Preparation of Standard Drawing of Pedestrian Bridge Span: 45 M Load Calculations
Preparation of Standard Drawing of Pedestrian Bridge Span: 45 M Load Calculations
Preparation of Standard Drawing of Pedestrian Bridge Span: 45 M Load Calculations
Span: 45 m
Load Calculations
Dead Load Calculation
Taking a steel deck of 6 mm,
UDL on a cross-beam = 2.0*0.01*78 = 1.56 kN/m
Nodal Loads kN
End Points No. Depth Length Area
Upper Lower
Chord Members
U 1 200 45000 9.00 18.00
L 1 200 45000 9.00 18.00
Vertical Members
U1 L1 2 200 3000 1.20 1.20 1.20
U2 L2 2 200 3000 1.20 1.20 1.20
U3 L3 2 200 3000 1.20 1.20 1.20
U4 L4 2 200 3000 1.20 1.20 1.20
U5 L5 2 200 3000 1.20 1.20 1.20
U6 L6 2 200 3000 1.20 1.20 1.20
U7 L7 2 200 3000 1.20 1.20 1.20
U8 L8 2 200 3000 1.20 1.20 1.20
U9 L9 2 200 3000 1.20 1.20 1.20
U10 L10 2 200 3000 1.20 1.20 1.20
U11 L11 2 200 3000 1.20 1.20 1.20
U12 L12 1 200 3000 0.60 0.60 0.60
Diagonals
U1 L2 2 200 3600 1.44 1.44 1.44
U2 L3 2 200 3600 1.44 1.44 1.44
U3 L4 2 200 3600 1.44 1.44 1.44
U4 L5 2 200 3600 1.44 1.44 1.44
U5 L6 2 200 3600 1.44 1.44 1.44
U6 L7 2 200 3600 1.44 1.44 1.44
U7 L8 2 200 3600 1.44 1.44 1.44
U8 L9 2 200 3600 1.44 1.44 1.44
U9 L10 2 200 3600 1.44 1.44 1.44
U10 L11 2 200 3600 1.44 1.44 1.44
U11 L12 2 200 3600 1.44 1.44 1.44
Also,
Frontal Area = 135.00
Solidity ratio = 0.40
Then, Drag Cofficient (Refer IRC 6 Annex C) = 1.7
Gust Factor = 2
Shielding Factor = 0.5
Therefore,
For Windward truss,
Load on each upper node = 8.01 kN
Load on each lower node = 8.07 kN
And,
For Leeward truss,
Load on each upper node = 4.60 kN
Load on each lower node = 4.60 kN
Chord Members
U -422.0685 -422.0685 -422.0685
L 419.517 572.23 386.361
Vertical Members
U1 L1 -160.671 -160.671 -160.671
U2 L2 -125.8635 -125.8635 -125.8635
U3 L3 -99.3615 -99.3615 -99.3615
U4 L4 -78.141 -78.141 -78.141
U5 L5 -60.564 -60.564 -60.564
U6 L6 -45.4755 -45.4755 -45.4755
U7 L7 -32.214 -32.214 -32.214
U8 L8 -20.16 -20.16 -20.16
U9 L9 -9.1875 -9.1875 -9.1875
U10 L10 0.483 0.483 0.483
U11 L11 3.675 3.675 3.675
Diagonals
U1 L2 185.304 185.304 185.304
U2 L3 144.0495 144.0495 144.0495
U3 L4 114.03 114.03 114.03
U4 L5 90.7725 90.7725 90.7725
U5 L6 71.778 71.778 71.778
U6 L7 55.692 55.692 55.692
U7 L8 41.6955 41.6955 41.6955
U8 L9 28.9695 28.9695 28.9695
U9 L10 17.2515 17.2515 17.2515
U10 L11 5.691 5.691 5.691
Design of RC Slab
Span between Cross-beams = 1.25 m
Thickness of Slab = 125 mm
Design of Section:
Basic Permissible Stresses of Concrete as per IRC : 21-2000:
Concrete Grade = M25
Reinforcement Used = Fe500
Characteristics Strengthof Concrete fck = 25 MPa
Permissible direct compressive stress, sc = 6.25 MPa
Permissible flexural compressive stress, scbc = 8.33 MPa
Maximum Permissible shear stress, tmax ( 0.07*fck) = 1.9 MPa
m σ cbc
Neutral Axis Factor = = 0.258
m σ cbc σ st
Lever Arm Factor = k = 0.914
1
3
M
d= 50.9 mm
Rb
Ast =
M = 145 mm2
st j d
b d
Provided percentage area of tensile steel = 0.49 %
Design of Cross-beam
Design moment = 2.98 kNm
Design shear = 8.70 kN
Design of Section
Modulus of Section, Zreqd = M/σb
Where, σb - Permissible stress in bending = 0.66*fy= 165 N/mm2
The size of angle section should be sufficient to resist the maximum shear stress τv
τv = 1 .5 * R < 0.45 fy
2 * t * d
Design Data
Length of member = 2.000 m
Effective length of member = 2.000 m
Design Axial Force on member = 529.00 kN
B.M. is considered to be neglected
Assuming the permissible direct compression stress
σac = 141.00 N/mm2
Sectional area required = 3751.77 mm2
Let us take a double channel section with 2-ISMC200
a = 2850 mm2
h = 200 mm
bf = 11.4 mm
Cy = 2.34 mm
Ixx = 1840 cm4
Iyy = 156 cm4
For double channel separated 25 mm back to back,
A = 5700 mm2
Ixx = 3680 cm4
Iyy = 370.8 cm4
So,
rmin = 36.07 mm
Slenderness ratio = 55.44
From IS 800, Table 5.1 σac = 141 N/mm2
Then,
Compressive Strength of the member = 803.70 kN OK
Adopt thickness = 6 mm OK
Again,
Transverse shear force, V = 2.5% of Axial Load = 13.23 kN
Adopt d = 150 mm
Design Data
Length of member = 2.00 m
Effective length of member = 2.00 m
Design Axial Force on member = 680.00 kN
B.M. is considered to be neglected
Assuming the permissible direct tensile stress
σac = 150.00 N/mm2
Sectional area required = 4533.33 mm2
Let us take a double channel of ISMC250
Area of one channel a = 2850 mm2
tw = 6.2 mm
Then,
Total Area = 5700 mm2
Tensile Strength of the member = 855.00 kN OK
= 𝐿/600 = 75 mm OK
= 31 mm OK
= 𝐿/800