A simple supported beam of Span Gm is
uniformly distributed load of
over the left half of the span
find the maximum bending moment & draw
bending
moment & shear forced diagram
Carrying
4 kn/m
Answers
Answered by
0
Answer:
Mx = - 800x - 500(x- 0.8) Bending moment between B and C also varies by a straight line law. B.M. at B is obtained by substituting x = 1.5 m in equation (i). MB = -800 X 1.5 - 500 (1.5 - 0.8) = 1200 – 350 = 1550 Nm.
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Answered by
16
Answer:
Step-by-step explanation:
Mx = - 800x - 500(x- 0.8) Bending moment between B and C also varies by a straight line law. B.M. at B is obtained by substituting x = 1.5 m in equation (i). MB = -800 X 1.5 - 500 (1.5 - 0.8) = 1200 – 350 = 1550 Nm.
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