A concentrated load P is applied at the end of a cantilever as shown in figure. The cross section of the beam is a square of side ‘a’ with a hole of a dia ‘a/2’. The deflection at the tip of the cantilever is given by

\(\frac{{1024P}}{{\left( {256 - \frac{\pi }{{64}}} \right)E}}\frac{{{L^3}}}{{{a^4}}}\)

\(\frac{{3P}}{E}\frac{{{L^3}}}{{{a^4}}}\)

\(\frac{{1024P}}{{\left( {256 - 3\pi } \right)E}}\frac{{{L^3}}}{{{a^4}}}\)

\(\frac{{256P}}{{\left( {1025 - 3\pi } \right)}}\frac{{{L^3}}}{{{a^4}}}\)

A concentrated load P is applied at the end of a cantilever as shown i

1.	A concentrated load P is applied at the end of a cantilever as shown in figure. The cross section of the beam is a square of side ‘a’ with a hole of a dia ‘a/2’. The deflection at the tip of the cantilever is given by
A.	\(\frac{{3P}}{E}\frac{{{L^3}}}{{{a^4}}}\)
B.	\(\frac{{1024P}}{{\left( {256 - 3\pi } \right)E}}\frac{{{L^3}}}{{{a^4}}}\)
C.	\(\frac{{1024P}}{{\left( {256 - \frac{\pi }{{64}}} \right)E}}\frac{{{L^3}}}{{{a^4}}}\)
D.	\(\frac{{256P}}{{\left( {1025 - 3\pi } \right)}}\frac{{{L^3}}}{{{a^4}}}\)
Answer» C. \(\frac{{1024P}}{{\left( {256 - \frac{\pi }{{64}}} \right)E}}\frac{{{L^3}}}{{{a^4}}}\)

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A concentrated load P is applied at the end of a cantilever as shown in figure. The cross section of the beam is a square of side ‘a’ with a hole of a dia ‘a/2’. The deflection at the tip of the cantilever is given by

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