Question

Select shaft material and confirm 1.2 factor of safety for static failure for the intermediate shaft. (20pts)

81.5 lb 1.26" TA The 19616 377 PA 0.87" 248 3016 60.9716

use labels from this picture for gear info. Do not use forces

Answer 1

-0.87 I 87.5 16 168 19616 * 0.87% 1.26 li 377 3016 248 66.97 16 Vertical plane :- 0.87" 0.87" 1961b 66.9716 ברב 248 Lb 1 58.26 170.52 Horizontal plane - 1875 168 3016 78.72. 26.1

T = Ssy 0.5 (40o) - 166.66 N/mm2 Let select carbon steel :- Yield tensile stress is 400 N/mm2 Factar of safety is 3 1.2 0.5 Syt fo = 24171.98 lb/im? Torsional moment Bending moment Mo = √ 170.5 2 2 + 125.725 = 186.5759 to in 1.2 MT Mb MT = tuxuro = 4a9.5016 in d = 16 + T I max (186.5157494.5 XX 2471.98 d =0.9825 choose dia more than 0.482910 otherwise it will static failure. We have to

12216 -0.5 1686 37715 273 lb 248 144 lb vertical farce - 248 377 1446 273 72 136.5 Harizontal farce - D 111 168.6 122 64 16 61 32 1.2 Let select carbon steel shaft :- 400 N/mm2 FoS = T = 0.J SYE O.TX 600 = 166.66 Nforma ts = 2451.98 16/12 √ 136.52 +66.57 149.51 lb in MT = luxuso = uaq. 50lbin de 11 Jim 2 cm 73 d = 0.4789" Mb