DESIGN OF COMPOSITE COMPOSITE STEEL GIRDER B RIDGE(WELDED)
6500 1676 1000
200 0 5 0 2 5 0 1
75
0 0 2
2100
2300
2100
CROSS SECTION OF BRIDGE INPUT DATA :
DESIGN CRITERIA :
1.0
Length of the Girder
=
20.0
M
Ultimate Concrete Stress,
f 'C =
25.00 N/mm2
2.0
Girder depth
=
2500
mm
Ultimate Steel Stress,
f 'y
=
415.0 N/mm2
3.0
EUDL for Moment
=
2065.5 kN
Allow. stress stress of Concrete Concrete in Compression Compression
f C
=
11.25 N/mm
4.0
EUDL for Shear
=
2272.4 kN
Modulus of Elasticity of Concrete
Ec
=
24000.00
N/mm
2
Modulus of of El Elasticity of of St Steel
Es
=
200000
N/mm
2
f c
=
10.00 N/mm2
f s
=
m
=
166.7 N/mm2 8
r
=
16.67
k j R
= = =
0.32 0.89 1.45
5.0
Coefficent o off Dy Dynamic Au Augument (C (CDA)
=
0.458 kN
6.0
Design EUDL for Moment / Girder
=
1505.4 kN
=
75.272 kN / m
=
1656.2 kN
=
82.812 kN / m
=
318.5 kN
= = = = = =
3.0 53.083 735.0 552.0 24.0
7.0 8.0
Design EUDL for Shear / Girder Axle Load Load of Train
(
32.5
)
t
9.0
Distribution width of L Lo oad Dispersion Design wheel Load per meter width Tractive effort Braking force 13.0 Unit weight of Concrete 14.0 Area of Steel Steel Girder 10.0 11.0 12.0
Bs s 0 m d 0 2 m
0 4 5 1
=
2300
M kN/m kN kN 3
Kn/m 2 88000 mm
Bs
mm
288 mm mm
0 5 1
=
N
=
=
m m 1 7 9
0 9
2
25
217
n
A
s
y
C.G of Steel Section
COMPOSITE SECTION
EQUIVEL A NT STEEL SECTION
Effective width of the Slab :
i) One - third of the effective Span ii) Distance Distance between between centres centres of of the rib iii)
= =
12 times the thickness ckness of the the slab + breadth of the the rib
6666.7 mm 2300 mm
=
2410
mm
Area of Steel Steel Girder
A S
=
88000 mm2
Area of composite composite Section Section
A C
=
145500 mm2
Moment of inertia of the Steel section
IS
=
1.1504E+11 mm4
Equivalent moment of inertia of the composite section
Ic
=
1.8735E+11 mm4
=
5.000 kN/m
2.0 Self wt. of the Slab/m
=
21.840 kN/m
3.0 Rail, Sleeper and Ballast load on top of the Slab
=
23.417 kN/m
4.0 EU EUDL / Girder for Live load with impact
=
75.272 kN/m
1.0 Self wt. of the Steel Girder/m
=
(200L + 1000) N
s 0 d 0 2
Fatigue considerration:
Total DL Shear per Girder
Shear (mini)
=
502.57 KN
Total DL + LL Shear per Girder
Shear (maxi)
=
2158.8 KN
=
0.2328
Ratio of Minimum Shear & Maximum Shear 2,000,000 Cycle
K
=
a) Bending Moment due to Self wt. of the Steel section
MS
=
268.82 kN -m
b) Bending Moment due to Dead load of Concrete
MC
=
1174.2 kN -m
c) Bending Moment due to Rail, Sleeper, Ballast
MSU =
1259.0 kN -m
d) Bending Moment due to EUDL (Applied load)
M A
=
4046.9 kN -m
Over all depth of the Composite Section,
D
=
Section modulus of Steel Section at bottom
Zb
=
1E+08 mm3
Section modulus of Composite Section at bottom
Z bc
=
2E+08 mm3
Section modulus of Composite Section at bottom
Z tc
=
2E+08 mm3
Considering
Maxi. Stress in Steel at bottom
σbc = =
Maxi. Stress in Concrete at top
σc t = =
Vartical Shear at support
0.93
2700
MS Zb
+
mm
MC
+ MSU +
42.832 N / mm2 MC
<
+ MSU +
165.00 N / mm2
OK
11.25 N / mm2
OK
M A
Z bc 4.1978 N / mm2
<
F
= 1430.85 kN
