ⎞
⎛ KL ⎞
⎜ ⎟
⎜
⎟ = ⎜
⎟ + 0.82
(1 + α 2 ) ⎝ rib ⎠
⎝ r ⎠m
⎝ r ⎠o
rib = 0.711”
α=
2
(6.9.4.3.11)
h = 9” – (2)(0.634”) = 7.73”
h
7.732"
=
= 5.44
2rib (2)(0.711")
⎛ KL ⎞
⎜
⎟ =
⎝ r ⎠m
⎛ (5.44) 2 ⎞ ⎛ 30" ⎞ 2
⎟
= 71.70
( 61.07 ) + 0.82 ⎜
⎜ (1 + (5.44) 2 ) ⎟ ⎜ 0.711" ⎟
⎝
⎠
⎝
⎠
2
Now we can see that after the Y axis slenderness ratio is modified, Y axis buckling
actually governs over X axis buckling.
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #4
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 117 
123.
Column Design Capacity:
2
2
ksi
(6.9.4.13)
⎞
⎛ KL ⎞ Fy ⎛ 71.70 ⎞ ⎛ 36
λ=⎜
=⎜
= 0.6466
⎟
⎟ ⎜
ksi ⎟
⎝ r π ⎠ y E ⎝ π ⎠ ⎝ 29, 000 ⎠
Since λ ≤ 2.25, Inelastic Buckling Governs
(
)
Pn = 0.66λ Fy As = 0.658( 0.6466) ( 36ksi )( 8.96 in 2 ) = 246.1kip
(6.9.4.11)
φPn = (0.90)(246.1kip)
φPn = 221kip
Since φPn > ΣγQ, the column is adequate.
Batten Design:
Assume that there are inflection points half way between the battens and design for a shear equal
to 2% of the compressive design strength (AISC Section E6. Pg 16.139)
Vu = (0.02)(221kip) = 4.42kip
2.21 kip
4.42kip
kip
= 2.21 channel
2
ΣM
2Mu,Batten
I Batten
Mu,Batten = 33.15kin
t (6")3
=
= 18t
12
S Batten =
2.21 kip
18t
= 6t
3
for first yield, My = Fy S
Let φFy S ≥ Mu,Batten
33.15kin
t≥
(1.00)(36ksi )(6 in 3 )
t ≥ 0.153”
use t = 5/16”
(Min Thickness)
use PL6 x 8 x 5/16 Battens
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #4
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 118 
124.
AASHTO Compression Member Example #5:
Problem:
Find the design strength of a WT15x146 made of M27050 steel. KL = 24’ for buckling in all
directions. Use the provisions in the AISC Specification to determine the FlexuralTorsional
Buckling strength of the column.
Solution:
Check Local Buckling:
Flange:
15.3"
b bf
=
=
= 4.14
t 2t f (2)(1.85")
b?
E
29, 000ksi
≤k
= 0.56
= 13.5 OK
t
Fy
50ksi
Web:
b h
= = 15.7 (Tabulated)
t tw
b?
E
29, 000ksi
≤k
= 0.75
= 18.1
t
Fy
50ksi
OK
Calculate the buckling load for Flexural Buckling about the XAxis:
2
ksi
in
⎛ KL ⎞ Fy ⎛ ( 24 ') (12 ft ) ⎞ ⎛ 50
λX = ⎜
=⎜
⎟ ⎜
⎟
ksi
⎝ r π ⎠ X E ⎝ (4.48")(π) ⎠ ⎝ 29, 000
2
⎞
⎟ = 0.7219
⎠
Since λX < 2.25, Inelastic Buckling Governs
(
)
Pn = 0.66( 0.7219) ( 50ksi )( 42.9 in 2 ) = 1,586kip
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #5
(6.9.4.11)
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 119 
125.
Calculate the Critical Stress for FlexuralTorsional Buckling about the Yaxis:
Fcrft =
Fcr ,Y + Fcr , Z ⎛
⎜1 − 1 − 4 Fcr ,Y Fcr , Z H
2
⎜
2H
( Fcr ,Y + Fcr ,Z )
⎝
⎞
⎟
⎟
⎠
(AISC E42)
2
ksi
in
⎞
⎛ KL ⎞ Fy ⎛ ( 24 ') (12 ft ) ⎞ ⎛ 50
λY = ⎜
=⎜
= 1.131
⎟ ⎜
⎟
ksi ⎟
⎝ r π ⎠Y E ⎝ (3.58")(π) ⎠ ⎝ 29, 000 ⎠
2
Since λY < 2.25, Inelastic Buckling Governs
Fcr ,Y =
(
)
Pn
= 0.66(1.131) ( 50ksi ) = 31.15ksi
As
2
2
r o2 = xo + yo +
Ix + I y
…… yo = 3.62"−
Ag
(6.9.4.11)
1.85"
= 2.695"
2
(AISC E47)
(861 in 4 + 549 in 4 )
r = (0.00) + ( 2.695 ) +
= 40.13 in 2
2
42.9 in
2
o
2
Fcr , Z =
2
GJ
(11, 200ksi )(37.5 in 4 )
=
= 244.0ksi
2
2
2
(42.9 in )(40.13 in )
Aro
(AISC E43)
2
2
xo + yo
(0.000") 2 + (2.695") 2
H = 1−
= 1−
= 0.8190
40.13 in 2
r o2
Fcrft
⎛ 31.15ksi + 244.0ksi
=⎜
(2)(0.819)
⎝
(AISC E48)
⎛
⎞
⎞⎜
(4)(31.15ksi )(244.0ksi )(0.819) ⎟
= 30.37 ksi (AISC E42)
⎟ ⎜1 − 1 −
ksi
ksi 2
⎟
⎠
( 31.15 + 244.1 )
⎝
⎠
Pn = AsFcrft = (42.9 in2)(30.37ksi) = 1,303kip
Since 1,303kip < 1,586kip, FlexuralTorsional Buckling Governs
φPn = (0.90)(1,303kip) = 1,170kip
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #5
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 120 
126.
AASHTO Compression Member Example #6:
Problem:
y
Find the design strength of a C12x30 made of A36 steel.
KLy = 7’ and KLx = KLz = 14’.
x
Solution:
Check Local Buckling:
b?
E
29, 000ksi
≤k
= 0.56
= 15.89 OK
36ksi
t
Fy
Flange:
3.17"
b bf
=
=
= 6.327
t t f 0.501"
Web:
b h d − 2t f 12.0"− (2)(0.501")
= =
=
= 21.56
0.510"
t tw
tw
b?
E
29, 000ksi
≤k
= 1.49
= 42.29
36ksi
t
Fy
OK
Since both the flange and the web are nonslender, local buckling is OK.
Buckling Strength:
Note that the axes of the channel are not arranged properly for the equations in the AISC
Specification. These axes need to be rearranged so that the y axis is the axis of symmetry.
Using this modified set of axes, note that KLx = 7’ and KLy = KLz = 14’.
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #6
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 121 
127.
Calculate the buckling load for Flexural Buckling about the XAxis:
2
ksi
in
⎞
⎛ KL ⎞ Fy ⎛ ( 7 ' ) (12 ft ) ⎞ ⎛ 36
λx = ⎜
=⎜
= 1.528
⎟ ⎜
⎟
ksi ⎟
⎝ r π ⎠ x E ⎝ (0.762")(π) ⎠ ⎝ 29, 000 ⎠
2
Since λx < 2.25, Inelastic Buckling Governs
(
)
Pn = 0.66(1.528) ( 36ksi )( 8.81 in 2 ) = 167.3kip
(6.9.4.11)
Calculate the Critical Stress for FlexuralTorsional Buckling about the Yaxis:
For Singly symmetric Sections:
Fe =
Fey =
Fey + Fez ⎛
⎜1 − 1 − 4 Fey Fez H
2H ⎜
(Fey + Fez )2
⎝
(π2 )(29, 000ksi )
⎛ (14 ')(12 ) ⎞
⎜ 4.29" ⎟
⎝
⎠
in
ft
2
⎞
⎟
⎟
⎠
(AISC E45)
= 186.6ksi
(AISC E410)
⎡ (π2 )(29, 000ksi )(151 in 6 )
⎤
1
Fez = ⎢
+ (11, 200ksi )(0.861 in 4 ) ⎥
2
2
2
in
⎢
⎥ (8.81 in )(4.54") (AISC E411)
( (14 ')(12 ft ) )
⎣
⎦
Fez = 61.54ksi
⎛ (186.6ksi + 61.54ksi ) ⎞ ⎛
(4)(186.6ksi )(61.54ksi )(0.919") ⎞
Fe = ⎜
⎜1 − 1 −
⎟
⎟⎜
⎟
(2)(0.919")
(186.6ksi + 61.54ksi ) 2
⎝
⎠⎝
⎠
(AISC E45)
Fe = 59.30ksi
λ=
Fy
Fe
=
36ksi
= 0.6071
59.30ksi
Pn = ( 0.66(0.6071) )( 36ksi )( 8.81 in 2 ) = 246.0kip
(6.9.4.11)
since 167.3kip < 246.0kip, Flexural Buckling about the x Axis Governs
φPn = (0.90)( 167.3kip) = 151kip
ODOTLRFD Short Course  Steel
AASHTO Compression Member Example #6
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 122 
128.
AASHTO Compression Members Example #7:
Problem:
A pair of L4x4x1/2 angles are used as a compression member. The angles are made of
M27036 steel and have an effective length of 12’. The angles are separated by 3/8” thick
connectors.
Y
4"
3
/8"
4"
Solution:
X
Check Local Buckling:
b 4.0"
=
= 8.0
1 "
t
2
Fully
Tensioned
b?
E
29, 000ksi
≤k
= 0.45
= 12.77
t
Fy
36ksi
Local Buckling is OK
Check the Connector Spacing:
⎛ (12 ') (12 in ) ⎞
⎛ KL ⎞
ft
=⎜
⎟ = 119.0
⎜
⎟
⎝ r ⎠ X ⎝ (1.21") ⎠
ry = 1.83” from AISC 2L
Table 115, Pg 1104.
in
⎛ KL ⎞ ⎛ (12 ') (12 ft ) ⎞
=⎜
⎟ = 78.69
⎜
⎟
⎝ r ⎠Y ⎝ (1.83") ⎠
⎛ 3 ⎞ ⎛ KL ⎞
a ≤ ri ⎜ ⎟ ⎜
⎟
⎝ 4 ⎠ ⎝ r ⎠ max
⎛3⎞
a ≤ (0.776") ⎜ ⎟ (119.0 ) = 69.26"
⎝4⎠
Use 5 connectors….. a = 36”
ODOTLRFD Short Course – Steel
AASHTO Compression Member Example #7
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 123 
129.
Check Flexural Buckling about the XAxis: (Y axis is the axis of symmetry)
2
2
ksi
⎛ KL ⎞ Fy ⎛ 119.0 ⎞ ⎛ 36
λX = ⎜
=⎜
⎟
⎟ ⎜
ksi
⎝ r π ⎠ X E ⎝ π ⎠ ⎝ 29, 000
(
⎞
⎟ = 1.781
⎠
Since λX < 2.25, Inelastic Buckling Governs
)
Pn = 0.66(1.781) ( 36ksi )( 7.49 in 2 ) = 127.9kip
(6.9.4.11)
Check FlexuralTorsional Buckling about the YAxis:
For Tees and Double Angles where the Y axis is the Axis of Symmetry:
Fcrft
Fcr ,Y + Fcr , Z ⎛
⎜1 − 1 − 4 Fcr ,Y Fcr , Z H
=
2
⎜
2H
( Fcr ,Y + Fcr ,Z )
⎝
⎞
⎟
⎟
⎠
(AISC E42)
Since the section is builtup and the connectors will be in shear for Yaxis buckling, we
must consider a modified slenderness ratio…
Calculate Modified Slenderness and Yaxis Flexural Buckling Stress:
2
α2 ⎛ a ⎞
⎛ KL ⎞
⎛ KL ⎞
= ⎜
+ 0.82
⎜ ⎟
⎜
⎟
⎟
(1 + α 2 ) ⎝ rib ⎠
⎝ r ⎠m
⎝ r ⎠o
α=
h
2rib
2
(6.9.4.3.11)
h = ( 2 )(1.18") + ( 3 8 ") = 2.735"
rib = ry for a single angle = 1.21”
α=
2.735"
= 1.130
(2)(1.21")
⎛ KL ⎞
⎜
⎟ =
⎝ r ⎠m
2
(0.82)(1.130) 2 ⎛ 36" ⎞
( 78.69 ) +
⎜
⎟ = 81.24
(1 + (1.130) 2 ) ⎝ 1.21" ⎠
2
ODOTLRFD Short Course – Steel
AASHTO Compression Member Example #7
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 124 
130.
Compute the Yaxis Flexural Buckling Stress, Fcry:
2
2
ksi
⎞
⎛ KL ⎞ Fy ⎛ (81.24) ⎞ ⎛ 36
λY = ⎜
=⎜
= 0.8301
⎟ ⎜
⎟
ksi ⎟
⎝ r π ⎠Y E ⎝ (π) ⎠ ⎝ 29, 000 ⎠
Since λY < 2.25, Inelastic Buckling Governs
Fcr ,Y =
(
)
Pn
= 0.66( 0.8301) ( 36ksi ) = 25.43ksi
As
(6.9.4.11)
Calculate Torsional Buckling Stress, Fcr,Z:
ro = 2.38" (AISC Table 115, Pg 1104)
J = 0.322 in4 for a single angle
∴ r o2 = 5.664 in 2
(AISC Table 17, Pg 142)
∴ Jtotal = (2)(0.322 in4) = 0.644 in4
Fcr ,Z =
GJ (11, 200ksi )(0.644 in 4 )
=
= 170.0ksi
2
2
2
Ar o (7.49 in )(5.664 in )
(AISC E43)
H = 0.848 (AISC Table 115, Pg 1104)
Fcrft
⎛ 25.43ksi + 170.0ksi
=⎜
(2)(0.848)
⎝
⎛
⎞
⎞⎜
(4)(25.43ksi )(170.0ksi )(0.848) ⎟
= 24.79ksi
⎟ ⎜1 − 1 −
ksi
ksi 2
⎟
⎠⎜
( 25.43 + 170.0 ) ⎟
⎝
⎠
Pn = AsFcrft = (7.49 in2) (24.79ksi) = 185.7kip
Since 127.9kip < 185.7kip,
Flexural Buckling Governs
φPn = (0.90)(127.9kip ) = 115kip
ODOTLRFD Short Course – Steel
AASHTO Compression Member Example #7
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 125 
132.
AASHTO Flexure Example #1:
Problem:
Determine the plastic moment of the steel section shown below.
Solution
Since the section is made up of components of different materials,
the location of the PNA must be determined by equating the force
above the PNA to the force below the PNA.
Pc = (16")(1") ( 50ksi ) = 800.0kip
Pw = (22") ( 3 4 ") ( 36ksi ) = 594.0kip
Pt = ( 8")( 2") ( 70ksi ) = 1,120kip
Since Pc + Pw > Pt
(800.0
kip
+ 594.0kip = 1,394kip > 1,120kip ) , the
PNA must lie in the web. Define q as the fraction of the web that
lies above the PNA.
Pcompression = Ptension
Pc + qPw = (1 − q ) Pw + Pt
(800.0 ) + q ( 594.0 ) = (1 − q ) ( 594.0 ) + (1,120 )
kip
kip
kip
kip
q = 0.7694
I.e., 76.94% of the web lies above the PNA (acts in compression assuming a positive moment).
