Question

Use the energy method to solve for the deflection and slope at point x=a. Assume that a=1m, b=1m, q=4kN/m, P=6kN, and EI=5MN*m^2. Please also draw the FBDs for both global and local cut-out equilibriums. The answers should be , and $\Theta = -1.93*10^{-3}$ .

S = -1.1 * 10-3

Answer 1

Sigma F_y = 0: A_y = 4(1) + 6 = 10kn. Sigma M_A = 0: M_A = 6(2) + 4(1) (0.5) m_A = 14 KN.M. M_AC = 10x -14 -4x^2/2 M_AC = 10x -2x^2 -14 (0 less than or equal to x less than or equal to1). M_BC = 6(x): (0 less than or equal to x less than or equal to1). Apply unit vertical load @ 'c' m_AC = 1(x) -1 m_BC = 0. delta_cv = -sigma interal^2_0 mmdx/EI = -integral^1_0 (10x -2x^2 -14) (x -1) dx/EI + 0 = -1/EI [10x^3/3 -2x^4/4 -14x^2/2 -10x^2/2 + 2x^3/3 + 14x]^1_0 = -5.5/EI = -5.5 times 10^3/5 times 10^6 = -1.1 times 10^- 3. delta_cv = -1.1 times 10^- 3 m. Apply unit moment @ 'c'. m_AC = -1. m_BC = 0. Theta = -integral^2_0 mmdx/IE theta = -integral^1_0 (10x -2x^2 -14) (- 1)dx/EI + 0. theta -1/EI [- 10x^2/2 + 2x^3/3 + 14x]^1_0 theta = -1.9933 * 10^- 3 rad.