Steel strucure lec # (10)

Prof. Dr. Zahid Ahmad Siddiqi
In case of buildings, the maximum service live load
deflection is usually limited to L /360.
This limit is considered invisible not damaging the
surface finishes.
The deflections may be limited to L/1500 or L /2000
for structures supporting delicate machinery.
In case of bridge, deflections due to live and impact
loads are restricted to L /800.

During initial proportioning of steel beams, it is
customary to indirectly control deflections by limiting
the span-over-depth ratio (L /d ratio) for the
members.
When these conditions are satisfied it is more likely
that the deflection check, to be performed later on,
will be satisfied eliminating the need for greater
number of trials to get a reasonable section.
Typical span-over-depth ratios used for various
types of members are as under:

1- For buildings, L/d ratio is usually limited to a
maximum of 5500 / Fy.
L /d £ 5500 / Fy
dmin = Fy L /5500 (L /22 for A36 steel)
2- For bridge components and other beams
subjected to impact or vibratory loads,
L /d £ 20
3- For roof purlins,
L /d £ 6900 / Fy (27.5 for A36 steel,
sometimes relaxed to a value equal to 30)

The actual expected deflections may be calculated
using the mechanics principles.
However, results given in Manuals and Handbooks
may also be used directly.
Some of the typical deflection formulas are
reproduced here.
1- For uniformly loaded and simply supported
beams,
Dmax =
EI
w
384
5 4
ll

2- For uniformly loaded continuous beams,
Dmidspan = [Mc – 0.1(Ma + Mb)]
Where Mc = magnitude of central moment,
Ma, Mb = magnitude of end moments.
EI48
5 2
l
3- For simply supported beams subjected to
point load (refer to Figure 4.16),
Dmidspan = where a £ L/ 2( )22
4
3
12
a
EI
aP
-l

a b
P
4- For overhanging part of beam subjected to
UDL,
a
No Load
w per meter
l
Dmax @ ( )a
EI
aw
34
24
3
+l

5. For the above case, with UDL also present
within supports,
Dmax @ ( )332
34
24
aa
EI
aw
+-ll
6- For overhanging part of beam subjected to
point load,
P
No load
al
Dmax =
( )
EI
aaP
3
2
+l

FLOW CHART FOR DESIGN OF BEAMS
Write Known Data
Estimate self-weight of the member.
a. The self-weight may be taken as 10 percent
of the applied dead UDL or dead point load
distributed over all the length.
b. If only live load is applied, self-weight may
be taken equal to 5 percent of its magnitude.
c. In case only factored loads are given, self-
wt. may be taken equal to 3 % of the given loads.

Calculate Factored Loads
Draw B.M. and S.F. Diagrams
Calculate Cb For Each Unbraced Segment
Find Mu,max, Vu,max, Lb for each segment and guess
which segment is the most critical.
Design this segment first and then check for
others.

Assume the section to be compact without LTB in
the start and calculate Zx accordingly.
Assumed Zx,req = fb = 0.9 ;
Fy = 250 MPa for A36 steel
yb
u
F
M
f
6
10´
Zsel ³ Zreq
Selection of Section Minimum weight
d ³ dmin
dmin = For Dmax = L/360
5500
lyF
= L /22 for A36 steel and simply supported beams

For Dmax required to be lesser than L /360, like L /500
or L /800, find (Ix)req from the deflection formula, with
only the live load acting, and select section such that Ix
³ (Ix)req.
Method 1 Use Of Selection Tables
1. Enter the column headed Zx and find a value
equal to or just greater than the plastic
section modulus required.
2. The beam corresponding to this value in the
shape column and all beams above it have
sufficient flexural strength based on these
parameters.

3. The first beam appearing in boldface type
(top of a group) adjacent to or above the
required Zx is the lightest suitable section.
4. If the beam must have to satisfy a certain
depth or minimum moment of inertia
criterion, proceed up the column headed
“Shape” until a beam fulfilling the
requirements is reached.
5. If Cb > 1.0, use Lm in place of Lp for the
approximate selection.
6. If Lb is larger than Lm of the selected section,
use the unbraced design charts.

