BRIDGES

Railway truss bridges
Layout
1/15-LA
I – RAILWAY BRIDGES
General Remarks for all kinds of bridges:
1- The deck bridge is the best type because the compression chord of the
truss is braced by cross girders (X.G). [Lb out of upper compression chord
will be distance between cross girders].
2- There must be horizontal wind bracing at the level of the lower flange
of X.G.
7.2-7.5m
sec 3
sec 2
sec 1
Stringer Br.
3.5m
B/7
-
B/9
H
5.3 m
1.8m
H
S/10
12.3-12.5 m
1.7m
1.8m 1.8m
B/7
-
B/9
S/10
1.7m
1.8m
0.3-0.4m
Min 5.5m
1.8m 1.7m 1.8m
3- There must be stringer bracing to carry lateral shock
 There must be braking force bracing to carry braking force.
 In Roadway there was no lateral shock & the braking force was
carried by slab to main girders to supports. But in case of Railway
Bridge, if there is no braking force bracing, there will be M y on X-
girders.
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Layout
2/15-LA
M.G.
M.G.
X.G.
stringer
S=4.0-7.0m
B
Plan of floor beams
Plan of bracing
X.G.
Wind bracing
Braking force bracing
Stringer bracing
4- What is the difference between pony & through bridge?
1 - The X.G. in both lies at the level of the lower chord.
2 – Also both of them have lower wind bracing.
3 – Only the through bridge has
upper bracing
We use through bridge only if h 
5.5 m, h is measured from the top of
the rail level h = H MG – h XG – 0.20
m (sleeper) – 0.15 m (rail) – 0.3
(height of upper member)
 The through bridge is preferable than pony bridge because the
compression chord is supported by bracing.
 In case of using through bridge, a closed frame or inclined
frames at both ends of the bridge must be provided to transfer
load from upper lever of upper bracing to bearings.
12.3-12.5 m
1.7m
1.8m 1.8m
B/7
-
B/9
H
S/10
1.7m
1.8m
0.3-0.4m
Min 5.5m
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Layout
3/15-LA
Elevation of M.G.
roller bearing
Elevation of M.G.
4-7m
H
4-7m
L=70-80m
H
Box portal frame
Elevation of M.G.
4-7m
H
Inclined portal frame
Box portal frame
Box portal frame (2)
L=70-80m
L=70-80m
Intermediate closed frame may be added if span is too big.
5- To draw plans, we have to take many sections:
a- The pony bridge has 2 plans, one shows the stringers and the stringer
bracing, while the other shows the wind bracing with braking force
bracing.
b- Deck and through bridges have 3 plans. Two of them are the same as
that of Pony Bridge, while the third is either lower bracing for Deck
Bridge or upper bracing for through bridge.
We can draw the first 2 plans in one plan only as shown below:
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Layout
4/15-LA
M.G.
M.G.
stringer
S=4.0-7.0m
B
Plan of floor beams
Plan of bracing
X.G.
X.G.
6- The width of the bridge "B" is nearly constant because the width of
train is constant and is determined as follows:
- 3.5 m in deck bridges & 5.3 in pony & through bridges.
- For each additional track in any kind of bridges add 3.5 m
i.e. Deck bridge:
Single track = 3.5 m
Double track = 3.5 + 3.5 = 7 m
Triple track = 3.5 + 2 *3.5 = 10.5 m
Pony or through bridge:
Single track = 5.3 m
Double track = 5.3 + 3.5 = 8.8 - 9 m
Triple track = 5.3 + 2 *3.5 = 12.5 m
Note:
 For long span bridges, L > 40 m, we use the main system as a
truss with height "
H" =
10
8 
L
.
 The shape of the M.G. is as follows the upper
chord is in the shape of " "- section while the
lower chord is in the shape of "Double T"
Upper chord
Lower chord
Vertical
Haunch
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Layout
5/15-LA
 It's preferable that number of panels is even
‫ﻋﺪد‬
‫اﻟﻤﺴﺎﻓﺎ‬
‫ال‬ ‫ﺑﯿﻦ‬ ‫ت‬
XG
‫اﻟﺸﻜﻞ‬ ‫ﻟﺘﻤﺎﺛﻞ‬ ‫زوﺟﻲ‬ ‫ﯾﻜﻮن‬
7- The spacing between X.G "S" is taken mg
H
L

10
so the angle of
diagonal of main truss will be equal 450
.
The number of panels will be 8 or 10 or maximum 12 ( )
‫زوﺟﻰ‬ .
8- There is no concrete slab so we cannot make composite action.
9- There must be a stringer nearly under
each rail.
10- The spacing between stringers is 1.8 m
(constant) [the distance is not 1.5 m as
distance between rails for construction
requirement.[ ‫ﻟﺴﮭﻮ‬
‫اﻟﺘﺮﻛﯿﺐ‬ ‫ﻟﺔ‬ ]
11- Distance between center lines of any 2 successive tracks is 3.5m.
12- There is no side walk (unless mentioned).
13- The braking force bracing is added at both ends & if span > 6 0 m we
add another bracing at mid span.
M.G.
X.G.
M.G.
S=4.0-7.0m
B
Plan of wind and braking force bracing
7.2-7.5m
sec 3
sec 2
sec 1
Stringer Br.
3.5m
B/7
-
B/9
H
1.8m 1.8m
1.7m
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Layout
6/15-LA
14- Drawing of lateral shock bracing (stringer bracing), the bracing is a
horizontal truss to carry horizontal loads. It consists of stringer as upper
& lower chord, verticals and diagonals are added.
M.G.
X.G.
stringer
Vl. of stringer
Bracing
Dl. of stringer
Bracing
M.G.
X.G.
stringer
S=4.0-7.0m
Plan of floor beams(sec 1)
B
15- The distance between verticals of stringer bracing is 1.5 – 2.0 m (so
the diagonal is inclined at nearly 45 o
)
M.G.
Dl. of stringer
Bracing
X.G.
Vl. of stringer
Bracing
Important: So the shape of stringer bracing is as shown (according to
spacing between X-G "S".
M.G.
X.G.
S=4.5-6.0m
S=4.5-6.0m
3 spacings
M.G.
X.G.
S=3-4m
S=3-4m
2 spacings
M.G.
X.G.
S=6.0-8.0m
4 spacings
S=6.0-8.0m
16- The level of braking force bracing is at the level of the lower
flange of X-G. The braking force is transmitted from level of Rail (upper
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Layout
7/15-LA
flange of XG) to level of braking force bracing (lower flange of XG) by
one of the following 2 methods : -
Either Full depth stringer or bracket ( as discussed in plate girder)
‫ھﺎﻣﺔ‬ ‫ﻣﻠﺤﻮﻇﺔ‬
:
‫ﻋﻠﻲ‬ ‫اﻟﻔﺮاﻣﻞ‬ ‫ﺣﻤﻞ‬ ‫ﻧﻘﻞ‬ ‫ﻛﯿﻔﯿﺔ‬ ‫ﺗﻮﺿﯿﺢ‬
braking force bracing
17- In Deck Bridge, the X.F is added each (15 – 18 m)
18- The X.F. is either X-system or V-system according to the angle (35o
-
55o
).
3.5 m
1.8m H
S/10
7.2-7.5m
sec 3
sec 2
sec 1
Stringer Br.
3.5m
1.7m 1.8m
1.8m
B/7
-
B/9
H
B/7
-
B/9
19- The upper bracing in through bridge and lower bracing in Deck
Bridge can be in the shape of either x-system or rhombic system.
S=4.0-7.0m S=4.0-7.0m
20- In case of using semi-deck bridge, the cross-girder will be nearly at
midpoints of the vertical members which give concentrated loads at these
points. This means that half the member will be tension and the other will
be compression. It is preferable to use K-truss.
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Layout
8/15-LA
roller bearing
upper bracing
lower bracing
elevation of M.G.
X-bracing
X.G.
L=40-70m
H
roller bearing
upper bracing
lower bracing
elevation of M.G.
X-bracing
X.G.
L=40-70m
H
4-7m
4-7m
21- There must be end bracket because there is no cantilever slab.
22- In case of single track deck bridges, the
bridge may compose of 2 main girders and
cross girder without stringers. (The main
girders are used at the position of stringers.)
1.50 ms
2.60 ms
H
1.7-1.8m
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Layout
9/15-LA
23- The main dimensions of the train envelope ( ‫اﻟﺠﺒﺎرﯾﺖ‬ ) is as shown:
4.62 ms
5
.
0
0
m
s
3.16 ms 0.80 m
2.31ms
3.16 ms
2.31 ms
5
.
0
0
m
s
3.16 ms
3.495 ms
0.15 m
5
.
0
0
m
s
0.15m
0.80m
24- In some cases (when needed) especially in through bridge, we can
lower the stringer by maximum of 15cm in order to decrease the height
of construction and increase distance between rail & upper beam.
Min 5cm
1.8m
X.G.
12.3-12.5 m
1.7m
1.8m 1.8m
B/7
-
B/9
H
S/10
1.7m
1.8m
0.3-0.4m
Min 5.5m
Min 5cm
1.8m
1.8m1.7m
8.8-9.2 m
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Layout
10/15-LA
25- The stringer may be added over the X.G. The actual height of
construction in this case will be hstringer + hx.g.+0.45
1.8m 1.7m 1.8m
8.8-9.2 m
26- When to use Subdivided trusses
For long spans (70 m or more) with narrow bridge (single track)
In this case the span of stringer will be 7m or more while span of
X.G. will be 3.5m in case of deck or 5.5m in case of pony or through.
i.e. the depth of stringer will be bigger than depth of X.G.
So use subdivided to reduce span of stringer
7-8m
70-80m
Deck bridge
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Layout
11/15-LA
7-8m 70-80m
Pony or through
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Layout
12/15-LA
Very Very Important:
How to determine the type of bridge whether it is deck, pony,
through?
The choice of the type of bridge depends mainly on:
1- Whether the compression chord is supported or not.
From the point of supporting compression chord, Deck Bridge is the
best choice then through bridge & finally pony is the most expensive.
2- The relation between the available height of construction "H"
and the construction height of the bridge "Hc".
a- The construction height of the bridge "Hc" is measured from the top of
the bridge floor (the rail level) to the level of the bottom flange of the
lower chord.
‫اﻟ‬ ‫ﻣﻨﺴﻮب‬ ‫ﻣﻦ‬ ‫ﯾﻘﺎس‬
‫ﻘﻄﺎرات‬
)
‫أﻟ‬
‫ﻘﻀﯿﺒﺎن‬
(
‫اﻟﻰ‬
‫اﻟﻜﻮﺑﺮى‬ ‫ﻓﻰ‬ ‫ﻧﻘﻄﺔ‬ ‫أﺳﻔﻞ‬ ‫ﻣﻨﺴﻮب‬
- Height for deck bridge is htruss + 0.2(sleeper) + 0.15 (rail) +
(0.1 is the deflection)
- Height of construction for pony and through bridge is h XG +
0.45.
b- Available construction height "H" may be given or calculated as
follows:
H = Bank level – H.W.L – Δ required for navigation
H.W.L.
Navigation
H
Bank level
The actual height of construction "Hc" must be less than or equal the
available height of construction "H". (HC ≤ H).
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Layout
13/15-LA
 If Htruss + 0.45 =
12
10 
L
+0.45 ≤ H So use deck bridge.
 If Htruss + 0.45 =
12
10 
L
+0.45 > H So use through or pony.
We try first to make it through bridge, if not so pony bridge
How to determine whether the bridge is through or not?
If the height of the truss
(
12
10 
L
) – hxg – 0.35 (sleeper + rail) – 0.3 (height of upper beam) 
5.5m
So we can use through bridge.
12.3-12.5 m
1.7m
1.8m 1.8m
B/7
-
B/9
H
S/10
1.7m
1.8m
0.3-0.4m
Min 5.5m
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Layout
14/15-LA
Examples to determine the type of bridge:
Example 1:
2 tracks railway bridge crosses a river with level of H.W.L (10.00) and
L.W.L (8.00). The Rail level is (15.00) and the height required for
navigation is 3.00. The width of the river is 50 m.
Solution:
 Width of river = span of bridge
 H = Available height of construction = 15 – 10 – 3 = 2 m
H C deck =
12
10
50

+ 0.45 = 5.45
 4.6 m > 2 m
So we cannot make Deck Bridge.
Since we will use pony or through bridge, so B = 5.3+3.5 = 9.0m
H XG =
9
7
9
8 

B
= 1.0m
For through bridge: h = 5- 1 – 0.35 -0.3 = 3.35m  5.5m
So we cannot use through Bridge
So the bridge must be pony
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Layout
15/15-LA
Example 2: Important
3 tracks railway bridge crosses a river with level of H.W.L (10.00) and
L.W.L (8.00). The Rail level is (18.00) and the height required for
navigation is 3.00. The width of the river is 50 m.
Solution:
 H = Available height of construction = 18 – 10 – 3 = 5 m
 H C = for deck =
12
10
50

+ 0.45 = 4.60  5.45 m
O.K. use Deck Bridge with H truss = 5- 0.45
= 4.55m
Example 3:
One track railway bridge crosses a river with level of H.W.L (10.00)
and L.W.L (8.00). The Road level is (15.00) and the height required
for navigation is 3.00. The width of the river is 70 m.
Solution:
H= 15 – 10 – 3 = 2 m
H C for deck =
12
10
70