Fh
= F * Bs*ds*(n - ds/2)
Design of Shear Connector :The horizontal Shear at the top of the Steel section
m IC Total
Or
=
382.47 N/mm
=
109958.79 N Fh
Horizontal force resisted by the compressive area of the concrete slab
= o.85 * f c' *Ac 2
4887500.00
= Fh =
Design horizontal force Let us considered H H D
= 150 mm =
6.8
>
D 4
=
22
N
N
4887500.00
150 mm Long
Ø MS Rod as Shear connector
22 Ø
OK
Allowable Shear taken by one Shear connector
q
= 1.518 * H * d √ f CK
n
=
14025 N
= Total nos. of studs
349
Nos
Let 'S' be the size of the welding all arround the bar with the top flange We know shear strength
P
= KLS ζ va
382.47
=
20900 S
S
=
0.0183
Pitch of the Shear connector, ρ
≤
n * Z r
Vsr
=
V f * Q
≈
3.0 mm
Vsr I
Where,
n is the number of Shear connector in Cross - section Z r Shear fatigue resistance of an individual Shear connector
=
4
=
α * d2
V f Vertical Shear force under Fatigue load combination
=
2861.7 kN
Q Area moment about N . A I C Moment of inertia of composite section
=
50078350.5 mm 3
=
1.8735E+11 mm4
d Dia of Shear connector
=
22
N Number of Cycle
= Vsr
=
Z r
Pitch of the Shear connector
ρ
764.93
=
=
= =
4*d L 2
=
x
2.5 d s
2
2
Ø
-
52.12
29.5
≥
= 132 mm
Longitudinal Spacing
38 * d
N = 238
Minimum Pitch of Shear connector
Maximum Pitch of Shear connector
≥
2000000
= α*d2 25225.8937 = 1 n/4 =
88 mm = 115 ≈ 115 mm 500
≤
M A
Z bc
600
mm
log(N)
19.00
DESIGN OF RCC DECK SLAB : 53.08 kN 6.24 kN
53.08 kN 1676
312
312
6.24 kN
46.833 kN/m 100
4.8 kN/m 350
A
C
2100 R1
=
138.93 kN
1.0 Self wt. of deck slab
=
2.0 Self wt. of Ballast, Sleeper & Rail
=
46.833 kN/m
3.0 Load per wheel
=
53.083 kN/m
4.0 Parapet Load
=
Maxi. (-) Moment at A
=
Maxi. (+) Moment at C
=
B
4.80
6.24
2100 R2
kN/m
kN/m
-25.93 kN - m 21.292 kN - m
DESIGN FOR CANTILIVER PART : Maxi. (-) Moment at A
=
25.93 Kn-M
(Top Tension)
Mcr = 1.2 x Mcr
Minimum flaxural strength
Mcr = 19.70 x √ f 'c x Z b Section modulus,
0.0017 m
Mcr =
5.1914 Kn-M
=
6.2297 Kn-M
=
25.93 Kn-M
MD
Design Moment,
d reqd. =
Use Y
16
3
Zb =
Minimum flaxural strength
133.9 mm
d avil. =
142.00 mm OK
As =
1228.5 mm2 / m @ 177 mm c/c
mm Ø
DESIGN FOR MID SECTION : Maxi. (+) Moment at C
=
21.29 Kn-M
Minimum flaxural strength
Mcr = 1.2 x Mcr
Section modulus,
Mcr = 19.70 x √ f 'c x Z b Z b = 0.0017 m Mcr =
5.1914 Kn-M
=
6.2297 Kn-M
Minimum flaxural strength MD
Design Moment,
=
(Bottom Tension)
21.29 Kn-M
d reqd. =
121.33 mm
d avil. =
142.00 mm
OK
1008.7 mm2 / m mm Ø @ 199 mm c/c
As = Use Y
16
DISTRIBUTION REINF : Distribution reinforcement is
67 % A dist. =
Use Y
12 @
350
2300
of the main reinforcement 675.8 mm 167
mm c/c
=
138.93
kN
100
DESIGN OF PARAPET WALL : 200
1.47
kN/m
0 5 0 . 1
Moment at base
= 1.5435 = 32.668 = 144.00 72.11 = @ 12 mm Ø
M d reqd. d avil. As Use Y
Kn-M mm mm
OK
2
mm / m 1568 mm c/c
MINIMUM REINF. REQUIRMENT :
A st Use Y
12 mm Ø
500 mm2 = 214 mm c/c ( Both Way )
12
Ø @ 12
214 mm c/c Ø @ 167 mm c/c 16 Ø @
177 mm c/c
16
Ø @
12
Ø @
214 mm c/c
12
Ø @
214 mm c/c
199 mm c/c
REINF. DETAILS OF DECK SLAB:
