SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Question : 169 of 200

Marks: +1, -0

A prismatic, straight, elastic, cantilever beam is subjected to a linearly distributed transverse load as shown below. If the beam length is L, Young’s modulus E, and area moment of inertia I, the magnitude of the maximum deflection is

${{\qL}^{4}}/{15 {\EI}}$
${{\qL}^{4}}/{30 {\EI}}$
${{\qL}^{4}}/{10 {\EI}}$
${{\qL}^{4}}/{60 {\EI}}$

Validate

Solution:

Explanation:

To find maximum deflection expression.
Students are advised to memorise the standard results. This

\δ={{\qL}^{4}}/{30 {\EI}}

is a standard result for cantilever beam with UVL.
For derivation of formula refer this

Here

{\w}={{\p\x}}/{{\L}}

per unit length at a distance

{\x}

from

{\O}

∴

{\EI} {{\d}^{4} {y}}/{{\dx}^{4}}=-{{\px}}/{{\L}}

{\EI} {{\d}^{3} {\y}}/{{\dx}^{3}}=-{{\px}^{2}}/{2 {\L}}+{\C}_{1}

S.F. is zero at x = 0, giving

\C_1

= 0
∴

{\EI} {{\d}^{3} {\y}}/{{\dx}^{3}}=-{{\px}^{2}}/{2 {\L}}

∴

{\EI} {{\d}^{2} {\y}}/{{\dx}^{2}}=-{{\px}^{3}}/{6 {\L}}+{\C}_{2}

B.M. is zero at x = 0, giving

\C_2

= 0

{\EI} {{\d}^{2} {\y}}/{{\dx}^{2}}=-{{\px}^{3}}/{6 {\L}}

Integrate again

{\EI} {{\dy}}/{{\dx}}=-{{\px}^{4}}/{24 {\L}}+{\C}_{3}

But

{\d \y}/{\d \x}=0

\x=\L

, giving

0=-{{\pL}^{4}}/{24 {\L}}+{\C}_{3}

i. e,

{\C}_{3}={{\pL}^{3}}/{24}

∴

{\d \y}/{\d \x}={1}{\E \I}(-{\p \x^{4}}/{24 \L}+{\p \L^{3}}/{24})={\p}/{24 \L \E\I}(\L^{4}-\x^{4})

\y={\p}/{24 \L \E \I}(\L^{4} \x-{x^{5}}/{5})+\C_{4}

Now y = 0 at x = L, giving

{\C}_{4}=-{4 {\pL}^{5}}/{5 ×24 {\EIL}}=-{{\pL}^{4}}/{30 {\EI}}

∴

\y={\p}/{24 \L \E \I}(\L^{4} \x-{\x^{5}}/{5})-{\p \L^{4}}/{30 \E \I}

\y=-{\p}/{120 \E \I \L}(4 \L^{5}-5 \L^{4} \x+\x^{5})

Since the deflection will be maximum at the end O i.e. at x = 0, put this in the above
obtained expression to get

\δ={{\pL}^{4}}/{30 {\EI}}

, where negative indicates the deflection is towards negative

{\y}

axis.

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