7. Apply moment capacity, shear capacity,
deflection and all other checks.
8. The column headed fbMp may also be used
in place of the Zx column in the above
method.
Method 2: Use Of Unbraced Design Charts
This method is applicable in cases where the
above method is not fully applicable and Lb ³ Lp.
The design charts are basically developed for
uniform moment case with Cb = 1.0.

Following notation is used to separate full plastic,
inelastic LTB, and elastic LTB ranges:
Solid Circle represents Lp
Hollow Circle represents Lr
1. According to Mu in kN-m units and Lb in
meters, enter into the charts.
2. Any section represented by a curve to the
right and above ( ) the point selected
in No. 1 will have a greater allowed unbraced
length and a greater moment capacity than
the required values of the two parameters.

3. A dashed line section is not an economical
solution.
If dashed section is encountered while
moving in top-right direction, proceed
further upwards and to the right till the first
solid line section is obtained.
Select the corresponding section as the trial
section, and it will be the lightest available
section for the requirements.
4. If Cb > 1.0, use Mu,req = Mu / Cb but check
that the selected section has fbMp > Mu.

Check the three conditions of compact section for
internal stability, namely,
1. web continuously connected with flange,
2. flange stability criterion, and
3. web stability criterion.
If any one out of the above three is not satisfied,
revise the section.
Either calculate Lp, Lr, and Lm or find their values
from beam selection tables.

Lp = (mm)
= 0.05 ry (m) for A36 steel
yfy FEr76.1
Lr =
2
7.0
76.611
7.0
95.1 ÷÷
ø
ö
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è
æ
++
cJ
hS
E
F
hS
cJ
F
E
r oxy
oxy
ts
Mr = 0.7Fy Sx/106 kN-m
= =
for doubly symmetric I-sections
2
tsr
x
wy
S
CI
x
oy
S
hI
2
c = 1.0 for a doubly symmetric I-shape
ho = d – tf

BF =
pr
rp
LL
MM
-
-
Lm = Lp + £ Lr÷÷
ø
ö
çç
è
æ -
b
bp
C
C
BF
M 1
Calculate design flexural strength:
1- If Lb £ Lm Mn = Mp = Zx Fy / 106 (kN – m)
2- If Lm < Lb £ Lr Mn = Cb [Mp – BF(Lb – Lp)]
£ Mp (kN – m)
3- If Lb > Lr Mn = CbFcrSx £ Mp

where Fcr =
2
2
2
078.01 ÷÷
ø
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è
æ
+
÷÷
ø
ö
çç
è
æ ts
b
ox
ts
b
b
r
L
hS
cJ
r
L
EC p
Bending strength check:
Mu £ fbMn OK
If not satisfied revise the trial selection.
Shear check:

For £ 2.24 (= 63.4 for A36 steel)
Cv = 1.0
wt
h
ywFE /
fv Vn = Fyw Aw Cv (kN)
1000
6.09.0 ´
If not satisfied, revise the section.
Deflection check:
Find Dact due to service live loads.
Dact £ L/360 or other specified limit OK

Check self-weight:
Calculated self weight £ 1.2 ´ assumed self
weight OK
Otherwise, revise the loads and repeat the
calculations.
Write final selection using standard designation.

Example 4.1: BEAMS WITH CONTINUOUS
LATERAL SUPPORT
Design a 7m long simply supported I-section beam
subjected to service live load of 5 kN/m and
imposed dead load of 6 kN/m, as shown in Figure
4.18. The compression flange is continuously
supported.
Use (a) A36 steel and (b) steel with fy = 345 MPa
and permissible live load deflection of span / 450.

Solution:
In beams with continuous lateral support,
unbraced length is not applicable or it may be
assumed equal to zero in calculations.
Assumed self weight = 10% of
superimposed DL
= 0.6 kN/m
wu = 1.2D + 1.6L
= 1.2 ´ 6.6 + 1.6 ´ 5
= 15.92 kN/m

97.51 kN-m
B.M.D.
15.92 kN/m
55.72 kN55.72 kN
7 m
55.72 kN
55.72 kN
S.F.D.

