Question

Problem #2 (55pts): In the torsion bar made of the same material fixed against rotation at its ends, a torque (I) of 45kN-m and an axial compressive load (P) of 200KN are applied, a) (35) determine maximum shear stress in the bar b) (20) determine factor of safety (FOS) against yielding using Energy Distortion Criteria (Consider 0=0) di 6cm; d 12cm x-0.5m; 1-1.5m Sy-300MPa

Answer 1

The fixed end at A and B must resist the torque so summation of these 2 resistive torque must be 45 KN-m

also angle of twist at application of torque must be same with respect to both sides of bar

after determining resistive torques we can calculate shear stress in both sides

and then apply Distortion energy theorem after finding the priciple stresses due to compressive force and shear stress

i have calculated all in my solution

Thanks
+10. 5mm TA+TB =T=45 PAC - OBC There 20.5TA - To

| 05 A 8. from md D To + To = 45 - To - 40 kv.m TA = 5 kNm TAC = The 8 - 5x10 x 3 x 102 P X 0.067 & TAC = 117.89 x103 kN/m2 = 11.789 KM/Com BC BC XY - 450103x8x102 x(0.1234 TBC 13.26 KM/m2 so, maximum to shear stress exist in and burie, of beside a max = 13.26 KM/C m² 5.6 - - 2002 -1.77 KM/m² Oy -2002- 7.08% 200 TX192 02-01 86 side 200 -=-7.08 KN 2.

so for AC side 25151-7.081/7.00 44X1,2017] 62 = (-7.08 124.62) 20,2 8:77-15.85 accadley to max" distortion energy therg 2 x (x = 6,676-5)+ (626) 2 7 Expo XTO) 76.71415-8378.774 GA-300 MP = 300x100 m - 3. kom 2x (30) 2 (6.774 15.85² + 8 7 7 7 (-15.85² - 30 = 2161 F08 -1.35 Fo8 for BC now section. 6,2 = 21-177 [1.77² + 4x13262) 362[-1.77€ 26.58) 5, --14.7.12.40 2x (30) ² - E14.17 - 12 40 ) + (-14.12² 12 40) 2 FOB - 1.30 (lower than 1.39) so we have to take Fos = 1.39.