+ 0.45
= 6.3m  7.45 m > 3 m
So it is either pony or through
Since this bridge is through or pony
So B of X.G = 5.3m
H XG =
8
3
.
5
= 0.65 m Take HMG = 0.65 + 0.45 + 5.5 + 0.3 = 7.0 m
5.3m
H
1.8m
0.3
Min 5.5m
0.65
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Loads and design of Stringer
1/13L.S
Design of stringer
Loads on stringer:
1- Live load + Impact:
- Each track is a main lane and there are no adjacent lanes.
- The wheel loads are given below ( )
The train is Locomotive + tender + locomotive + tender + any number of
wagons.
1) This train is [type D]
2) The Previous loads are [wheel loads], Wheel load =
2
1 axle load
2*6.25t
1.5 1.75 1.8
3*12.5t
2.0
2.0
1.75 2.0 1.5 1.5
1.8 1.8
4*10t
1.5
4*10t
5.5
1.75 1.75 1.5
Locomotive Tender Any number of wagons
+ Tender
Locomotive
• The length of locomotive is 10.5m and of weight 100t (wheel loads
50t).
• The length of tender is 8.40m and of weight 80t (wheel loads 40t).
• The length of tender is 12.0m and of weight 80t (wheel loads 40t).
• Always use wheel loads.
Important note: In the exam, axle loads may be given so we have to
divide it / 2.
- Impact factor = I =
L
+
24
24
( ) 0.25 ≤ I ≤ 0.75
I = 0.25 → 0.75
Where: "L" in meters is the loaded length of the main track in direction of
motion.
Impact factor shall be taken for the loads in all the main lanes.
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2/13L.S
Note that: If I > 0.75, take I = 0.75
If I < 0.75, take I = 0.25
Loads on stringer (Railway Bridge)
1- Dead load:
- O.W. of stringer. = 150 – 200 kg / m/
- Weight of sleepers, rail = 600 kg / m/
- Weight of stringer bracing = 50 kg / m/
W dead = 0.2 +
2
05
.
0
2
6
.
0
+ = ……… t / m/
(We divide by 2 because we have two tracks)
M d = W d *
8
2
S = ---- m t Q d = W d *
2
S = ---- t
2- Live load:
Not less than 2 trails to get M L.L. & Q L.L.
There is no strips, we can calculate the moment and shear directly.
• Calculate ML.L max :
Both cases, we calculate the
moment at mid span under
the middle concentrated
load.
If we want to draw another case of loading: Use 2 wheel loads of 12.5t
and third with 6.25t. We have to get the position of resultant to get the
position middle concentrated load with respect to middle of beam.
3*12.5t
2.0
2.0
1.8 1.8
3*10t
Case 2 Case 1
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3/13L.S
Calculate resultant = 12.5*+6.25 = 31.25t
X =
25
.
31
75
.
3
*
25
.
6
2
*
5
.
12 +
= 1.55m
Middle concentrated load at distance
= m
225
.
0
2
55
.
1
0
.
2
=
−
Get the moment under the concentrated load
Calculate maximum moment
• Calculate QL.L max :
Both cases, we calculate the
moment at mid span under
the middle concentrated
load.
We may make third case
Calculate maximum shear
• Calculate Impact:
I =
L
+
24
24
0.25 ≤ I ≤ 0.75
Where "L" is the length of the stringer "S", whatever the number
of tracks are.
1.75
6.25
0.45
2
2.0
R=31.25
12.5
12.5
x=1.55
3*12.5t
2.0
2.0
1.8 1.8
3*10t
Case 1
Case 2
R R
Case 3
12t 12t
2.0 1.75
6.25t
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4/13L.S
• Total straining actions:
Mtotal = ML(1+I) + MD Qtotal = QL(1+I) + QD
Design the section (as rolled section):
• Design as simple beam:
Fsr: number of cycles (ECP 40, table (3.1b), Class I, Longitudinal flexural
member with L < 10m. n = over 2'000'000. Rolled section "detail A".
From table (3.2) ECP 41 Fsr = 1.68 t/cm2
Estimation of section: F max =
max
min
1
68
.
1
M
M
−
≤ 0.64Fy
If F max > 0.64Fy Take Fmax = 0.64Fy
Important note: Mmax (for fatigue) = ML (1+I) + MD
Not 0.6ML (1+I) + MD see ECP 37 (for railway, take full LL+I)
Sx =
max
F
M I
L
D +
+
Checks:
1-
y
f F
t
c 9
.
16
≤ ,
y
w
w
F
t
d 127
≤
2- Lu act = distance between vertical members
of stringer bracing.
Note that: we considered that sleeper is not
supporting the stringer
Lu max = least of
y
f
F
b
20
or b
y
f
C
dF
A
1380
Take Cb = 1.35 (concentrated load) and
calculate Fbcx.
Lu act
Stringer
X.G.
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5/13L.S
3- fmax = ≤
+
+
x
I
L
D
S
M
Fbcx (mostly 0.64 Fy)
4- fmax fatigue =
x
S
M
M min
max −
= ≤
+
x
I
L
S
M
1.68 t/cm2
5- qact = ≤
+
+
w
I
L
D
ht
Q
0.35 Fy
6- Check deflection using conjugate
beam
800
L
EI
M
≤
= ε
δ (ECP 132)
• Design as continuous beam:
We prefer not to make the connection continuous because we will use
special sleeper over the connection.
M0 = M Simple beam
0.9Mo 0.8Mo 0.8Mo 0.8Mo
0.75Mo 0.75Mo 0.75Mo 0.75Mo Sec 1
Sec 2
Sec 3
1- Calculate bending moment for stringer simply supported over X.G.
2- Calculate M1 = 0.75Mo, M2 = 0.8Mo and M3 = 0.9Mo for both dead and
(live and impact).
Sec 1-1: M1-D = 0.75 Mo-D M1-(L+I) = 0.75 Mo-(L+I)
3*12.5 or 3*10
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6/13L.S
3- For choice of section, assume that the section is compact section.
• For sec (1-1):
For rolled sec. with holes of pre-tensioned bolts Detail "B", n = over
2'000'000 Fsr = 1.12 t / cm2
Calculate; −
−
−
=
+
−
=
+
−
−
− )
1
(
26
.
1
)
(
1
1
1
max
I
L
d
d
M
M
M
f
IF fmax > 0.64 Fy then take f max = 0.64 Fy and get the section from;
fmax =
1
)
(
1
1
x
I
L
d
S
M
M +
−
− +
get S X1 = ---- cm3
[We choose the required section using Mmax flexural]
• For sec (2 – 2): for rolled sec "Detail "A", Fsr = 1.68 t / cm2
)
−
−
−
=
+
−
=
+
−
−
− )
1
(
68
.
1
)
(
2
2
2
max
I
L
d
d
M
M
M
f
IF fmax > 0.64 Fy then take f max = 0.64 Fy and get the section from;
fmax =
2
)
(
2
2
x
I
L
d
S
M
M +
−
− +
get S X 2 = ---- cm3
From bigger S X 1, 2 check I.P.E. or I.P.N. (S.I.E.) No ……., and check
safety of the section.
:
1
2
check
3
1) Check compactness of the section:
The section is compact if:
)
76
127
(
&
10
9
.
16
( =
>
=
>
y
F
w
t
w
h
y
F
f
t
C
For st. 44
)
67
127
(
&
9
.
8
9
.
16
( =
>
=
>
y
F
w
t
w
h
y
F
f
t
C
For st. 52
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

7/13L.S
Check L u act for negative sections = 0.2S and L u act for positive sections =
distance between verticals of stringer bracing.
2) Check maximum stresses: Mmax = MD + ML+I for required section
For sec 3:
x
I
L
d
S
M
M )
(
3
3 +
−
− +
= ------ t / cm2
< 0.64 FY for compact
< 0.58 FY for non-compact
3) Check stress range:
Mmax = MD +ML+I Mmin = MD (for required section)
For sec 1: Fsr =
x
S
M
M min
max −
= ------ t / cm2
< 1.12 t / cm2
For sec 3: Fsr =
x
S
M
M min
max −
= ------ t / cm2
< 1. 68 t / cm2
4) Check shear stress:
w
t
w
h
I
L
d
Q +
+ = ------ t / cm2
< 0.35 FY
5) Check deflection:
800
8
.
0
S
simple
continuous <
∆
=
∆
NOTE:
If the section is 3 unsafe, take the same section without increase
dimension and strengthen ( ) the first and the last panel only by
adding cover plate to increase the inertia of the section. We will add the
plate to the lower flange only because there are sleepers on the upper
flange.
• The cover plate is welded to the tension flange with fillet weld
parallel to direction of stress (assume weld is automatic with no
start stop position, detail B with F sr = 1.12 t / cm 2
. Then we must
check section 2 as rolled section.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

8/13L.S
i.e. Check flexural stresses:
For section 2 as rolled and section 3 as rolled with cover plates.
And check fatigue stresses:
For section 2 as detail "A" and section 3 as detail "B".
Note:
)
2
(
*
/
)
(
3
3
PL
X
I
L
D
t
h
I
M
M
+
+ +
−
−
< 0.64Fy t / cm2
)
2
(
*
/
)
(
3
PL
X
I
L
t
h
I
M
+
+
−
< 1. 12 t / cm2
Where: I/
x = I X-beam + 2 A PL* 2
)
2
2
( PL
t
h
+ =cm4
Example:
Design the Stinger as rolled section twice (Simple and continuous). St 44
5.7
5.7
Solution:
1) Dead Load
Track w =
2
1 * 600 = 300 kg/m'
St. Bracing w =
2
1 *40 = 20 kg/m'
O.w stringer ≅ 150 kg/m' ∑w = 470 kg/m'
M d =
8
7
.
5
*
47
.
0 2
= 1.909 mt Q d =
8
7
.
5
*
47
.
0
= 1.34 mt
t=10-20mm
f
b -4
Sec 3
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

9/13L.S
2) Live Loads: ( not less than 2 trails )
Calculate ML.L max :
M 1 = 28.44 mt
M 2 = 24.75 mt
Calculate resultant
= 12.5*+6.25 = 31.25t
X =
25
.
31
75
.
3
*
25
.
6
2
*
5
.
12 +
= 1.55m
Middle concentrated load at distance
= m
225
.
0
2
55
.
1
0
.
2
=
−
Get the moment under the concentrated load
Calculate maximum moment
R = 14.39t
M3 = 26.84mt
ML-max = 28.44 mt
• Calculate QL.L max :
Q1 = 24.34 t Q2 = 21.05 t Q3 = 22.97 t
QL-max = 24.34 mt
3*12.5t
2.0
2.0
1.8 1.8
3*10t
Case 1
Case 2
0.85
0.85
1.05
1.05
1.75
6.25
x=1.55
0.225
2.0
R=31.25
12.5
12.5
1.775 1.075
1.75
0.875
12.5
12.5
6.25
3*12.5t
2.0
1.8
Case 1
Case 2
R R
1.8 1.8 0.3
4*10t
2.0 1.7
2.0 1.75
Case 3
6.25t 6.25t
1.75 0.2
12.5t 12.5t
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

10/13L.S
3) Impact:
I =
7
.
5
24
24
+
= 0.808 > 0.75 ∴ I = 0.75
M d + L + I = 1.909 + (28.44 * 1.75) = 51.676 mt
Q d + L + L = 1.34 + (24.34 *1.75) = 43.939 t
• Design as simple beam:
Fsr: number of cycles (ECP 40, table (3.1b), Class I, Longitudinal flexural
member with L < 10m. n = over 2'000'000. Rolled section "detail A".
From table (3.2) ECP 41 Fsr = 1.68 t/cm2
676
.
51
909
.
1
1
68
.
1
−
= 1.74 t/cm2
< 0.64*2.8 = 1.79 t/cm2
F max = 1.74 t/cm2
Sx = 2970
74
.
1
100
*
676
.
51
= cm3
Use IPE 600 Sx = 3070 cm3
(Note: hstringer = cm
Span
57
10
≈ )
Checks:
1- 1
.
10
8
.
2
9
.
16
8
.
5
9
.
1
2
/
22
=
<
=
=
f
t
c
9
.
75
8
.
2
127
8
.
46
2
.
1
9
.
1
*
2
60
=
<
=
−
=
w
w
t
d
The Section is Compact
To get L u act divide spacing between 2 stringers as
shown 3 * 1.9 m, so angle is ≈ 45 0
2 – L u act = 1.9 m
L u max =
8
.
2
22
*
20
= 263 cm
Stringer
X.G.
1.9m
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

11/13L.S
=
60
*
8
.
2
9
.
1
*
22
(
*
1380
* 1.35 = 463.5 cm
L u act < L u max
∴ F bcx = 0.64 * 2.8 = 1.79 t / cm2
3 -
3070
100
*
676
.
51
max =
f = 1.68 t / cm2
< 1.79 t / cm2
4 -
3070
100
*
)
909
.
1
676
.
51
( −
=
fatique
f = 1.62 t / cm2
< 1.68 t / cm2
5 -
2
.
1
*
60
939
.
43
=
q = 0.61 t / cm2
< 0.35 * 2.8 = 0.98 t / cm2
6 – Check deflection:
ε
M = 51.15 * 2.85 – 12.5 *
3
2
*2 – 31.88 * 1 – 6.77 * (2 +
3
2
* 0.85) =
79.85 m3
t
800
570
41
.
0
92080
*
2100
10
*
85
.
79 6
<
=
= cm
δ
= 0.71 cm O.K.
• Design as continuous beam:
M0 = M Simple beam =
0.9Mo 0.8Mo 0.8Mo 0.8Mo
o 0.75Mo 0.75Mo 0.75Mo
Sec 3 Sec 2
Sec 1
0.75M
M1 = 0.75*51.676 = 38.76mt, M2 = 0.8*51.676 = 41.34 mt and M3 =
0.9*51.676 = 46.51mt.
Sec 1-1: M1-D = 1.43mt M1-(L+I) = 37.33mt
Sec 2-2: M2-D = 1.53mt M2-(L+I) = 39.82mt
Sec 3-3: M3-D = 1.72mt M3-(L+I) = 44.79mt
12.5 12.5 12.5
12.5
31.88
6.77
52.15
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

12/13L.S
Sec. (1-1): Fsr = 1.12 t/cm2
, n = over 2'000'000, Detail "B" HSB
f max =
676
.
51
909
.
1
1
12
.
1
−
= 1.16 t/cm2
S x =
16
.
1
100
*
76
.
38
= 3341cm3
Sec. (2-2): Fsr = 1.68 t/cm2
, n = over 2'000'000, Detail "A" rolled
f max =
676
.
51
909
.
1
1
68
.
1
−
= 1.74 t/cm2
< 0.64*2.8 = 1.79 t/cm2
S x =
74
.
1
100
*
34
.
41
= 2376cm3
S x-required = 3341cm3
> IPE 600 which means that the stringer in the
continuous connections is more critical. Choose (S.I.B 550)
Checks:
1- 1
.
10
8
.
2
9
.
16
33
.
3
3
2
/
20
=
<
=
=
f
t
c
9
.
75
8
.
2
127
8
.
25
9
.
1
3
*
2
55
=
<
=
−
=
w
w
t
d
2 – Lu-act for positive sections = 1.9m, Lu-act for negative sections =
0.2*5.7 = 1.14m
L u max =
8
.
2
20
*
20
= 239 cm
=
55
*
8
.
2
)
3
*
20
(
*
1380
* 1.35 = 725.8 cm
L u act < L u max ∴ F bcx = 0.64 * 2.8 = 1.79 t / cm2
1- f max 3 =
3610
100
*
51
.
46
= 1.29 t/cm2
< 0.64*2.8 = 1.79 t/cm2
o.k.
2- f max - f min 1 =
3610
100
*
33
.
37
= 1.034 t/cm2
< 1.12 t/cm2
o.k.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

13/13L.S
3- f max 3 =
3610
100
*
79
.
44
= 1.24 t/cm2
< 1.68 t/cm2
o.k.
4- q max =
55
*
9
.
1
939
.
43
= 0.42 t/cm2
< (0.35*2.8=0.98 t/cm2
) o.k.
5-
800
570
33
.
0
41
.
0
*
8
.
0
8
.
0
. <
=
=
= cm
simple
cont δ
δ = 0.71 cm O.K.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Loads and design of X.G.
1/20XG
Design of cross girder
Loads on X.G.:
1- Live load + Impact:
- Each track is a main lane and there are no adjacent lanes.
The train is Locomotive + tender + locomotive + tender + any number of
wagons.
• Always use wheel loads.
Important note: In the exam, axle loads may be given so we have to
divide it / 2.
- Impact factor = I =
L
+
24
24
( ) 0.25 ≤ I ≤ 0.75
L = n*2S where "n" is the number of tracks
X.G.
stringer
X.G.
stringer
X.G.
stringer
X.G.
stringer
X.G.
stringer
X.G.
stringer
Loaded length (shaded)
Required X.G.
Loaded length (shaded)
Required X.G.
Example: Calculate the impact factor of design of X.G for a triple track
pony bridge with spacing between X.Gs = 5.7m
I = 41
.
0
)
7
.
5
*
2
(
*
3
24
24
=
+
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

2/20XG
Loads on X.G. (Railway Bridge)
1- Dead load: Reaction of stringer = 2 QD-stringer
- O.W. of stringer. = 150 – 200 kg / m /
- Weight of sleepers, rail = 600 kg / m /
- Weight of stringer bracing = 50 kg / m /
W dead for stringer = (0.15-0.2) +
2
05
.
0
2
6
.
0
+ ≈ 0.5 t / m/
(We divide by 2 because we have two tracks)
RD = wdead * S
O.w. of X.G. ≅ 300→ 350 kg/m'
Or
Single track
D
R
D
R
Double track
D
R
D
R
D
R
D
R
2- Live load: Not less than 2 trails to get R L.L.
We have to take one strip (the strip is the stringer), so we load the
stringer and get the maximum reaction. Then put the maximum reaction
over X.G. and calculate maximum moment and shear without moving
the reactions. The position of the reaction is the position of rail.
It is preferable to use the wheel loads only without impact and
multiply both the moment and the shear of X.G. * impact later.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