Y = 1"+ ( 0.7694 )( 22") = 17.93" from the top of steel
Find the moment arms from the resultant forces to the PNA.
tc
1"
= 17.93"− = 17.43"
2
2
1
1
= ( 2 ) qh = ( 2 )( 0.7694 )( 22") = 8.463"
dc = Y −
d wc
d wt = ( 1 2 )(1 − q ) h = ( 1 2 )(1 − 0.7694 )( 22") = 2.537"
dt = d − Y −
tt
2"
= 25"− 17.93"− = 6.074"
2
2
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #1
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 127 
133.
PL16 x 1, 50ksi
50ksi
Pc
36ksi
Y
Pwc
dc
PL22 x 3/4, 36ksi
dwc
36ksi
PNA
Pwt
dwt
dt
PL8 x 2, 70ksi
70ksi
Pt
Compute the plastic moment by summing the moments about the PNA.
M p = ∑ Pdi = Pc d c + Pwc d wc + Pwt d wt + Pdt
i
t
= ( 800kip ) (17.43") + ( 457 kip ) ( 8.463") + (137 kip ) ( 2.537") + (1,120kip ) ( 6.074")
= 24,960kin = 2, 080kft
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #1
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 128 
134.
AASHTO Flexure Example #2:
Problem:
Determine the plastic moment capacity for the composite beam shown below. The section is a
W30x99 and supports an 8” concrete slab. The dimensions are as shown. Use Fy = 50ksi and
f’c = 4ksi. Assume full composite action.
100"
8"
Solution:
Determine the Controlling Compression Force:
Ps = 0.85 f c'bets = ( 0.85 ) ( 4ksi ) ( 8")(100") = 2720kip
PSteel = Ast Fy = ( 29.1 in 2 )( 50ksi ) = 1455kip
Assuming full composite action, the shear
connectors must carry the smallest of Ps and Psteel.
W30 x 99:
A = 29.1 in2
d = 29.7"
bf = 10.5"
tf = 0.670"
tw = 0.520"
Zx = 312 in3
Ix = 3,990 in4
Iy = 128 in4
rx = 11.7"
ry = 2.10"
Since Ps > Psteel, the PNA must lie in the slab.
Determine the Location of the PNA:
The PNA location is determined by equating the compressive force in the slab, acting over a
depth ac, with the tensile force in the steel section.
0.85 f c'be ac = Ast Fy
PConc = PSteel
ac =
Ast Fy
0.85 f c'be
( 29.1 in )( 50 ) = 4.279" (measured from the top of slab)
=
( 0.85) ( 4 ) (100")
2
ksi
ksi
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #2
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 129 
135.
Determine the Plastic Moment:
The plastic moment is calculated by summing the tension and compression forces about any
point. In general, the moments are summed about the PNA. In this case (where the PNA is in
the slab) it is simplest to sum moments about either force PSteel or the force Pconc. Note that the
tension force in the concrete is ignored.
100"
0.85f’c
Pconc
ac
8"
PNA
a1
Psteel
Fy
M p = ( PConc ) (a1 ) = ( PSteel ) (a1 )
a1 =
d st
a
29.7"
4.279"
+ ts − c =
+ 8"−
= 20.71"
2
2
2
2
M p = (1455kip ) (20.71") = 30,130kin = 2,511kft
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #2
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 130 
136.
AASHTO Flexure Example #3:
Problem:
Determine the plastic moment capacity for the composite beam shown below. The section is a
W30x99 and supports a 6” thick concrete slab. The dimensions are as shown. Use Fy = 50ksi and
f’c = 4ksi. Assume full composite action.
Solution:
Determine the Controlling Compression Force:
Ps = 0.85 f c'bs ts = ( 0.85 ) ( 4ksi ) ( 6")( 50") = 1020kip
PSteel = Ast Fy = ( 29.1 in 2 )( 50ksi ) = 1455kip
Assuming full composite action, the shear connectors must
carry the smallest of Ps and Psteel.
Since Ps < Psteel, the PNA must lie in the steel. When this
occurs, it is simplest to use the aids in Appendix D of the
AASHTO Specification to determine the location of the
PNA and plastic moment.
Referring to Table D6.11 in Appendix D6.1, Page 6290:
Determine the forces in the components of the cross section. The forces in the rebar will be
conservatively taken as zero (we don’t know what size the rebar is any ways…)
Ps = 0.85 f c'bs ts = ( 0.85 ) ( 4ksi ) ( 6")( 50") = 1020kip
Pc = (0.670")(10.5") ( 50ksi ) = 351.8kip
Pw = ⎡ 29.1 in 2 − (2)(0.670")(10.5") ⎤ ( 50ksi ) = 751.5kip
⎣
⎦
kip
Pt = Pc = 351.8
In this case, I took Aw = Asteel  2Af. Otherwise,
Pc+Pt+Pw ≠ Psteel. If you take Aw = D tw where
D = d  2tf, the plastic moment changes by ~2%
?
Check Case I Pt + Pw ≥ Pc + Ps
?
351.8kip + 751.5kip ≥ 351.8kip + 1020kip
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #3
NO
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 131 
137.
?
Check Case II Pt + Pw + Pc ≥ Ps
?
351.8kip + 751.5kip + 351.8kip ≥ 1020kip
YES  PNA in Top Flange
50"
0.85f’c
Ps
Pc1
6"
Fy
ds
PNA
Pc2
dw
Pw
dt
Pt
Fy
First, the location of the PNA within the top flange is determined.
⎞
⎛ t ⎞ ⎛ P + P − Ps
Y = ⎜ c ⎟⎜ w t
+ 1⎟
Pc
⎝ 2 ⎠⎝
⎠
kip
kip
kip
⎞
⎛ 0.670" ⎞ ⎛ 751.5 + 351.8 − 1, 020
=⎜
+ 1⎟ = 0.2368"
⎟⎜
kip
351.8
⎝ 2 ⎠⎝
⎠
Next, the distances from the component forces to the PNA are calculated.
6"
+ 0.2368" = 3.237"
2
29.7"
dw =
− 0.2368" = 14.61"
2
0.670"
dt = 29.7"−
− 0.2368" = 29.13"
2
ds =
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #3
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 132 
138.
Finally, the plastic moment is computed.
⎛P ⎞
2
M p = ⎜ c ⎟ ⎡Y 2 + ( tc − Y ) ⎤ + [ Ps d s + Pw d w + Pdt ]
t
⎦
⎝ 2tc ⎠ ⎣
⎛ 351.8kip ⎞ ⎡
2
2
=⎜
⎟ ⎣( 0.2368") + ( 0.670"− 0.2368") ⎤ + ...
⎦
⎝ (2)(0.670") ⎠
(
... + ⎡(1, 020kip ) ( 3.237") + ( 751.5kip ) (14.61") + ( 351.8kip ) ( 29.13") ⎤
⎣
⎦
)
= 262.5 kip ⎡ 0.2437 in 2 ⎤ + ⎡ 24,530kin ⎤
in ⎣
⎦ ⎣
⎦
= 24,590kin = 2, 049kft
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #3
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 133 
139.
AASHTO Flexure Example #4:
Problem:
Determine the plastic moment capacity for the composite beam shown below for negative
flexure. The section is a W30x99 and supports an 8” concrete slab. The dimensions are as
shown. Use Fy = 50ksi and f’c = 4ksi. Assume full composite action. The grade 60 reinforcement
in the slab is made up of #4 bars, with a clear cover of 17/8”.
100"
8"
Solution:
The concrete slab will be in tension,
therefore none of the concrete is assumed
to be effective.
Slab Reinforcement:
Top Layer:
#4 bars @ 6" cc
Bottom Layer: #4 bars @ 12" cc
Referring to Table D6.12 in Appendix D6.1, Page 6291:
⎛ (π)(0.5") 2 ⎞
kip
Prt = Fyrt Art = ( 60 ) ( 8 ) ⎜
⎟ = 94.25
4
⎝
⎠
2
⎛ (π)(0.5") ⎞
kip
Prb = Fyrb Arb = ( 60ksi ) ( 4 ) ⎜
⎟ = 47.12
4
⎝
⎠
ksi
W30 x 99:
A = 29.1 in2
d = 29.7"
bf = 10.5"
tf = 0.670"
tw = 0.520"
Sx = 269 in3
Zx = 312 in3
Ix = 3,990 in4
Iy = 128 in4
rx = 11.7"
ry = 2.10"
Pt = (0.670")(10.5") ( 50ksi ) = 351.8kip
Pw = ⎡ 29.1 in 2 − (2)(0.670")(10.5") ⎤ ( 50ksi ) = 751.5kip
⎣
⎦
kip
Pc = Pt = 351.8
?
Check Case I: Pc + Pw ≥ Pt + Prb + Prt
?
351.8kip + 751.5kip ≥ 351.8kip + 47.12kip + 94.25kip
⎛ D ⎞ ⎡ P − P − Prt − Prb ⎤
Y = ⎜ ⎟⎢ c t
+ 1⎥
Pw
⎝ 2 ⎠⎣
⎦
⎛ 28.36" ⎞ ⎡ 351.8
Y =⎜
⎟⎢
⎝ 2 ⎠⎣
kip
YES  PNA is in Web
Take D as d − 2t f = 29.7"− ( 2 )( 0.670") = 28.36"
− 351.8kip − 94.25kip − 47.12kip ⎤
+ 1⎥
751.5kip
⎦
Y = 11.51" (measured from the bottom of the top flange)
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #4
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 134 
140.
d rt = 11.51"+ 0.670"+ 8"− ⎡1 7 8 "+ ( 1 2 )( 1 2 ") ⎤ = 18.06"
⎣
⎦
d rb = 11.51"+ 0.670"+ ⎡1 7 8 "+ ( 1 2 )( 1 2 ") ⎤ = 14.31"
⎣
⎦
dt = 11.51"+ ( 1 2 )( 0.670") = 11.85"
d wt = ( 1 2 )(11.51") = 5.755"
d wc = ( 1 2 )( 28.36"− 11.51") = 8.425"
Not needed when using Table D6.12
d c = ( 28.36"− 11.51") + ( 1 2 )( 0.670") = 17.19"
⎛ P ⎞
M p = ⎜ w ⎟ ⎡ y 2 + ( D − y ) 2 ⎤ + [ Prt d rt + Prb d rb + Pdt + Pc d c ]
t
⎣
⎦
⎝ 2D ⎠
751.5kip
⎡ (11.51") 2 + (28.36"− 11.51") 2 ⎤ + [(94.25kip )(18.06") + ...
=
⎦
(2)(28.36") ⎣
... + (47.12kip )(14.31") + (351.8kip )(11.85") + (351.8kip )(17.19")]
= 18,110kin = 1,509kft
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #4
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 135 
141.
The plastic moment can also be computed from “first principles” as well, though it is a bit more
involved. What follows is an example of how this would be completed.
Determine the Location of the PNA:
Since Pc + Pw ≥ Pt + Prb + Prt , the PNA is in the web of the section.
The location of the PNA within the web is determined by equating the tensile force acting
above the PNA with the compressive force acting below it. Assume the PNA lies at a
depth Y below the bottom of the top flange.
Pc + Pwc = Pwt + Pt + Prb + Prt
351.8kip + ( 50ksi ) ( 0.520") ( 28.36"− Y ) = ( 50ksi ) ( 0.520") (Y ) + 351.8kip + 47.12kip + 94.25kip
Y = 11.46 ''
Pwt = (50ksi )(11.46") ( 0.520") = 298.0kip
Pwc = (50ksi )(28.36"− 11.46") ( 0.520") = 439.4kip
Determine the Plastic Moment:
The plastic moment is calculated by summing the moments of the tensile and
compressive forces about any point. In general, the moments are summed about the
PNA. In this case (where the PNA is in the web) note that the tension force in the
concrete is ignored.
d rt = 11.46"+ 0.670"+ 8"− ⎡1 7 8 "+ ( 1 2 )( 1 2 ") ⎤ = 17.88"
⎣
⎦
d rb = 11.46"+ 0.670"+ ⎡1 7 8 "+ ( 1 2 )( 1 2 ") ⎤ = 14.38"
⎣
⎦
dt = 11.46"+ ( 1 2 )( 0.670") = 11.80"
d wt = ( 1 2 )(11.46") = 5.730"
d wc = ( 1 2 )( 28.36"− 11.46") = 8.450"
d c = ( 28.36"− 11.46") + ( 1 2 )( 0.670") = 17.24"
M p = (94.25kip )(17.88'') + (47.12kip )(14.38") + (351.8kip )(11.80 '') + ...
... + (298.0kip )(5.730") + (439.4kip ) ( 8.450") + ( 351.8kip ) (17.24")
= 18, 000kin = 1,500kft
The minor difference in between the two answers can be attributed to the fillet area.
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #4
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 136 
142.
AASHTO Flexural Example #5a:
Problem:
A noncomposite W30x99 made of M27050 steel is used to span 48’. The beam is braced
laterally at 12’0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes
the self weight of the beam. Check to see if the section is adequate considering flexural failure
modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Section
6.10.8 to determine capacity.
Solution:
Determine Classification of the Section:
Check
2 Dc ?
E
≤ 5.7
tw
Fyc
(6.10.6.2.31)
Take D = d  2tf = 29.7”  (2)(0.670”) = 28.36”
Dc =
D 28.36"
=
= 14.18"
2
2
?
(2)(14.18")
29, 000ksi
= 54.54 ≤ 5.7
= 137.3
(0.520")
50ksi
Check
I yc
I yt
OK, ∴ web is nonslender
?
≥ 0.3
(6.10.6.2.32)
Since Section is doubly symmetric, Iyc = Iyt
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5a
OK
AASHTOLRFD 2007
Created July 2007: Page 1 of 4
 137 
143.
Since the web is nonslender and Eq 6.10.6.2.32 is satisfied, we have the option of using either
AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member.
For this example, the provisions of 6.10.8 will be used and we will work with stresses.
The following failure modes must be investigated:
• Flange Local Buckling of the Compression Flange Fnc(FLB)
• Compression Flange Lateral Buckling
Fnc(LTB)
• Yielding of Tension Flange
Fnt
Fnc
Investigate Compression Flange Local Buckling:
λf =
λ pf
b fc
2t fc
=
10.5"
= 7.836
(2)(0.670")
(6.10.8.2.23)
E
29, 000ksi
= 0.38
= 0.38
= 9.152
Fyc
50ksi
(6.10.8.2.24)
Since λf < λp, the flange is compact and,
Fnc ( FLB ) = Rb Rh Fyc
(6.10.8.2.21)
Rb = 1.00 (since the web is nonslender)
Rh = 1.00 (since the section is rolled and is ∴ nonhybrid)
Fnc ( FLB ) = (1.00)(1.00) ( 50ksi ) = 50ksi
Investigate Compression Flange LateralTorsional Buckling:
The unbraced length of the beam is Lb = 12’0” = 144.0”.
rt =
b fc
⎛ 1 Dc tw ⎞
12 ⎜1 +
⎟
⎜
⎟
⎝ 3 b fc t fc ⎠
L p = 1.0rt
=
(10.5")
⎛ ⎛ 1 ⎞ (14.18")(0.520") ⎞
12 ⎜ 1 + ⎜ ⎟
⎟
⎝ ⎝ 3 ⎠ (10.5")(0.670") ⎠
E
29, 000ksi
= (1.0)(2.609")
= 62.84"
Fyc
50ksi
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5a
= 2.609"
(6.10.8.2.39)
(6.10.8.2.34)
AASHTOLRFD 2007
Created July 2007: Page 2 of 4
 138 
144.
