Question

Use slope-deflection method to analyze the frame shown below. Segments AB and BD of the frame have moment of inertia I. Segment BC has moment of inertia 2/. Modulus of elasticity E is constant throughout the frame. The frame is supported by fixed-supports at A and D, and by a roller-support at C. Joint B is rigid. A downward point load of 20 kN is applied at mid-span of AB. Uniformly distributed load of intensity 2 kN/m acting downwards is applied along BC. Consider flexural response only a) Calculate the rotation at B b) Calculate the support moments at A and D. c) Sketch the deflected shape. Hint: MN 20 KN 2 kN/m BI 21 8 m 4 m 4 m 4 m

Answer 1

Based on the figure A and D are fixed so reaction theta_A = 0: theta_A = 0 �C� is hinged so there is no moment M_CB = 0: M^bar_BC = M^bar_Cb = 0 Now, using slope and deflection equation M_AB = M^bar_AB + 2EZ/l (2 theta_A + theta_B - 3 delta/l) M^bar_AB = -omega t/8 = -2 times 8/8 = -2 kN xi - m M^bar_BA = omega t/8 = 2kn xi - m M^bar_BC = 0: M^bar_BD = M^bar_DB = 0 M_AB = -2 + 2Ez/8 (2(0) + theta_B - 0) M_AB = -2 + 2EJ/8 (Theta_B) M_BA = 2 + 2EJ/8 (2 theta_B) M_BA = 2 + 2EJ/8 (2 theta_B) = 2 + EJ/4 theta_B M_BC = M^bar_BC + 2EJ/4 (2 theta_B + theta_C - 3 delta/2)_detla = 0 M_BC = -M^bar_CB + 2EJ/4 (2 theta_B + theta_C) = -8/8 + (2 theta_B + theta_C) EI