Question

20 KN 3 kN/m 3 2 KN- y, 2 kN-m 12.5 mm 200 mm? Z B 150 mm 121.43 mm 1.5 m 1.5 m 2.2 m 12.5 mm Part [1] (a.) Construct shear and bending moment diagrams. Show all work. Label completely. (b.) Determine the maximum value of the transverse shear force (in magnitude) and where it occurs. 'Box' answers. (c.) Determine the maximum values of the bending moment (both positive and negative) and where each occurs. 'Box' answers. Part 21 The cross-sectional geometry is shown for this same beam and loading above. The location of the centroid of the cross-section is shown (121.43 mm from the bottom). The second moment of area ('moment of inertia') about the z axis is: Iz = 10.621 x 106 mm4 = 10.621 x 10-6 m (a.) Determine the maximum compressive normal stress due to bending and indicate where it occurs (location along the length, plus top or bottom). Show all work. 'Box' answers. (b.) Determine the maximum tensile normal stress due to bending and indicate where it occurs (location along the length, plus top or bottom). Show all work. 'Box' answers.

Answer 1

Part - 1 (a) S . F . D & B . M . D Reactions - sigma V = 0 R_A + R_C = 20 + (3 times 2.2) R_A + R_C = 26.6 KN sigma M_A = 0 - (3 times 2.2 times 4.1) + (R_C times 3) - 2 - (20 times 1.5) = 0 R_C times 3 = 59.06 [R_C times 3 = 59.06 [R_C = 19.68 KN] [R_A = 26.6 - 19.68 = 6.92 KN] \r\n shear force - F_D = 0 KN F_CR = 3 times 2.2 = 6.6 KN F_C_L = 6.6 - 19.68 = -13.08 KN F_B = -13.08 + 20 = + 6.92 KN F_A = 6.92 - 6.92 = 0 KN Bending Moment - M_D = 0 KN.m M_C = 3 times 2.2 times 1.1 = 7.26 KN.m M_B_L = (3 times 2.2 times 2.6) - (12.68 times 1.5) = -12.36