3/20XG
• Calculate RL.L max :
R of case 1 will be mostly
critical.
• Calculation of ML and QL:
R
L
R
L
R
L
R
L
R
L
R
L
Single track Double track
R
L
R
L
R
L
R
L
Triple track
R
L
R
L
Calculate ML and QL.
ML+I = ML * (1+I)*F QL+I = QL * (1+I)*F
Where I =
S
n 2
*
24
24
+
, n is number of tracks and "S" is span of stringer
L = 2 S for single track = 4 S for double tracks
= 6S for triple track
"F" is a reduction factor depends on number of tracks
F = 1 Single track F = 0.9 Double tracks
F = 0.8 Triple tracks F = 0.75 4 tracks or more
This is because the case of maximum reaction for all tracks cannot
happen.
3.0
2.0
1.75 1.75
Case 1
2.0
10 10
1.8
6.25
6.25
6.25
10
2.0
12.5
12.5
Case 2
1.75
1.75 2.0 3.0
12.5
1.0
I.L. of reaction
6.25
S S
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

4/20XG
• Total straining actions:
Mtotal = ML(1+I) + MD Qtotal = QL(1+I) + QD
Design the section:
The X.G. may be rolled for single track and B.U. of double tracks
Important note: In design of X.G. and especially in single track deck
bridge, and if the stringer is given. The depth of X.G. must be at least of
the same depth of stringer.
• Design as rolled beam:
Fsr: number of cycles (ECP 40, table (3.1b), Class III, Transverse
floor beams. n = over 2'000'000 for single track and 500'000 for double
tracks. Rolled section "detail A".
Note: For more than 2 tracks, we can take 500'000 cycles as case of
double tracks.
From table (3.2) ECP 41: Fsr = 1.68 t/cm2
(single) or Fsr = 2.52 t/cm2
(double tracks or more)
I
L
D
D
sr
M
M
F
+
+
−
1
≤ 0.64Fy
If F max > 0.64Fy Take F max = 0.64Fy
Important note: Mmax (for fatigue) = ML (1+I) + MD
Sx =
max
F
M I
L
D +
+
Checks:
1-
y
f F
t
c 9
.
16
≤ ,
y
w
w
F
t
d 127
≤
2- Lu act = distance between stringers = 1.8m.
Stringer
X.G.
1.80
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

5/20XG
Lu max = least of
y
f
F
b
20
or b
y
f
C
dF
A
1380
Take Cb = 1.35 (concentrated load) and calculate Fbcx.
3- fmax = ≤
+
+
x
I
L
D
S
M
Fbcx (mostly 0.64 Fy)
4- fmax fatigue =
x
S
M
M min
max −
= ≤
+
x
I
L
S
M
Fsr t/cm2
5- qact = ≤
+
+
w
I
L
D
ht
Q
0.35 Fy
6- Check deflection using conjugate beam
R
L
R
L
R
L
R
L L
R
L
R
800
*
L
F
EI
M
≤
= ε
δ (ECP 132) ("F" is the reduction factor)
• Design as built-up section:
As will be seen in Main girder
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

6/20XG
• Lateral shock: (47 )
The train makes lateral shock in the direction
perpendicular to stringer. The lateral shock is taken
constant and equals to 6t. The 6t is to be added in
order to produce the maximum effect in the studied
member.
The lateral shock is constant = 6t whatever the
number of tracks. Do not take impact factor.
• Braking force: (47 )
The braking force is taken as ∑ 7
/
P "P is the wheel load".
6.25
12.5
Loc.(50t)
Tender(40t) Loc.(50t) Tender(40t)
9m
1.5
10.5
8.4m 10.5m 8.4m
Wagon(40t)
12.0m
10.0
1.5 1.75
The wagons are added in length = L - 36.3 of the main girder
In case of one track, we take total braking force.
In case of double tracks, we take total braking force on one track
and half the braking force on the other track.
In case of triple tracks, we take braking force on 2 tracks only as
follows: total braking force on one track, half the braking force on the
second track in which this combination gives maximum straining actions
on the studied member.
Stringer
X.G.
1.80
6t
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

7/20XG
B
B
B
B
B/2
B/2
B
B
B/2
B/2
The braking force is distributed over the braking force bracings of
the bridge. The number of braking force bracings may be "2" if we put 2
bracings at both ends only or "4" if we put at mid span.
B =
.
B.F.B
of
Number
7
/
∑P
(B.F.B. is braking force bracing)
M.G.
X.G.
M.G.
S=4.0-7.0m
B
4 bracings
2 bracings
B
S=4.0-7.0m
M.G.
X.G.
M.G.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

8/20XG
Very important note: If there is no braking force bracing, there will be
My on X.G. The braking force will be distributed over the total number of
X.G.
B =
.
X.G.
of
Number
7
/
∑P
then calculate My and design the X.G.
Number of X.G = Number of spacing + 1
• We always make braking force bracing except if it is mentioned in
the exam not to use B.F.B.
Illustrative example:
Through bridge triple track of span 72 m the X.G. is spaced 6 m.
Calculate braking force in the following cases:
1 – There is only B.F.B at both ends
2 - There is only B.F.B at both ends & at mid span
3 - There is no B.F.B
Solution:
Length of loc + tender + loc + tender = 10.5 * 2 + 2 * 8.4 – 1.5 = 36.3 m
Length of wagons = 12 m
12
7
.
35
∴ = 2.975 Means 3 wagons
Loc.(50t)
1.5 0.3
8.4m
9m
10.5
10.5m 8.4m 12.0m 12.0m 11.7
60.3
10.0 10.0
Wagon(40t) Wagon(40t)
Loc.(50t)
12.5
6.25 Tender Tender
40t 40t
∑P (Wheel loads) = 2 * 50 (2 locomotives) + 2 * 40 (2 tenders) + 3 * 40
(3 wagons) = 300 t
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

9/20XG
Case 1:
2
7
/
300
=
B = 21.43 t
M.G.
X.G.
M.G.
B
2 bracings
B
B
B/2
B/2
B
B
B/2
B/2
B
R
Case 2:
4
7
/
300
=
B = 10.7t
Note for long spans > 50 m , we have to use 2 B.F.B at mid span
M.G.
X.G.
M.G.
B
4 bracings
B
B
B/2
B/2
B
B
B/2
B/2
B
R
B
B
B/2
B/2
B
B
B/2
B/2
Case 3: Number of X.G. = 12 + 1 = 13
13
7
/
300
=
B = 3.3 t
M.G.
X.G.
M.G.
B
B
B
B/2
B/2
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

10/20XG
• Madd for pony bridges:
If the maximum compression force in the upper chord is given, we have
to calculate Madd on X.G. and check as case II
5.3 m
1.8m
C
100
h
C
100
M
M
add
Example1:
Design the X.G. of a single track deck bridge as rolled section. Steel used
is 37. The spacing between cross girders is 5.7m.
Solution:
B = 3.5 m (single track, deck bridge)
1) Dead load:
o.w of stringer = 0.175 t / m/
Stringer bracing + rail + sleeper on each stringer =
2
65
.
0
= 0.325 t / m/
DL on stringer = 0.5 t / m/
R D = 0.5 * 5.7 = 2.85 t
Assume o w = 300 kg / m/
M D = 2.88 m t
Q D = 3.375 t
2.85 2.85
0.3
0.85 0.85
1.8
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

11/20XG
2) Live load:
R max = 12.5 (1 + 2 * 0.65)
+ 10 * 0.12
+ 6.25 (0.34 + 0.035)
= 32.3 t
Make another case of loading
I =
7
.
5
*
2
*
1
24
24
+
= 0.68 < 0.75
Q L = 32.3 t , Q L+I = 32.3 * 1 * 1.68 = 54.26 t
M L = 27.45 m t, M L+I = 27.45*1*1.68 = 46.11 m t
3) Maximum Straining actions:
Q max = 3.375 + 54.26 = 57.64 t M max = 2.88 + 46.11 = 49 m t
For fatigue: Class III , one track, over 2000000
Rolled: Detail "A" à F sr = 1.68 t / cm2
Estimation of section:
49
88
.
2
1
68
.
1
max
−
=
F = 1.78 t / cm2
< 0.64 * 2.8 = 1.79 t / cm2
78
.
1
100
*
49
=
x
S = 2753 cm4
Use IPE no. 600
For single track deck bridge, we must check stringer since Lstringer = 5.7 m
The section will be IPE 600 ( ≈
= cm
57
10
570
60 cm) . So use IPE 600 for
X.G section (minimum section)
2.0
Case 1
1.0
I.L. of reaction
6.25
0.2 1.75 1.75 2.0
10
3.0 0.7
0.035
0.34
0.65
0.65
0.12
5.7 5.7
6.25
32.3 32.3
0.85 0.85
1.8
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

12/20XG
Check:
5
.
6
8
.
2
9
.
10
8
.
5
9
.
1
2
/
22
=
<
=
=
f
t
c
76
8
.
2
127
8
.
46
2
.
1
9
.
1
*
2
60
=
<
=
−
=
w
w
t
d
L u act = 180 cm
L u max =
8
.
2
22
*
20
= 263 cm Or = 35
.
1
*
8
.
2
*
60
9
.
1
*
22
*
1380
= 463 cm
∴The section is Compact and no LTB
1)
3070
4900
max =
f = 1.6 t / cm2
< 0.64 * 2.8 = 1.79 t / cm2
2)
3070
4611
=
fatique
f =1.5 t / cm2
< 1.68 t / cm2
3) Check shear =
2
.
1
*
60
64
.
57
= 0.8 t / cm2
< 0.35 * 2.8 = 0.98 t / cm2
4) Check deflection:
32.3 32.3
0.85 0.85
24.7
11.67
27.45
36.36
0.9 0.9
92080
*
2100
)
3
85
.
0
9
.
0
(
*
67
.
11
2
9
.
0
*
7
.
24
75
.
1
*
36
.
36 +
−
−
=
=
EI
Mξ
δ * 106
= 0.2 cm <
800
350
= 0.44 cm
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

13/20XG
Example 2:
Design the X.G of double track through bridge as built up section. The
distance between X.G is 5.7 m. steel used is st. 44
Solution:
From previous: R D = 2.85 m t & R L = 32.3 t
1) Dead load:
B = 5.3 + 3.5 = 8.8 m
Q D = 5.7 t & M D = 15.1 m t
2) Live load:
Q L = 64.6 t M L = 171.2 m t
7
.
5
*
2
*
2
24
24
+
=
I = 0.51 < 0.75
Q L+I = 64.6 * 1.51 = 97.5 t , M L+I = 171.2*1.51 = 258.5 m t
Q max = 97.5 + 5.7 = 103.2 t , M max = 258.5 + 15.1 = 273.6 m t
8
880
=
w
h = 110 cm
t w = 8 mm or
110
*
8
.
2
*
35
.
0
2
.
103
= 0.96 cm or
830
8
.
2
*
110
= 0.37 cm
h w < 1.5 m no h z stiffeners
8
.
2
190
110
=
w
t
t w = 0.97 cm take t w = 10 mm
n = 500'000 from class III, B.U. detail "B"
6
.
273
1
.
15
1
0
.
2
max
−
=
F = 2.1 t / cm2
> 0.58 * 2.8 = 1.6 t / cm2
∴ F max = 1.6 t / cm2
T = C =
110
*
97
.
0
100
*
6
.
273
= 256.4 t
2.85 2.85
0.3
32.3
1.75 1.8
32.3
64.6
0.85
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

14/20XG
Area of upper flange = 1
*
110
*
6
1
6
.
1
4
.
256
− = 142 cm2
Assume b = 20 t 20 t2
= 142
t u = 2.8 cm b u = 55 cm
2
3
)
2
8
.
2
2
110
(
*
55
*
8
.
2
*
2
12
110
*
1
+
+
=
x
I = 1'090'652 cm4
Checks:
8
.
2
21
6
.
9
8
.
2
2
/
)
1
55
(
≤
=
−
=
f
t
c
= 12.55
Note that
8
.
2
3
.
15
>
f
t
c
= 9.14 so the flange is non-compact
L u act = 180 cm
L u max =
8
.
2
55
*
20
= 657 cm or = 35
.
1
*
8
.
2
*
110
8
.
2
*
55
*
1380
= 931 cm
∴ There is no LTB ∴ F bcx = 0.58 * 2.8 = 1.6 t / cm2
)
8
.
2
2
110
(
*
1090652
100
*
6
.
273
max +
=
f = 1.45 t / cm2
< 1.6 t / cm2
)
8
.
2
2
110
(
*
1090652
100
*
5
.
258
+
=
fatique
f = 1.37 t / cm2
< 2 t / cm2
1
*
110
2
.
103
=
q = 0.94 t / cm2
< 0.35 * 2.8 = 0.98 t / cm2
Weld: )
4
.
4
*
2
.
0
(
2
1090652
)
2
8
.
2
2
110
(
*
)
8
.
2
*
55
(
*
2
.
103
s
=
+
s = 0.47 cm take smin = 6mm
55*2.8
110*1
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

15/20XG
ε
M = 500.26 * 4.4 – 145.52 *
2
85
.
0
– 52.33 (0.85+
2
8
.
1
)
– 203.49 (0.85 +
2
8
.
1
)
– 98.92 (0.85 + 1.8 +
3
75
.
1
)
= 1387.5 m3
t
800
880
55
.
0
9
.
0
*
1090652
*
2100
10
*
5
.
1387 6
<
=
= cm
δ
= 1.1 cm
Example 3:
Design the X.G of triple track Pony Bridge as built up section. The
distance between X.G is 5.7 m. steel used is st. 52. The maximum
compression force upper chord is 1800 t. H MG = 6 m
Solution:
As before R D = 2.85 m t & R L = 32.3 t
B = 5.3 + 7m = 12.5 m
32.3 32.3 32.3
1.85 1.8 1.7 1.8 1.7 1.8 1.85
2.85 2.85 2.85
1.85 1.8 1.7 1.8 1.7 1.8 1.85
2.675
1.825
3.125
0.925
Q D = 2.85*3 = 8.55 t & M D = 2*2.85(0.925+1.825+2.675) = 30.92 m t
Q L = 32.3*3 = 96.9 t & M L = 2*32.3(0.925+1.825+2.675) = 350.4 m t
32.3
1.75 1.8 0.85
32.3
171.2
113.05
145.52
203.49
98.92
52.33
32.3 32.3
500.26
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

16/20XG
7
.
5
*
2
*
3
24
24
+
=
I = 0.41 > 0.25
Q L+I = 96.9*1.41*0.8 = 109.3 t , M L+I = 350.4*1.41*0.8 = 395.25 m t
Q max 8.55+109.3 = 117.85 t , M max = 395.25+30.92 = 426.17 m t
8
5
.
12
: =
XG
add h
M = 1.60 m
2
6
.
1
6
2
−
=
−
=
∴ XG
MG
h
h
h = 5.2 m
100
1800
100
=
c
= 18 t M add = 18 * 5.2 = 93.6 m t
5.2
93.6
1800
100
93.6
1800
100
5.2
1800
100
Note that: The vertical number of truss also is subjected to moment
h w = 160 cm
t w = 8 mm or
160
*
6
.
3
*
35
.
0
85
.
117
= 0.58 cm
Or
6
.
3
830
160
=
w
t
à t w = 0.59 cm
Or
6
.
3
190
160
=
w
t
à t w = 1.6 cm too big
Use vertical stiffener & horizontal stiffener at
5
h
from compression
flange
6
.
3
320
160
=
w
t
à t w = 0.94 cm Take t w = 10 mm
For class III, BU. Detail "B" , n = 500'000
Note that: Case of triple tracks is not mentioned in the code but we can
take n = 500'000 (conservative)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

