SSC JE Civil Engineering 2018 Set 1 Solved Paper

Question : 190 of 200

Marks: +1, -0

A simply supported beam carries a varying load from zero at one end and w at the other end. If the length of the beam is a, the maximum bending moment will be
(A)

{wa}/27

(B)

{wa}^2/27

(C)

{w^2a}/√27

(D)

{wa}^2/{9√3}

A only
B only
C only
D only

Validate

Solution:

ΣY = 0

R_A+ R_B=1/2 Wa

ΣM_A = 0

R_B× Q +1/2W.\l{.a×2}/.3 =0

R_B={wa}/3

R_B={wa}/6

ΣBM_x = 0

-R_A× x+1/2{wx}/a .x. x /3+[x]=0

M_x=-R_A× x+1/2{wx}/a\ x\ {xx }/3

.... (i)

{dMx}/{dx}= 0

{ for max imum BM}

0={dMx}/{dx}=-{wa}/6+{3wx^2}/{6Q}

{wa}/6={ wx^2}/{2Q}

{wa^2}/3={ wx^2}

x=√{a^2/3}

[x={a /√3}]

Put value of x in equation (i) we go →

M_x=R_A=Q/√3+1/2 w/a(a/√3)^3×1/3

M_x=- {wa}/6 a/3+1/2 w/a(a/√3)^3×1/3

[M_x={wa^2}/{9√3}]

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