Question

solve with explanasion please

2. The 20 mm diameter stainless steel shaft shown is fixed at A and B. Calculate the maximum shear stress in the shaft. (15 pts) 200 Nm 1 m 500 N·m 1.5 m

Answer 1

* mm Diameter D = 20 be Let TA and To depe developed the resisting torque 500 200 TB TA A im 1.5m D IMB on torque balancing - TATTB = 500+200 TA + TB = 700 Nm 1. Torque be as diagram will follows: TA TL GI A TB-200 (angle of tulist TA OAB = 0 as both ends are fixed O A/C & Oclo too/B = 0 TAC (LAC) + TCD Lost TDB LDB GI > -TA (1) + To -200 (115) + TB (1) » – (700-TB) + 3 (TB-200) + TB 0. 2 » -2(700 - TB) + 3 (TB - 200) + TB = 0

Data Page -1400 +2TB + 3 TB - 600 + TB =0. 6TB - 2000 To 1000 Nm. = 333-33 Nm. - TA= 700-TB 700 -333-33 366.67 Nm. - TA = 366.67 Nm TB -333.33 Nm . Maximum torque any where in shaft = 366.67 Nm = Imax 16T Max shear stress AD3 Imax = 164 366 67 N/m² t(0.02) 233 429.37 N/m² Tmax = 233.43 kN/m²