17/20XG
17
.
426
92
.
30
1
0
.
2
max
−
=
F = 2.16 t / cm2
> 0.58 * 3.6 = 2.1 t / cm2
T = C =
160
*
97
.
0
100
*
17
.
426
= 274.6 t
Area of flange = 1
*
160
*
6
1
1
.
2
6
.
274
− = 104 cm2
= 104 t = 2.4 cm b = 48 cm
Checks:
6
.
3
21
8
.
9
4
.
2
2
/
)
1
48
(
≤
=
−
=
f
t
c
= 11.1
L u act = 180 cm
L u max =
6
.
3
48
*
20
= 506 cm
Or = 35
.
1
*
6
.
3
*
160
4
.
2
*
48
*
1380
= 276 cm > 180 cm
The section is non- compact & No L.T.B.
F bcx = 0.58 * 3.6 = 2.1 t / cm2
2
3
)
2
4
.
2
2
160
(
*
)
4
.
2
*
48
(
2
12
160
+
+
=
I = 1860462 cm4
)
4
.
2
2
160
(
*
1860462
100
*
17
.
426
max +
=
f = 1.89 t / cm2
< 2.1 t / cm2
Check with M add (case B)
)
4
.
2
2
160
(
*
1860462
100
*
)
6
.
93
17
.
426
(
max +
+
=
f =2.3 t / cm2
< 1.2*2.1=2.52t /
cm2
)
4
.
2
2
160
(
*
1860462
100
*
25
.
395
+
=
fatique
f = 1.75 t / cm2
< 2 t / cm2
1
*
160
85
.
117
=
q = 0.73 t / cm2
< 0.35 * 3.6 = 1.26 t / cm2
160*1
48*2.4
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

18/20XG
Weld: )
2
.
5
*
2
.
0
(
2
1860462
)
2
4
.
2
2
160
(
*
)
4
.
2
*
48
(
*
85
.
117
s
=
+
s = 0.28 cm take smin = 6mm
Check deflection: as before
800
1250
<
δ = 1.56 cm
Example 4:
Redesign Question 2 assume NO Braking force bracing is used & L Bridge
= 57 m
Solution:
Since No Braking force Bracing, so the X-girder will carry M y
L = 57 m no of X.G = 1
7
.
5
57
+ = 11 X.G.
6.25
12.5
Loc.(50t)
Tender(40t) Loc.(50t) Tender(40t)
9m
1.5
10.5
8.4m 10.5m 8.4m
Wagon(40t)
12.0m
10.0
1.5 1.75 5.45<5.5
∑P (Wheel loads) = 2 * 50 + 2 * 40 + 40 + 20 = 240 t
11
7
/
240
=
B = 3.1 t
1.75 1.8 0.85
A
5.27
3.1 3.1 1.56 1.56
8
.
8
)
05
.
7
25
.
5
(
1
.
3
)
55
.
3
75
.
1
(
56
.
1 +
+
+
=
B
R = 5.27 t
M = 13.14 m t (At point "A" not at mid span)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

19/20XG
F bcx = 0.58 F y F bcy = 0.58 F y = 1.6 t / cm2
Check:
2
*
12
55
*
8
.
2
3
=
y
I = 77642 cm4
6
.
1
2
55
*
77642
100
*
14
.
13
6
.
1
)
8
.
2
2
110
(
*
1090652
100
*
6
.
273
max +
+
=
f = 1.2 ≤1.2 case B
Example 5:
Redesign Question 3 assume NO Braking force bracing L = 68.4 m (12
panel * 5.7 m)
Solution:
Loc.(50t)
1.5
10.0
8.4m
9m 10.5m 8.4m 12.0m 12.0m
60.3
10.5
12.5
6.25
1.5 1.75
4.85<5.5
Tender
40t
Loc.(50t) Wagon(40t) Wagon(40t)
Tender
40t
∑P (Wheel loads) = 2 * 50 + 2 * 40 +2 * 40 + 20 = 280 t
13
7
/
280
=
B = 3.1 t (Number of X. girders = Number of panels + 1)
We have to arrange braking force on 2 lanes in order to get maximum
effect (maximum straining actions on X.G.)
1.85 1.8 1.7 1.8 1.7 1.8 1.85
3.1 3.1 1.54 1.54
zero
A
10.65
8.85
7.15
5.35
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

20/20XG
M y = 21.45 m t (at point "A")
12
48
*
4
.
2
*
2
3
=
y
I = 44237 cm4
1
.
2
2
48
*
44237
100
*
45
.
21
1
.
2
3
.
2
max +
=
f = 1.65 > 1.2
Unsafe Increase dimensions specially of flange & recheck.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
Inf line and Loads 2009
1/29
I-Drawing of influence lines
a- Upper and lower chord:
• To draw influence line of upper or lower chord, we have to determine
the pole.
•
pole of upper
chord
pole of lower
chord
• If the member is in the lower chord, so the pole is in the upper chord
and vise versa.
• The ordinate under the pole is
Lh
ab
For marked member from the upper chord:
h
bupper
aupper
L=span of truss
a*b
L*h
I.L of upper
Chord (u)
pole(u)
M= a b
L
M
h
=
a b
L h
=
Force
Upper
(comp.)
(-)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
2/29
For marked member from the lower chord:
pole(L)
h
blower
a
lower
L=span
of truss
a*b
L*h
Lower
(tens.) M= a b
L
M
h
=
a b
L h
=
Force
I.L of lower
Chord (L)
(+)
Important note: The chord member (whether upper or lower) subjected
to either tension always or compression always.
b- Diagonal:
• Steps of drawing of I.L. of diagonal:
• If α is the angle between the diagonal and the horizontal,
1
sin
1
sin
1
sin
1
sin
1
sin
1
sin
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
3/29
•
a , b
diagonal
1
sin
1
sin
a
b
•
a, b
a, b
a
b
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
4/29
For dead load: We have to load both parts (positive and negative)
For LL: We have to load twice
a
b DL
a
b LL
a
b LL
• To get the sign of diagonal (tension or compression), first we have to
recognize the main sign of the diagonal.
• The diagonal is tension if tangent and compression if perpendicular to
B.M.D.
(tens.) diagonal
(comp.) diagonal
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
5/29
Important note: Always loading the bigger part of the I.L. gives the
main sign of the member, while loading the smaller part gives the other
sign.
.
diagonal
So the influence line of diagonal is as shown below
sin
1
* I.L of diagonal (D )
1.00
* I.L of Reaction (R)
h
sin
1
a
b
L L2
(+)
(-)
sin
1
sin
a
1
1
D
1
D
2
2
1
b
(+)
Very important:
Always for N-truss, the main sign is tension for all diagonals.
Always for W-truss, the main sign is one tension and the other is
compression.
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
6/29
c- Vertical:
• Always use the joint away from the position of X.G., so we can use
method of joint to calculate the vertical. (∑ = 0
y ).
Deck N-truss:
• For the first and last verticals (V1), it is always = reaction
• For the middle vertical (V5). Force = Reaction of XG
• For other verticals, V = D sinα and with opposite sign.
i.e. V4 = D3 sinα (from the lower joint away from X.G.)
1
* I.L of V
1.00
* I.L of V
h
1
a
b
L1 L2
V1
V2
D1
D2 V4
D3
4
(+)
(-)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
7/29
Vertical:
y ).
• For the middle vertical, it is zero member
• For all other verticals, V = D sinα and with opposite sign.
i.e. V3 = D3 sinα (from the upper joint away from X.G.)
1
* I.L of V
h
1
a
b
L1 L2
(-)
(+)
V1
V2
D1
D2 V
4
D3
4
V
5
V3
--------------------------------------------------------------------------------------------------
2- Deck W-truss (warren):
a- Upper and lower chord: Exactly as N-truss deck bridge
For marked member from the upper chord:
h
bupper
aupper
L=span of truss
a*b
L*h
I.L of upper
Chord (u)
pole(u)
M= a b
L
M
h
=
a b
L h
=
Force
Upper
(comp.)
(-)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
8/29
Note that: Each 2 adjacent members in the upper chord are equal in
magnitude and both are compression.
For marked member from the lower chord:
pole(L)
h
blower
a
lower
L=span
of truss
a*b
L*h
Lower
(tens.) M= a b
L
M
h
=
a b
L h
=
Force
I.L of lower
Chord (L)
(+)
Note that: Each 2 adjacent members in the lower chord are equal in
magnitude and both are tension.
The first and the last lower chords are zero members. These members
cannot be removed for stability.
b- Diagonal:
• The main sign of the diagonal varies according to its direction of
inclination. Tangent to moment is tension, while perpendicular is
compression.
h
T T T
T
C C C C
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
9/29
sin
1
h
sin
1
a
b
L L2
(+)
(-)
sin
1
sin
a
1
1
D
1
D
2
2
1
b
(-)
Tension
Zone
Comp
Zone
D
3
sin
1
b
(+) Tension
Zone
3
Important
c- Vertical:
y ).
• In W-deck truss, half the verticals are zero members, while the other
half are compression members with I.L. as shown below
h
1.0
1.0
1.0
1.0
1.0
2
1
V 2
V
4
V 6
V
8
V
I.L. of V
4
I.L. of V
8
I.L. of V
6
I.L. of V
1
I.L. of V
(-)
(-)
(-)
(-)
(-)
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
10/29
From previous:
• The first and last diagonals carry the reaction, so they have the same
I.L. of reaction.
• Verticals V2, V4, V6 and V8 which support X.G. carry loads only when
X.G. carry loads.
• Verticals V3, V5 and V7 are always zero members and can be removed
without affecting the stability of the truss. They only decrease the
buckling inside of the lower chord.
----------------------------------------------------------------------------------------------------
3- Pony or through W-truss: (beginning with tension diagonal)
a- Upper and lower chord: Exactly as N-truss deck bridge
b- Diagonal: As previous in case of W-deck truss
sin
1
h
sin
1
a
b
L L2
(+)
(-)
sin
1
sin
a
1
1
D
1
D
2
2
1
b
(-)
Tension
Zone
Comp
Zone
c- Vertical:
• In W-pony truss, half the verticals are zero members, while the other
half are tension members with I.L. as shown below
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
11/29
h
1.0
1.0
1.0
3
1
V
2
V
4
V 6
V 8
V
I.L. of V
5
I.L. of V
7
I.L. of V
1
I.L. of V
(+)
1
D
1.00
(-)
----------------------------------------------------------------------------------------------------
4- Pony or through W-truss: (beginning with compression diagonal)
a- Upper and lower chord: Exactly as W-truss deck bridge
b- Diagonal: As previous in case of W-pony truss
sin
1
h
sin
1
a
b
L2
(+)
1
Tension
Zone
1
D
2
D
1
L
(-)
Zone
2
b
1
sin
(-)
Comp
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
12/29
c- Vertical:
• In W-pony truss, half the verticals are zero members, while the other
half are tension members with I.L. as shown below
h
V V V V
1
D 1 3 5 7
(+)
I.L. of V
I.L. of V
I.L. of V
1.0 1.0
1.0
1
3
5
ENG_KHALED(BHIT)
ENG_KHALED(BHIT)