DESIGN OF STEEL GIRDER : 45
` 0 0 5 2
45
20.00 M
375
375
LONG SECTION OF GIRDER
STRUCTUREAL DESIGN OF STEEL I GIRDER. 1.0
Length of the Girder
=
20.0
DESIGN CRITERIA :
M
Yield stress of steel
mm
Allowable shear stress,
f y
=
250.00 N/mm2
=
100.00 N/mm2
=
165.00 N/mm2
=
187.50 N/mm2
=
150.00 N/mm2
=
70.00 N/mm 108.00 N/mm2
2.0
Girder depth
=
3.0
EUDL for Moment
=
2065.5 kN
4.0
EUDL for Shear
=
2272.4 kN
Permissible bearing stress,
5.0
Coefficent of Dynamic Augument (CDA)
=
0.458 kN
Allowable stress for tension,
= = = = =
1505.4 kN 1656.2 kN 735.0 kN
Allowable stress for compre.
ζ va бbc бp бt бC
Shear stress in the weld
ζ vf
Weight of Rail Weight of Rail Length of Sleeper
=
60.00
= =
0.589
Design EUDL for Moment / Girder Design EUDL for Shear / Girder 8.0 Tractive effort 9.0 Braking force 10.0 Unit wt. of Timber 6.0 7.0
2500
552.0 kN 3 12.0 kN/m
Allowable bending stress,
2750
2
kg/m kn/m mm
(Mpa)
700 mm t f =
45mm
b m m
N
A
tw = 10 mm
0 0 5 2
= = = =
Weight of Rail, Sleeper over a girder Load per meter over a girder
=
67.987 kN
=
3.276 kN/m
= 1 d
t f =
Self wt. of Plate Girder = Self wt. of Plate Girder/m = Total uniform load, w' = Maxi. Shear force V = Maxi. Bending moment, M = Let the thickness of web, tw =
Width of Sleeper Thickness of Sleeper Spacing of Sleeper No.of Sleeper on top Slab
45mm
2065.50 kN/m 103.28 kN/m 240.65 kN/m 2000.94 kN 12938.1 kN - m ( directly exposed to weather and unpainted ) 10 mm = d1
Economical depth of Plate Girder
2500 mm 3080.3 mm ≈ x 10 mm Minimum web thickness of web from Shear considaration t w (mini) = 8.0038 < tw OK
d1
Selected web Plate size
=
=
2500
DESIGN OF FLANGE :
A f
Approximate flange area required
=
31365 mm
The flange outstand should not be greater than 12 x t b Area of flange
= ( 2b + t w ) x
=
f
12 x t f
t f
= ( 2 x 12 x t f + 8 ) x t f 24 t f
2
24 t f
2
t f
+
8 t f
=
31365
+
8 t f
-
31365
35.984 mm ≈
=
=
0
40 mm
≈ 600 mm 2 2 24000 mm < 31365 mm NOT OK t f = 45 mm Increase flange thickness, Let Width of flange = 697 ≈ 700 mm 2 2 Area of flange = 31500 mm > 31365 mm OK Width of the flange Plate
=
Calculated area of flange
=
784.13
Hance provide a flange Plate of size
=
700
x
45 m m
CHECK BY MOMENT INERTIA METHOD :
I xx
3
= t wd 12 =
3
+
w f * t f
2*
12
+ (w f * t f )* (d/2 + t f ) 2
4
1.1504E+11 mm
Maxi. bending stress
бbc = M y I
145.64 N / mm2 <
=
165.0 N / mm2 OK
CONNECTION BY WELDING :
Welds are designed for horizontal Shear V * Ay Horizontal shear per unit length, = Ixx Vertical Shear force, V = 2000.9 kN Area moment of the flange about N - A, A y I xx
= 2 * A f * (d/2)
+
=
mm
40083750
3
tw*d 12
= ( A f +
A w
6 Horizontal Shear / mm
2
)*
d 2 =
=
4
1. 11 5E+1 1
mm
719.6 N / mm
SIZE OF WELD :
Welding is done on both side of the web The strength of well is given by P =
KLS ζ
va
Where, S K L ζ vf
= = = =
Size of th e wel d in mm Constant = 0.7 Effective length of the weld in mm 2 Shear stress in the weld in N / mm 719.6
=
2 x ( 0.7
x
S
2
= 108 N / mm x
1
x 108 )
220 250 445 48
mm mm mm c/c nos.
