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Find the maximum moment of the above beam which is subjected to triangular vertical load.

**SOLUTION:**

ΣΜ_{Β} = 0 → Α_{Υ} –(1/2)ql(l/3)=0 → Α_{Υ} = (1/6)ql

ΣF_{Y} = 0 → ql/6 + B_{Υ} –(1/2)ql=0 → B_{Υ} = (1/3)ql

Assuming the equilibrium in distance x from the A end:

ΣΜ_{X} = 0 → (1/6)qlx – (1/2)(x/l)qx(x/3) - M_{x }= 0 → M_{x} = (ql/6)x(1 - x^{2}/l^{2})

ΣF_{Y} = 0 → (1/6)ql –(1/2)(x/l)qx - Q_{x}=0 → Q_{x} =(ql/6)(1 - 3x^{2}/l^{2})

Maximum M means zero Q:

Q_{X} = 0 → (ql/6)(1 - 3x^{2}/l^{2}) = 0 → x =* l*/ √* 3 *

maxM = (ql/6)(*l*/ √* 3 *)(1 - l^{2}/(3l^{2})) → maxM = (q*l*^{2}*/ 9*√* 3 *)

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