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Member AB can now be cut member at a-a. At the cut, both the shear and moment internal forces are unknowns. The vertical and horizontal components of the force FAB are
FAB-x = FAB-y = FAB / cos45 = 150 N
Vertical and horizontal distances are shown on diagram. Summing moments at the cut gives,
ΣMcut = 0
M + 150(10)(1-cos25) - 150(10)(sin25) = 0
M = 493.4 N |