S
=
4.7592 mm
But minimum size of a fillet weld for a Let us provide an intermittent fillet weld Th e ef fect ive we ld le ngt h 4 S = 40 Mini effective weld length, L = 40 Provide in termit tent fillet welds of 40 Pitch of welds
45 mm thick Plate is 10 mm
mm mm mm long
Strength of weld on both face of web Horizontal shear / mm length
=
=
84.047 mm
≈ Clear spacing between the welds should not be more than Maxi. Clear spacing
=
120 mm,
12 t w
C/C spacing of welds
=
mm 200 mm
130
or
160.00 mm
130.00 mm (Provide continuous welding)
Provide 40 mm L ong fillets welds at a Pitch of BEARING STIFFENER :
Maxi. Shear force V = 2000.94 kN 2 Permissible bearing stress, бp = 187.5 N/mm 2 The bearing area required, A = 10672 mm Let the thickness of bearing stiffener = t bs Out stand bbs should not be more than
12 t bs
Use 2 bearing stiffener of b bs mm wid e 24 t bs
2 * (t bs x 12 bs )
=
A
= 10671.7
t bs = 21.087 mm ≈ 30 mm Th er ef or e, out st an d = 3 60 mm ≈ 200 mm 2 2 Bearing area provided = 12000 mm > 10671.7 mm
OK
As per I S specification the bearing stiffener should be designed as column with a distance of the web 20 times the thickness of web ( t w ) on both side as a part of the stifferers. Length of web as stiffener =
2
x
20
tw
x
=
400
mm
30
0 0 2
1 0
0 0 2
20 t w
20 t w
= 200 mm
Eff ecti ve ar ea of st if fe ner
=
200 mm 2
A I xx
= =
16000 mm 172300000 mm
r
=
103.77 mm
Radius of gyration ,
Effective length of the web ℓ = 1750 mm Slenderness ratio of the bearing stiffener λ = 16.864 бac = 148.63 N / mm2 Permissible axial compression stress OK > 2000.9 kN Safe load = 2378.04 kN Provide 200 mm x
30 mm P late as stiffeners
CONNECTION BY WELDING :
Let us provide 6 mm size intermittent fillet weld Strength of weld / mm = 453.6 N / mm Required strength of weld / mm = 200.09 N / mm The le ngth of intermittent fillet weld is taken as 10 t bs C / C spacing of fillet weld = 680.08 mm Maxi. c / c spacing of fillet weld = 16 t bs
<
=
453.6 N / mm OK 300 mm
= 480 mm ≈ 300 mm
Provide 300 mm long fillet welds at a spacing of 300 mm INTERMEDIATE STIFFENERS :
For an unstiffened web the minimum required thickness is as below t w (mini) 1.0 = d √ ζ va
ζ va =
2.0
816 80.04 N / mm
t w (mini)
=
t w (mini)
= d√
27.409 mm f y
1344 3.0
t w (mini)
=
d 85
=
29.411 mm
=
29.412 mm
The web thickness provided is 8 mm, which is quite less than required if it to be unstiffened. So, intermidiate stiffenrs will be required Vertical stiffeners to be provided if the web thickness as calculated below is more than the web thickness provided. 1.0
t w (mini)
= d2√
f y
3200 2.0
t w (mini)
=
d2 200
=
12.353 mm
=
12.50 mm
The web thickness provided is 8 mm which is less than 12.50 mm Therfore, Vertical stiffeners will be provided A horizontal stiffener will be required at a distance of 2/5 of the distance from the compression flange to the N . A, if the web thickness as calculated below is more than that provided. 1.0 2.0
t w (mini) t w (mini)
= d2√ =
f y
4000 d2
=
9.8821 mm
= 10.00 mm 250 Therefore, a horizontal stiffener will be provided at