Main Truss
13/29
II-Calculation of loads
The members of the truss are subjected to either tension or compression
forces due to dead and live loads.
1- Dead load:
(1)Weight of tracks:
O.W. of one track (rail + sleeper) = 600 kg / m/
(2) O.W. of steal structure:
(Stringer + X.G. + truss + bracing + ------ )
• For single track (S.T.) (for spans ≤ 60 m)
W S = 0.5 + 0.05 L t / m/
à (deck)
W S = 0.75 + 0.05 L t / m/
à (through or pony)
• For double tracks (D.T.) (for spans ≤ 60 m)
W S = (0.5 + 0.05 L) * 1.9 t / m/
à (deck)
W S = (0.75 + 0.05 L) * 1.9 t / m/
à (through or pony)
Where
L = span of the Bridge (truss span)
W S is for the whole bridge width
Note: for spans > 60.0 m, the loads are reduced by 10%
)
05
.
0
75
.
0
(
9
.
0
)
05
.
0
5
.
0
(
9
.
0
L
W
L
W
S
S
+
=
+
=
à Single track (S.T.)
)
05
.
0
75
.
0
(
9
.
1
*
9
.
0
)
05
.
0
5
.
0
(
9
.
1
*
9
.
0
L
W
L
W
S
S
+
=
+
=
à Double tracks (D.T.)
2
)
2
∑
+
=
ack
weightoftr
w
W s
d = ----- t / m/
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Example: Calculate the dead load of (S.T.) deck bridge with span = 62 m
WS = 0.9 (0.5 + 0.05 * 62) = 3.24 t / m/
2
6
.
0
2
24
.
3
+
=
d
W = 1.92 t / m/
Example: Calculate the dead load of (D.T.) pony bridge with span = 48
m
WS = 1.9* (0.75 + 0.05 * 48) = 5.985 t / m/
2
2
*
6
.
0
2
985
.
5
+
=
d
W = 3.59 t / m/
2- Live load:
We have to use I.L. to calculate the maximum forces in different
members.
For diagonals and verticals, we must calculate both cases (tension and
compression for fatigue)
Important: Always calculate the force without impact load using wheel
load. The force will be F = f*m*n*(1+I)
Where I =
nL
+
24
24
"n" is number of tracks. 1 for single, 2 for double, 3 for triple…..
And "m" is reduction factor: 1 for single, 0.9 for double, 0.8 for triple.
• The reduction factor is used because in double tracks, we consider
the 2 trains reach the maximum case of loading at the same time
which is not the real case.
• In impact we use "nL" because each lane is considered as main
lane. So we have to load both of them to get maximum forces in
truss members.
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Example 1:
Pony bridge double track L = 54 m, S = 5.4 m Warren truss and N-truss,
Find maximum forces in truss members. (h = 6.0
m)
10*5.4=54
V1 V2 V3
V
4 V5
V6
U1
U2
U3
U2
U1
L1
L2
L3
L3
L2
L1
d1
d2 d3
d4 d5
L1
U1
Solution: Dead load for the bridge:
WS = 1.9 (0.75 + 0.05 * 54) = 6.56 t / m/
2
2
*
6
.
0
2
56
.
6
+
=
d
W = 3.88 t / m/
Force in member (U 1)
1
10t
12.5t
6.25t
0.42
0.81
0.78
0.65
0.54
0.46
0.33
0.22
0.1
0.02
h =6 m
pole (U )
1
10t
10t
10t
10t
10t
12.5t
6.25t
L*h
a*b
=0.81=
54*6
5.4*48.6
48.6 m
5.4
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Dead load: )
2
54
*
81
.
0
(
*
88
.
3
=
d
F = 84.8 t
Live load
f L = (2*6.25*0.42) + (3*12.5*0.78) + (4*10*0.65) + (2*6.25*0.54) +
(3*12.5*0.46) + (4*10*0.33) + (2*10*0.22) + (2*10*0.1) + (2*10*0.02)
= 104.5 t
I =
)
54
*
2
(
24
24
+
= 0.18 Not less than 0.25
FL+I = 0.9*2*1.25*104.5 = 235.1t (comp.)
1
max
min
26
.
320
8
.
84
U
t
F
t
F
→
−
=
−
=
Force in member (U 2):
12.5t
10t
maximum ordinate
37.8 m
0.9
0.21
0.01
16.2*37.8
54*6 =1.89
a*b
L*h
h=
6 m
0.57
1.28
1.52
1.75
1.89
1.68
1.10
0.56
=
16.2 m
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t
6.25t
2
pole (U )
2
U2
pole (U )
2
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Dead load: )
2
54
*
89
.
1
(
*
88
.
3
=
d
F = 197.87 t
Live load
fL = (2*6.25*0.56) + (3*12.5*1.1) + (10*1.68) + (10*1.89) + (2*10*1.75)
+ (2*6.25*1.52) + (3*12.5*1.28) + (4*10*0.9) + (2*10*0.57) +
(2*10*0.21) + (10*0.01) = 237.7 t
I =
)
54
*
2
(
24
24
+
FL+I = 0.9*2*1.25*237.7 = 535 t (comp.)
F d + L + I = 197.87 + 535 = 732.87 t (comp.) 2
max
min
87
.
732
87
.
197
U
t
F
t
F
→
−
=
−
=
Force in member (U 3):
27.0 m
27*27
54*6
=2.25
a*b
L*h
h=
6 m
=
2.03
1.48
0.84
0.46
2.25
2.08
1.44
0.89
0.23
27.0 m
U3
U3
pole (U )
3
pole (U )
3
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t
6.25t
Dead load: F d = w d * area of I.L. ∴ )
2
54
*
25
.
2
(
*
88
.
3
=
d
F = 235.56 t
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Live load
f L = (2*6.25*0.46) + (3*12.5*0.84) + (4*10*1.48) + (2*6.25*2.03) +
(3*12.5*2.08) + (4*10*1.44) + (2*10*0.89) + (2*10*0.29) = 281.4t
I as before = 0.25
FL+I = 0.9*2*1.25*281.4 = 633.3t (comp.)
F d + L + I = 235.56 + 633.3 = 868.82 t (comp.) 3
max
min
82
.
868
56
.
235
U
t
F
t
F
→
−
=
−
=
Force in member (L1): L1 is zero member under vertical loads.
F max = F min = zero à L1 because there is no hl. loads.
Force in member (L2):
43.2 m
10.8*43.2
54*6
=1.44
a*b
L*h
h=
6 m
=
0.77
0.16
1.44
1.35
1.13
0.98
0.72
0.5
0.26
0.1
10.8 m
pole (L )
2
L2
L1
L1 L2
pole (L )
2
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t
6.25t
10t
Dead load: )
2
54
*
44
.
1
(
*
88
.
3
=
d
F = 150.76 t
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Main Truss
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Live load
f L = (2*6.25*0.16) + (3*12.5*0.77) + (4*10*1.35) + (2*6.25*1.13) +
(3*12.5*0.98) + (4*10*0.72) + (2*10*0.5) + (2*10*0.26) + (2*10*0.1)
= 181.7t
I = 0.25 FL+I = 0.9*2*1.25*181.7 = 408.8t (Tension)
F d + L + I = 150.76 + 408.8 = 559.6t (tens.) 2
max
min
6
.
559
76
.
150
L
t
F
t
F
→
+
=
+
=
Force in member (L3):
32.4 m
21.6 m
1.12
0.66
1.89
2.16
1.9
1.59
1.08
0.64
0.16
pole (L )
3
L3
L3
pole (L )
3
h=
6 m
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t
6.25t
21.6*32.4
54*6
=2.16
a*b
L*h =
Dead load: )
2
54
*
16
.
2
(
*
88
.
3
=
d
F = 226.14 t
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Live load: fL = (2*6.25*0.66) + (3*12.5*1.12) + (4*10*1.89) +
(2*6.25*1.9) + (3*12.5*1.59) + (4*10*1.08) + (2*10*0.64) +
(2*10*0.16) =268.5 t (tension)
I = 0.25 FL+I = 0.9*2*1.25*268.5 = 604.1t (Tension)
F d + L + I = 226.14 + 604.1 = 830.3 t 3
max
min
3
.
830
14
.
226
L
t
F
t
F
→
+
=
+
=
Force in member (D 2): α = tan-1
(6/5.4) = o
48 35
.
1
sin
1
=
α
48.0m
10*5.4=54 m
(+)
(-)
6m
0.49
1.08
1.03
0.83
0.67
0.56
0.36
0.20
0.04
0.13
0.13
0.09
0.05
1.35
D2
D2
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t
6.25t
1.35
12.5t
h=
6 m
x
0.13
=
4.5-x
1.08 x = 0.6m
Dead load:
2
1
)
( =
−ve
A * 48 * 1.08 = 25.83
2
1
)
( =
+ve
A * 6 * 0.13 = 0.4
Fd = 3.88*(25.83-0.4) = 98.6t
Live load:
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F L max = (2*6.25*0.49) + (3*12.5*1.03) + (4*10*0.83) + (2*6.25*0.67) +
(3*12.5*0.56) + (4*10*0.36) + (2*10*0.2) + (10*0.04) = 126.1
I max =
)
48
*
2
(
24
24
+
= 0. 2 Imin= 0.25
FL+I(max) = 2*0.9*1.25*126.1= 283.8t
FL min = 12.5 * (0.05 + 0.09 + 0.13) = 3.4t
I min =
)
6
*
2
(
24
24
+
= 0.67 < 0.75 FL+I(min) = 2*0.9*1.67*3.4= 10.22t
F max = 98.6 + 283.8 = 382.4 t F min = 98.6 – 10.22 = 88.4 t
Note that: For case of N-truss, the maximum tension in the diagonal is
382.4t and the minimum tension is 88.4t. For case of W-truss, the
maximum compression in the diagonal is 382.4t and the minimum
compression is 88.4t.
Force in member (D 3): α = o
48 35
.
1
sin
1
=
α
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42.0m
12.0 m
0.89
0.70
0.54
0.42
0.23
0.07
0.27
1.35
0.34
0.94
1.35
0.01
0.11
0.23
0.27
(+)
(+)
D
D3
h=
6 m
12.5t
6.25t 10t
10t
10t
10t
12.5t
6.25t
12.5t
6.25t
10t
3
Dead load
2
1
)
( =
−ve
A * 12 * 0.27 = 1.61
2
1
)
( =
+ve
A * 42 * 0.94 = 19.78
F d = 3.88 * (19.78 – 1.61) = 70.5 t
Live load
F L max = (2*6.25*0.34) + (3*12.5*0.89) + (4*10*0.7) + (2*6.25*0.54) +
(3*12.5*0.42) + (4*10*0.23) + (2*10*0.07) = 98.7 t
I max =
)
42
*
2
(
24
24
+
= 0. 22 Imin =0.25
FL+I(max) = 2*0.9*1.25*98.7= 222.1t
F L min = (3*12.5*0.23) + (2*6.25*0.11) + (10*0.01) = 10.1t
I min =
)
12
*
2
(
24
24
+
= 0.5 FL+I (min) = 2*0.9*1.5*10.1= 27.3t
F max = 70.5 + 222.1= 292.6 t F min = 70.5 –27.3 = 43.2 t
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Note that: For both cases of N-truss and W-truss, the maximum tension
in the diagonal is 292.6 and the minimum tension is 43.2.
Force in member (V 1)
h=
6 m
U1
D1
1
V
1
V
1
D
1
U
For both cases of N and W-truss, the vertical is D1 sinα
So we have to calculate the force of D1
We can calculate D1 by 2 methods:
1- D1 cosα = U1 D1 =
α
cos
1
U
α = 0
48
48
cos
26
.
320
max
1 =
D = 478.74 t (tens.) 7
.
126
48
cos
8
.
84
min
1 =
=
D t (tens.)
V 1 = D 1 sinα =
α
cos
1
U
sinα = U 1 tan α
V 1 max = 320.26 tan α = 355.84 t (comp.)
V1 min = 84.8 tan α = 94.2(comp.)
OR using I.L. of diagonal D1
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48.6 m
10t
12.5t
6.25t
0.63
1.21
1.17
0.98
0.81
0.69
0.5
0.33
0.15
0.03
h =6 m
5.4
10t
10t
10t
10t
10t
12.5t
6.25t
1.35
1.21
1
D
Dead load: )
2
54
*
21
.
1
(
*
88
.
3
=
d
F = 126.7 t
Live load
f L = (2*6.25*0.63) + (3*12.5*1.17) + (4*10*0.98) + (2*6.25*0.81) +
(3*12.5*0.69) + (4*10*0.5) + (2*10*0.33) + (2*10*0.15) + (2*10*0.03)
= 156.75 t
I =
)
54
*
2
(
24
24
+
FL+I = 0.9*2*1.25*156.75 = 352.7 t (tension)
1
max
min
4
.
479
7
.
352
7
.
126
7
.
126
D
t
F
t
F
→
=
+
=
=
V 1 = D 1 sinα )
(
3
.
356
48
sin
*
4
.
479
1
.
94
48
sin
*
7
.
126
1
max
min
Comp
V
t
F
t
F
→
−
=
=
−
=
=
Gives the same answer as before
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V2
V2
h=
6 m
D2
• V2 for N-truss: V2 = D2*sin48
From previous: D 2-max = 382.4 t D 2-min = 88.4 t (tension)
V2-max = 382.4*sin48=284.2 t V2-min = 88.4*sin48=65.7 t
(compression)
• V2 for W-truss for Pony Bridge is zero member.
h=
6 m
D
V3 3
V3
• V3 for N-truss: V3 = D3*sin48
From previous: D 3-max = 292.6 t D 3-min = 43.2 t (tension)
D max = 292.6*sin48=217.4 t D min = 43.2*sin48=32.1 t (compression)
• V3 of W-truss for Pony Bridge is tension
)
2
8
.
10
*
1
(
*
88
.
3
=
d
F = 20.94 t (tens.)
Live load
F L = (6.25*0.31) + (12.5*0.63*2)
+ (12.5*1) + (10*0.07) = 30.9 t
0.07
0.63
1.00
0.63
0.31
10t
12.5t
6.25t
V3
10.8m
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I =
)
8
.
10
*
2
(
24
24
+
= 0.526
F L + I = 2*0.9*1.526*30.9= 84.9t
F d + L + I = 20.94 + 84.9 = 105.84 t (tens.) for pony
7
5
3
max
min
,
,
81
.
105
94
.
20
V
V
V
t
F
t
F
→
+
=
+
=
Example 2:
Find the maximum forces in the marked truss members for the shown
Deck bridge triple track L = 54 m, S = 5.4 m in the case of Warren truss
and the case of N-truss. (h=6.0m)
10*5.4=54
Solution: Dead load for the bridge:
WS = 2.7 (0.5 + 0.05 * 54) = 8.64 t / m/
2
3
*
6
.
0
2
64
.
8
+
=
d
W = 5.22 t / m/
V1 in both W-truss and N-truss is the same and equals to the reaction of
the truss. So we draw the I.L. of reaction of truss
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Main Truss
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54.0 m
R
R
0.96
0.82
0.70
0.61
12.5t
6.25t 10t
10t
10t
10t
10t
12.5t 10t 10t
V1
V1
1.0
0.35
0.21
0.13
0.01
0.47
Dead load: F d = (
2
1
* 1 * 54) * 5.22 = 140.9 t
Live load:
F L = (3*12.5*0.96) + (4*10*0.82) + (2*6.25*0.7) + (3*12.5*0.61) +
(4*10*0.47) + (2*10*0.35) + (2*10*0.21) + (2*10*0.13) + (10*0.01)
= 133.2 t
I =
)
54
*
3
(
24
24
+
= 0.13 I min= 0.25
F L + I = 3*0.8*1.25*133.2 = 399.6t
F d + L + I = 140.9 + 399.6 = 540.5 t
F min = 140.9 t F max =540.5 t
This is the same for the reaction.
i.e. the force used to design bearing is 540.5t
a- For W-truss:
V
2
10.8m
6.25t
12.5t 10t
0.31
0.63
1.00
0.63
0.07
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The reaction is the load on the X.G. and the member is always
compression.
)
2
8
.
10
*
1
(
*
22
.
5
=
d
F = 28.2 t (tens.)
Live load
F L = (6.25*0.31) + (12.5*0.63*2)
+ (12.5*1) + (10*0.07) = 30.9 t
I =
)
8
.
10
*
3
(
24
24
+
= 0.42 F L + I = 3*0.8*1.42*30.9= 105.3t
F d + L + I = 28.2 + 105.3 = 133.5 t (Comp.) for deck
8
6
4
max
min
,
,
5
.
133
2
.
28
V
V
V
t
F
t
F
→
−
=
−
=
b- For N-truss:
10*5.4=54
V
D 2
1
From the unloaded joint (lower joint)
V2 = D1sinα , so we have to calculate the force in the diagonal D1
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Main Truss
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48.6 m
10t
12.5t
6.25t
0.63
1.21
1.17
0.98
0.81
0.69
0.5
0.33
0.15
0.03
5.4
10t
10t
10t
10t
10t
12.5t
6.25t
1.35
1.21
1 2
D V
Dead load: )
2
54
*
21
.
1
(
*
22
.
5
=
d
F = 170.5 t
Live load
f L = (2*6.25*0.63) + (3*12.5*1.17) + (4*10*0.98) + (2*6.25*0.81) +
(3*12.5*0.69) + (4*10*0.5) + (2*10*0.33) + (2*10*0.15) + (2*10*0.03)
= 156.75 t
I =
)
54
*
3
(
24
24
+
FL+I = 0.8*3*1.25*156.75 = 470.25 t (tension)
1
max
min
75
.
640
25
.
470
5
.
170
5
.
170
D
t
F
t
F
→
=
+
=
=
V 2 = D 1 sinα )
(
2
.
476
48
sin
*
75
.
640
7
.
126
48
sin
*
5
.
170
1
max
min
Comp
V
t
F
t
F
→
−
=
=
−
=
=
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<15
Y
Y
X
X
/ o
Butt Weld
Gusset plate
Diagonal
member
Vertical
member
5
:
1
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Design of truss members in Bridges
(As welded members)
b
b
t
t
b
1
2
2
3 b
h
t
t
Upper
Chord
Lower
Chord
Vertical
or
diagonal
t
t
h
G.PL
G.PL
10-20cm
fl
w
Gusst
plate
fl
b
Buckling lengths of members: ECP 58, 59 table 4.5
Some notes for table 4.5
• Single triangulated web system is W or N-truss.
• Compression chord effectively braced is for all verticals and
diagonals of Deck Bridge.
• Compression chord un-braced is for all verticals and diagonals of
Pony Bridge.
• Compression chord may be braced or un-braced for verticals and
diagonals of through bridge according to the suggested shape of
the upper bracing.
• For all buckling lengths see page 40
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We have to study how to design truss member subjected to:
1- Compression only 2- Tension only 3- Moment & tension
4- Moment & Comp 5- Zero member 6- M only
(I) Design procedure of upper chord (compression member):
Assume stress for steel 44 f ≈1.2 à 1.5 t / cm2
Assume stress for steel 52 f ≈1.5 à 1.9 t / cm2
Calculate area =
f
Fmax ------ cm2
= 2 h t + b/
t
Take h =
15
12
15
12
length
→
=
→
S
Panel
ECP 129
Note: Panel length is the length of the member
Take b = (0.75 à 1) h for deck or through ECP 129
b = (1 à 1.25) h for pony bridge
b/
= b + 2 * (10 à 20 cm )
Applying the equation of area à get (t) = ---- cm
Note: If t ≥ 4 cm, use reduced Fy t≥1.2cm
Minimum thickness is 1.2cm for upper and lower chord because this
plate will be used as gusset plate at the position of connection. The
minimum thickness of gusset plate is 12 mm ECP 133.
• If the thickness "t" is very big, we can use box section or use stiffener
10-20 cm
t
t
h
b
C=10-20cm
t
b
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(1) Check local buckling:
We must be sure that the section is not slender by applying the non-
compact conditions in the code.
y
F
t
b 64
≤ (Table 2-1 b) code p. 10 ( plate)
y
F
t
c 21
≤ (Table 2-1 c) code p. 11 (as cantilever of L – section)
y
F
t
h 30
≤ (Table 2-1 d) code p. 12 (as cantilever of T – section)
Note: IF
y
F
t
h 30
> so we can use "box section"
Then check that
y
F
t
h 64
≤
(1) Check Global buckling:
The limit of buckling "λ " ≤ 90 for railway ECP 51
x
in
in
r
l
=
λ = ---- ≤ 90
y
out
out
r
l
=
λ = ---- ≤ 90
To calculate r x , r y :
Calculate
t
b
ht
t
h
t
b
h
ht
y
/
/
2
)
5
.
0
(
)
5
.
0
*
2
(
+
+
+
=
v
Calculate I X = 2 *
12
3
th
+ 2 t h ( y
r
- 0.5 h) 2
+ b/
t ( y
r
- h – 0.5 t) 2
= --- cm4
I y = 2 t h (0.5 b + 0.5 t) 2
+
12
3
/
tb
= ---- cm4
Calculate A = b/
t + 2 h t &
A
I
r x
x = &
A
I
r
y
y =
Y
t
h
b
X X
Y
Y
b
t
t
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(2) Check compressive stresses
Actual stresses =
A
F
f I
L
d
ca
+
+
= = ---- t / cm2
≤ F C
Allowable stresses = F C = 1.6 – 8.5 * 10-5 2
max
λ for st. 44
F C = 2.1 – 13.5 * 10-5 2
max
λ for st. 52
Important note: There is no check fatigue in the upper chord because
both the maximum and the minimum forces are compression.
II- Design procedure of lower chord (tension member):
Fatigue in tension members must be considered
How to consider fatigue?
)
1
(
max
min
max
T
T
F
F sr
−
= So we have to get Fsr from ECP 41
The member will be welded, so detail "B"
To calculate number of cycles Table 3.1b ECP40
Member description No Fs r
- Chord members (Class I) (L<10m) > 2 * 10 6
1.12
Double track
(D.T.)
2 * 10 5
2.80
web members
(vl.& diag.)
(Class II) single track (S.T.) 5 * 10 5
2.00
Double track
(D.T.)
5 * 10 5
2.00
web members
(carry load of
X.G. only)
(Class III)
single track (S.T.) > 2 * 10 6
1.12
Notes:
• For chord member: "L" is length of member and always < 10m
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• If more that 2 tracks, we can use "n" of double tracks
• "n" = 2*105
is not in the table, so we have to calculate it from the
chart ECP 42.
