SSC JE Mechanical Engineering 27 Sep 2019 Shift 1 Paper

Question : 190 of 200

Marks: +1, -0

Maximum bending moment for simply supported beam with udl over entire length of beam, if W = weight of beam and L = length of beam, is:

Solution:

Net weight of the UDL = wL = W

{R}_{{A}}+{R}_{{B}}={wL}

Due to symmetry,

{R}_{{A}}={{wL}}/{2} ; {R}_{{B}}={{wL}}/{2}

Taking moment about B

M=R_{A} x-{w x^{2}}/{2}={w L}/{2} x-{w x^{2}}/{2}

For maximum B.M,

{{dM}}/{{dx}}=0

i. e.

{{wL}}/{2}-{{w} \.2 {x}}/{2}=0

{x}={{L}}/{2}

{M}_{\max }={{wL}}/{2} \×{{L}}/{2}-{{wL}^{2}}/{8}={{wL}^{2}}/{8}

{W}={wL} \⇒{M}_{\max }={{WL}}/{8}

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