2 2500 = 500 mm from the x 5 2 compression flange. Another horizontal stiffener will be required at the N . A if the web thickness as calculated below is more than the provided one. t w (mini) 1.0 = d2√ f y = 6.176 mm 6400 t w (mini) d2 2.0 = = 6.25 mm 400 Web thickness provided is more than 6.25 mm. So, horizontal stiffener at N . A is not required. VERTICAL STIFFENERS :
> 180 t w
The smaller clear panel dimenssion of vertical stiffener
=
1800
mm
Therefore, Vertical stiffener should be provided at a spacing less than 1800 mm Number of stiffeners = 11 Nos. 1800 mm Actual smaller clear panel dimenssion = 1818.2 mm ≈ Provide vertical stiffeners at spacing of 1800 mm. A horizontal stiffener will be provided at 500 mm from the compression flange. d = 2000 mm Spacing of stiffeners c = 0.9 d > 0.7 d = 1400 mm c/c Provide vertical stiffeners at a spacing of 1400 mm c/c d 200 tw = Shear stress ζ va = From Table 6.6A of I. S : 800 - 1984, for
d tw
2
80.038 N / mm =
200
and c =
0.7 d
ζ va = 100.00 N / mm2 OK Minimum required thickness of vertical stiffener for shear consideration. t vs = 8.0038 mm ≈ 10 mm Maxi. Outstand of vertical stiffener Required moment of inertia
>
12 t vs = 3
120
mm ≈
mm
110
3
1.5 d t 4 = 3139106.2 mm 2 c I x x = 4436666.67 mm4 Moment of inertia of the vertical stiffener about the face of the web Use I. S. F
110
=
mm
x
>
3139106.2 mm4
OK
10 mm
Connection :
Connection are designed to rasist a shear of
2
125 t Where, kN / m h Shear force = 113.64 kN / m = 113.64 N / mm Let us provide a 3 mm size of the intermittent filled weld Strength of weld / mm = 226.80 N / mm > 113.64 N / mm OK The length of the intemittent fillet weld is taken as 10 t = 100 mm C / C Spacing of welds = 399.17 mm But the c / c spacing of the weld 16 t or 300 mm which ever is less C / C spa cing of weld = Provide 3 mm size weld
16 0 m m 100 mm long welds at c /c spacing of
h =
Outstand of stiffener
160 mm on both sides of the web.
A horizontal stiffener be provided at 500 mm from the compression flange. Let the thickness of horizontal stiffener be t = 10 mm ( minimum thickness of the web ) Maxi. Outstand = 12 t = 120 mm Consider a filet section = 100 x 10 mm 4
Moment of inertia required = 4 C t = 2871235.8 mm C = Actual distance between vertical stiffeners = 1400 mm t = Minimum required thickness = 8.0038 mm 4 4 Moment of Inertia provided = 3333333.33 mm > 2871235.8 mm
OK
Connection :
Provide
3 mm size
8 mm lo ng f ill et w eld at a c / c spa cing o f 1 28 mm
110
mm
200 L + 1000 N / m
HORIZONTAL STIFFENERS :
3
=
(
16 t
= 16 0 mm )
(Provide continuous welding)
3 mm size continuous fillet welding
VERTICAL STIFFENERS : CONNECTION : I SF 1 00
x
1 0 m m ( H or iz on ta l S ti ff en er )
0 0 5
IS F 110
1400
x
10 m m ( Ve rt ic al Sti ff ener )
mm
HORIZONTAL STIFFENERS:
DESIGN OF LATERAL BRACING :
Lateral braching is provided at the compression flange level. The diagonals in tension will be effective only. The influence line diagram for the bracing system show the diagonals are in tention. The force in the members L 0U1 and L12U9 will be maximum and also equal. Therefore, L0U1 or L0U9 is designed and the same section is provided for the other diagonals. Lateral force is condiderd
