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Structures under Combined Loading

Example: (continued)

Use equilibrium (net force and net moment must sum to zero) along with the FBD
(free body diagram) of the pipe cut at x = a (designated by O) to find the resisting
forces, R, and resisting moments, M, at this section.

Define forces: F₁ = 150 i lb, F₂ = 200 k lb, F₃ = - 50 k lb, F₄ = -150 k lb

and define position vectors: r₁ = 10 i + 10 j in, r₂ = 4 i in, r₃= 10 i in

r₁ x F₁ = (10 i + 10 j ) x 150 i = - 1500 k lb in = M₁

r₁ x F₂ = (10 i + 10 j ) x 200 k = 2000 i - 2000 j lb in = M₂

r₂ x F₄ = 4 i x (- 150) k = 600 j lb in = M₃

r₃ x F₃ = 10 i x (- 50) k = 500 j lb in = M₄

For equilibrium Σ F = 0 and Σ M = 0

So Σ F = R + F₁ + F₂ + F₃ + F₄ = 0 which gives R = - 150 i lb

and for moment equilibrium,

ΣM = M + r₁ x F₁ + r₁ x F₂ + r₂ x F₄ + r₃ x F₃ = 0 which gives

M = M_xi + M_yj + M_zk = - 2000 i + 900 j + 1500 k lb in

Click here to continue with this example.

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