Fyr = (0.7) ( 50ksi ) = 35ksi
Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc
Lr = π rt
(Pg 6110)
E
29, 000ksi
= (π)(2.609")
= 235.9"
Fyr
35ksi
(6.10.8.2.35)
Since L p = 62.84" < Lb = 144" < Lr = 235.9" , Inelastic LTB must be investigated.
⎡ ⎛
F
Fnc ( LTB ) = Cb ⎢1 − ⎜1 − yr
⎜ Rh Fyc
⎢ ⎝
⎣
⎞⎛ Lb − Lp
⎟⎜
⎟⎜ Lr − Lp
⎠⎝
⎞⎤
⎟ ⎥ Rb Rh Fyc ≤ Rb Rh Fyc
⎟⎥
⎠⎦
(6.10.8.2.32)
⎡ ⎛
⎞ ⎛ 144"− 62.84" ⎞ ⎤
35ksi
⎥ (1.0 )(1.0 ) ( 50ksi ) ≤ (1.0 )(1.0 ) ( 50ksi )
⎟
Fnc ( LTB ) = Cb ⎢1 − ⎜1 −
⎜ (1.0 ) ( 50ksi ) ⎟ ⎜ 235.9"− 62.84" ⎟ ⎥
⎢ ⎝
⎠
⎠⎝
⎣
⎦
Fnc ( LTB ) = ( Cb )( 0.8593) ( 50ksi ) = ( Cb ) ( 42.97 ksi ) ≤ 50ksi
Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.
2
⎛ f ⎞
⎛ f ⎞
Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3
⎝ f2 ⎠
⎝ f2 ⎠
(6.10.8.2.37)
C
M 2 = M c = 1, 080kft →
f 2 = 48.18ksi
M o = M B = 810.0kft →
f o = 36.13ksi
M BC ,mid = 1, 013kft
f mid = 45.19ksi
→
B
fmid
f1
Since the BMD is not concave,
f1 = 2 f mid − f 2 ≥ f o = (2) ( 45.19ksi ) − ( 48.18ksi ) = 42.20ksi ≥ 36.13ksi
2
⎛ 42.20 ⎞
⎛ 42.20 ⎞
Cb = 1.75 − 1.05 ⎜
⎟ + 0.3 ⎜
⎟ = 1.061 ≤ 2.3
⎝ 48.18 ⎠
⎝ 48.18 ⎠
= 1.061
Fnc ( LTB ) = (1.061) ( 42.97 ksi ) = 45.57 ksi ≤ 50ksi
= 45.57 ksi
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5a
AASHTOLRFD 2007
Created July 2007: Page 3 of 4
 139 
f2
145.
The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB):
Since Fnc ( LTB ) = 45.57 ksi < Fnc ( FLB ) = 50.00ksi , LTB governs the strength of the
compression flange.
Fnc = Fnc ( LTB ) = 45.57 ksi
φFnc = (1.00) ( 45.57 ksi ) = 45.57 ksi
Check fbu +
1
f l ≤ φ f Fnc
3
(6.10.8.1.11)
fbu = f C = 48.18ksi
Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0
Since fbu +
1
f l = 48.18ksi > φ f Fnc =45.57 ksi , the compression flange is not adequate.
3
Investigate the Strength of the Tension Flange:
Since the tension flange is discretely braced,
Fnt = Rh Fyt = (1.0) ( 50ksi ) = 50ksi
(6.10.8.31)
φFnt = (1.00) ( 50ksi ) = 50ksi
Check fbu +
1
f l ≤ φ f Fnt
3
(6.10.8.1.21)
fbu = f C = 48.18ksi
Since fbu +
1
f l = 48.18ksi < φ f Fnt =50.00ksi , the tension flange is adequate.
3
Since the compression flange is not adequate, the section is not adequate for flexure.
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5a
AASHTOLRFD 2007
Created July 2007: Page 4 of 4
 140 
146.
AASHTO Flexural Example #5b:
Problem:
A noncomposite W30x99 made of M27050 steel is used to span 48’. The beam is braced
laterally at 12’0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes
the self weight of the beam. Check to see if the section is adequate considering flexural failure
modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6
to determine capacity.
Solution:
Determine Classification of the Section:
Check
2 Dc ?
E
≤ 5.7
tw
Fyc
(6.10.6.2.31)
Take D = d  2tf = 29.7”  (2)(0.670”) = 28.36”
Dc =
D 28.36"
=
= 14.18"
2
2
?
(2)(14.18")
29, 000ksi
= 54.54 ≤ 5.7
= 137.3
(0.520")
50ksi
Check
I yc
I yt
OK, ∴ web is nonslender
?
≥ 0.3
(6.10.6.2.32)
Since Section is doubly symmetric, Iyc = Iyt
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
OK
AASHTOLRFD 2007
Created July 2007: Page 1 of 6
 141 
147.
Since the web is nonslender and Eq 6.10.6.2.32 is satisfied, we have the option of using either
AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member.
For this example, the provisions of A6 will be used and we will work with moments.
The following failure modes must be investigated:
• Flange Local Buckling of the Compression Flange Mnc(FLB)
Mnc(LTB)
• LateralTorsional Buckling
Mnt
• Yielding of Tension Flange
Mnc
Compute Web Plasticity Factors, Rpc and Rpt (Section A6.2):
Investigate the classification of the web.
Check
2 Dcp
tw
?
≤ λ pw( Dcp )
λ pw( Dcp ) =
(A6.2.11)
E
Fyc
⎛ Dcp ⎞
≤ λ rw ⎜
⎟
⎛ 0.54M p
⎞
⎝ Dc ⎠
− 0.09 ⎟
⎜
⎜ RM
⎟
h
y
⎝
⎠
(A6.2.12)
2
λ rw = 5.7
E
= 137.3
Fyc
(A6.2.13)
Rh = 1.00 (since the section is rolled and is ∴ nonhybrid)
M y = S x Fy = ( 269 in 3 )( 50ksi ) = 13, 450kin = 1,121kft
M p = Z x Fy = ( 312 in 3 )( 50ksi ) = 15, 600kin = 1,300kft
λ pw( Dcp ) =
29, 000ksi
50ksi
⎛ 14.18" ⎞
= 83.76 ≤ 137.3 ⎜
⎟ = 137.3
⎝ 14.18" ⎠
⎛ ( 0.54 ) (15, 600kin )
⎞
⎜
− 0.09 ⎟
⎜ (1.0 ) (13, 450kin )
⎟
⎝
⎠
2
λ pw( Dcp ) = 83.76
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
AASHTOLRFD 2007
Created July 2007: Page 2 of 6
 142 
148.
2 Dcp
tw
=
?
(2)(14.18")
= 54.54 ≤ λ pw( Dcp ) = 83.76
0.520"
OK, ∴ web is compact
Since the web is compact,
R pc =
R pt =
M yc
Mp
M yt
=
15, 600kin
= 1.160
13, 450kin
(A6.2.14)
=
Mp
15, 600kin
= 1.160
13, 450kin
(A6.2.15)
Investigate Compression Flange Local Buckling:
Investigate the compactness of the compression flange.
λf =
b fc
2t fc
=
λ pf = 0.38
10.5"
= 7.836
(2)(0.670")
(A6.3.23)
E
29, 000ksi
= 0.38
= 9.152
50ksi
Fyc
(A6.3.24)
Since λf < λpf, the flange is compact and,
M nc ( FLB ) = R pc M yc = (1.160 ) (13, 450kin ) = 15, 600kin
(A6.3.21)
Investigate LateralTorsional Buckling:
The unbraced length of the beam is Lb = 12’0” = 144.0”.
rt =
b fc
⎛ 1 Dc tw ⎞
12 ⎜1 +
⎜ 3b t ⎟
⎟
fc fc ⎠
⎝
L p = 1.0rt
=
(10.5")
⎛ ⎛ 1 ⎞ (14.18")(0.520") ⎞
12 ⎜ 1 + ⎜ ⎟
⎟
⎝ ⎝ 3 ⎠ (10.5")(0.670") ⎠
E
29, 000ksi
= (1.0)(2.609")
= 62.84"
50ksi
Fyc
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
= 2.609"
(A6.3.310)
(A6.3.34)
AASHTOLRFD 2007
Created July 2007: Page 3 of 6
 143 
149.
E
Lr = 1.95 rt
Fyr
⎛ Fyr S xc h ⎞
1 + 1 + 6.76 ⎜
⎟
S xc h
⎝ E J ⎠
J
2
(6.10.8.2.35)
⎛
⎞
S
Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5Fyc Fyr = (0.7) ( 50ksi ) = 35ksi
S xc
⎝
⎠
(Pg 6222)
Sxc = 269 in3
h = d − ( 1 2 ) ( t fc + t ft ) = 29.7"− 0.670" = 29.03"
J=
3
3
D tw b fc t fc
+
3
3
⎛
t fc
⎜1 − 0.63
⎜
b fc
⎝
⎞ b ft t 3
ft
⎟+
⎟
3
⎠
⎛
t ft
⎜ 1 − 0.63
⎜
b ft
⎝
⎞
⎟
⎟
⎠
(A6.3.39)
⎡ (10.5")(0.670")3 ⎛
(28.36")(0.520")3
⎛ 0.670" ⎞ ⎞ ⎤
4
J=
+ (2) ⎢
⎜1 − ( 0.63) ⎜ 10.5" ⎟ ⎟ ⎥ = 3.350 in
3
3
⎝
⎠ ⎠⎦
⎝
⎣
J = 3.350 in4
J = 3.77 in4 from AISC Manual….use J = 3.77 in4
E 29, 000ksi
=
= 828.6
Fyr
35ksi
3
S xc h ( 269 in ) ( 29.03")
=
= 2, 071
3.77 in 4
J
Lr = (1.95 )( 2.609")( 828.6 )
2
1
⎛ 2, 071 ⎞
1 + 1 + 6.76 ⎜
⎟ = 254.9" = 21.24 '
2, 071
⎝ 828.6 ⎠
This value of Lr = 21.24’ agrees well with the value published in AISC on Page 315
Since L p = 62.84" < Lb = 144" < Lr = 254.9" , Inelastic LTB must be investigated.
⎡ ⎛
F S
M nc ( LTB ) = Cb ⎢1 − ⎜1 − yr xc
⎢ ⎜ R pc Fyc
⎣ ⎝
⎞⎛ Lb − Lp
⎟⎜
⎟⎜ Lr − Lp
⎠⎝
⎞⎤
⎟ ⎥ R pc Fyc ≤ R pc Fyc
⎟⎥
⎠⎦
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
(A6.3.32)
AASHTOLRFD 2007
Created July 2007: Page 4 of 6
 144 
150.
⎡ ⎛
( 35ksi )( 269 in3 ) ⎞ ⎛ 144"− 62.84" ⎞⎤ (1.160 ) 13, 450kin ≤ (1.160 ) 13, 450kin
⎟⎜
M nc ( LTB ) = Cb ⎢1 − ⎜1 −
(
)
(
)
⎟⎥
⎢ ⎜ (1.16 ) (13, 450kin ) ⎟ ⎝ 254.9"− 62.84" ⎠ ⎥
⎠
⎣ ⎝
⎦
= ( Cb )( 0.9656 ) (15, 600kin ) = ( Cb ) (12,990kin ) ≤ 15, 600kin
Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.
2
⎛M ⎞
⎛M ⎞
Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3
⎝ M2 ⎠
⎝ M2 ⎠
(A6.3.37)
M 2 = M c = 1, 080kft
M o = M B = 810.0kft
M BC ,mid = 1, 013kft
Since the BMD is not concave,
M 1 = 2M mid − M 2 ≥ M o = (2) (1, 013kft ) − (1, 080kft ) = 946kft ≥ 810kft
2
⎛ 946 ⎞
⎛ 946 ⎞
Cb = 1.75 − 1.05 ⎜
⎟ + 0.3 ⎜ 1, 080 ⎟ = 1.061 ≤ 2.3
⎝ 1, 080 ⎠
⎝
⎠
= 1.061
M nc ( LTB ) = (1.061) (12,990kin ) = 13, 780kin ≤ 15, 600kin
M nc ( LTB ) = 13, 780kin
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
AASHTOLRFD 2007
Created July 2007: Page 5 of 6
 145 
151.
The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB):
Since M nc ( LTB ) = 13, 780kin < M nc ( FLB ) = 15, 600kin , LTB governs the strength of the
compression flange.
M nc = M nc ( LTB ) = 13, 780kin = 1,148kft
φM nc = (1.00) (1,148kft ) = 1,148kft
Check M u +
1
f l S xc ≤ φ f M nc
3
(A6.1.11)
M u = M C = 1, 080kft
Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0
Since M u +
1
fl S xc = 1, 080kft < φ f M nc =1,148kft , the compression flange is adequate.
3
Investigate the Strength of the Tension Flange:
Since the tension flange is discretely braced,
M nt = R pt M yt = (1.160) (13, 450kin ) = 15, 600kin = 1,300kft
(A6.41)
φM nt = (1.00) (1,300kft ) = 1,300kft
Check M u +
1
fl S xt ≤ φ f M nt
3
(6.10.8.1.21)
M u = M C = 1, 080kft
Since M u +
1
f l S xt = 1, 080kft < φ f M nt =1,300kft , the tension flange is adequate.
3
Since both flanges are adequate, the section is adequate for flexure.
Note that the benefits of using Appendix A6 are illustrated here since the section was found
to be not adequate when the provisions in Section 6.10.8 were used to compute capacity.
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #5b
AASHTOLRFD 2007
Created July 2007: Page 6 of 6
 146 
152.
AASHTO Flexural Example #6a:
Problem:
A noncomposite builtup girder made of M27050 steel is used to span 48’. The beam is braced
laterally at 12’0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes
the self weight of the beam. Check to see if the section is adequate considering flexural failure
modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Section
6.10.8 to determine capacity.
PL16 x 3/4
Solution:
PL38 x 3/8
Ix = 10,730 in4
Iy = 513.2 in4
Sx = 543.1 in3
Sy = 64.15 in3
PL16 x 3/4
Determine Classification of the Section:
Check
2 Dc ?
E
≤ 5.7
tw
Fyc
(6.10.6.2.31)
Take D = 38”
Dc =
D 38"
=
= 19"
2
2
?
(2)(19")
29, 000ksi
= 101.3 ≤ 5.7
= 137.3
( 38 ")
50ksi
Check
I yc
OK, ∴ web is nonslender
?
≥ 0.3
I yt
Since Section is doubly symmetric, Iyc = Iyt
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6a
(6.10.6.2.32)
OK
AASHTOLRFD 2007
Created July 2007: Page 1 of 5
 147 
153.
Since the web is nonslender and Eq 6.10.6.2.32 is satisfied, we have the option of using either
AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member.
For this example, the provisions of 6.10.8 will be used and we will work with stresses.