If the beam is continuously braced, Cb value is not
applicable but may be considered equal to 1.0 in case it is
required in the formulas.
Mu = Mmax = 97.51 kN-m
Vu = Vmax = 55.72 kN
Compression flange continuously braced.
(a) A36 Steel
Assuming the section to be internally compact and
knowing that there is no LTB,

(Zx)req = = = 433.4 ´ 103 mm3
yb
u
F
M
f
6
10´
2509.0
1051.97 6
´
´
dmin = = 318 mm
22
7000
Zsel ³ Zreq
Selection of section Min. weight section
d ³ dmin
Consulting beam selection tables (Reference-1),
following result is obtained:
W310 ´ 32.7 provides sufficient strength but
depth is little lesser.

Select W360 ´ 32.9; Zx = 544 ´ 103 mm3
Check internal compactness of section as under:
1. web is continuously connected OK
2. = 7.5 < lp = 10.8 OK
3. = 53.3< lp =107 OK
f
f
t2
b
wt
h
The section is internally compact.
Continuously braced conditions imply: Lb < Lp
and Cb = not included in formulas

fbMn = fbMp
= 0.9 ´ 544 ´ 103 ´ 250 / 106 @ 121 kN-m
Mu = 97.51 kN-m < fbMp OK
h/tw =53.3 < 63.4
Þ shear yield formula is applicable
fvVn = FywAwCv
= ´ 250 ´ 349 ´ 5.8 ´ 1.0
= 273.27 kN
1000
6.09.0 ´
1000
6.09.0 ´

Vu = 55.72 (kN) < fvVn OK
Dact due to service live load =
= = 9.44 mm
EI
w 4
384
5 ll
´
( )
4
4
108280200000
70005
384
5
´´
´
´
Dall = L/360 = 7000 / 360 = 19.44 mm
Dact < Dall OK
Self-weight = = 0.32 kN/m
1000
81.99.32 ´

This is lesser than assumed self-weight of 0.6 kN/m in
the start. OK
Final Selection: W360 ´ 32.9
(b) Steel With Fy = 345 MPa
(Zx)req = = = 314.0 ´ 103 mm3
yb
u
F
M
f
6
10´
3459.0
1051.97 6
´
´
dmin = = 438 mm
16
7000

This dmin is much larger, so the direct deflection criteria
will be used.
EI
wL
4
384
5
450
ll
=
=
= 5024 ´ 104 mm4
E
w
I L
req
3
384
4505 l´
=
200000
70005
384
4505 3
´´
Zsel ³ Zreq
I ³ Imin

Consulting beam selection tables (Reference-1),
following result is obtained:
W310 ´ 28.3 provides sufficient strength and
moment of inertia.
Ix = 5410 ´ 104 mm4 ; Zx = 405 ´ 103 mm3
Check internal compactness of section as under:
1. web is continuously connected OK
2. bf / 2tf = 5.7 < lp = 9.1 OK
3. h / tw= 46.2 < lp =90.5 OK

Continuously braced conditions imply: Lb < Lp
and Cb = N.A.
fbMn = fbMp
= 0.9 ´ 405 ´ 103 ´ 345 / 106
@ 125.8 kN-m
Mu = 97.51 kN-m < fbMp OK
h/tw =53.3 < = 53.9
Þ shear yield formula is applicable
yFE /24.2

fvVn = FywAwCv
= ´ 345 ´ 309 ´ 6.0 ´ 1.0
= 345.40 kN
1000
6.09.0 ´
1000
6.09.0 ´
Vu = 55.72 (kN) < fvVn OK
Dact < Dall Already satisfied OK
Self-weight = = 0.28 kN/m
1000
81.93.28 ´

This is lesser than assumed self-weight of 0.6 kN/m in
the start. OK
Final Selection: W310 ´ 28.3
Example 4.2: BEAMS WITH COMPRESSION
FLANGE LATERALY SUPPORTED (BRACED)
AT REACTION POINTS ONLY
Design the beam of Figure 4.19 with the lateral
bracing only provided at the reaction points.

wD = 8 kN/m
wL = 10 kN/m
3.5m3.5m
PD = 60 kN
PL = 50 kN

Solution:
Self load = 0.1
= 1.7 kN/m
÷
ø
ö
ç
è
æ
+
7
60
8
wu = 1.2 ´ 9.7 + 1.6 ´ 10
= 27.64 kN/m
Pu = 1.2 ´ 60 + 1.6 ´ 50
= 152 kN
The factored loading and the shear force and bending
moment diagrams are shown in Figure 4.20.

