EIy = -(5/6)x³ + C₁x + (5/2)(x - 6)³ is the equation of deflection obtained by using Macaulay's method for the beam shown in the following figure [Simply supported beam AB of 6m with overhang BC of 3m, 10 kN downward load at C, RA = 5 kN downward, RB = 15 kN upward]. Find the slope at A. Given EI = 10 × 10¹³ N·mm². Where x = horizontal distance measured from support
- AA. 0.00015 radians
- B0.0003 radiansCorrect
- C0.003 radians
- D0.0015 radians
Solution
Concept
Find C1 from the support condition y = 0 at B. Slope at A is then C1/EI.
Formula
EI y = -(5/6)x3 + C1 x + (5/2)(x - 6)3 ; θA = C1/EI
Step-by-step solution
1. At A, x = 0 and y = 0. The bracket term (x - 6) is not active, so C2 = 0. 2. At B, x = 6 m and y = 0: -(5/6)(6)3 + 6C1 = 0. 3. -180 + 6C1 = 0, so C1 = 30 kN·m². 4. Slope: EI (dy/dx) = -(5/2)x2 + C1 + (15/2)(x - 6)2. At x = 0, EI θA = C1 = 30 kN·m². 5. EI = 10 × 1013 N·mm² = 10 × 1013 × 10-6 N·m² = 108 N·m² = 105 kN·m². 6. θA = 30 / 105 = 0.0003 rad.
Explanation
Correct answer: B) 0.0003 radians