The following failure modes must be investigated:
• Flange Local Buckling of the Compression Flange Fnc(FLB)
• Compression Flange Lateral Buckling
Fnc(LTB)
• Yielding of Tension Flange
Fnt
Fnc
Investigate Compression Flange Local Buckling:
λf =
λ pf
b fc
2t fc
=
16"
= 10.67
(2)( 3 4 ")
(6.10.8.2.23)
E
29, 000ksi
= 0.38
= 0.38
= 9.152
Fyc
50ksi
(6.10.8.2.24)
Since λf < λp, the flange is non compact and,
⎡ ⎛
F ⎞⎛ λ − λ pf
Fnc ( FLB ) = ⎢1 − ⎜ 1 − yr ⎟⎜ f
⎢ ⎜ Rh Fyc ⎟⎜ λ rf − λ pf
⎠⎝
⎣ ⎝
λrf = 0.56
⎞⎤
⎟ ⎥ Rb Rh Fyc
⎟⎥
⎠⎦
E
Fyr
(6.10.8.2.22)
(6.10.8.2.25)
Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc
(Pg 6109)
Fyr = (0.7) ( 50ksi ) = 35ksi
λrf = 0.56
29, 000ksi
= 16.12
35ksi
Rb = 1.00 (since the web is nonslender)
Rh = 1.00 (since the section is nonhybrid)
Fnc ( FLB )
⎡ ⎛
⎞ ⎛ 10.67 − 9.152 ⎞ ⎤
35ksi
ksi
ksi
= ⎢1 − ⎜1 −
⎟ ⎥ (1.00)(1.00) ( 50 ) = 46.74
ksi ⎟ ⎜
⎣ ⎝ (1.00)(50 ) ⎠ ⎝ 16.12 − 9.152 ⎠ ⎦
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6a
AASHTOLRFD 2007
Created July 2007: Page 2 of 5
 148 
154.
Investigate Compression Flange LateralTorsional Buckling:
The unbraced length of the beam is Lb = 12’0” = 144.0”.
rt =
b fc
⎛ 1 Dc tw
12 ⎜1 +
⎜ 3b t
fc fc
⎝
L p = 1.0rt
⎞
⎟
⎟
⎠
=
(16")
⎛ ⎛ 1 ⎞ (19")( 38 ") ⎞
12 ⎜1 + ⎜ ⎟
⎟
⎝ ⎝ 3 ⎠ (16")( 3 4 ") ⎠
E
29, 000ksi
= (1.0)(4.220")
= 101.6"
Fyc
50ksi
Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc
Lr = π rt
= 4.220"
(6.10.8.2.39)
(6.10.8.2.34)
Fyr = (0.7) ( 50ksi ) = 35ksi
E
29, 000ksi
= (π)(4.220")
= 381.6"
Fyr
35ksi
(Pg 6110)
(6.10.8.2.35)
Since L p = 101.6" < Lb = 144" < Lr = 381.6" , Inelastic LTB must be investigated.
⎡ ⎛
F
Fnc ( LTB ) = Cb ⎢1 − ⎜1 − yr
⎜ Rh Fyc
⎢ ⎝
⎣
⎞⎛ Lb − Lp
⎟⎜
⎟⎜ Lr − Lp
⎠⎝
⎞⎤
⎟ ⎥ Rb Rh Fyc ≤ Rb Rh Fyc
⎟⎥
⎠⎦
(6.10.8.2.32)
⎡ ⎛
⎞ ⎛ 144"− 101.6" ⎞ ⎤
35ksi
ksi
ksi
⎟⎜
Fnc ( LTB ) = Cb ⎢1 − ⎜1 −
⎟ ⎥ (1.0 )(1.0 ) ( 50 ) ≤ (1.0 )(1.0 ) ( 50 )
⎜ (1.00 ) ( 50ksi ) ⎟ ⎝ 381.6"− 101.6" ⎠ ⎥
⎢
⎠
⎣ ⎝
⎦
Fnc ( LTB ) = ( Cb )( 0.9546 ) ( 50ksi ) = ( Cb ) ( 47.73ksi ) ≤ 50ksi
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6a
AASHTOLRFD 2007
Created July 2007: Page 3 of 5
 149 
155.
Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.
2
⎛ f ⎞
⎛ f ⎞
Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3
⎝ f2 ⎠
⎝ f2 ⎠
(6.10.8.2.37)
C
M 2 = M c = 1, 080kft →
f 2 = 23.86ksi
M o = M B = 810.0kft →
f o = 17.90ksi
M BC ,mid = 1, 013kft
f mid = 22.38ksi
→
B
f2
fmid
f1
Since the BMD is not concave,
f1 = 2 f mid − f 2 ≥ f o = (2) ( 22.38ksi ) − ( 23.86ksi ) = 20.90ksi ≥ 17.90ksi
2
⎛ 20.90 ⎞
⎛ 20.90 ⎞
Cb = 1.75 − 1.05 ⎜
⎟ + 0.3 ⎜
⎟ = 1.061 ≤ 2.3
⎝ 23.86 ⎠
⎝ 23.86 ⎠
→ Cb = 1.061
Fnc ( LTB ) = (1.061) ( 47.73ksi ) = 50.64ksi ≤ 50ksi
→ Fnc ( LTB ) = 50ksi
The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB):
Since Fnc ( LTB ) = 50ksi > Fnc ( FLB ) = 46.70ksi , FLB governs the strength of the compression
flange.
Fnc = Fnc ( FLB ) = 46.70ksi
φFnc = (1.00) ( 46.70ksi ) = 46.70ksi
Check fbu +
1
f l ≤ φ f Fnc
3
(6.10.8.1.11)
fbu = fC = 23.86ksi
Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0
Since fbu +
1
f l = 23.86ksi < φ f Fnc =46.70ksi , the compression flange is adequate.
3
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6a
AASHTOLRFD 2007
Created July 2007: Page 4 of 5
 150 
156.
Investigate the Strength of the Tension Flange:
Since the tension flange is discretely braced,
Fnt = Rh Fyt = (1.0) ( 50ksi ) = 50ksi
(6.10.8.31)
φFnt = (1.00) ( 50ksi ) = 50ksi
Check fbu +
1
f l ≤ φ f Fnt
3
(6.10.8.1.21)
fbu = fC = 23.86ksi
Since fbu +
1
f l = 23.86ksi < φ f Fnt =50.00ksi , the tension flange is adequate.
3
Since both the compression flange and tension flange are adequate, the section is adequate
for flexure.
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6a
AASHTOLRFD 2007
Created July 2007: Page 5 of 5
 151 
157.
AASHTO Flexural Example #6b:
Problem:
A noncomposite builtup girder made of M27050 steel is used to span 48’. The beam is braced
laterally at 12’0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes
the self weight of the beam. Check to see if the section is adequate considering flexural failure
modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6
to determine capacity.
Solution:
Determine Classification of the Section:
Check
2 Dc ?
E
≤ 5.7
tw
Fyc
(6.10.6.2.31)
Take D = 38”
Dc =
D 38"
=
= 19"
2
2
?
(2)(19")
29, 000ksi
= 101.3 ≤ 5.7
= 137.3
( 38 ")
50ksi
Check
I yc
I yt
OK, ∴ web is nonslender
?
≥ 0.3
(6.10.6.2.32)
Since Section is doubly symmetric, Iyc = Iyt
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
OK
AASHTOLRFD 2007
Created July 2007: Page 1 of 7
 152 
158.
Since the web is nonslender and Eq 6.10.6.2.32 is satisfied, we have the option of using either
AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member.
For this example, the provisions of A6 will be used and we will work with moments.
The following failure modes must be investigated:
• Flange Local Buckling of the Compression Flange Mnc(FLB)
• LateralTorsional Buckling
Mnc(LTB)
• Yielding of Tension Flange
Mnt
Mnc
Compute Web Plasticity Factors, Rpc and Rpt (Section A6.2):
Investigate the classification of the web.
Check
2 Dcp
tw
?
≤ λ pw( Dcp )
λ pw( Dcp ) =
(A6.2.11)
E
Fyc
⎛ Dcp ⎞
≤ λ rw ⎜
⎟
⎛ 0.54M p
⎞
⎝ Dc ⎠
− 0.09 ⎟
⎜
⎜ RM
⎟
h
y
⎝
⎠
(A6.2.12)
2
λ rw = 5.7
E
= 137.3
Fyc
(A6.2.13)
Rh = 1.00 (since the section is rolled and is ∴ nonhybrid)
M y = S x Fy = ( 543.1 in 3 )( 50ksi ) = 27,160kin = 2, 263kft
M p = Z x Fy = ( 600.4 in 3 )( 50ksi ) = 30, 020kin = 2,502kft
λ pw( Dcp ) =
29, 000ksi
50ksi
⎛ 19" ⎞
= 93.68 ≤ 137.3 ⎜
⎟ = 137.3
⎝ 19" ⎠
⎛ ( 0.54 ) ( 30, 020kin )
⎞
⎜
− 0.09 ⎟
⎜ (1.0 ) ( 27,160kin )
⎟
⎝
⎠
2
λ pw( Dcp ) = 93.68
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
AASHTOLRFD 2007
Created July 2007: Page 2 of 7
 153 
159.
2 Dcp
tw
=
?
(2)(19")
= 101.3 > λ pw( Dcp ) = 93.68
3 "
8
∴ web is non compact
Since the web is non compact,
⎡ ⎛ Rh M yc
R pc = ⎢1 − ⎜ 1 −
Mp
⎢ ⎜
⎣ ⎝
⎞ ⎛ λ w − λ pw( Dc ) ⎞ ⎤ M p M p
≤
⎟⎥
⎟⎜
⎟⎜ λ − λ
⎟
M yc
rw
pw ( Dc ) ⎠ ⎥ M yc
⎠⎝
⎦
(A6.2.24)
Where,
⎛D ⎞
λ pw( Dc ) = λ pw( Dcp ) ⎜ cp ⎟ ≤ λ rw
⎝ Dc ⎠
⎛ 19 '' ⎞
= 93.68 ⎜
⎟ ≤ 137.3
⎝ 19 '' ⎠
= 93.68 ≤ 137.3
(A6.2.26)
⎡ ⎛ (1.00)(27,160kin ) ⎞ ⎛ 101.3 − 93.68 ⎞ ⎤ (30, 020kin ) (30, 020kin )
R pc = ⎢1 − ⎜ 1 −
≤
⎟⎜
⎟⎥
(30, 020kin ) ⎠ ⎝ 137.3 − 93.68 ⎠ ⎦ (27,160kin ) (27,160kin )
⎣ ⎝
= 1.087 ≤ 1.105
= 1.087
⎡ ⎛ Rh M yt
R pt = ⎢1 − ⎜1 −
Mp
⎢ ⎜
⎣ ⎝
⎞ ⎛ λ w − λ pw( Dc ) ⎞ ⎤ M p M p
≤
⎥
⎟⎜
⎟ ⎜ λ rw − λ pw( D ) ⎟ ⎥ M yt M yt
⎟
⎠⎝
c ⎠⎦
(A6.2.25)
⎡ ⎛ (1.00)(27,160kin ) ⎞ ⎛ 101.3 − 93.68 ⎞ ⎤ (30, 020kin ) (30, 020kin )
R pt = ⎢1 − ⎜1 −
≤
⎟⎜
⎟⎥
(30, 020kin ) ⎠ ⎝ 137.3 − 93.68 ⎠ ⎦ (27,160kin ) (27,160kin )
⎣ ⎝
= 1.087 ≤ 1.105
= 1.087
Investigate Compression Flange Local Buckling:
Investigate the compactness of the compression flange.
λf =
b fc
2t fc
=
λ pf = 0.38
16"
= 10.67
(2)( 3 4 ")
(A6.3.23)
E
29, 000ksi
= 0.38
= 9.152
Fyc
50ksi
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
(A6.3.24)
AASHTOLRFD 2007
Created July 2007: Page 3 of 7
 154 
160.
Since λf >λpf, the flange is non compact and,
⎡ ⎛
F S ⎞⎛ λ − λ pf ⎞ ⎤
M nc ( FLB ) = ⎢1 − ⎜1 − yr xc ⎟⎜ f
(A6.3.22)
⎟ ⎥ R pc M yc
⎜
⎟⎜
⎟
⎢ ⎝ R pc M yc ⎠⎝ λ rf − λ pf ⎠ ⎥
⎣
⎦
⎛
⎞
S
Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5 Fyc ; Fyr = (0.7) ( 50ksi ) = 35ksi (Pg 6222)
S xc
⎝
⎠
Ekc
λrf = 0.95
(A.6.3.25)
Fyr
kc =
4
=
D
tw
4
38''
= 0.3974
(A6.3.26)
3 ''
8
( 29, 000 ) ( 0.3974 ) = 17.24
λ = 0.95
( 35 )
⎡ ⎛
⎤
⎞
( 35 )( 543.1 in ) ⎟ ⎛ 10.67 − 9.152 ⎞⎥ (1.087 ) 27,160
= ⎢1 − ⎜1 −
(
)
⎟
⎜
⎟⎜
⎢ ⎝ (1.087 ) ( 27,160 ) ⎠ ⎝ 17.24 − 9.152 ⎠ ⎥
⎣
⎦
ksi
rf
ksi
ksi
3
kin
M nc ( FLB )
kin
= 27,550kin = 2, 296kft
Investigate Compression Flange LateralTorsional Buckling:
The unbraced length of the beam is Lb = 12’0” = 144.0”.
rt =
b fc
⎛ 1 Dc tw
12 ⎜1 +
⎜ 3b t
fc fc
⎝
L p = 1.0rt
⎞
⎟
⎟
⎠
=
(16")
⎛ ⎛ 1 ⎞ (19")( 38 ") ⎞
12 ⎜ 1 + ⎜ ⎟
⎟
⎝ ⎝ 3 ⎠ (16")( 3 4 ") ⎠
= 4.220"
E
29, 000ksi
= (1.0)(4.220")
= 101.6"
Fyc
50ksi
E
Lr = 1.95 rt
Fyr
⎛ Fyr S xc h ⎞
1 + 1 + 6.76 ⎜
⎟
S xc h
⎝ E J ⎠
J
(A6.3.310)
(A6.3.34)
2
(6.10.8.2.35)
⎛
⎞
S
Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5 Fyc Fyr = ( 0.7 ) ( 50ksi ) = 35ksi (Pg 6222)
S xc
⎝
⎠
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
AASHTOLRFD 2007
Created July 2007: Page 4 of 7
 155 
161.
h = D + ( 1 2 ) ( t fc + t ft ) = 38"+ 3 4 " = 38.75"
J=
3
3
t fc
D tw b fc t fc ⎛
+
⎜1 − 0.63
3
3 ⎜
b fc
⎝
⎞ b ft t 3
ft
⎟+
⎟
3
⎠
⎛
t ft
⎜1 − 0.63
⎜
b ft
⎝
⎞
⎟
⎟
⎠
(A6.3.39)
⎡ (16")( 3 4 ")3 ⎛
(38")( 3 8 ")3
⎛ 34 " ⎞ ⎞⎤
4
+ (2) ⎢
J=
⎜1 − ( 0.63) ⎜ 16" ⎟ ⎟ ⎥ = 5.035 in
3
3
⎝
⎠ ⎠⎦
⎝
⎣
E 29, 000ksi
=
= 828.6
Fyr
35ksi
3
S xc h ( 543.1 in ) ( 38.75")
=
= 4,180
J
5.035 in 4
Lr = (1.95 )( 4.220")( 828.6 )
2
1
⎛ 4,180 ⎞
1 + 1 + 6.76 ⎜
⎟ = 396.8" = 33.06 '
4,180
⎝ 828.6 ⎠
Since L p = 101.6" < Lb = 144" < Lr = 396.8" , Inelastic LTB must be investigated.