172.74 kN
172.74 kN S.F.D.
76
76
B.M.D.
169.30 kN-m
266.00 kN-m
wu = 27.64 kN/m
172.74 kN172.74 kN 3.5m3.5m
Pu = 152 kN

Calculation of Cb:
MB = Mmax = 435.3 kN-m
MA = MC = 172.4 ´ 1.75 – 27.64 ´ 1.752/2
= 259.97 kN-m
Cb = Rm = 1.24
£ 3.0 (where Rm = 1.0)
97.259630.4355.6
30.4355.12
´+´
´
Mmax = 435.3 kN-m : Vmax = 172.74 kN : Lb = 7 m

Assuming the section to be internally compact with no LTB,
(Zx)req =
= = 1935 ´ 103 mm3
dmin. = L/22 = 7000/22 = 318 mm
yb
u
F
M
f
6
10´
2509.0
103.435 6
´
´
Zsel ³ Zreq
d ³ dmin
Using beam selection tables of Reference-1,

Trial section: W610 ´ 82
Mp = 549.00 kN-m, Mr = 327.25 kN-m,
BF = 64.99, Lr = 5.10 m
Lp = 1.67 m : Lm = 1.67 + = 3.33 m÷
ø
ö
ç
è
æ -
24.1
124.1
99.64
00.549
Lb >> Lm Þ revise the section using beam
selection charts of Reference-1
For Lb = 7m and Mu,eq = 435.3 / 1.24 = 351.05 kN-
m, from design charts of Reference-1,

Select W410 ´ 100
fbMp = 479.25 kN-m > Mu OK
d > dmin OK
Check internal compactness of section:
1- web is continuously connected OK
2- = 7.7< lp = 10.8 OK
3- = 35.9 < lp = 107 OK
f
f
t
b
2
wt
h

Lp = 3.11 m
Lr = 10.00 m
BF = 28.53 kN
Mp = 532.50 kN-m
Lm = Lp + £ Lr
Lm = 3.11 + = 6.72 m
÷÷
ø
ö
çç
è
æ -
b
bp
C
C
BF
M 1
÷
ø
ö
ç
è
æ -
24.1
124.1
53.28
50.532
Lm < Lb £ Lr:
fbMn = Cb ´ fb [Mp – BF(Lb - Lp)] £ fbMp
= 1.24 ´ 0.9 [532.50 – 28.53(7.00 - 3.11)]
= 470.44 kN-m

Mu = 435.3 kN-m £ fbMn OK
h/tw = 35.9 < 63.4 Þ
fvVn = Fyw Aw Cv
= ´ 250 ´ 415 ´ 10.0 ´ 1.0
= 560.25 kN
1000
6.09.0 ´
1000
6.09.0 ´
Vu = 172.74 kN < fvVn OK

Dact = ( ) ( ) ( )( )22
44
4
3500700075.0
103970020000012
350050000
1039700200000
700010
384
5
-´
´´´
´
+
´´
´
´
= 3.94 + 4.50 = 8.44 mm
L/360 = 19.44 mm
Dact < L/360 OK
Self weight = = 0.981 kN/m
< 1.7 kN/m assumed in the start OK
1000
81.9100´
Final Selection: W410 ´ 100

Example 4.3: BEAMS WITH COMPRESSION
FLANGE LATERALLY SUPPORTED AT
INTERVALS
Design the beam of Figure 4.21, ignoring self-weight
and deflection check. Compression flange is laterally
supported at points A, B and C.
Solution:
The factored loading and the shear force and bending
moment diagrams are shown in Figure 4.21.

7m 8m 6m
A C
B
D
PD =35 kN
PL =30 kN
PD =60 kN
PL =50 kN
wD =3 kN/m
wL =2 kN/m

Pu =90 kNPu =152 kN
wu =6.8 kN/m
208.04 kN81.56 kN
9090
118.04118.04
33.96
81.56
S.F.D.
(-)
(+)
540.00 kN-m
404.32 kN-m
B.M.D.

