A girder that carries a uniformly distributed dead load of 1.7 k/ft plus its self-weight and three concentrated live loads of 15
kips at the quarter points of the 36 ft span is to be sized. Using A992 steel, determine the lightest Wshape to carry the load with lateral supports provided at the supports and load points. Use the correct Cb, check shear, and limit deflection to 1/360 of span.
1 answer:
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Answer:
Maximum shear stress= 5.66 ksi
Maximum tensile stress= 5.66 ksi
Maximum compressive stress=-5.66 ksi
Maximum shear strain=0.000943
Maximum tensile strain= 0.0004715
Maximum compressive strain= -0.0004715
Explanation:
For acircular bar, the maximum shear stress will be given by
where d is the diameter and T is torque.
By substituting 30 kip-in for torque and 3 in for d then
Maximum shear stress=
Also, the maximum tensile and compressive stresses will be 5.66 ksi and -5.66 ksi respectively.
The maximum shear strain will be given by stress divided by modulus of elasticity, in this case 6000 G
Maximum shear strain will be
The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence
Maximum compressive strain will be
Answer:
The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.
Explanation:
Let us assume that the total mass of composite is 100 lbm So as per the given conditions
- 15 lbm is copper and 85 lbm is rubber.
- Density of rubber is 70 lbm/ft3
- Specific gravity of Copper is 9
So As per the formula of specific gravity
here density of water is 62.4 lbm/ft3
Solving for Density of Copper gives
For composition on volume basis, volume of individual components and composite are calculated as
The composition is given as
So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.