⎡ ⎛
F S
M nc ( LTB ) = Cb ⎢1 − ⎜1 − yr xc
⎢ ⎜ R pc M yc
⎣ ⎝
M nc ( LTB )
⎞⎛ Lb − Lp
⎟⎜
⎟⎜ Lr − Lp
⎠⎝
⎞⎤
⎟ ⎥ R pc M yc ≤ R pc M yc
⎟⎥
⎠⎦
(A6.3.32)
⎡ ⎛
( 35ksi )( 543.1 in 3 ) ⎞ ⎛ 144"− 101.6" ⎞⎤ (1.087 ) 27,160kin ≤ (1.087 ) 27,160kin
⎢1 − ⎜1 −
⎟⎜
= Cb
(
)
(
)
⎟⎥
⎢ ⎜ (1.087 ) ( 27,160kin ) ⎟ ⎝ 396.8"− 101.6" ⎠ ⎥
⎠
⎣ ⎝
⎦
= ( Cb )( 0.9488 ) ( 29,520kin ) = ( Cb ) ( 28, 010kin ) ≤ 29,520kin
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
AASHTOLRFD 2007
Created July 2007: Page 5 of 7
 156 
162.
Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.
2
⎛M ⎞
⎛M ⎞
Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3
⎝ M2 ⎠
⎝ M2 ⎠
(A6.3.37)
M 2 = M c = 1, 080kft
M o = M B = 810.0kft
M BC ,mid = 1, 013kft
Since the BMD is not concave,
M 1 = 2M mid − M 2 ≥ M o = (2) (1, 013kft ) − (1, 080kft ) = 946kft ≥ 810kft
2
⎛ 946 ⎞
⎛ 946 ⎞
Cb = 1.75 − 1.05 ⎜
⎟ + 0.3 ⎜ 1, 080 ⎟ = 1.061 ≤ 2.3
⎝ 1, 080 ⎠
⎝
⎠
= 1.061
M nc ( LTB ) = (1.061) ( 28, 010 kin ) = 29, 720kin ≤ 29,520kin
M nc ( LTB ) = 29,520kin = 2, 460kft
The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB):
Since M nc ( LTB ) = 2, 460kft > M nc ( FLB ) = 2, 296kft , ∴ FLB governs the strength of the
compression flange.
M nc = M nc ( FLB ) = 2, 296kft
φM nc = (1.00) ( 2, 296kft ) = 2, 296kft
Check M u +
1
f l S xc ≤ φ f M nc
3
(A6.1.11)
M u = M C = 1, 080kft
Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0
Since M u +
1
fl S xc = 1, 080kft < φ f M nc =2,296kft , the compression flange is adequate.
3
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
AASHTOLRFD 2007
Created July 2007: Page 6 of 7
 157 
163.
Investigate the Strength of the Tension Flange:
Since the tension flange is discretely braced,
M nt = R pt M yt = (1.087) ( 27,160kin ) = 29,520kin = 2, 460kft
(A6.41)
φM nt = (1.00) ( 2, 460kft ) = 2, 460kft
Check M u +
1
fl S xt ≤ φ f M nt
3
(6.10.8.1.21)
M u = M C = 1, 080kft
Since M u +
1
f l S xt = 1, 080kft < φ f M nt =2,460kft , the tension flange is adequate.
3
Since both flanges are adequate, the section is adequate for flexure.
Note that the benefits of using Appendix A6 are illustrated here. Even though the capacity
was found to be adequate in both Examples #6a and #6b, using Appendix A6, the capacity
was found to be 16% greater than the capacity found using the provisions in Section 6.10.8.
ODOTLRFD Short Course  Steel
AASHTO Flexure Example #6b
AASHTOLRFD 2007
Created July 2007: Page 7 of 7
 158 
164.
AASHTO Shear Strength Example #1:
Problem:
Check the beam shown below to see if it has adequate shear strength and web strength to resist
the factored loads shown. The beam is a W27x94 made of M27050 steel.
75kip
12'
30kip
12'
12'
Solution:
Draw the shear force diagram.
60kip
SFD(kip)
15kip
45kip
kip
From the diagram, Vu = 60 .
Referring to Section 6.10.9.2 of the Specification, Check Design Shear Strength:
Vn = CVp
(6.10.9.21)
V p = 0.58Fy Dtw
(6.10.9.22)
D = d − 2t f = 26.9"− (2)(0.745") = 25.41"
V p = (0.58)(50ksi ) [ (25.41")(0.490") ] = 361.1kip
ODOTLRFD Short Course  Steel
AASHTO Shear Strength Example #1
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 159 
165.
Since the web is unstiffened, k = 5.00.
Ek
=
Fy
( 29, 000 ) ( 5.00 ) = 53.85
( 50 )
ksi
ksi
D 25.41"
=
= 51.86
tw 0.490"
1.12
Ek
= (1.12 )( 53.85 ) = 60.31
Fy
Since
D
Ek
= 51.86 < 1.12
= 60.31 , shear yielding governs and,
tw
Fy
C = 1.00
(6.10.9.3.24)
Vn = CVp = (1.00)(361.1kip) = 361.1kip
φVn = (1.00)(361.1kip) = 361.1kip > Vu = 60kip
ODOTLRFD Short Course  Steel
AASHTO Shear Strength Example #1
O.K.
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 160 
166.
AASHTO Shear Strength Example #2:
Problem:
A builtup section made of M27050 steel, is used as a beam. Determine the design shear
capacity of the beam and determine if the beam can sustain a factored shear force of 242kip.
Solution:
Referring to Section 6.10.9.2 of the Specification:
Vn = CVp
(6.10.9.21)
V p = 0.58Fy Dtw
(6.10.9.22)
V p = (0.58)(50ksi ) [ (38")( 3 8 ") ] = 413.3kip
Since the web is unstiffened, k = 5.00.
Ek
=
Fy
( 29, 000 ) ( 5.00 ) = 53.85
( 50 )
ksi
ksi
1.12
Ek
= (1.12 )( 53.85 ) = 60.31
Fy
1.40
Ek
= (1.40 )( 53.85 ) = 75.39
Fy
Since
D
Ek
= 101.3 > 1.40
= 75.39 , Elastic shear buckling governs and,
tw
Fy
D 38"
=
= 101.3
tw 38 "
⎛ Ek ⎞ (1.57 ) ⎛ ( 29, 000ksi ) ( 5.00 ) ⎞
⎜
⎟ = 0.4437
C=
⎟=
2 ⎜
2
⎟
( 50ksi )
( D / tw ) ⎜ Fy ⎟ (101.3) ⎜
⎝
⎠
⎝
⎠
1.57
(6.10.9.3.26)
Vn = CVp = (0.4437) (413.3kip) = 183.4kip
φVn = (1.00)(183.4kip) = 183.4kip < Vu = 242kip
ODOTLRFD Short Course  Steel
AASHTO Shear Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 161 
167.
Try adding transverse stiffeners to the web to increase the shear strength. A panel aspect ratio of
1.25 to 1.50 looks good…
do
D
→ do
1.25
1.25 D = (1.25)(38") = 47.5" say 48" ∴
d o 48"
=
= 1.263
D 38"
For stiffened webs,
k = 5+
Ek
=
Fy
5
5
= 5+
= 8.134
2
(d o / D)
(1.263) 2
( 29, 000 ) (8.134 ) = 68.68
( 50 )
ksi
ksi
1.12
Ek
= ( 68.68 )(1.12 ) = 76.92
Fy
1.40
Since
(6.10.9.3.27)
Ek
= ( 68.68 )(1.40 ) = 96.15
Fy
D
Ek
= 101.3 > 1.40
= 96.15 , Elastic shear buckling governs and,
tw
Fy
⎛ Ek ⎞ (1.57 ) ⎛ ( 29, 000ksi ) ( 8.134 ) ⎞
⎜
⎟ = 0.7218
C=
⎟=
2 ⎜
2
⎟
( 50ksi )
( D / tw ) ⎜ Fy ⎟ (101.3) ⎜
⎝
⎠
⎝
⎠
1.57
(6.10.9.3.26)
Vn = CVp = (0.7218) (413.3kip) = 298.3kip
φVn = (1.00)(298.3kip) = 298.3kip > Vu = 242kip
ODOTLRFD Short Course  Steel
AASHTO Shear Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 162 
168.
The previous calculations were based on the buckling strength of the web. For interior panels
where:
2 Dtw
≤ 2.5
( b fct fc + b ft t ft )
(6.10.9.3.21)
(2)(38")( 38 ")
28.5
=
= 1.188 ≤ 2.5
[(16")( 3 4 ") + (16")( 3 4 ")] 24.0
OK
Tension Field Action can be developed:
⎡
⎤
⎢
⎥
0.87(1 − C ) ⎥
⎢
Vn = V p ⎢C +
2 ⎥
⎛ do ⎞ ⎥
⎢
1+ ⎜ ⎟
⎢
⎝D⎠ ⎥
⎣
⎦
(6.10.9.3.22)
⎡
⎤
⎢ (0.7218) + (0.87)(1 − 0.7218) ⎥ = 360.4kip
Vn = (413.3 )
2
⎢
⎥
1 + (1.263)
⎣
⎦
kip
φVn = (1.00)(360.4kip) = 360.4kip
ODOTLRFD Short Course  Steel
AASHTO Shear Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 163 
170.
AASHTO Web Strength Example #1:
Problem:
Check the beam shown below to see if it has adequate shear strength and web strength to resist
the factored loads shown. The beam is a W27x94 made of M27050 steel.
75kip
12'
30kip
12'
12'
Solution:
Draw the shear force diagram.
60kip
SFD(kip)
15kip
45kip
Referring to Section D6.5 of the Specification, Check the End Reactions
for Web Yielding and Web Crippling: (Assume that the bearing length, N, is 31/4”.)
Check Web Yielding
Since the supports are likely to be at a distance less than or equal to d from the end of the
member:
Rn = (2.5k + N ) Fywtw
(D6.5.23)
Rn = ((2.5)(1.34") + 3.25")(50ksi )(0.490") = 161.7 kip
φRn = (1.0)(161.7 kip ) =161.7 kip > 60kip
O.K.
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #1
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 165 
171.
Check Web Crippling
Since the supports are likely to be at a distance less than or equal to d/2 from the end of
the member.
Check
N
:
d
3.25"
= 0.1208 ≤ 0.20
26.9"
Therefore, (D6.5.33) controls: Rn = 0.40t
2
w
1.5
⎡
⎤
⎛ N ⎞ ⎛ tw ⎞ ⎥ EFywt f
⎢1 + 3 ⎜ ⎟ ⎜ ⎟
tw
⎢
⎝ d ⎠⎜ tf ⎟ ⎥
⎝ ⎠ ⎦
⎣
1.5
ksi
ksi
⎡
⎛ 3.25" ⎞⎛ 0.490" ⎞ ⎤ (29, 000 )(50 )(0.745")
Rn = 0.40(0.490") 2 ⎢1 + 3 ⎜
⎟⎜
⎟ ⎥
0.490"
⎝ 26.9" ⎠⎝ 0.745" ⎠ ⎥
⎢
⎣
⎦
Rn = (0.09604 in 2 )(1.193)(1485ksi ) =170.2kip
φRn = (0.80)(170.2kip ) =136.2kip > 60kip
O.K.
Check the Interior Concentrated Loads for Web Yielding and Web Crippling:
(Assume that the bearing length, N, is 3.25”)
Check Web Yielding
Since the applied load is located at a distance greater than d from the end of the member.
Rn = (5k + N ) Fywtw
(D6.5.22)
Rn = ((5)(1.34") + 3.25")(50ksi )(0.490") = 243.8kip
φRn = (1.0)(243.8kip ) = 243.8kip > 75kip
O.K.
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #1
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 166 
172.
Check Web Crippling
Since the applied load is located at a distance greater than d/2 from the end of the
member.
2
Therefore, (D6.5.32) controls: Rn = 0.80tw
1.5
⎡
⎤
⎛ N ⎞ ⎛ tw ⎞ ⎥ EFywt f
⎢1 + 3 ⎜ ⎟ ⎜ ⎟
tw
⎢
⎝ d ⎠⎜ tf ⎟ ⎥
⎝ ⎠ ⎦
⎣
1.5
ksi
ksi
⎡
⎛ 3.25" ⎞⎛ 0.490" ⎞ ⎤ (29, 000 )(50 )(0.745")
Rn = 0.80(0.490") ⎢1 + 3 ⎜
⎥
⎟⎜
⎟
0.490"
⎝ 26.9" ⎠⎝ 0.745" ⎠ ⎥
⎢
⎣
⎦
2
Rn = (0.1921 in 2 )(1.193)(1485ksi ) = 340.3kip
φRn = (0.80)( 340.3kip ) = 272.2kip > 75kip
O.K.
The Web Yielding and Web Crippling Strengths are Satisfactory
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #1
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 167 
173.
AASHTO Web Strength Example #2:
Problem:
A builtup section made of M27050 steel, is used as a beam. It was determined in AASHTO
Shear Strength Example #2 that intermediate stiffeners were required to develop adequate shear
strength in the web. Determine the required size of these intermediate stiffeners. And check the
web to see if an end reaction of 128kip can be supported.
Solution:
Design the intermediate stiffeners that were added to increase
the shear strength:
The moment of inertia of the intermediate stiffeners should satisfy the smaller of:
3
I t ≥ bt w J
(6.10.11.1.31)
and
1.5
D 4 ρt1.3 ⎛ Fyw ⎞
It ≥
⎜
⎟
40 ⎝ E ⎠
(6.10.11.1.32)
where:
It  Moment of inertia of the stiffener pair about the midthickness of the web.
2
⎛ D ⎞
2.5
− 2.0 ≥ 0.5
J = 2.5 ⎜
⎟ − 2.0 ≥ 0.5 use… J =
2
( do / D )
⎝ do / D ⎠
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
(6.10.11.1.33)
AASHTOLRFD 2007
Created July 2007: Page 1 of 6
 168 
174.
J=
2.5
( 48"/ 38")
2
− 2.0 = −0.4332 in 2 ≥ 0.5
b = smaller of do and D,
take J = 0.50
b = 38”
ρ = larger of Fyw / Fcrs and 1.00
0.31E
Fcrs =
⎛ bt
⎜
⎜t
⎝ p
⎞
⎟
⎟
⎠
2
≤ Fys
Fcrs =
(0.31)(29, 000ksi )
⎛ 6" ⎞
⎜ 3 "⎟
⎝ 8 ⎠
ρ = 50 / 35.11 ≥ 1.00 ρ =1.424
3
btw j = (38")( 3 8 ")3 (0.5) = 1.002 in 4
1.5
D 4 ρt1.3 ⎛ Fyw ⎞
⎜
⎟
40 ⎝ E ⎠
It =
2
= 35.11ksi ≤ 50ksi
Using a 6” wide stiffener is based on the
assumption that 6” bar stock can be
used, which should be readily available.
Base the width of 3/8” on engineering
judgment of minimum thickness of
stiffener.
1.5
(38") 4 (1.424)1.3 ⎛ 50 ksi ⎞
=
⎜
ksi ⎟
40
⎝ 29, 000 ⎠
(ts )(2bs + tw )3
12
= 5.909 in 4
take bs = 6”
(t ) [ (2)(6") + 38 "]
= (ts )157.9 in 3
It = s
12
3
tp ≥
3
btw J
1.002 in 4
=
= 0.006345"
157.9 in 3 157.9 in 3
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
…say ts = 3 8 "
AASHTOLRFD 2007
Created July 2007: Page 2 of 6
 169 
175.
