Question

Problem 2: A circular shaft transmits power as shown with pulley loads. The shaft carries a torque, bending, shear and axial loads. Draw LVM diagram to find Mmax and Vmax. Show all loads (moments and forces) on the circular x-section of the shaft below. Use double arrows for moments. Compute shear and normal stresses and show them on the same section. Create stress elements for points A and B of the section. Combine the stresses and compute Tmax (Tresca) and om (von Mises) for the stress elements. Zero credit if no values shown on the section below. (Work in N and mm) 3.2 kN 4.2 kN -0.15 m 0.30 m -0.15 m BU Х С 2 kN Z A B 2.5 kNm D 25 mm dia.

Answer 1

Given! A circular shaft transmits power os shown with pulley loods 3-2119 42KN А 0.15m 0.3m 0.15m ។ 1. MC A 2.5Kim A 2KN B D 25mm TAY B c Dy force balance, 7.4 Ay + by = 3.2 + 4 2 = MAzo 3.2 x 0.15 + 4.2% 0.45 by x 0.6 by = '3.95 kn, ay = 3.45 KYN Shear force diagrom- 3.45 KN 1.25 KN C D A B 13.95KN bending moment diagrom:. 592.5 kNmm 517.5kN mm A c D How here, Uman = 3.95 KN ar point c (pulleye) Mman = 592.5 kN mm now axial stress same as point Agc, - 2x103 x Tar 4074 mpa

now Shear Strecess. てっ (both at point 2B) 165 таз 16x2.5x10' ITX 253 TE てっ 815 mpa how bending stress at point B is zero (because at heutral plane) and bending stress occurs at point A only = -32% 542.54103 -386 MPa ITX 253 (66) A Oba 32m T2² - 386mpa now at point A compressive stress, a = -3867 -4.074 39ompa Shear stres. 2 = 815 mpa how maximum principle stress om= 2 + +22 300 + 13,992 +815? om = -1038mpa Iman = 838mpal at point normal stress - 4.07ampa > shear stress, t 815mpa maximum normal stress, n = E)+ VE)+22 818 MPa maximum shear stress, Eman = 816mpa