SSC JE Civil Engineering 25-Sep-2019 Shift 1 Solved Paper

Question : 147 of 200

Marks: +1, -0

For the beam having a uniformly distributed load of w per unit length as shown in the given figure, the maximum positive bending moment is equal to the maximum negative bending moment. The value of L

_1

to L for this condition is:

$1 /√2$
$√ 2$
$1 /2$
$1 /{2√ 2}$

Validate

Solution:

As per given figure:
Let us consider

(L_1 = a)

Max. positive

B.M = W /8[ L^2-4a^2 ]

...(i)
Max Negative

B.M = {Wa^2}/ 2

...(ii)
By equating eq (i) and (ii)

B.M = W /8[ L^2-4a^2 ]={Wa^2}/ 2

L^2 - 4a^2 = 4a^2

L^2 = 8a^2

a^2=L^2/8

a=L/√8=L/{2√2}

Put the value of (a = L

_1

)
Hence,

[L_1=L/{2√2}]

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