Question

AISI 1018 HR

Section #2 (80p) 035 CD 1018 HR Q1. The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of AISI 1035 CD steel. Using both conservative failure theory and distortion energy theory with a design factor of 2.5, determine the minimum shaft diameter to avoid yielding. dete n-2.5 127lbt 8y=61& psi 3 22166 300 lbf 50 lbf 392 lb D por 8 8

Answer 1

- 300 + 59 7392 ksis & sin in consider forces in vertical direction EMAFO Rox22 4 350X8 = 0 1. RD=-27.3 lbf RA+RO-350 RA = 350 – Rp --350 +227.3 =-172.7462

طلهم M BV=RAX8 A Solb TROV Part sin I --172.7x8 --1382 lb-in sin MAVE RB X6 . = -127-3x6 MA =-764 lb-in 1382 YS1lbf to TRBH RAHL 8 m Bin to in the consider forces in pai harisontal direction EMAH=O ROH X22 – 451x16=0 - Rank 8 in [ROH = 328 lbf) RAH = 451-328 = 1231bf MBH - 123x8 =984 lb-in

Пон - Кон х 6 - 326х6 Мен = 1966 16- Mc = √ Men't Mc - / 446,4ч - нс — 2 ш 16 - ) В = І Ізва навч = 646 lb. . - 32 м - 12х2)11 'd3 та акая 21S09 ча

T= (1,-12) r r = diameter or radius of pulley B24 is =( 300-8034 Ti=300 lb Tz =sodb = 1000 16-imi T = 161 – 16X1000 - 5093 - Tid Tod do for AISI 1035 steel Sy + = 66 700 PSI Sot tyl907 77 F2 2 15093 66700 - 21502.43 + 1/ 21502.4714 , 2d3 | 2d374

33 350 = + (22647.7) 3335 j? = 0.67909 d=0.8789 in d=22.3mm (d=23mm distortion energy they :- (sut) = [orto, 2010 a = a + Eur 5. november 9 = 22647.7 d3 or= a - Elita 3 215092 - pas ng isang 2/3 =-1145

Swar): कम (suneey- a nd- 11 2222500 = 488 297723.74 16 d' = 0 4390 d = 0.8718 d - 22 14mm d -23Mm