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Determine the fixed end moment FEMBC with a settlement of 0….

Determine the fixed end moment FEMBC with a settlement of 0.6 in. at support B.   As in Chapter 16, include the effect of settlement in the FEM calculation.  Let w = 2.9 kip/ft, L = 30 ft, E = 29,000 ksi and I = 1,650 in.4.

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Determine the reaction force at B. Let w = 17 lb/in., a = 68…

Determine the reaction force at B. Let w = 17 lb/in., a = 68 in., and EI = 273 × 106 lb·in.2.

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Use Robot to determine the force in member AD. Use a consist…

Use Robot to determine the force in member AD. Use a consistent material and cross section for the truss members. Delete the self-weight of the members. Let P = 14 kN, L1 = 5 m, and L2 = 4 m.

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Determine the fixed end moment FEMBC. Let w = 1.1 kip/ft and…

Determine the fixed end moment FEMBC. Let w = 1.1 kip/ft and L = 28 ft. Assume EI = constant.

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Determine the fixed end moment FEMAB. Let P = 20 kips, L1 =…

Determine the fixed end moment FEMAB. Let P = 20 kips, L1 = 30 ft, and L2 = 15 ft. Assume EI = constant.

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Use the portal method to determine the magnitude of the appr…

Use the portal method to determine the magnitude of the approximate shear in column AE. Let P1 = 36 kN, P2 = 54 kN, L1 = 8 m, L2 = 6 m, L3 = 7 m, and L4 = 5 m.

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Use the cantilever method to determine the magnitude of the…

Use the cantilever method to determine the magnitude of the approximate axial force in column FI. Let P1 = 22.4 kN, P2 = 37.6 kN, L1 = 10 m, and L2 = 7 m.

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Determine the distribution factor DFCE. Let P1 = 15 kips, P2…

Determine the distribution factor DFCE. Let P1 = 15 kips, P2 = 16 kips, L1 = 21 ft, L2 = 11 ft, and L3 = 16 ft. Assume EI = constant.

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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 = 18,600 lb·in., a = 70.44 in., b = 51.56 in., and EI = 52 × 106 lb·in.2. Note that b = (a + b)[1 – sqrt(3)/3].

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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 = 10,000 lb·in., a = 70.44 in., b = 51.56 in., and EI = 55 × 106 lb·in.2. Note that b = (a + b)[1 – sqrt(3)/3].

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