A concentrated load P is applied at the end of a cantilever as shown in Fig. The cross section of the beam is a square of side 'a' with a hole of dia 'a/2'. The deflection at the tip of the cantilever is given by
- A. 3P L³ / E a^4
- B. 1024P L³ / (256 - 3\pi)E a^4
- C. 1024P L³ / (256 - \pi/64)E a^4
- D. 256P L³ / (1024 - 3\pi)E a^4
Answer: B. 1024P L³ / (256 - 3\pi)E a^4
Explanation: Moment of Inertia I = I_square - I_circle = a^4/12 - \pi(a/2)^4/64 = a^4/12 - \pi a^4 / 1024 = a^4(256 - 3\pi) / 3072. Deflection \delta = PL³ / 3EI = PL³ / (3E * a^4(256-3\pi)/3072) = 1024 PL³ / ((256-3\pi)E a^4).