Mu = Mmax = 540 kN-m
Mu = 404.32 kN-m for portion AC
Mu = 540.00 kN-m for portions CB and BD
Vu = Vmax = 118.04 kN
Lb for portion AC = 7 m
Lb for portion CB = 8 m
Lb for portion BD = 6 m
Calculation of Cb:
i) Portion AC

M = 81.56 x – 6.8 x2/2
Mmax = 404.32 kN-m
MA = 132.32 kN-m (x = 1.75 m)
MB = 243.81 kN-m (x = 3.50 m)
MC = 334.48 kN-m (x = 5.25 m)
Cb = Rm
£ 3.0 (where Rm = 1.0)
= 1.49
48.334381.243432.132332.4045.2
32.4045.12
´+´+´+´
´
ii) Portion CB
M = 540 – 118.04 x (x starts from the end B)

Mmax = 540.00 kN-m
MA = 303.92 kN-m (x = 2 m)
MB = 67.84 kN-m (x = 4 m)
MC = 168.24 kN-m (x = 6 m)
Cb = Rm
£ 3.0 (where Rm = 1.0)
= 2.22
24.168384.67492.303300.5405.2
00.5405.12
´+´+´+´
´
iii) Portion BD
Cb = 1.0

Portion (Mu)max
(kN-m)
(Vu)max
(kN)
Lb
(m)
Cb
AC 404.32 81.56 7 1.49
CB 540.00 118.04 8 2.22
BD 540.00 90.00 6 1.00

Greater value of Mu makes a segment more
critical.
Similarly, a longer unbraced length reduces the
member capacity.
However, smaller value of Cb is more critical for a
particular segment.
The effect of these three factors together
determines whether a selected segment is actually
more critical than the other segments.

The decision about which part of the beam is
more critical for design depends on experience
but may not always be fully accurate.
If a wrong choice is made for this critical section,
the procedure will correct itself.
Only the calculations will be lengthy in such cases
with the end result being always the same.
For this example, assume that portion BD is more
critical and first design it. Later on, check for the
other two portions.

dmin » = = 682 mm
22
l
22
15000
Depth for portion AB may be relaxed a little due to the
presence of cantilever portion.
However, the portion BD may have its own larger depth
requirements.
Portion BD
Assuming the section to be internally compact
with no LTB,
Assumed (Zx)req = =
= 2400 ´ 103 mm3
yb
u
F
M
f
6
10´
2509.0
10540 6
´
´

From beam selection tables, the trial section is:
W610 ´ 92
Lp = 1.75 m, Lr = 5.33 m, Lb = 6 m,
Lm = Lp (as Cb = 1.0)
Lb >> Lm Þ use beam selection charts
For Mu = 540 kN-m and Lb = 6m
From unbraced beam curves Þ W610 ´ 125
is the first choice but select W690´125 to
satisfy dmin.

Trial Section: W690´125
Actual d = 678 (still a little lesser but within
permissible limits)
Lp = 2.62 m ; Lr = 7.67 m ; BF = 77.0 kN ; Lm = Lp ;
Mp = 999.50 kN-m ; Mr = 610.75 kN-m
Compactness Check:
1- web is continuously connected OK
2- bf/2tf = 7.8 £ lp = 10.8 OK
3. h/tw = 52.7 £ lp = 107 OK
Section is internally compact.

Lm < Lb £ Lr
fbMn = Cb ´ fb [Mp – BF(Lb – Lp)] £ fbMp
= 1.0 ´ 0.9 [999.50 – 77.0(6 – 2.62)]
= 665.22 kN-m £ 899.55 kN-m
Mu = 540 kN-m < fbMn OK
h/tw = 52.7 £ 63.4 Þ
fvVn = ´ 250 ´ 678 ´ 11.7 ´ 1.0
= 1070.9 kN
1000
6.09.0 ´