Referring to SectionD6.5 of the Manual, Check the End Reactions
for Web Yielding and Web Crippling:
The bearing length, N, is 9” and we’ll assume that 3/8” fillet welds connect the flanges and web.
This gives an effective “k” distance of 3/4” + 3/8” = 1.125”
Check Web Yielding
Since the supports are likely to be at a distance less than or equal to d from the end of the
member.
Rn = (2.5k + N ) Fywtw
(D6.5.23)
Rn = ((2.5)(1.125") + 9")(50ksi )( 3 8 ") = 221.5kip
φRn = (1.0)(221.5kip ) = 221.5kip > 128kip
O.K.
Check Web Crippling
Since the supports are likely to be at a distance less than or equal to d/2 from the end of
the member.
N
:
d
9"
= 0.2278 > 0.20
38"+ (2)( 3 4 ")
Check
Therefore, (D6.5.34) controls: Rn = 0.40t
2
w
1.5
⎡
⎤
⎛ 4N
⎞ ⎛ tw ⎞ ⎥ EFywt f
⎢1 + ⎜
− 0.2 ⎟ ⎜ ⎟
tw
⎢ ⎝ d
⎠⎜ tf ⎟ ⎥
⎝ ⎠ ⎦
⎣
1.5
⎡ ⎛
⎞ ⎛ 3 8 " ⎞ ⎤ (29, 000ksi )(50ksi )( 3 4 ")
(4)(9")
Rn = (0.40)( 8 ") ⎢1 + ⎜
− 0.2 ⎟ ⎜
⎟ ⎥
( 3 8 ")
⎠ ⎝ 34 " ⎠ ⎥
⎢ ⎝ (38"+ (2)( 3 4 ")
⎣
⎦
3
2
Rn = (0.05625 in 2 )(1.252)(1, 703ksi ) =119.9kip
φRn = (0.80)(119.9kip ) = 95.92kip <128kip
No Good.
The web strength is satisfactory with regard to web yielding but not for web crippling.
Bearing stiffeners will need to be added. (Technically speaking, stiffeners are required by
AASHTO at all bearing locations on builtup sections anyways…)
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 3 of 6
 170 
176.
Design the bearing stiffeners that need to be added to increase the web crippling strength:
Check local buckling of the bearing stiffener:
bt ≤ 0.48t p
bt ≤ 0.48t p
tp ≥
E
Fy
take Fy = 50ksi
29, 000ksi
= 11.56t p
50ksi
bt
7"
=
= 0.6055"
11.56 11.56
(6.10.11.2.21)
take bs = 7”
take ts = 5 8 "
Taking Fy = 50ksi here is a conservative
assumption since I am not sure what
material will actually be used.
(7” x 5/8” bar stock may be used)
Check the bearing stiffeners as an effective column section:
3
3
I = ( 112 ) ⎡( 5 8 ")( 7"+ 3 8 "+ 7") + ( 6.75"− 5 8 ")( 3 8 ") ⎤
⎣
⎦
4
= 154.7 in
A = ⎡( 5 8 ")( 2 )( 7") + ( 6.75")( 3 8 ") ⎤ = 11.28 in 2
⎣
⎦
r=
I
154.7 in 4
=
= 3.704"
A
11.28 in 2
⎛ KL ⎞ ⎛ (0.75)(38") ⎞
⎜
⎟=⎜
⎟ = 7.695
⎝ r ⎠ ⎝ 3.704" ⎠
2
2
ksi
⎞
⎛ KL ⎞ Fy ⎛ 7.695 ⎞ ⎛ 36
λ=⎜
=⎜
= 0.007448
⎜
⎟
⎟
ksi ⎟
⎝ r π ⎠ E ⎝ π ⎠ ⎝ 29, 000 ⎠
Pn = 0.66λ Fy As
(
Inelastic Buckling (6.9.4.13)
Taking Fy = 36ksi here is a conservative
assumption since I am not sure what
material will be used.
)
= 0.66( 0.007448) ( 36ksi )(11.28 in 2 ) = 404.8kip
(6.9.4.11)
φPn = ( 0.90 ) ( 404.8kip ) = 364.3kip
In this solution, it is assumed that the bearing stiffener is located at the middle of the 9”
wide plate. Thus, there is at least 4.5” of web between the stiffener and the end of the
girder, which is greater than 9tw.
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 4 of 6
 171 
177.
Check bearing stress between the end of the bearing stiffeners and the loaded flange:
Rn = 1.4 Apn Fys
(6.10.11.2.31)
The corners of the stiffeners are clipped 1”
horizontal and 2 1/2” vertical to provide clearance
for the flangetoweb welds
Apn = (2)(7"− 1") ( 5 8 ") = 7.50 in 2
Rn = (1.4)(7.50 in 2 )(36ksi ) = 378.0kip
φRn = (1.00)(378.0kip ) = 378.0kip
The capacity of the bearing stiffeners is governed by the “equivalent column”
capacity. φRn = 364kip, which is greater than the reaction of 128kip.
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 5 of 6
 172 
178.
Just for fun ☺, lets check the capacity of 2 pairs of 7” x 5/8” interior bearing stiffeners:
The local buckling check will be the same as for the single pair of bearing stiffeners.
Check the bearing stiffeners as an effective column section:
3
3
I = ( 112 ) ⎡ (2) ( 5 8 ")( 7"+ 3 8 "+ 7") + ( (2)(3.375") + 7"− (2) ( 5 8 ") ) ( 3 8 ") ⎤ = 309.5 in 4
⎣
⎦
A = ⎡( 5 8 ")( 4 )( 7") + ⎡(2) ( 3.375") + 7"⎤ ( 3 8 ") ⎤ = 22.66 in 2
⎣
⎦
⎣
⎦
r=
I
309.5 in 4
=
= 3.696"
A
22.66 in 2
2
2
ksi
⎛ KL ⎞ Fy ⎛ 7.711 ⎞ ⎛ 36
λ=⎜
=⎜
⎟
⎟ ⎜
ksi
⎝ r π ⎠ E ⎝ π ⎠ ⎝ 29, 000
(
⎛ KL ⎞ ⎛ (0.75)(38") ⎞
⎜
⎟=⎜
⎟ = 7.711
⎝ r ⎠ ⎝ 3.696" ⎠
⎞
⎟ = 0.007479
⎠
)
Pn = 0.66λ Fy As = 0.66( 0.007479) ( 36ksi )( 22.66 in 2 ) = 813.2kip
Inelastic Buckling (6.9.4.13)
(6.9.4.11)
φPn = ( 0.90 ) ( 813.2kip ) = 731.9kip
The bearing stress between the end of the bearing stiffeners and the loaded flange would be twice
that calculated for a single pair of stiffeners: φRn = (2)(378.0kip) = 756.0kip.
The strength is governed again by the “equivalent column” capacity, φRn = 732kip.
ODOTLRFD Short Course  Steel
AASHTO Web Strength Example #2
AASHTOLRFD 2007
Created July 2007: Page 6 of 6
 173 
180.
AASHTO Connection Example #1:
Problem:
A C12x30 is used as a tension member as is shown in the sketch below. The channel is made of
M27036 material and is attached to the gusset plate with 8, 7/8” diameter M164 (A325) bolts.
The gusset is 5/8” thick and made of M27036 steel. Calculate the design capacity, φPn, of the
connection considering the failure modes of bolt shear, bolt bearing, and block shear. Also
compute the load which will cause slip of the connection.
Solution:
Section AA
Shear Strength of the Bolts:
Assume that the threads are included in the
shear plane of the connection.
3"
Rn = 0.38 Ab Fub N s
6"
3"
1.5"
(6.13.2.72)
3"
3"
2
⎛π⎞
Ab = ⎜ ⎟ ( 7 8 ") = 0.6013 in 2
⎝4⎠
3"
For A325 bolts, Fub = 120ksi
Bolts are in single shear so Ns = 1
(
)(
A
A
)
kip
Rn = (0.38) 0.6013 in 2 120ksi (1) = 27.42 bolt
(
C12 x 30
)
kip
kip
φRn = (0.80) 27.42 bolt = 21.94 bolt
(
For all 8 bolts, φRn = (8 bolts) 21.94
kip
bolt
) = 175.5
ODOTLRFD Short Course  Steel
AASHTO Connection Example #1
Pu
kip
AASHTOLRFD 2007
Created July 2007: Page 1 of 4
 175 
182.
Since the channel web is thinner than the gusset plate and they’re made of the same material,
block shear of the channel will govern over block shear of the gusset plate.
Check Block Shear in the Channel Web:
Atg = (6")(0.510") = 3.060 in 2
Atn = ⎡(6") − (2) ( 1 2 )( 7 8 "+ 18 ") ⎤ (0.510") = 2.550 in 2
⎣
⎦
Avg = (2) [ (1.5") + (3)(3") ] (0.510") = 10.71 in 2
Avn = (2) ⎡ (1.5") + (3)(3") − (3.5) ( 7 8 "+ 18 ") ⎤ (0.510") = 7.140 in 2
⎣
⎦
?
Atn ≥ 0.58 Avn
?
2.550 in 2 ≥(0.58)(7.140 in 2 ) = 4.141 in 2
NO!
∴ Rn = 0.58Fu Avn + Fy Atg
(
)(
(6.13.42)
) (
)(
)
Rn = (0.58) 58ksi 7.140 in 2 + 36ksi 3.060 in 2 = 350.3kip
(
)
φRn = (0.80) 350.3kip = 280.3kip
The Shear Strength of the Bolts Governs, φRn = 176kip
ODOTLRFD Short Course  Steel
AASHTO Connection Example #1
AASHTOLRFD 2007
Created July 2007: Page 3 of 4
 177 
183.
Check the Slip Capacity of the Connection:
Rn = K h K s N s Pt
(6.13.2.81)
Kh = 1.00 (for standard holes)
Ks = 0.33 (assume Class A surface)
Ns = 1
Pt = 39kip (from Table 6.13.2.81 for M164 Bolts)
(
)
kip
Rn = (1.00)(0.33)(1) 39kip = 12.87 bolt
For All 8 Bolts:
(
)
kip
Rn = (8 bolts) 12.87 bolt = 103.0kip
ODOTLRFD Short Course  Steel
AASHTO Connection Example #1
AASHTOLRFD 2007
Created July 2007: Page 4 of 4
 178 
184.
AASHTO Connection Example #2:
Problem:
An 8” long WT 10.5 x 66 is attached to the bottom flange of a beam as is shown below. This
hanger must support a factored load of 120kip. Given that 4, 1” diameter M164 (A325) bolts are
used to attach the hanger to the beam, investigate the adequacy of the bolts and tee flange.
Solution:
Prying must be investigated in this situation:
⎛ 3b t 3 ⎞
Qu = ⎜ − ⎟ P
⎝ 8a 20 ⎠
b=
(6.13.2.10.41)
gt
7"
− k1 = − 1 1 8 " = 2.375"
2
2
(k1 = 11/8” for W21 x 132)
a = ( 1 2 ) ( b f − gt ) = ( 1 2 )(12.4"− 7") = 2.700"
⎛ (3)(2.375") (1.04")3 ⎞
Qu = ⎜
−
⎟ Pu = 0.2736 Pu
20 ⎠
⎝ (8)(2.700")
Tu = Qu + Pu = 1.274 Pu = (1.274) (120kip ) = 152.8kip
Tensile Resistance of the Bolts:
Tn = 0.76 Ab Fub
(6.13.2.10.21)
2
⎛π⎞
Ab = ⎜ ⎟ (1") = 0.7854 in 2
⎝4⎠
Fub = 120ksi
kip
Tn = (0.76) ( 0.7854 in 2 )(120ksi ) = 71.63 bolt
For All 4 Bolts:
(
)
kip
φTn = (4 bolts) 57.30 bolt = 229.2kip
ODOTLRFD Short Course  Steel
AASHTO Connection Example #2
(
)
kip
kip
φTn = (0.80) 71.63 bolt = 57.30 bolt
OK
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 179 
185.
Check the Strength of the Flange of the WT:
Assume that a plastic mechanism forms between the bolt lines and stem.
Q
T
T
Q
Moment Equilibrium about at the fillet:
∑ M → 2M
u
⎛ Pb⎞
=⎜ u ⎟
⎝ 2 ⎠
⎛ (120kip ) (2.375") ⎞
⎟
Mu = ⎜
⎜
⎟
4
⎝
⎠
kin
= 71.25
For Safety, φM p ≥ M u .
Tu
Mu
Mu
Pu/2
b
Pu
Lt 2
⎛ (8")(1.04") ⎞
ksi
kin
Fy = ⎜
⎟ ( 50 ) = 108.2
4
4
⎝
⎠
kin
φM p = (1.00) (108.2 ) = 108.2kin
OK
Mp =
ODOTLRFD Short Course  Steel
AASHTO Connection Example #2
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 180 
186.
AASHTO Connection Example #3:
Problem:
Assuming an unfactored fatigue load of 60kip, determine the fatigue life of the tension bolts in the
previous example.
Solution:
For Safety, γ ( Δf ) ≤ ( ΔF )n
(6.6.1.2.21)
kip
γ ( ΔP ) (0.75) ⎡(1.274 ) ( 60 ) ⎤
⎣
⎦ = 18.24ksi
γ ( Δf ) =
=
Abolts
(4) ( 0.7854 in 2 )
⎛ A ⎞ 3 ( ΔF )TH
=⎜ ⎟ ≥
2
⎝N⎠
1
( ΔF )n
For infinite life,
( ΔF )n =
( ΔF )TH
2
31.0ksi
=
= 15.5ksi
2
Since γ ( Δf ) = 18.24ksi > ( ΔF )n = 15.5ksi , the bolts will have a finite life
1
For finite life, γ ( Δf ) ≤ ( ΔF )n
N≤
A
( γ ( Δf ) )
3
⎛ A ⎞3
=⎜ ⎟
⎝N⎠
=
17.1× 108 ksi3
(18.24 )
ksi 3
= 281,800 cycles
Note that if prying is not included, γ ( Δf ) = 14.32ksi and the calculations
would incorrectly show that the bolts have an infinite fatigue life.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #3
AASHTOLRFD 2007
Created July 2007: Page 1 of 1
 181 
187.
AASHTO Connection Example #4:
Problem:
Suppose that the hanger depicted in Examples #2 and #3 is subjected to a force that is applied at
an angle as is shown below. Determine if the connection is adequate in this configuration.
Solution:
Vu =
2 Pu
= 0.8944 Pu = 107.3kip
5
Vu =
107.3kip
kip
= 26.83 bolt
4 bolts
Tu =
1.274 Pu
= 0.5694 Pu = 68.33kip
5
Tu =
68.33kip
kip
= 17.08 bolt
4 bolts
Assume that the threads are included in the shear plane of the connection.
Vn = Rn = 0.38 Ab Fub N s
(6.13.2.72)
(
)(
)
kip
Vn = (0.38) 0.7854 in 2 120ksi (1) = 35.81 bolt
kip
Vu 26.83 bolt
=
= 0.7491 ,
kip
Vn 35.81 bolt
Tn = (0.76) ( 0.7854 in
(
2
∴Tn = 0.76 Ab Fub
)(120 )
ksi
⎛ V ⎞
1− ⎜ u ⎟
⎝ φVn ⎠
2
(6.13.2.112)
⎛ ( 26.83 kip bolt ) ⎞
kip
1− ⎜
= 25.11 bolt
⎜ (0.80) ( 35.81 kip bolt ) ⎟
⎟
⎝
⎠
2
)
kip
kip
φTn = (0.80) 25.11 bolt = 20.09 bolt
kip
kip
Since φTn = 20.09 bolt > Tu = 17.08 bolt , the bolts are OK for the loading shown.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #4
AASHTOLRFD 2007
Created July 2007: Page 1 of 3
 182 
188.
