Question

Problem # 3: Using the Virtual work method (Unit load method) method, calculate following quantities for the beam shown below. The bending moment diagram due to the applied loads is provided below. Also, note that the "moment of inertia" value is different for part of the beam (for AC: it is 2I and for CF: it is I) a) Vertical deflection at Point F b) Rotation at point D E is constant for the beam 40 k 15 k oft Bending Moment diagram CM) 200 A: 150 200

Answer 1

V_A + V_0 + V_E = 55 k V_A times 20 = 40 times 10 = 20 k V_0 times 10 + V_0 times 25 = 15 times 35 V_0 + V_E = 35 Rightarrow V_E = 11.67 k V_D = 23.33 k Apply load at B = 1k V_A 0.5 k V_A + V_B + V_C = 1 m_C = 0 V_B times 10 + 25 times V_E V_B = -2.5 V_E V_p 0.5 -2.5 V_E + V_E = 1 delta = 1/E integral^L_0 dx V_C = -0.33 k V_D = 0.833 k \r\n delta_B = 1/EI [integral^10_0 (20x)(0.5x)/2 + integral^20_10 [20x -40(x -10)/2 [0.5x -(x -10) + integral^30_20 [20x -40(x -1 -0)][0.5x -(x -10)] + integral^45_30 m x_2 mx + integral^55_45 mx_2 m_x' For AB mx = 10x^2 For BC rightarrow mu = (400 -20x)(1.0 -0.5 x) = 400 -200 x -200 x + 10x^2 For BC = 10x^2 -400 x + 400 CD = 10x^2 -400 x + 400 DE = (20 x -40 x + 400 + 23.33 -700)(1.0 -0.5 x + 0.833 x -28) DE = mx = (3.33 x -300) (0.33 x -15) DE = 1.11 x^2 -148.45 x + 4500