Question

The torsional spring shown is clamped at both ends and subjected to torque T=1500 lbf in, at a distance x=1.5 in from one end. The length is L=10 in, and d1=0.5,d2=0.45 in. The material properties are E=29.9,G=11.5 MPsi and ν=0.3. Calculate: ∙ The reaction T1.

(10 points)

Answer 1

T= 1500 lbf en Ito G=1115 Mpsi di=0.5 in I Į da=0.45 in > Liel. 51 - L2= 3.5in - Ti = Reachon turque at polar moment of, Bechon 02 x 07 - 20.54194 ual Torgu in section 0 Ti = Ti angle of twist en section ( 0,- Til = TXI15 - GITI 11.5x 27 x0:54 poter moment of Snertia cf section ® 12- 024 = *0.454 m4 Torque in section ② T2 = T-1500 angle of twist in section ® 02 = T2lz (T, -1500) X8.5 GJ 2 - 11,50 2 x 0.454 but angle or of twist & +02=0 because shaft is fixed from both ends 0, +02-0 Tixlos_ +(T-1500 )x8.5 11.5*32*0.54 11.5**0.454 20

L |-F , - 1500X805 2.5 8 7 0.454 J - 0.54 0.45€ Ti = 1344.348202 lbf in Answer