=
9000
N/m
U0
U1
U2
U3
U4
L0
L1
L2
L3
L4
U5
U6
U7
U8
U8
U9
L5
L6
L7
L8
L9
L10
M 3 . 2
20
m
@ 2.3 m e ach panal
TOP VIEW:
U0
U1
U2
U3
U4
L1
L2
L3
L4
U5
U6
U7
U8
U8
U9
L5
L6
L7
L8
L9
L10
Mx 3 . 2
L0
x
20 m
2
@ 2.3 m each panal
2
√
√
x
x
1
9 . 0
I L D For L 0U1 :
Consider the section x - x and taking the moment about L 10 Sin
45
L0 U1
=
o
L0 U1
x
1 sin
45
20.0
=
o
= 1 1
(Unit load place at L 0)
x 20.0 1 x √2
1
=
√2 Force in L0U1
=
114551.299 N 2
Area A = 763.68 mm Since the angle will not be accessible for clearing and painting thickness <
8 mm provide ISA 100 x 65 Cx x = 32.80 mm
is tried from IS Handbook No. 1. The relevant properties of the section, Shear center A 1 .= ( 100
-
4
) x 8
=
768
mm
A 2 .= ( 65
-
4
) x 8
=
488
mm
k
=
3 A1 3 A1 + A 2
=
0.8252
x
8 mm
Net area provided Load carrying capacity
= =
2
1121.8 mm 168265 N >
114551 N
OK
PB
CONNECTION :
5 4 0 8 . 2 3
114551.30
A
B
N 31.0
Le he force resised by he weld near he ou er edge of flange be P B. Taking momen abou A PB
x
100 PB
= =
114551.30
x
32.80
37572.8 N
Force resisted by the weld at A P A
=
76978.47
Let us provide a size of weld S Length of welds 37572.8
ℓB
= =
0.7 x 99.399
=
0.7
=
203.65
=
A
≈
N 5
mm weld for making the connection.
x S x 108.00 100 mm
And 76978.47
ℓ A
x
B
L0 U0
Force in the end strut Area of Angle required Let us try
90
x
90
x
≈
x
S
204
x
108.00
mm
=
90000 N
=
1285.7 mm
8
mm from IS Hand book No. 1. The releveant properties are
2
2
A = 1379 mm Radius of gyration r = 17.5 mm The strut will be welded to the flange plate girder Area
PB
CONNECTION :
5 4
A
1 . 5 2
B
0 9
90000
N 25.1
Slenderness ratio of the strut
λ
=
For λ
111.71 and f y
=
250.00 N/mm2
бac
= =
82.86 N/mm2 114263.94 N
=
Axial comprassive stress Load carrying capacity
111.71
>
90000 N
OK
Let the force resisted by the weld near the outer edge of flange be P B. Taking moment about A PB
x
90 PB
= =
90000
x
Force resisted by the weld at A P A = 64900.0 N Let us provide a size of weld S = Length of welds
5
mm weld for making the connection.
x
ℓ A
x
=
66.402
≈
70 mm
=
0.7
x
ℓB
x
=
171.69
≈
25100.0 =
ℓB
25.10
25100.0 N
0.7
S
x 108
And 64900.00
ℓ A
S
175
x 108 mm
DESIGN CROSS - FRAME :
U0 90000 N
L0
θ 0 0 5 2
Bottom strut A
B 2300
Two end frames one at each end of the plate girder are provided. In cross - frames double diagonals are provided of which one in tension is assumed to be effective. θ = 45 0 Froce in one diagonal U 0 A Cos 45 o = 90000 N U0 A = 127280 N OK Provide ISA 100 x 65 x 8 mm, Load carrying capacity of the Angle = 168265 N > 127280 N as design for the lateral bracing. The connection will also be the same as designed in for the section in lateral bracing system. 90 x 90 x 8 mm for the bottom strut. Provide a nominal section say Intermediate cross frames at every 4 m are provided. Since the lateral forae = 36000 N which is very le ss so, provide 75 x 75 x 8 mm angles for diagonals as well as for struts.