• Class III means that the force in the vertical member calculated
from one X.G. and there is no affect of diagonals. (See marked
members)
Sub-divided W-Truss
Design procedure:
Take "b, h" as the upper chord
b/
= (1/2 2/3 ) * b
Assume
)
1
(
max
min
max
T
T
F
F sr
−
=
For st 44 Take smaller from F max or 1.6 t / cm2
=
Area
Tmax
For st 52 Take smaller from F max or 2.1 t / cm2
=
Area
Tmax
Get A = -- cm2
= 2 h t + 2 b/
t
Get t = -- cm tmin = 12mm t≤ 4 cm or use reduced Fy
There is no check local buckling for
pure tension members.
b
1
2
2
3 b
h
t
t
Lower Chord
1
2
2
3 b
b =
y
X
X
y
Stiffener
h
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x
in
in
r
l
=
λ = ---- ≤160
y
out
out
r
l
=
λ = ---≤ 160
Where
A
I
r x
in =
A
I
r
y
out =
And we have to calculate y to get Ix
(3) Check stresses:
Max. Stress à
A
Tmax = --- ≤ 1.6 t / cm2
or 2.1 t / cm2
Stress range à
A
F
F
A
T
T d
I
L
d −
=
− +
+
min
max = ---- Fsr
Note that: For case of railway, Tmax = Fd+L+I .
For case of roadway, Tmax = Fd +0.6 FL+I.
Note that: For lower chord, Fmin = Fd
(4) Check deflection: (roadway)
35
&
(railway)
30
d
l
≤ ECP 129
Note that: for lower chord: d = h + t
(III)- Design procedure of web members (compression members):
Assume f ≈1.2 à 1.5 t / cm2
st. 44
1.5 à 1.9 st. 52
Get Area =
f
force
Max
= --- cm2
= 2 bfl * tfl + b t w
Take "b" as chord members exactly
Get t w from
y
w F
t
b 64
≤ & t w ≥1.0 cm
t
t
b
Y
C
Y
x
x
Flange
Web
f
fl
b W
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Applying the equation of area à Get bFL * tF L = --- cm2
Assume fl
t
20)
(10−
=
fl
b Get bFL& tFL
)
20
(
)
10
(
flange
.
It is preferable that b > bfl
y
fl F
t
C 21
≤
y
w
fl
F
t
s
t
b 64
2
2
≤
−
−
Calculate Iout = 2
3
)
2
2
(
2
12
)
2
( fl
fl
fl
fl
w
x
t
b
t
b
t
b
t
I −
+
−
= (Buckling outside
is about the stronger axis)
12
*
2
fl
fl
y
b
t
I =
A = (b – 2 fl
t ) t w + 2 fl
flt
b
A
I
r x
x =
A
I
r
y
y =
y
in
in
r
l
=
λ = --- ≤ 90 & 90
≤
−−
=
=
y
out
out
r
l
λ
(y - y axis with inside plane) & (x-x axis with outside plane)
(3) Check stresses:
2
max
5
10
*
5
.
8
6
.
1 λ
−
−
≤
−−
=
=
Area
Force
fca Or 2.1 – 13.5 * 10-5 2
max
λ
(4) Check deflection:
30
d
l
≤ For web member d = bfl
No check deflection for vertical members
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(IV) Design procedure of web members (as tension members):
Assume −
−
−
=
−
=
max
min
max
1
T
T
F
F sr
y
F
58
.
0
≤ Get area =
max
max
F
T
--- cm2
Get b, t w, fl
t , fl
b as before. (No check local buckling)
If Area of flange is negative, so use minimum flange which is:
tfl = 10mm bfl = 6φ + tw (assume using M24 or M27)
- Check global buckling ≤
λ 160
- Check maximum stress = y
F
A
I
L
d
F
T
F 58
.
0
max
max ≤
+
+
=
=
- & stress range (fatigue). = sr
I
L
F
A
F
≤
+
Check deflection: 30
d
l
≤ For web member d = bfl
(V) Design procedure of zero members (web members):
Take
y
w F
t
b 64
≤ get t w = -- cm
Take t w = fl
t = --- 1.0 cm
Get fl
b by assuming that there is only one
row of bolts each side of the flange taking
bolts M24 , So fl
b 6d + t w + 2 s
Check local buckling
y
w
fl
F
t
s
t
b 64
2
2
≤
−
−
&
y
fl
w
fl
F
t
s
t
b 21
2
/
2
/
≤
−
−
Check global buckling:
y
in
in
r
l
=
λ = --- ≤ 90 & 90
≤
−−
=
=
y
out
out
r
l
λ
b
l
fl
≤
b
<1.5d
bolt
Gusset
plate
tW
b
fl
t
f
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(VI) Design procedure of zero members (lower chord):
Take "t" the bigger value of
y
F
t
h 64
≤
y
F
t
b
t
c 21
2
/
/
/
≤
≈
Where h, b, b/
are taken as other
chords.
- Check local buckling
- Check global buckling
y
in
in
r
l
=
λ ≤ 90 & 90
≤
=
y
out
out
r
l
λ
- Check deflection: (roadway)
35
&
(railway)
30
d
l
≤ ECP 129
Note that: for lower chord: d = h + t
(VII) Design vertical or diagonal subjected to tension and
compression at the same time:
Calculate Area from both tension and compression
a) Assume f ≈1.2 à 1.5 t / cm2
st. 44
1.5à 1.9 st. 52
Acomp =
f
force
n
compressio
Max
= --- cm2
b) Aten =
1
.
2
6
.
1
force
n
Max tensio
or
= --- cm2
Take bigger of Acomp and Aten
b is the same as the truss
Take
y
w F
t
b 64
≤ get t w = -- cm
Stiffener
b
c
h
c
b =(1
2
2
3
)b
t
X
X
y
y
10cm
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A2 flanges = A – b*tw Assume bf = 20tf get bf and tf
Checks: 1) As compression member:
Check local buckling
y
w
fl
F
t
s
t
b 64
2
2
≤
−
−
&
y
fl
w
fl
F
t
s
t
b 21
2
/
2
/
≤
−
−
y
in
in
r
l
=
λ = --- ≤ 90 & 90
≤
−−
=
=
y
out
out
r
l
λ
Check stresses: c
F
A
≤
C
2) As tension member:
Check stresses: y
F
A
58
.
0
T
≤
b
l
fl
≤
Check fatigue: fsr = sr
F
A
c
T
≤
−
− )
(
i.e. If the forces in the diagonal is as following:
Fd = 40 t(Comp) FL+I(max) = 120t (Comp) FL+I(min) = 50t (tens)
So Fmin = 50-40 = 10t (tens) Fmax = 120+40 = 160t (comp)
We have to design the member as compression member (160t) and check
as tension member (10t).
10/A
60/A
fsr
ft
fc
Check fatigue: fsr = sr
F
A
A
A
c
T
<
=
−
−
=
−
− 170
)
160
(
10
)
(
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Summary of buckling lengths of members:
Table 4-5 ECP 58 and 59
1- Deck bridge:
• All upper chords: Lin = 0.85L and Lout = 0.85L
• All verticals and diagonals = Lin = 0.7L and Lout = 0.85L
• For lower chord: Lin = 0.85L and Lout = 0.85L or 0.85(2L)
According to the shape of the lower bracing
L
1 L
3
Lower bracing of deck bridge
• For L1: Lin = 0.85L and Lout = 0.85L
• For L3: Lin = 0.85L and Lout = 0.85(2L)
2- Pony bridge
• For lower chord: Lin = 0.85L and Lout = 0.85L
• All upper chords: Lin = 0.85L and Lout = 2.54 δ
a
EI y
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4
5
.
2 δ
a
EI
l y
out = a is spacing between X.G.
E = 2100 t / cm2
& a = X.G spacing
I y = y – y inertia of upper chord
=
δ Flexibility of U- frame (cm / t)
If not given
2
2
2
1
3
1
2
3 EI
d
EI
d
+
=
δ
I 1 = I X - X of vertical member I2 = IX – X of X.G
x
x
x
x
X.G.
Vertical
member
Important note: Incase of using U – Frame every X.G, all the verticals
will be subjected to moment due to C/100 and due to wind in addition to
the force from vertical loads which is tension or compression.
d
B
2
d1
y
y
Upper
Chord
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3- Through bridge
a) If the upper bracing is as shown:
Upper bracing of through bridge
• All upper chords: Lin = 0.85L and Lout = 0.85L
• For lower chord: Lin = 0.85L and Lout = 0.85L
b) If the upper bracing is as shown:
For All lower chords: Lin = 0.85L and Lout = 0.85L
Upper chord: For U1 : Lin = 0.85L and Lout = 0.85L
For U2 : Lin = 0.85L and Lout = 0.85(2L)
M.G.
Plan of upper bracing
V
1
V
2
V
3 V
4 V
5
D
1
D
2
D
3
D
4
D
5
u1
u2
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Verticals and diagonals
For V1, D1, V2, D2, V4, D4 the compression chord is
effectively braced Lin = 0.7L and Lout = 0.85L
For V3, D3, the compression chord is not braced
Lin = 0.7L and Lout = 1.2L
Design of members subjected to moment
Design procedure of Verticals of pony bridge:
M Only
All verticals
(M+N)
(M+T)
Always all verticals of Pony Bridge subjected to moment in addition
to the main force from cases of loading.
Mx of pony = h
F
*
100
max where h = hmg -
2
xg
h
Mx of wind = (0.1S*0.5)*
2
2
h
where "S" is spacing between X.G.
F
100
h
max F
100
max
C.L. of X.G.
VL
of
pony
bridge
C.L. of X.G.
VL
of
pony
bridge
M1
M
2
The moment is outside the plan.(about x-axis of the vertical member)
Fmax is the maximum compression force in the upper chord. (The force in
the upper chord at mid span of truss)
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Note that: The effect of h
F
*
100
max and wind is case "B"
Procedures:
(1) If the member is subjected to M + N:
a- Estimate section: "b" of the whole truss or ≈
15
12 −
S
For web:
y
w F
t
b 64
=
For flange: Force =
3
N
b
M x
+ (comp)
Assume F max = 1.5 t / cm2
for st. 44 & 2.0 t / cm2
for st.52)
Af =
max
F
Force
= bfl tfl complete as before
Note that: We can estimate the section as previous in case of web
member subjected to pure compression & decrease the assumed stress.
b- Check member as pure compression member for case "A"
- Local buckling
- Global buckling ≤ 90
- C
ca F
f <
c- Check member as beam – column (subjected to moment & normal)
for case "B"
2
.
1
1 ≤
+ A
F
f
F
f
bcx
bcx
C
ca
Where C
ca F
&
f from previous step "b"
y
I
M
f
out
bcx
max
= [
2
b
y
&
I
is
I X
out = ]
N/2 M
M
N
x
b
N/2
b tW
t
f
b
Y
C
fl
Y
x
x
x
M
b
x
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in
b
l
L −
≈
act
u of member (important)
Because the moment is outside plan, so L.T.B is inside plan.
Calculate L u max then F bcx
bigger
smaller
M
M
=
α = zero C b = 1.75
If 15
.
0
1
≤
C
ca
F
f
A 1 = 1
If 15
.
0
1
>
C
ca
F
f
EX
ca
F
f
A
−
=
1
85
.
0
1
Where
2
7500
x
EX
F
λ
= out
is
x
λ
Note: The moment rotates about the stronger axis
- Check using interaction equation
2
.
1
1 ≤
+ A
F
f
F
f
bcx
bcx
C
ca
(2) If the member is subjected to M + T:
a - Estimate section as before (web subjected to pure tension) using
y
sr
F
T
T
F
F 58
.
0
1
max
min
max ≤
−
=
Or Force in the critical flange =
3
T
b
M x
+ (tens) Afl =
max
max
F
T
b - Check member as pure Tension
- Global buckling ≤ 160
- ft ≤ 0.58 F y
- fsr ≤ Fsr
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No check deflection because it is vertical member
c -Check using interaction equation as member subjected to M & T
2
.
1
≤
+
btx
btx
t
t
F
f
F
f
ft as before =
Area
Tmax
, F t = 0.58 F y
y
I
M
f
X
btx = Where I X = I out & y =
2
b
Fbtx = 0.58 F y (always because there is no. L.T.B in tension side)
d- Check fatigue:
Fmax – Fmin ≤ Fsr
Fmax = y
I
M
A
T
out
*
max
+ M is Mw + h
C
*
100
max
Cmax = CD+L+I
Fmin = y
I
M
A
T
out
*
min
+ M is h
C
*
100
min
Cmin = CD
CD and CD+L+I are of upper chord at mid span of truss.
DO Not take the effect of wind in studying fatigue
(3) If the member is subjected to M only
a - Estimate section as web zero member (zero member) use b of
truss, t w = 1 cm, b f (min.) = 6φ + 2s + t w (s is size of weld ≈1cm)
t f (min.) = 1 cm.
b - Check as zero member
- Local buckling
- Global buckling < 90
- No check deflection because it is vertical member
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c – Check y
I
M
X
X
(I X = I out & y =
2
b
) ≤ F bcx * 1.2 where F bcx is
calculated using in
b
act
u l
≈
L & L u max (C b = 1.75).
Design procedure of Vertical of through bridge forming the closed
portal frame:
The member is subjected to
either tension and moment
or compression and
moment.
a- Estimate the section as before (section subjected to moment and
compression or moment and tension).
b- Check the member as pure compression or pure tension (case A)
- To calculate buckling lengths:
Buckling inside: From table of code for web members lin = 0.7l
Buckling outside: The vertical member forms portal frame outside plan,
so lout is calculated using GA and GB.
GA =
xg
xg
v
v
l
I
l
I
/
/
GB =
beam
beam
v
v
l
I
l
I
/
/
Lxg is the width of the bridge = Lbeam
Ibeam is the Ix of the horizontal beam.
Note that: Buckling inside and outside of all other verticals and
diagonals is from table of code.
The frame is allowed to sway: Cmx = 0.85
First vertical
(M+N) (M+T)
CL of upper beam
CL of XG
Vl
of
truss
2h/3
h/3
R
u
M
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5
.
0
+
=
=
bigger
smaller
M
M
α Cb = 1.75 + 1.05*0.5+0.3*(0.5)2
= 2.35, taken
2.3
c- Check as beam column (case B)
Design procedure of diagonal of through bridge forming the open
portal frame:
The member is subjected to compression
and moment.
a- Estimate the section as before
using the same section of the upper chord.
b- Check the member as pure compression (case A)
a. To calculate buckling lengths:
Buckling inside: From table of code for web members lin = 0.7l
Buckling outside: The diagonal forms portal frame outside plan, so lout is
calculated using GA and GB.
GA = 10 (hinged base)
GB =
beam
beam
dl
dl
l
I
l
I
/
/
Lbeam is the width of the bridge
Ibeam is the Ix of the horizontal
beam.
Note that: Buckling inside and outside of all other verticals and
diagonals is from table of code.
The frame is allowed to sway: Cmx = 0.85
First diagonal
(M+N)
CL of upper beam
Vl
of
truss
L(inclined distance)
R
u
R /2
u R /2
u
M
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0
=
=
bigger
smaller
M
M
α Cb = 1.75
c- Check as beam column (case B)
Summary of all previous:
1. For deck bridge:
- All the members of the truss are subjected to either tension
or compression only (NO moment)
- lin & lout of members are from table in Code ECP 58&59
2. For pony bridge:
- All members of the truss are subjected to either tension or
compression except verticals.
- All verticals are subjected to moment in addition to actual
tension or compression.
M = 0.1*0.4 S * h2
/2 + h
C
*
100
max
where h = hMG -
2
XG
h
- Buckling length of all members are from ECP 58&59 except
upper chord.
- Buckling length of upper chord is 4
5
.
2 δ
S
EI y
3. For through bridge:
• If there is inclined frame, so the section of the first inclined
diagonal is the same as the upper chord.
- All the members subjected to either tension or compression
except vertical or diagonal forming the portal frame.
Vertical: M = Ru/2 * 2/3 h Ru of loaded case
Horizontal: M = Ru/2 * l Ru of loaded case
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Lin & lout for all members are from ECP 58&59 except for vertical
or diagonal forming portal frame. We use GA & GB
For vertical: GA =
xg
xg
v
v
l
I
l
I
/
/
GB =
beam
beam
v
v
l
I
l
I
/
/
For diagonal: GA = 10 (hinged base) GB =
beam
beam
dl
dl
l
I
l
I
/
/
Very important note;
b is constant for whole of the truss. So before design we must check if
any section is given in the exam, we must take "b" from it
Solved Example 1:
Deck bridge as shown, h of truss = 4m, St.44, single track
U4 U
3
U
1
V
1
d
1
V
2
d
2
V3
d
3
V
4
d4
V5
L
1
L
2
L
3
R=241t 5.0m
40.0m
L
4
2
U
The forces in the shown members are given in the following table:
Member Maximum Minimum Member Maximum Minimum
V2 -96 -15 D1 +352 +78
U3 -515 -113 D2 -260 -46
L2 +418 +90 D3 +181 -2
L4 +553 +119 D4 -110 0
It is required to design U3, U4, L4, V1 and D3
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Design members 2009
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Solution:
Design (U 3 &, U4 ): They have the same force
h =
15
12
500
→
= 41.7 ≈
→ 3
.
33 40 cm
For deck bridge b ≈(0.75→1)* 40
= 30 →40 cm ≈36 cm
Assume f = 1.2 à 1.5 for st. 44
A =
2
.
1
515
= 429.2 cm2
Taking b/
≈36 + 2*(10 à 20) = 56 à76 ≈60 cm
So A = 429.2 = (2 * 40 + 60) * t t = 3.07 cm ≈3.2 cm
• Check local buckling
Taking weld size ≈1 cm
8
.
2
64
6
.
10
2
.
3
1
*
2
36
≤
=
−
=
t
b
= 38.2
8
.
2
21
4
.
2
2
.
3
1
2
.
3
12
≤
=
−
−
=
t
c
= 12.5
8
.
2
30
2
.
12
2
.
3
1
40
≤
=
−
=
t
h
= 17.9
• Check global buckling
)
2
.
3
*
60
(
)
2
*
2
.
3
*
40
(
)
6
.
41
*
2
.
3
*
60
(
)
2
*
20
*
2
.
3
*
40
(
+
+
=
y = 29.3 cm
]
2
*
12
40
*
2
.
3
[
3
=
x
I + [3.2*40*(29.3-20)2
*2] + [3.2*60*(29.3-41.6)2
]
= 85322 cm4
]
12
60
*
2
.
3
[
3
=
y
I + [2*3.2*40*(18+1.6)2
] = 155945 cm4
A = (3.2 * 2 * 40) + (60 * 3.2) = 448 cm2
3.2 cm
y
3.2
40
36
X X
Y
29.3
cm
12
60
Y
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Design members 2009
24/46
488
85322
=
x
r = 13.8 cm &
488
155945
=
y
r = 18.7 cm
Deck bridge (upper) in
l = 0.85 l & out
l = 0.85 l
90
8
.
30
8
.
13
500
*
85
.
0
≤
=
=
in
λ 90
7
.
22
7
.
18
500
*
85
.
0
≤
=
=
out
λ
• Check stress
16
.
1
448
515
=
=
ca
f t / cm2
≤ F C = 1.6 – 8.5*10-5
* 30.82
= 1.52 t / cm2
Safe but waste Try smaller thickness 2.6 cm
Design (L 4 ) Lower tension
h = 40 cm & b = 36 cm
36
*
)
3
2
2
1
(
/
→
=
b = 18 à 24 cm ≈ 20 cm
Chord, L<10m n=over 2,000,000
Detail B Fsr = 1.12 t / cm2
553
119
1
12
.
1
max
−
=
F = 1.43 t / cm2
< 1.6 t / cm2
1.43 =
A
553
get A = 386.7 cm2
= (2*20 + 2*40) * t
Get t = 3.22 cm ≈ 3.4 cm
• Check global buckling
)
4
.
3
*
40
*
2
(
)
4
.
3
*
20
*
2
(
)
4
.
23
*
4
.
3
*
40
*
2
(
)
7
.
1
*
20
*
4
.
3
*
2
(
+
+
=
y = 16.2 cm
A = 408 cm2
]
2
*
12
40
*
4
.
3
[
3
=
x
I + [3.4*40*2*(16.2-1.7)2
] + [3.4*20*2(16.2- 23.4)2
]
= 100505 cm4
36
40
t
3.4
Y
Y
X X
20
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]
2
*
12
20
*
4
.
3
[
3
=
y
I + [3.4*20*2*(
2
4
.
3
2
36
+ ) 2
] + [3.4*40*2(18+1.7)2
]
= 162874 cm4
408
100505
=
x
r = 15.7 cm &
408
162874
=
y
r = 20 cm
Take in
l = 0.85 l & out
l (from lower bracing)
Assume the lower bracing is as shown below:
Plan of suggested lower bracing
L 3 4
and L
160
1
.
27
7
.
15
500
*
85
.
0
≤
=
=
in
λ 160
5
.
42
20
500
*
85
.
0
*
2
≤
=
=
out
λ
• Check stress
6
.
1
36
.
1
408
553
.
max
≤
=