Check Bearing of the Flange of the WT:
It is given that the WT is 8” long. Since the minimum spacing is 3”, we’ll assume that an
end distance of 2” is provided resulting in a spacing of 4” bolttobolt.
Interior Bolts:
Lc = 4"− ( 2 )( 1 2 )(1"+ 116 ") = 2.938"
Since Lc = 2.938” > 2d = 2”,
Rn = 2.4dtFu
(6.13.2.91)
(
)
kip
Rn = (2.4) (1") (1.04") 65ksi = 162.2 bolt
End Bolts:
Lc = 2"− ( 1 2 )(1"+ 116 ") = 1.469"
Since Lc = 1.469” < 2d = 2”,
Rn = 1.2 Lc tFu
(6.13.2.92)
(
)
kip
Rn = (1.2) (1.469") (1.04") 65ksi = 119.1 bolt
For all 4 Bolts:
(
)
(
)
kip
kip
Rn = (2 bolts) 162.2 bolt + (2 bolts) 119.1 bolt = 562.8kip
(
)
φRn = (0.80) 562.8kip = 450.1kip
Since φRn = 450kip > Vu = 107 kip , the flange of the WT will be OK in bearing.
Note that since the flange thickness of the W24x176 is greater than that of the
WT10.5x66 and they are made of the same material, bearing of the WT will govern
over bearing of the W24x176.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #4
AASHTOLRFD 2007
Created July 2007: Page 2 of 3
 183 
189.
Check Shear in the Stem of the WT:
Rn = 0.58 Ag Fy
(6.13.5.32)
Rn = (0.58) [ (8")(0.650") ] ( 50ksi ) = 150.8kip
φRn = (1.00) (150.8kip ) = 150.8kip
Since φRn = 150.8kip > Vu = 107.3kip , the stem will be satisfactory in shear.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #4
AASHTOLRFD 2007
Created July 2007: Page 3 of 3
 184 
190.
AASHTO Connection Example #5:
Problem:
An L6 x 4 x 1/2, M27036, is welded to a 3/8” thick gusset plate made of M27050 steel. The
long leg of the angle is attached using two, 8” long fillet welds. The capacity of the angle was
previously computed as φPn = 163kip based on Gross Yielding. Determine the weld size required
to develop the full capacity of the member.
Solution:
Design the Welds:
Use an E70 Electrode since the
gusset has a strength of Fu = 65ksi.
The maximum weld size is 1/2”  1/16” = 7/16”.
Since the gusset and angle are both less than 3/4” thick, the
minimum weld size that can be used is 1/4”.
φRn , weld = 0.6φe 2 Fexx Aw ≥ φPn , member
φRn, weld = ( 0.6 )( 0.80 ) ( 70ksi ) × ⎡(0.7071) ( w ) ⎤ ( 2 )( 8") ≥ 163kip
⎣
⎦
w≥
163kip
= 0.4288"
kip
380.1 inch
→ Say
ODOTLRFD Short Course  Steel
AASHTO Connection Example #5
7
16
"
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 185 
191.
Check Tension for the Gusset:
Check tension on the Whitmore
section: Since the overall width of
the gusset is not given, I’ll check the
Whitmore width assuming that it
governs. If the overall width is less
than the Whitmore width, these
calculations will be unconservative.
Compute the width of the Whitmore section:
Lw = 6"+ ( 2 )( 8") Tan ( 30° ) = 15.24"
Gross Section Yielding:
(
)
(
)
φPn = φFy Ag = (0.95) 50ksi ⎡(15.24")( 3 8 ") ⎤ = 271.4kip
⎣
⎦
Net Section Fracture:
φPn = φFu Ae = (0.80) 65ksi ⎡(15.24")( 3 8 ") ⎤ (1.00 ) = 297.1kip
⎣
⎦
(Taking U = 1.00)
Check Block Shear in the Gusset Plate:
Atg = Atn = (6") ( 3 8 ") = 2.250 in 2
?
Atn ≥ 0.58 Avn
Avg = Avn = (2) ( 8")( 3 8 ") = 6.000 in 2
?
→
2.250 in 2 ≥(0.58)(6.000 in 2 ) = 3.480 in 2
∴ Rn = 0.58Fu Avn + Fy Atg
(
)(
(6.13.42)
) (
Rn = (0.58) 65ksi 6.000 in 2 + 50ksi
(
NO!
)( 2.250 in ) = 338.7
2
kip
)
φRn = (0.80) 338.7 kip = 271.0kip
Use 7/16” x 8” Fillet Welds made with an E70 Electrode
ODOTLRFD Short Course  Steel
AASHTO Connection Example #5
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 186 
192.
AASHTO Connection Example #6a:
Use the elastic vector method to compute the maximum force on any bolt in the
eccentrically loaded bolt group shown in the figure below. The bolts are all the
same size. (Example 4.12.1 from Salmon & Johnson)
Problem:
4"
P
3"
B
C
D
E
F
3"
3"
A
Solution:
τ=
Tr
J
→ T = ( 24kip ) ( 3"+ 2") = 120kin
J = ∑ Ad 2 = A∑ d 2
(
⎡
J = ⎢( 4 ) (2")2 + (3") 2
⎣
J = 47.12 in 4
) + ( 2)( 2") ⎤⎥⎦ ⎛⎜⎝ π ⎞⎟⎠ (1")
4
2
Tr (120
Corner Bolts: τ =
=
J
τ = 9.182ksi
2
kin
)(
2
(2") 2 + (3") 2
47.12 in
)
4
2
⎛π⎞
V = ( 9.182ksi ) ⎜ ⎟ (1") = 7.211kip
⎝4⎠
ODOTLRFD Short Course  Steel
AASHTO Connection Example #6a
AASHTOLRFD 2007
Created July 2007: Page 1 of 2
 187 
193.
Force acts perpendicular to line drawn from bolt to C.G.
Breaking force into horizontal and vertical
components…
⎛ 2 ⎞
⎛ 2 ⎞
kip
kip
Vy = ⎜
⎟ (V ) = ⎜
⎟ ( 7.211 ) = −4.000
⎝ 3.606 ⎠
⎝ 3.606 ⎠
⎛ 3 ⎞
⎛ 3 ⎞
kip
kip
Vx = ⎜
⎟ (V ) = ⎜
⎟ ( 7.211 ) = 6.000
3.606 ⎠
3.606 ⎠
⎝
⎝
Add evenly distributed vertical force to the vertical,
torsional force for Bolt B…
Vy = −4.000
kip
−24kip
+
= −8.000kip
6
Vx = 6.000kip
Vtotal = Vx 2 + Vy 2 = (−8.000kip ) 2 + (6.000kip ) 2 = 10.00kip
Check These Results with a Spreadsheet Solution:
Px:
0
xCG:
0
Py:
24
yCG:
0
ex:
5
ey:
0
Σd:
60.00
T=
120.0
Vmax =
10.0
Bolt
A
B
C
D
E
F
x
2.00
2.00
2.00
2.00
2.00
2.00
2
y
3.00
3.00
0.00
0.00
3.00
3.00
2
d
13.00
13.00
4.00
4.00
13.00
13.00
Vx
6.0
6.0
0.0
0.0
6.0
6.0
Vy
Vtotal
4.0
4.0
4.0
4.0
4.0
4.0
6.0
10.0
0.0
8.0
6.0
10.0
Everything checks out OK.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #6a
AASHTOLRFD 2007
Created July 2007: Page 2 of 2
 188 
194.
AASHTO Connection Example #6b:
Use the simplified equations to solve the previous example problem. (Example
4.12.1 from Salmon & Johnson)
Problem:
4"
3"
A
B
C
D
E
P
F
Solution:
T = ( 24 kip ) ( −3"− 2") = −120 kin
∑d
∑d
2
= ( 4 ) ⎡ (2") 2 + (3") 2 ⎤ + ( 2 )( 2")
⎣
⎦
2
= 60.00 in 2
2
Looking at Bolt B:
VB , x
VB , y
⎛ ( − 120 kin ) ( 3") ⎞
⎟ = 6.000 kip
= −⎜
2
⎜ 60.00 in
⎟
⎝
⎠
kin
⎛ ( − 120 ) ( 2") ⎞
⎟ = − 4.000 kip
=⎜
⎜ 60.00 in 2 ⎟
⎝
⎠
VB ,total =
( 6.000 )
kip 2
⎡
⎛ − 24 kip
+ ⎢ ( − 4.000 kip ) + ⎜
⎝ 6
⎣
⎞⎤
⎟⎥
⎠⎦
2
VB ,total = 10.00 kip
ODOTLRFD Short Course  Steel
AASHTO Connection Example #6b
AASHTOLRFD 2007
Created July 2007: Page 1 of 1
 189 
195.
AASHTO Connection Example #7:
Problem:
Detail a splice between two noncomposite W30 x 99 M270 Gr 50. Take Mu at
the location of the splice as 810kft and take φMn as 1,300kft.
Solution:
The splice is designed for the larger of:
M u , Beam + φM n, Beam
2
=
810kft + 1,300kft
= 1, 055kft
2
(
)
0.75 φM n, Beam = ( 0.75 ) 1,300kft = 975.0kft
Since 1,055kft > 975.0kft, Mu,Splice = 1,055kft
A) Flange Splice:
In this case, it makes no difference which flange is the “controlling flange” and which one is the
“noncontrolling flange,” (Since the beam is noncomposite and we are assuming that moment
could be either positive or negative).
For the Controlling flange:
⎛ 1 ⎞ ⎛ f cf
Fcf = ⎜ ⎟ ⎜
+ αφ f Fyf
⎝ 2 ⎠ ⎜ Rh
⎝
(810 ) (12 )(
=
kft
f cf
⎞
⎟ ≥ 0.75 αφ f Fyf
⎟
⎠
in
ft
29.7"
2
4
(6.13.6.1.4c1)
− 0.670" )
2
3,990 in
= 35.36ksi
Rh = 1.00, since the beam is nonhybrid.
α = 1.00. Since we are assuming that φMn = φMp, Fn = Fyf
ksi
⎛ 1 ⎞ ⎛ 35.36
Fcf = ⎜ ⎟ ⎜
+ (1.00 )(1.00 ) 50ksi
2 ⎠ ⎜ 1.00
⎝ ⎝
(
= 42.68ksi ≥ 37.50ksi
⎞
) ⎟ ≥ ( 0.75)(1.00 )(1.00 ) ( 50 )
⎟
ksi
⎠
→ Fcf = 42.68ksi
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 1 of 20
 190 
196.
For the NonControlling Flange:
Fncf = Rcf
f ncf
≥ 0.75 αφ f Fyf
Rh
(6.13.6.1.4c3)
f ncf = f cf = 35.36ksi
Rcf =
Fcf
f cf
Fncf = (1.207 )
=
42.68ksi
= 1.207
35.36ksi
(
35.36ksi
≥ ( 0.75 )(1.00 )(1.00 ) 50ksi
1.00
= 42.68ksi ≥ 37.50ksi
)
→ Fncf = 42.68ksi
For the Compression Flange:
Pu ,Comp = Fcf Ae
In compression, Ae is taken as the gross area of the flange.
Ae = Ag = (10.5")( 0.670") = 7.035 in 2
(
)(
)
Pu ,Comp = 42.68ksi 7.035 in 2 = 300.3kip
For the Tension Flange:
Pu ,Ten = Fcf Ae
⎛φ F
In tension, Ae = ⎜ u u
⎜ φ y Fy
⎝
⎞
⎟ An ≤ Ag
⎟
⎠
(6.13.6.4c2)
For 1” diameter bolts, An = ⎡(10.5") − ( 2 )(1 18 ") ⎤ ( 0.670") = 5.528 in 2
⎣
⎦
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 2 of 20
 191 
197.
(
(
⎛ ( 0.80 ) 65ksi
Ae = ⎜
⎜ ( 0.95 ) 50ksi
⎝
) ⎞ 5.528 in ≤ 7.035 in
⎟(
)
)⎟
⎠
2
= 6.052 in 2 ≤ 7.035 in 2
(
)(
2
→ Ae = 6.052 in 2
)
Pu ,Ten = 42.68ksi 6.052 in 2 = 258.3kip
Proceed assuming that the flange splice will consist of plate on both the outside and inside of the
flange. Assume that the flange force will be equally distributed between in the inner and outer
plates (we’ll check the validity of this assumption later). Also assume that the outer splice plate
will be 10.5” wide (the same width as the flange) with two rows of 1” diameter M164 (A325)
bolts.
Pu ,Ten =
258.3kip
= 129.1kip
2
Pu ,Comp =
300.3kip
= 150.2kip
2
For the Outer Flange Splice Plate:
Gross Yielding (Tension):
φPn = φ y Fy Ag ≥ Pu ,Ten
(
(6.8.2.11)
)
φPn = ( 0.95 ) 50ksi (10.5") ( touter ) ≥ 129.1kip
touter ≥
129.1kip
( 0.95) ( 50ksi ) (10.5")
≥ 0.2589"
→ say
5
16 "
Net Section Fracture (Tension):
φPn = φu Fu AnU ≥ Pu ,Ten
(
(6.8.2.12)
)
φPn = ( 0.80 ) 65ksi ⎡(10.5") − ( 2 )(1"+ 18 ") ⎤ ( touter ) (1.00 ) ≥ 129.1kip
⎣
⎦
touter ≥
129.1kip
( 0.80 ) ( 65ksi ) ( 8.25")
≥ 0.3010"
→ say
5
16 "
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 3 of 20
 192 
198.
Gross Yielding (Compression):
φPn = φc Fy Ag ≥ Pu ,Comp
(
(6.13.6.1.4c4)
)
φPn = ( 0.90 ) 50ksi (10.5") ( touter ) ≥ 150.2kip
touter ≥
150.2kip
( 0.90 ) ( 50ksi ) (10.5")
≥ 0.3179"
→ say
3
8"
For the Inner Flange Splice Plates:
The widths of the inner splice plates will be roughly equal the flange width of the section minus
the thickness of the web and fillets.
Winner = b f − 2k1 = 10.5"− ( 2 )(1 116 ") = 8.375"
Take the width of each of the two inner plates as 4”.
Gross Yielding (Tension):
φPn = φFy Ag ≥ Pu ,Ten
(
φPn = ( 0.95 ) 50ksi
tinner ≥
(6.8.2.11)
) ( 2)( 4.00") ( t
Inner
) ≥ 129.1kip
129.1kip
( 0.95) ( 50ksi ) ( 2 )( 4.00")
≥ 0.3297"
→ say
3
8"
Net Section Fracture (Tension):
φPn = φFu AnU ≥ Pu ,Ten
(
φPn = ( 0.80 ) 65ksi
(6.8.2.12)
) ( 2) ⎡( 4.00") − (1"+
⎣
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
1
8"
)⎤ ( tInner ) (1.00 ) ≥ 129.1kip
⎦
AASHTOLRFD 2007
Created July 2007: Page 4 of 20
 193 
199.
tinner ≥
129.1kip
( 0.80 ) ( 65ksi ) ( 5.75")
≥ 0.4318"
→ say
7
16 "
Gross Yielding (Compression):
φPn = φc Fy Ag ≥ Pu ,Comp
(
φPn = ( 0.90 ) 50ksi
t Inner ≥
(6.13.6.1.4c4)
) ( 2)( 4.00") ( t
Inner
) ≥ 150.2kip
150.2kip
( 0.90 ) ( 50ksi ) ( 2 )( 4.00")
≥ 0.4172"
→ say
7
16 "
For a flange splice with inner and outer splice plates, the flange design force at the strength limit
state may be assumed divided equally to the inner and outer plates and their connections when
the areas of the inner and outer plates do not differ by more than 10% (Commentary, Page 6191).
