Topic: Strength of Materials
Consider the beam shown in figure. The beam is simply supported at its left end A and fixed at its right end B. It carries a load that varies in intensity from zero at support A to 'w' at support B according to the relation w_x = (x/L)w. The reaction component at A is
- A. wL/10
- B. 2wL/5
- C. 5wL/2
- D. wL/5
Answer: A. wL/10
Explanation: Using Macaulay's double integration method, the bending moment at a section x from support A is M(x) = R_A * x - (w * x³) / (6L). Equating this to EI * y'', we integrate to find the slope (y') and deflection (y). EI * y' = (R_A * x²)/2 - (w * x^4)/(24L) + C_1. EI * y = (R_A * x³)/6 - (w * x^5)/(120L) + C_1 * x + C_2. Applying boundary conditions: at x=0, y=0 gives C_2=0. At the fixed end x=L, y'=0 gives C_1 = (w * L³)/24 - (R_A * L²)/2. At x=L, y=0 gives 0 = (R_A * L³)/6 - (w * L^4)/120 + ((w * L³)/24 - (R_A * L²)/2) * L. Solving this for R_A gives R_A = wL/10.