Week 1 Challenge
1. Efficient truss system:
Truss-A is Pratt truss and Truss-B is Howe truss in use. Pratt truss develops tension in diagonal members and compression in vertical members. Howe truss develops compression in diagonal members and tension in vertical members. Technically, the longer members are critically risk in buckling during compression. (Diagonal members are longer in truss structure). So it is best to keep tension in longer members than that of shorter members for more safety against buckling (out of plane rotation). In view of this, Pratt truss is more efficient with gravity loads.
2. Deflection & qualitative BMD
3. Determination of MOI, Weight and cross section
Moment of inertia = bd^3/12 + Ah^2
I1 = 8x180^3/12 = 3888000
I2 = 80x10^3/12 + (80x10x95^2) = 6666 + 7220000 = 7226666
I3 = 80x10^3/12 + (80x10x95^2) = 6666 + 7220000 = 7226666
I = I1 + I2 + I3 = 1.83x10^7
Density of steel = 7850 kg/m3
Area = (8x180)+(80x10)+(80x10) = 3040 mm2
Weight = 3040x10^-6 x 1 x 7850 = 23.9 kg/m
(a)
I1 = 8x1000^3/12 = 666666666
I2 = 300x30^3/12 + (300x30x515^2) = 675000 + 2387025000 = 2387700000
I3 = 300x30^3/12 + (300x30x515^2) = 675000 + 2387025000 = 2387700000
I = I1 + I2 + I3 = 5.4x10^9
Density of steel = 7850 kg/m3
Area = (8x1000)+(300x30)+(300x30) = 26000 mm2
Weight = 26000x10^-6 x 1 x 7850 = 204 kg/m
(b)
As per Steel book, MOI of ISMB300 = 8.9x10^7
Weight = 46 kg/m
Area = 5860 mm2
After adding reinforced plate, The revised MOI will be
I2 = I3 = (160x12^3/12) + (160x12x156^2) = 23040 + 46725120 = 4.6x10^7
I = (8.9+4.6+4.6)x10^7 = 18.1x10^7
Weight of each plate = 160x12x10^-6x1x7850 = 15 kg/m
Total weight = 46+15+15 = 76 kg/m
Total area = 5860 + 1920 + 1920 = 9700 mm2
(c)
As per steel book, MOI of ISMB500 = 45.2x10^7
Area of ISMB500 = 11100 mm2
Weight of ISMB500 = 86.9 kg/m
As per steel book, Area of ISMC300 = 4630 mm2
Weight of ISMC300 = 36.3 kg/m
MOI of channel section = 0.3x10^7 + (4630x255^2) = 30.4x10^7
Total MOI = 75.6x10^7 mm4
Weight of total section = 123.2 kg/m
Area of total section = 15730 mm2
4. Calculation of beam loads
Interior beam:
Tributory width = 2m
Floor load = 10kn/m2
UDL of interior beam = 20 kN/m
Exterior beam:
Tributory width = 1m
UDL of exterior beam = 10 kN/m
Interior girder:
Tributory width = 8m
UDL of interior girder = 80 kN/m
Exterior girder:
Tributory width = 4m
UDL of exterior girder = 40 kN/m
5. Thermal elongation of steel bridge
Temperature difference = 40 - (-20) = 60 degree
Strain at 80 degree temperature difference = 12x10^-6 x 60 = 7.2x10^-4
Total length of bridge = Original length x (1+strain) = 2000 x (1.00072) = 2001.4 meter