AOuter = (10.5")( 3 8 ") = 3.938 in 2
AOuter − AInner
AAve
=
AInner = ( 2 )( 4.00")( 7 16 ") = 3.500 in 2
( 2 ) ( 3.938 in 2 ) − ( 3.500 in 2 )
( 3.938 in ) + ( 3.500 in )
2
2
= 11.76%
Since the difference area is greater than 10%, either (1) the assumption that the flange force is
evenly divided between the outer and inner plates must be modified, or (2) the inner plate
thickness must be increased to 1/2”, which would result in a difference in area between the outer
and inner plates of less than 2%. The second option will be selected for the case of this
example.
Outer Flange Splice Plate: 101/2” x 3/8”
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
Inner Flange Splice Plates: 4” x 1/8”
AASHTOLRFD 2007
Created July 2007: Page 5 of 20
 194 
200.
Check Bolt Shear in the Flange Splice:
Assume that the threads are included in the shear plane of the connection. Bolts are in double
shear since both inside and outside splice plates are used.
Rn = 0.38 Ab Fub N s
(6.13.2.72)
2
⎛π⎞
Ab = ⎜ ⎟ (1") = 0.7854 in 2
⎝4⎠
(
)(
For A325 bolts, Fub = 120ksi
)
kip
Rn = (0.38) 0.7854 in 2 120ksi (2) = 71.63 bolt
(
)
kip
kip
φRn = (0.80) 71.63 bolt = 57.30 bolt
Determine the number of flange bolts required:
300.3kip
n fb =
= 5.24 bolts
kip
57.30 bolt
→
say 6 bolts
2"
6 1/2"
10 1/2"
2"
24 1/2"
2 1/2"
3 1/2"
3 1/2"
2 1/2"
2 1/2"
1
W30 x 99
3 1/2"
3 1/2"
2 1/2"
/2" gap between ends of beams
PL 241/2” x 101/2" x 3/8"
W30 x 99
PL 241/2" x 4" x 1/2" Each Side
1" dia M164 Bolts (12 places)
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 6 of 20
 195 
202.
Check Slip of the Flange Splice:
Bolted connections for flange splices shall be designed as slipcritical connections for the flange
design force. As a minimum, for checking slip of the flange splice bolts, the design force for the
flange under consideration shall be taken as the Service II design stress, Fs, times the smaller
gross flange area on either side of the splice.
Take the Service II moment as 548kft
Pslip = Fs Ag
(
)
548kft (12 in )( 29.7" − 0.670" )
fs
ft
2
2
where Fs =
=
= 23.92ksi
4
Rh
(1.00 ) 3,990 in
(
(
)
(6.13.6.1.4c5)
)
Pslip = 23.92ksi (10.5")( 0.670") = 168.3kip
The slip resistance of a single bolt is taken as:
Rn = K h K s N s Pt
(6.13.2.81)
Kh = 1.00 (for standard holes)
Ks = 0.33 (assume Class A surface)
Ns = 2
Pt = 51kip (from Table 6.13.2.81 for M164 Bolts)
(
)
kip
Rn = (1.00)(0.33)(2) 51kip = 33.66 bolt
Determine the number of flange bolts required:
n fb =
168.3kip
= 5.00 bolts
kip
33.66 bolt
→
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
6 bolts will work
AASHTOLRFD 2007
Created July 2007: Page 8 of 20
 197 
204.
B) Web Splice:
The web splice is to be designed for the following actions at the Strength Limit:
1.
2.
3.
4.
Vuw Mvuw Muw Huw 
The direct shear force.
The moment on the web bolts caused by the eccentricity of Vuw
The portion of the bending moment in the beam that is carried by the web.
The horizontal force resulting from the relocation of the beam moment from the
ENA location to the midheight of the beam.
The shear force in the beam at the location of the splice is Vu = 45kip and the nominal shear
capacity of the beam is φVn = 427.7kip.
1. Determine the direct shear force acting on the web splice, Vuw:
?
Vu < 0.5φvVn
?
(
)
45kip < ( 0.5 )(1.00 ) 427.7 kip = 213.8kip
Since Vu < 0.5φvVn ,
(
)
Vuw = 1.5Vu = (1.5 ) 45kip = 67.5kip
(6.13.6.1.4b1)
2. Determine the moment, Mvuw, that is caused by the eccentricity of the direct shear, Vuw:
Assuming the arrangement of bolts shown on Page 12, the distance from the CG of the
bolt group on one side of the splice to the CL of the splice is,
e = ( 1 ) ( 3 1 2 ") + 2 1 2 "+ ( 1 ) ( 1 2 ") = 4.50"
2
2
(
M vuw = e Vuw = ( 4.50") 67.5kip
)
= 303.8kin = 25.31kft
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
I used a gap of 1/2” here to be
conservative. I understand that most
splices will use much narrower gaps
 these calculations will be conservative
for smaller gaps.
AASHTOLRFD 2007
Created July 2007: Page 10 of 20
 199 
205.
3. Determine the portion of the beam moment that is carried by the web splice, Muw:
M uw =
tw D 2
Rh Fcf − Rcf f ncf
12
(C6.13.6.1.4b1)
Fcf = 42.68ksi
Rcf = 1.207
(from Before)
f ncf = −35.36ksi
M uw
(Positive since it’s in tension)
(Negative since it’s in compression)
⎡ ( 0.520")( 28.36")2 ⎤
=⎢
⎥ (1.00 ) 42.08ksi − (1.207 ) −35.36ksi
12
⎢
⎥
⎣
⎦
(
(
)
(
)
)
= 34.85 in 3 85.36ksi = 2,975kin = 247.9kft
4.
Determine the horizontal force that results from moving the beam moment, Huw:
H uw =
tw D
Rh Fcf + Rcf f ncf
12
(C6.13.6.1.4b2)
⎡ ( 0.520")( 28.36") ⎤
ksi
ksi
H uw = ⎢
⎥ (1.00 ) 42.68 + (1.207 ) −35.36
12
⎣
⎦
(
(
)
(
)
)
= 1.229 in 2 0.000ksi = 0.00kip
In this case, the ENA is at the midheight of the beam. Since Huw is the horizontal force
that results from the eccentricity of the ENA relative to the midheight of the beam, it
makes sense that Huw is zero.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 11 of 20
 200 
206.
The total moment acting on the web splice is,
M Total = M vuw + M uw = 303.8kin + 2,975kin = 3, 279kin
The total actions acting on the web splice are as shown below on the left.
To determine the forces acting on the bolts using the Elastic Vector Method, tables in the AISC
Manual of Steel Construction will be used for preliminary investigations. These tables are set up
to account for the shear force, Vuw, but not the moment, MTotal. This can be accommodated by
computing a fictitious shear force, P, that when applied over the eccentricity, e, results in the
same actions as the actually applied shear and moment.
3, 279kin
P=
= 728.7 kip
4.50"
2 1/2"
3 1/2"
1
2 3/4"
2
A
B
C
D
E
F
P = 728.7kip
G
H
e = 41/2"
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 12 of 20
 201 
207.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 13 of 20
 202 
208.
From Table 78 on Page 738 of the 13th Ed. of the AISC Manual of Steel Construction,
Cmin
Pu
728.7 kip
=
=
= 12.72
kip
φrn 57.30 bolt
From the Table for e = 4.00”, S = 3.00”, and for 8 bolts in a row, C = 13.2
From the Table for e = 5.00”, S = 3.00”, and for 8 bolts in a row, C = 12.2
The average of these two values is 12.7. Although this is slightly smaller than 12.72, the
proposed configuration will probably still work since our horizontal spacing is 31/2” instead
of 3”.
Elastic Vector Method for the Web Splice:
(
) (
) (
) (
)
2
2
2
2
2
Σd 2 = ( 4 ) ⎡ ( 4 ) ( d x ) + d y , D + d y ,C + d y , B + d y , A ⎤
⎢
⎥
⎣
⎦
2
2
2
2
2
2
Σd = ( 4 ) ⎡( 4 )(1.75") + (1.5") + ( 4.5") + ( 7.5") + (10.5") ⎤ = 805 in 2
⎣
⎦
Examine Bolt A2:
(
)
(
)
VT , X
3, 279kin (10.5")
T y
=
=
= 42.77 kip
2
2
805 in
Σd
VT ,Y
3, 279kin (1.75")
T x
=
=
= 7.128kip
2
2
Σd
805 in
The direct shear force is,
( 67.5 ) = 4.219
=
kip
VD ,Y
VTotal =
16 bolts
( 42.77 ) + ( 7.128
kip 2
kip
kip
bolt
+ 4.219kip
)
2
= 44.25kip
From this, an actual value of C can be computed, which
will be useful when investigating slip resistance.
C Actual =
PTotal 728.7 kip
=
= 16.47
kip
Pbolt 44.25 bolt
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 14 of 20
 203 
210.
Check Flexural Yielding of the Web Splice Plates:
Take Muw = 3,262kin
σ=
M y ⎡
12
=⎢ 3
I
⎢ d p 2t p
⎣
( )( )
Solve for tp:
⎤⎛ M d
p
⎥⎜
2
⎥⎝
⎦
⎞ 3M
⎟ = 2 ≤ φFy
⎠ d pt p
(
)
( 3) 3, 262kin
3M
tp ≥ 2
=
= 0.2787"
d p φFy ( 26.5")2 (1.00 ) 50ksi
(
)
Use one PL261/2” x 171/2” x 5/16” each side of web.
Check Shear Yielding of the Web Splice Plates:
Take Vuw = 67.5kip
( )( 2t p )( Fy )
Vuw ≤ φVn = ( φ )( 0.58 ) d p
(
)
= (1.00 )( 0.58 )( 26.5") ⎡( 2 )( 516 ") ⎤ 50ksi = 480.3kip
⎣
⎦
OK
Check Shear Rupture of the Web Splice Plates:
Take Vuw = 67.5kip
(
Vuw ≤ φVn = ( φ )( 0.58 ) d p ,net
)( 2t p ) ( Fu )
(
)
= ( 0.80 )( 0.58 ) ⎡( 26.5") − ( 8 )(1 1 8 ") ⎤ ⎡( 2 )( 516 ") ⎤ 65ksi = 329.9kip
⎣
⎦⎣
⎦
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
OK
AASHTOLRFD 2007
Created July 2007: Page 16 of 20
 205 
211.
Check Bearing of the Bolts in the Web Splice:
Edge Bolts Bearing on the Beam Web:
Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"
Rn = 1.2 Lc tFu
Since Lc = 1.969” < 2d = 2.0”,
(
)
kip
Rn = (1.2) (1.969") (0.520") 65ksi = 79.85 bolt
(6.13.2.92)
Edge Bolts Bearing on the Splice Plates:
Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"
Rn = 1.2 Lc tFu
Since Lc = 1.969” < 2d = 2.0”,
(
)
kip
Rn = (1.2) (1.969")( 2 )( 516 ") 65ksi = 95.99 bolt
(6.13.2.92)
Summary of Splice Plate Bearing:
Bearing on the beam web governs.
kip
Rn = 79.85 bolt
(
)
kip
kip
φRn = ( 0.80 ) 79.85 bolt = 63.88 bolt
OK
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 17 of 20
 206 
212.
Check Slip of the Bolts in the Web Splice:
Take the Service II moment as 548kft. From the slip check on the flange splice, Fs and fs were
determined to be 23.92ksi and Pslip was determined to be 168.3kip. Take the Service II shear force
at the location of the splice to be Vsw = 30.4kip.
The web splice is to be designed for the following actions at the Service II Limit
1.
2.
3.
4.
Vsw  The direct shear force.
Mvsw  The moment on the web bolts caused by the eccentricity of Vsw
Msw  The portion of the bending moment in the beam that is carried by the web.
Hsw  The horizontal force resulting from the relocation of the beam moment from the
ENA location to the midheight of the beam.
1. The direct shear force acting on the web splice is given as, Vsw = 30.4kip:
2. Determine the moment, Mvsw, that is caused by the eccentricity of the direct shear, Vsw:
Assuming the arrangement of bolts shown on Page 12, the distance from the CG of the
bolt group on one side of the splice to the CL of the splice is,
e = ( 1 ) ( 3 1 2 ") + 2 1 2 "+ ( 1 ) ( 1 2 ") = 4.50"
2
2
(
M vsw = e Vsw = ( 4.50") 30.4kip
)
= 136.8kin = 11.40kft
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
AASHTOLRFD 2007
Created July 2007: Page 18 of 20
 207 
213.
3. Determine the portion of the beam moment that is carried by the web splice, Msw:
M sw =
M sw
tw D 2
f s − f os
12
(C6.13.6.1.4b1 mod)
⎡ ( 0.520")( 28.36")2 ⎤
=⎢
⎥ 23.92ksi − −23.92ksi
12
⎢
⎥
⎣
⎦
(
(
) (
)
)
= 34.85 in 3 47.84ksi = 1, 667 kin = 138.9kft
4. Determine the horizontal force that results from moving the beam moment, Hsw:
H sw =
tw D
f s + f os
12
(C6.13.6.1.4b2 mod)
⎡ ( 0.520")( 28.36") ⎤
ksi
ksi
H sw = ⎢
⎥ 23.92 + −23.92
12
⎣
⎦
(
(
) (
)
)
= 1.229 in 2 0.000ksi = 0.00kip
The total moment acting on the web splice is,
M Total = M vsw + M sw = 136.8kin + 1, 667 kin = 1,804kin
The fictitious shear force, P, that when applied over the eccentricity, e, results in the same
actions as the actually applied shear and moment is determined as,
P=
1,804kin
= 400.8kip
4.50"
The largest bolt force in the web splice due to the Service II combination can be determined as,
PBolt
P
400.8kip
kip
Total
=
=
= 24.34 bolt
16.47
C
kip
This force is well below the slip capacity of 33.66 bolt that was computed on Page 8.
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
OK
AASHTOLRFD 2007
Created July 2007: Page 19 of 20
 208 
214.
Final Splice Detail:
The final splice configuration is shown below.
Technically speaking, fatigue should also be checked for beam flanges and flange splice plates at
the location of the splice.
2"
6 1/2"
2"
Outer Flange Splice Plate: PL 241/2” x 101/2" x 3/8" Each Flange
3 1/2"
3 1/2"
2 1/2"
2 1/2"
3 1/2"
3 1/2"
2 1/2"
2"
2"
2 1/2"
2"
2"
Inner Flange Splice Plates: PL 241/2" x 4" x 1/2" Each Flange
1" dia M164 Bolts (12 Places Each Flange)
W30 x 99
W30 x 99
2 3/4"
7 Spaces @ 3"
1" dia M164 Bolts (32 Places)
2 3/4"
Web Splice Plates: PL 261/2” x 171/2" x 5/16" Each Side of Web
2 1/2"
ODOTLRFD Short Course  Steel
AASHTO Connection Example #7
3 1/2"
2 1/2"
2 1/2"
3 1/2"
2 1/2"
AASHTOLRFD 2007
Created July 2007: Page 20 of 20
 209 
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