 →
 t / cm2
12
.
1
06
.
1
408
119
553
.
≤
=
−


 →

fatique
t /cm2
30
52
.
11
4
.
3
40
500
≤
=
+



 →

Deflection
Design (V 1): Web member comp:
Assume f = 1.1 à 1.4 = 1.1 t / cm2
=
A
241
A = 219 cm2
b = 36 cm
8
.
2
64
36
≤
w
t
= 38.2
t w = 0.94 ≈1.0 cm
A = 219 = (36 * 1) + (2 fl
b fl
t ) fl
b fl
t = 91.5 cm2
28
1.0
3.2
36
29.6
Y
Y
X
X
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Design members 2009
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Take fl
b ≈ 10 fl
t à fl
t = 3.02 ≈ 3.2 cm fl
b =
2
.
3
5
.
91
≈ 28 cm
1. Check local buckling
Taking size of weld s ≈ 1.0 cm
8
.
2
21
9
.
3
2
.
3
1
2
/
1
2
/
28
≤
=
−
−
=
fl
t
c
= 12.5
8
.
2
64
6
.
27
0
.
1
1
*
2
2
.
3
*
2
36
≤
=
−
−
=
w
t
b
= 38
2. Check global buckling
12
6
.
29
*
1 3
=
x
I + 2*3.2*28*(18-1.6)2
= 50359 cm4
12
28
*
2
.
3
*
2
3
=
y
I = 11708 cm4
A = (1*29.6) + (2*28*3.2) = 209 cm2
209
50359
=
x
r = 15.5 cm &
209
11708
=
y
r = 7.5 cm
In deck bridge in
l = 0.70 l & out
l = 0.85 l
90
37
)
5
.
7
(
400
*
70
.
0
≤
=
=
y
in
r
λ 90
22
)
5
.
15
(
400
*
85
.
0
≤
=
=
x
out
r
λ
3. Check stress
15
.
1
209
241
=
=
ca
f t / cm2
≤ F c = 1.6 – 8.5 * 10-5
* 372
= 1.48 t / cm2
Design (D 3): Web member tension
Fmax= +181t Fmin= -2t
Web member (diagonal), do not carry floor reaction only (class II)
Single track n=500,000 Detail B Fsr = 2 t / cm2
Note that: If double tracks, so n=200,000 Fsr = 2.8 t / cm2
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181
2
1
0
.
2
max
−
−
=
F = 1.98 > 1.6 t / cm2
So 1.6 =
A
181
∴ A = 113 cm2
, b = 36 cm
8
.
2
64
36
≤
w
t
= 38.2 t w = 0.94 ≈ 1.0 cm
We used local buckling because this member will be subjected to
compression.
A = 113 = (36 * 1) + (2 fl
b fl
t ) ∴ fl
b fl
t = 38.5 cm2
Taking fl
b ≈ 10 fl
t à fl
t = 1.96 = 2.0 cm & fl
b =
2
5
.
38
≈ 20 cm
1 - Check global buckling
12
32
*
1 3
=
x
I + 2*20*2*(18-1)2
= 25851 cm4
12
20
*
2
*
2
3
=
y
I = 2667 cm4
& A = 1 * 32 + 2 * 20 * 2 = 112 cm2
112
25851
=
x
r = 15.2 cm &
112
2667
=
y
r = 4.9 cm
in
l = 0.70 l & out
l = 0.85 l (deck) & 2
2
5
4 +
=
l = 6.4 m
160
91
)
9
.
4
(
640
*
70
.
0
≤
=
=
=
y
in
r
λ 160
36
)
2
.
15
(
640
*
85
.
0
≤
=
=
=
x
out
r
λ
2 - Check stress
6
.
1
61
.
1
112
181
.
max
≈
=

 →
 t / cm2
20
1.0
2.0
36
32
Y
Y
X
X
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0
.
2
65
.
1
112
)
2
(
181
.
≤
=
−
−