Vu = 118.04 kN £ fvVn OK
Check for Portion AC
Lb = 7m ; Cb = 1.49
Lp < Lb £ Lr
fbMn = Cb ´ fb [Mp – BF(Lb – Lp)] £ fbMp
= 1.49 ´ 0.9 [999.50 – 77.0(7 – 2.62)]
= 887.92 kN-m £ 899.55 kN-m

fbMn = 887.92 kN-m
Mu = 404.32 kN-m < fbMn OK
For W690´125:
Sx = 3490´103 mm3
Mp = 999.50 kN-m ; Mr = 610.75 kN-m
Check For Portion CB

fbMp = 899.55 kN-m
J = 117 ´104 mm4 ; ry = 52.6 mm
Lb = 8m ; Cb = 2.22
Iy = 4410 ´104 mm4 ; ry = 52.6 mm
d = 678 mm ; tf = 16.3 mm
ho = d - tf = 678 - 16.3 = 661.7 mm
tsr
x
oy
S
hI
2
3
4
1034902
7.661104410
´´
´´
=
= = 64.66 mm

Lb > Lr
= 123.86
fbMn = fb ´ Sx / 106 £ fbMp
59.64
8000
=
ts
b
r
L
2
2
2
078.01 ÷÷
ø
ö
çç
è
æ
+
÷÷
ø
ö
çç
è
æ ts
b
ox
ts
b
b
r
L
hS
cJ
r
L
EC p
632
32
2
10/10349086.123
7.661103490
11170000
078.01
86.123
20000022.2
9.0 ´´´
´´
´
+
´´
´
p
=
fbMn = 1137.10 kN-m £ 899.55 kN-m
Hence fbMn = fbMp = 899.55 kN-m and portion BD
is most critical as expected earlier.

Mu = 540 kN-m < fbMn OK
Deflection Check:
Requires detailed calculations for Dactual that is not
asked for in this example. Very approximate
calculations are performed as under:
Dmax for cantilever »
where a is the cantilever length and l is the span
=
( )
EI
aPa
3
2
+l
( )
4
2
10000,119000,2003
000,6000,15000,6000,30
´´´
+´

Dmax = 31.76 mm (will be reduced by loads on span ‘l’)
Also Dmax considering one end of the overhang as
fixed
»
=
EI
Pa
3
3
4
3
10000,119000,2003
000,6000,30
´´´
´
= 9.08 mm
l/360 = 6000/360 = 16.67mm
Deflection may be critical, detailed
calculations are recommended.

Self weight = = 1.23 kN/m
(asked to be ignored in the problem statement)
1000
81.9125´
Final Selection: W690´125
Example 4.4: Select suitable trial section for a 8m long
simply supported W-section beam subjected to service
live load of 10 kN/m and imposed dead load of 4 kN/m.
The compression flange is continuously supported in the
lateral direction and the deflection is limited to
span/1500.

Solution:
Assumed self weight = 10% of superimposed
dead load
= 0.4 kN/m
say 0.8 kN/m as live load is greater and deflection
requirements are strict.
wu = 1.2D + 1.6L
= 1.2 ´ 4.8 + 1.6 ´ 10 = 21.76 kN/m
Mu = Mmax = = 174.08 kN-m
8
876.21 2
´

(Zx)req = =
= 773.7 ´ 103 mm3
dmin = = 364 mm
yb
u
F
M
f
6
10´
2509.0
1008.174 6
´
´
22
8000
Dact due to service live load =
EI
w 4
384
5 ll
´
= 8,000 / 1500
( )
min
4
000,200
000,810
384
5
I´
´
´

Imin = 50,000 ´ 104 mm4
Zsel ³ 773.7 ´ 103 mm3
Selection Minimum weight section
of section d ³ 364 mm
Imin = 50,000 ´ 104 mm4
From beam selection tables,
Trial section no.1: W360 ´ 44
d = 352 mm < dmin
Ix = 12,100 ´ 104 mm4

Moving upwards in the same tables,
Trial section no.2: W410 ´ 46.1
d = 403 mm > dmin
Ix = 15,600 ´ 104 mm4
d = 355 mm < dmin
Ix = 14,200 ´ 104 mm4
Ix = 21,200 ´ 104 mm4

Skipping W410 ´ 60 and W460 ´ 60,
Ix = 35,100 ´ 104 mm4
Ix = 41,000 ´ 104 mm4
Ix = 56,200 ´ 104 mm4
Trial Section: W610 ´ 82
After selection of the trial section, all the checks are
to be performed. This part of the exercise is left for
the reader to be completed.

End of Chapter 4

Steel strucure lec # (10)

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Steel strucure lec # (10)