Determine the deflection at B that would be caused by the distributed load if the middle support was not there. Let w = 9 lb/in., a = 64 in., and EI = 116 × 106 lb·in.2.
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Determine the distribution factor DFBC. Let w = 2.5 kip/ft,…
Determine the distribution factor DFBC. Let w = 2.5 kip/ft, L1 = 35 ft, and L2 = 21 ft. Assume EI = constant.
Draw the influence line for the moment at B. What is the lin…
Draw the influence line for the moment at B. What is the line’s minimum value? Let L1 = 4 m and L2 = 15 m.
Determine the vertical displacement of joint A if members AE…
Determine the vertical displacement of joint A if members AE and DE experience a temperature increase of 38°C. Let L = 8 m and α = 11.5E-6/°C.
Determine the beam slope at B. Let w = 1.5 kip/ft, L1 = 35 f…
Determine the beam slope at B. Let w = 1.5 kip/ft, L1 = 35 ft, and L2 = 28 ft. Assume EI = constant.
Determine the deflection at B that would be caused by the di…
Determine the deflection at B that would be caused by the distributed load if the middle support was not there. Let w = 9 lb/in., a = 66 in., and EI = 118 × 106 lb·in.2.
Determine the deflection at B that would be caused by the co…
Determine the deflection at B that would be caused by the concentrated moment if the middle support was not there. Let M = 10,800 lb·in., a = 65.82 in., b = 48.18 in., and EI = 84 × 106 lb·in.2. Note that b = (a + b)[1 – sqrt(3)/3].
Since indeterminate structures have more support reactions a…
Since indeterminate structures have more support reactions and/or members than required for static stability, the equilibrium equations alone are sufficient for determining the reactions and internal forces of such structures.
Determine the fixed end moment FEMBC. Let w = 1.2 kip/ft and…
Determine the fixed end moment FEMBC. Let w = 1.2 kip/ft and L = 30 ft. Assume EI = constant.
Determine the deflection at B that would be caused by the di…
Determine the deflection at B that would be caused by the distributed load if the middle support was not there. Let w = 9 lb/in., a = 53 in., and EI = 113 × 106 lb·in.2.