material of bridge design

1. Section 2: Notional Design
The diagram was made in auto cad and show below,
Steel foot pedestrian bridge figure
The above figure is the model of steel foot pedestrian bridge.
It is span is about 20 ft. and its 8-ft. wide.
Some mechanical properties of steel are,
 Modulus of elasticity, E = 210,000 N/mm²
 Shear modulus, G taken as 81,000 N/mm²
 Poisson's ratio, ν = 0.3.
 Coefficient of thermal expansion, α = 12 x 10-6/°C (in the ambient temperature
range).
The beam which is used in it in it are w8*31 and its properties are given below
1. Area 9.1 inch
2. Depth 8.00 inch
3. Thickness 0.285 inch
4. Width 7.995
5. Fy =50ksi
6. fu=32 ksi

2. The components are listed below,
Beam / Girder
Beam or girder is that part of superstructure structure which is under bending along
the span. it is the load bearing member which supports the deck. Span is the
distance between points of support (eg piers, abutment). Deck is bridge floor directly
carrying traffic loads. Deck transfers loads to the Girders depending on the decking
material.
Columns or shafts
A column or pillar in architecture and structural engineering is a structural element
that transmits, through compression, the weight of the structure above to other
structural elements below. In other words, a column is a compression member.
Stairs
“A Stair is a series of steps with or without landings or platforms, which is installed
between two or more floors of a building to bridge a large vertical distance”.
Stair Case:
A part of a building having series of steps is called stair case.
Stringer
The structural member that supports the treads and risers. It is the part of the stairs
that will hold all the weight.
This the steel foot pedestrian bridge and we took steel beam w8*31

3. Section 3: Load and Analysis
The loads are given below,
Assume self-weight of a beam =W sw=130 lb/ft
Dead load =30+130=160lb/ft=0.16kip/ft
Live load= 150lb/ft=0.15kip/ft
Wu=1.2(Dead load)+1.6(live load)
Wu=1.2(0.16)+1.6(0.15)
=0.432 kip/ft
Mu = wu L^2 / 8
=(0.432)(20)^2/8
=21.6 kip/ft
So we assume Mu=22 kip/ft
Step II.
Determine unsupported length Lb and Cb
Lb = 20 ft and cb=1.14
III. Select a wide-flange shape
- Mu/Cb = 22/1.14 = 19.29=20kip-ft.
Bf/2 tf <_0.38 sqrt 29000/50
9.2<_2.89
the flange is compact
Mn=MP=fyzx
=(50)(30.4)
=1520kip in =126.666-127 kip ft
φbMn=(0.9)(127)=114.3 kip ft
φbMn > Mu
so w8*31 is selected.

4. Shear
he factored load and shear are:
Wu=1.2(0.16)+1.6(0.15)= 0.432 kip/ft
Vu = wu L / 2 = 0.432 (20) / 2 = 4.32 kips
w8*31 - Section Properties taken from Part 1 of the AISC Manual
d = 8 in tw =0.285 in h / tw =28.070
2.45 sqrt E / Fyw = 2.45 29,000 /36 = 69.54
Since h / tw = 28.070 is < 69.54, the shear strength is governed by shear
yielding of the web
Vn = 0.60 Fyw Aw = 0.6(36)(7.995)(9.13) = 1576kips
φvVn = 0.90(1576) = 1419 > 52.8 kips (OK)
The section w8*31 is adequate in resisting the design shear.
Buckling
Check for local buckling.
λ = bf / 2tf = 7.87; Corresponding λp = 0.38 (E/Fy)0.5 = 9.192
Therefore, λ < λp - compact flange
λ = h/tw = 50.0; Corresponding λp = 3.76 (E/Fy)0.5 = 90.55
Therefore, λ < λp - compact web
Torsion:
The formulas for torsion is
Now the below diagrams are shown,

5. Torsion :
Joint reaction :
Axial force :

6. Shear force :

7. Moment :