 →

fatique
t / cm2
30
32
20
640
.
>
=



 →

Deflection
Use bfl = 22cm and recheck
• Since the minimum force in the member is compression, so we
must check this member on "2t"
Local buckling: Taking size of weld s ≈ 1.0 cm
8
.
2
21
3
.
4
2
1
2
/
1
2
/
20
≤
=
−
−
=
fl
t
c
=12.5
8
.
2
64
30
0
.
1
1
*
2
2
*
2
36
≤
=
−
−
=
w
t
b
= 38
• Global buckling 90
91 ≈
=
in
λ & 90
36 ≤
=
out
λ O.K.
02
.
0
112
2
=
=
ca
f t / cm2
≤ 1.6 – 8.5 * 10-5
* 912
= 0.9 O.K.
Solved Example 2:
Pony bridge as shown, h of truss = 5.5m, St.44, single track
5.5 m
10*5.5=55 m
3.9 m
J1
-945t -985t
+
1
4
4
t
-
7
6
t
-220t
-198t
Joint1
5.5 m
5.5 m
It is required to design the upper chord and vertical member at joint J1
Take b = 420 mm δ = 0.12 cm / t
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Design members 2009
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Solution:
(1) Upper chord: F = 985 t (comp.)
Assume F C = 1.3 t / cm2
3
.
1
985
=
g
A = 757.7 cm2
Choose 45
15
12
550
≈
→
=
h cm
Take b u = 42 + 20 = 62 cm
757.7 = (2 * 45 + 62) t t = 4.98 cm ≈ 5.0 cm
Check local buckling:
8
.
2
64
8
5
1
*
2
42
≤
=
=
−
t
b
= 38 O.K.
8
.
2
21
8
.
0
5
1
5
)
2
42
62
(
≤
=
−
−
−
= 22.5 O.K.
8
.
2
30
8
.
8
5
1
45
≤
=
−
=
t
h
= 17.9 O.K.
)
5
*
62
(
)
2
*
5
*
45
(
)
5
.
47
*
5
*
62
(
)
2
*
5
.
22
*
5
*
45
(
+
+
=
y = 32.7 cm
A = 62 * 5 + 45 * 2 * 5 = 760 cm2
12
45
*
5
[
3
=
x
I +5*45*(32.7-22.5)2
]*2 + [62*5*(47.5-32.7)2
] = 191304 cm4
]
12
62
*
5
[
2
*
]
)
2
5
2
42
(
*
5
*
45
[
3
2
+
+
=
y
I = 348753 cm4
760
191304
=
x
r = 15.87cm &
760
348753
=
y
r = 21.42 cm
bx
bin l
l = = 0.85 * 550 = 467.5 cm
4 12
.
0
*
550
*
348753
*
2100
5
.
2
=
= by
bout l
l = 1172.2 cm
5
45
42
Y
32.7
62
5
5
X X
y
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87
.
15
5
.
467
=
x
λ = 29.47
42
.
21
2
.
1172
=
y
λ = 54.72 < 90
t > 4cm F c = 1.5-7.5*10-5
54.722
= 1.275 t / cm2
760
985
=
c
f = 1.296 t / cm2
> 1.275 t / cm2
Unsafe
Increase dimensions and recheck
(2)Vertical member:
Since this is Pony Bridge, so the vertical is subjected to bending
moment
The maximum force of vertical can be calculated from diagonal.
.)
(
102
)
45
sin
144
(
sin
max
max comp
F
F d −
=
−
=
−
= α
.)
(
54
)
45
sin
76
(
sin
min
min tens
F
F d +
=
−
−
=
−
= α
For (S.T.), class (II) (do not carry X.G. only) n = 500.000
Fsr = 2.0 t / cm2
M add =
100
1
F max (max. comp. force in upper chord) * h/
(HM.G – HX.G/2)
h = 5.5 – 0.7/2 = 5.15m
M add (max) =
100
1
* 985 * 5.15 + 0.1*5.5*0.5*
2
15
.
5 2
= 53.6 m t
M add (min) =
100
1
* 220 * 5.15 = 11.33 m t
Since the member is subjected to both tension and compression, so we
have to check the vertical twice.
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Design members 2009
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• Design as M + N:
8
.
2
64
42
=
w
t
à t w = 1.1 cm t w = 1.2 cm
Force in flange =
42
5360
2
102
+ = 178 cm2
Assume f = 1.2 t / cm2
f
f t
b
∴ =
2
.
1
178
= 157 cm2
= 157
t = 4 cm & b = 40 cm
I y (inside) = 2 * 4 *
12
403
= 42667 cm4
I X (outside) = 1.2 *
12
343
+ 2 * 40 * 4 * 2
)
2
34
2
4
( + = 119450 cm4
A = 2 * 40 * 4 + 34 * 1.2 = 360.8 cm2
8
.
360
42667
=
in
r = 10.9cm &
8
.
360
119450
=
y
r = 18.2 cm
• Check local buckling:
8
.
2
64
7
.
26
2
.
1
2
34
<
=
−
=
w
w
t
h
= 38
8
.
2
21
6
.
4
4
1
2
/
2
.
1
2
/
40
<
=
−
−
=
f
t
C
= 12.5
• Check global buckling:
90
35
9
.
10
550
*
7
.
0
<
=
=
in
λ 90
3
.
36
2
.
18
550
*
2
.
1
<
=
=
out
λ
• Check stresses:
F C = 1.6 – 8.5 * 10-5
* (36.3)2
= 1.49 t / cm2
8
.
360
102
=
ca
f = 0.28 t / cm2
< 1.49 t / cm2
42
Y
Y
X
X
4.0
40
34*1.2
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• Check as beam-column
2
3
.
36
7500
=
EX
F = 5.69 t / cm2
49
.
1
28
.
0
=
C
ca
F
f
= 0.19 > 0.15
69
.
5
28
.
0
1
85
.
0
A 1
−
=
∴ = 0.89 take A 1 = 1
L U act = L in = 0.7 * 550 = 385 cm α = zero c b = 1.75
8
.
2
40
*
20
max
u =
L = 478 cm or 75
.
1
*
8
.
2
*
42
4
*
40
*
1380
= 3286 cm > 385
bcx
F
∴ = 0.58 F y = 1.6 t / cm2
1
*
6
.
1
2
42
*
119450
5360
49
.
1
28
.
0
+ = 0.78 < 1.2 Waste
Decrease b, try flange 30 * 3 cm (Better to assume f = 1.5 t / cm2
from
the estimation)
• Check for case of M + T
λ < 160
8
.
360
54
=
t
f = 0.15 t / cm2
< 1.6 t / cm2
(case A)
2
42
*
119450
5360
=
btx
f = 0.94 t / cm2
6
.
1
94
.
0
6
.
1
15
.
0
+ = 0.68 < 1.2 (case B)
Check deflection:
fl
b
l
=
40
550
= 13.75 < 30
Check fatigue: Mfatigue= 15
.
5
*
985
*
100
1
= 50.7 mt (without wind)
F max =
8
.
360
54
+
2
42
*
119450
5070
= 0.15+ 0.89 = 1.04 t / cm2
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Design members 2009
33/46
2
42
*
119450
1139
8
.
360
102
min +
−
=
F = - 0.08 t / cm2
F max – Fmin = 1.04 – (- 0.08)
= 1.12 t / cm2
< 2 t / cm2
Web member single track,
n = 500000, detail "B"
Solved Example 3:
Through bridge single track, with span 78 m. Spacing between X.G. is 6
m & of depth 0.8 m. Steel used is st 52. Width of bridge is 5.5 m.
It is required to design the marked members.
7.5
Inclined portal frame
6m
78m
M
M
4
M
M2
1
3
Solution:
D.L.: W S = 0.9 * (0.75 + 0.05 * 78) = 4.185 t / m/
2
6
.
0
2
185
.
4
+
=
DL
W = 2.39 t / m/
For Member M 1:
6
5
.
7
tan 1
−
=
α = 510
10
0.35
0.55
0.67
1.16
1.19
0.72
1.03
0.92
12.5
6.25
6.25
12.5
12.5
10
6.25
10
10
10
10
6.25
12.5
12.5
12.5
10
10
10
10
10
10
10
10
10
10
0.15
10
10
10
0.03
10
10
0.84
M
C
min
max
T
M
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1- Dead load: F D =
2
1 * 78 * 1.19 * 2.39 = 111 t
2- Live load:
F L = 6.25 * 2 * 0.67 + 12.5 * 3 * 1.16 + 40 * 1.03 + 6.25 * 2 * 0.92 + 3 *
12.5 * 0.84 + 40 * 0.72 + 40 * (0.55 + 0.35 + 0.13) + 10 * 0.03 = 205.2 t
78
24
24
+
=
I = 0.23 take I = 0.25
F L+I = 205.2 * 1.25 = 256.5 t
From Wind: The diagonal is subjected to
moment as a member in the inclined open
portal frame.
W upper (loaded) = 0.1 [0.4 *
2
5
.
7
+ 0.4 *
2.45] = 0.25 t / m/
R U (loaded) = 0.25 *
2
78
= 9.75 t
Note that: Loaded case is critical when design the diagonal while un-
loaded case is critical when design upper bracing & the horizontal beam.
2
2
6
5
.
7 +
=
l = 9.6 m
M =
2
75
.
9
* 9.6 = 46.8 m t (case B)
Design of Member M 1:
For the whole truss; assume b =
12
600
= 50 cm
F D = 111 t & F L+I = 256.5 t & Ftot = 367.5 t (compression)
M wind = 46.8 m t & l inclined = 9.6 m
7.5
1.8m
For Upper bracing
0.1
0.1
2.45
3.75
0.8
CL of upper beam
Vl
of
truss
9.6
9.75
M
4.875 4.875
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Design members 2009
35/46
The diagonal will be in the shape of π - section (same shape of the
section of upper chord)
(We do not assume f = 1.8 t / cm2
because the
member is subjected to moment with normal force)
0
.
1
5
.
256
111+
=
A = 367.5 cm2
b /
= 50 + 20 = 70 cm
Assume h = b = 50 cm (2 * 50 + 70) t = 367.5
t = 2.2 cm
Check local buckling =
6
.
3
30
7
.
22
2
.
2
50
>
= = 15.8
We can either increase thickness
t
50
=15.8 t = 3.2 cm
Or use stiffness as shown and take t = 2.2 cm
Check local buckling:
6
.
3
21
54
.
4
2
.
2
10
<
=
=
t
c
= 11.1
6
.
3
30
54
.
4
2
.
2
10
/
<
=
=
t
c
= 15.8
6
.
3
64
7
.
22
2
.
2
50
<
=
=
t
b
= 33.73
6
.
3
64
2
.
18
2
.
2
40
/
<
=
=
t
h
= 33.73
A = (70 + 2 * 50 + 2 * 10) * 2.2 = 418 cm2
418
)]
40
2
.
2
(
*
10
*
2
)
25
2
.
2
(
*
50
*
2
2
2
.
2
*
0
[7
.2
2 +
+
+
+
=
y
v
= 19.16 cm
40
50
70
50 10
t=22mm
t=22mm
b
b
h h
c
c
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Design members 2009
36/46








−
+
+
−
+
=
= 2
3
2
3
in )
16
.
19
2
.
27
(
*
50
*
2
.
2
12
50
*
2
.
2
2
)
2
2
.
2
16
.
19
(
*
2
.
2
*
70
12
2
.
2
*
70
I
Ix
+ 2








−
+ 2
3
)
16
.
19
2
.
42
(
*
2
.
2
*
10
12
2
.
2
*
10
= 133720cm4








+
+
+
+
=
=
12
10
*
2
.
2
)
2
2
.
2
25
(
*
2
.
2
*
50
12
2
.
2
*
50
2
12
70
*
2
.
2 3
2
3
3
out
y I
I
+ 2 [ ]
2
)
5
2
.
2
25
(
*
10
*
2
.
2 +
+ = 258826 cm4
r x =
418
133720
= 17.9 cm r y =
418
258826
= 24.9 cm
in
l = 0.7 * 9.6 = 6.72 m
out
l = Since the diagonal is a member in the portal frame, so out
l will
be calculated using GA and GB. So we have to know section of
horizontal beam.
- To estimate the beam of the upper bracing, the critical case is the
un-loaded case because this beam carries wind only.
W upper (unloaded) = 0.2 * 0.4 *
2
5
.
7
* 2 = 0.6 t / m/
R U = 0.6 *
2
78
= 23.4 t M un-loaded =
2
4
.
23
* 9.6 = 112.32 m t
CL of upper beam
Vl
of
truss
9.6
23.4
M
11.7 11.7
7.5m
1.8m
0.8
For Upper bracing
3.75
0.2 0.2
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Design members 2009
37/46
1
.
2
11232
=
∴ X
S = 5350 cm3
Choose H E B 600 or B.U.S Note
2
550
L act
U = = 275 cm
Or assume upper horizontal beam is HEB 600 (Always if open frame)
GA = 10 (hinged base) GB = 87
.
0
5
.
5
/
171000
6
.
9
/
258826
= K = 1.8
Note: we used I y because it is I out
90
5
.
37
9
.
17
672
<
=
=
in
λ 90
3
.
71
9
.
24
960
*
85
.
1
<
=
=
out
λ
(1) Check stresses:
F C = 2.1 – 13.5 * 10-5
* (71.3)2
= 1.41 t / cm2
418
5
.
367
=
ca
f = 0.88 t / cm2
< 1.41 t / cm2
O.K. Safe
• For effect of moment due to wind: (case B)
I out = 258826 cm4
rout = 24.9 cm out
λ = 79.3
For Π - Section There is no LTB bcx
F
∴ = 0.58Fy = 2.1 t / cm2
41
.
1
88
.
0
=
C
ca
F
f
= 0.62 > 0.15 &
2
y
E
3
.
71
7500
)
( =
out
F = 1.47 t / cm2
47
.
1
88
.
0
1
85
.
0
A 1
−
=
∴ = 2.12 (too large because fca is too large)
fbcx = 63
.
0
35
*
258826
4680
= t / cm2
Using interaction equation:
1
.
2
*
1
.
2
63
.
0
62
.
0 + 2 = 1.26 > 1.2 (case B) Use t = 2.4cm
If we want to check for the unloaded case: (NOT CRITICAL)
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Main Truss
Design members 2009
38/46
418
111
=
ca
f = 0.27 t / cm2
41
.
1
27
.
0
=
C
ca
F
f
= 0.19 > 0.15
2
y
E
3
.
71
7500
)
( =
out
F = 1.47 t / cm2
41
.
1
27
.
0
1
85
.
0
A 1
−
=
∴ = 1.05
M un-loaded = 112.32 mt fbcx = 52
.
1
35
*
258826
11232
= t / cm2
Using interaction equation:
05
.
1
*
1
.
2
52
.
1
19
.
0 + = 0.95 < 1.2 (case B) (Not critical)
• Design of Member M 2:
The member is subjected to tension only
F D =
2
1 * 12 * 1.0 * 2.39 = 14.34 t
F L = 12.5 (1 + 2 * 0.67) + 10 * 0.17
+ 6.25 (0.375 + 0.083) = 33.8 t (tension)
12
24
24
+
=
I = 0.67
F L+I = 33.9 *1.67 = 56.6 t
Ftotal = 14.34 + 56.6 = 70.95 t
b = 50 cm (as in diagonal)
Truss vertical carry floor beam reaction only (class III) Single track
n = Over 2*106
detail B F sr = 1.12 t / cm2
95
.
70
34
.
14
1
12
.
1
max
−
=
F = 1.4 t / cm2
> 2.1 t / cm2
(0.58 F y)
4
.
1
95
.
70
=
∴ A = 50.7 cm2
Take t w = 1 cm
50
Y
Y
X
X
20
1.0
48*1.0
0.17
0.083
0.67
0.67
1.00
0.375
6.25
12.5
12.5
6.25
12.5
10
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Main Truss
Design members 2009
39/46
f
f t
b
2
∴ = 50.7 – 50 * 1≈ zero
b min = 6 ϕ + t w + 2 S
For M27 and size of weld 1cm, bmin = 20cm
So take minimum f
b = 20 cm & f
t = 1 cm
in
I = 2 * 1*
12
203
= 1333.3 cm4
12
483
=
out
I *1 + 2 * 20 * ( 2
)
2
48
2
1
+ = 33226 cm4
A = 20 * 1 * 2 + 48 * 1 = 88 cm2
88
3
.
1333
=
in
r = 3.9 cm &
88
33226
=
out
r = 19.43 cm
l = 7.5 m in
l = 0.7 * 7.5 = 5.25 m
out
l = 0.85 * 7.5 = 6.375 m (Supported on upper bracing)
160
135
9
.
3
525
<
=
=
in
λ & 160
33
43
.
19
5
.
637
<
=
=
out
λ
Check stresses:
88
95
.
70
=
t
f = 0.81 t/cm2
< 2.1 t /cm2
64
.
0
88
34
.
14
95
.
70
=
−
=
sr
f t /cm2
< 1.12 t /cm2
For verticals: no check deflection
• Design of member M3:
For x = zero
F = tan-1
= 7.5/6 = 51.30
Fmin= 111*cos51.3=69.3t (Tension)
FL+I= 256.5*cos51.3=160t (Tension)
FL+I+D = 69.3+160=229.3 t
Stiffener
c
X
X
y
y
10cm
50
25
50 12
1.2
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Design members 2009
40/46
We may use stiffener or not
in
l = 0.85 * 6 = 510 m & out
l = 0.85* 6 = 510 m
Complete as before
• Design of member M4: (zero members)
(1)
6
.
3
64
50
=
w
t
t w = 1.5 cm
Take t w = t f = 1.5 cm
b min = 6 ϕ + t w + 2 S Assume M 27
b min = 6 * 2.7 + 1.5+ 2 * 1 = 19.7 cm,
take 20 cm
6
.
3
21
5
.
5
5
.
1
1
2
/
5
.
1
2
/
20
<
=
−
−
=
f
t
c
= 11.1
Check global buckling
in
l = 0.7 * 7.5 = 5.25 m
lout is according to the shape of the upper bracing
Assume out
l = 0.85 * 7.5 = 6.375 m
in
I = 2 * 1.5 *
12
203
= 2000 cm4
12
47
*
5
.
1
3
=
out
I + 2 * 20 * 1.5 ( 2
)
2
47
2
5
.
1
+
= 48262 cm4
A = 20 * 1.5 * 2 + 47 * 1.5 = 130.5 cm2
5
.
130
2000
=
in
r = 3.9 cm
5
.
130
48262
=
out
r = 19.2 cm
20
50
Y
Y
X
X
1.5
47*1.5
S S
Lout
S S
out
L
out
L
out
L
= 0.85*7.5 = 0.85*7.5
= 0.85*7.5 = 1.2*7.5
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Main Truss
Design members 2009
41/46
90
135
9
.
3
525
>
=
=
in
λ
So increase dimensions of flanges (main item in Iin)
Assume flange 300*20 mm
in
I = 2 * 2 *
12
303
= 9000 cm4
A = 30 * 2 * 2 + 46 * 1.5 = 189 cm2
189
9000
=
in
r = 6.9 cm 90
76
9
.
6
525
<
=
=
in
λ
Solved Example 4:
Through bridge single track, with span 78 m. Spacing between X.G. is 6
m. The XG composed of web plate 800*14 and 2 flange plates 300*30.
Steel used is st 52. Width of bridge is 5.5 m.
It is required to design the marked members.
7.5
6m
78m
M1
M2
M
4
M3
Solution:
• For Member M1 It is subjected to compression only.
Design of Member M 1:
For the whole truss; Take b =
12
600
= 50 cm (given for U1)
From previous example:
F D = 111 t & F L+I = 256.5 t & Ftot = 367.5 t (compression)
The diagonal will be in the shape of π - section (same section of upper
chord)
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