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Find the maximum moment of the above beam which is subjected to triangular vertical load.
SOLUTION:
ΣΜΒ = 0 → ΑΥ –(1/2)ql(l/3)=0 → ΑΥ = (1/6)ql
ΣFY = 0 → ql/6 + BΥ –(1/2)ql=0 → BΥ = (1/3)ql
Assuming the equilibrium in distance x from the A end:
ΣΜX = 0 → (1/6)qlx – (1/2)(x/l)qx(x/3) - Mx = 0 → Mx = (ql/6)x(1 - x2/l2)
ΣFY = 0 → (1/6)ql –(1/2)(x/l)qx - Qx=0 → Qx =(ql/6)(1 - 3x2/l2)
Maximum M means zero Q:
QX = 0 → (ql/6)(1 - 3x2/l2) = 0 → x = l/ √ 3
maxM = (ql/6)(l/ √ 3 )(1 - l2/(3l2)) → maxM = (ql2/ 9√ 3 )
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