A and B are done. please only anser c,d and e (note there is a typo, its asking

Question

A and B are done. please only anser c,d and e (note there is a typo, its asking for maximum shear strain of titanium shaft, not aluminum)

also note for C angle of twist free end=sum of all angle of twist (thats what my notes say but im not sure how to tell whether angle of twist is negative or positive which is why i got stuck in C)

data you may want

max shear stress steel=11.13ksi

max shear stress titanium=5.73ksi

G steel-11000ksi G titanium=6400ksi   

For the member with torsion loads and section properties shown below, determine: 2 in outer diameter steel (A-36) tube with 3/8 in thick walls 1.25 in diameter solid Titanium shaft T-26 kip-in T = 9 kip-in T = 2.2 kip-in 3 ft 2.5 ft 2 ft Steel tube section a. b. c. d. e. The maximum shear stress in the steel tube. The maximum shear stress in the solid aluminum shaft. The maximum angle of twist at the free end. The maximum shear strain in the steel tube. The maximum shear strain in the aluminum shaft.

Explanation / Answer

Let us determine the polar moment of inertia of steel shaft and titanium shaft

Outside diamater of steel shaft = 2in

Inside diameter of steel shaft = 2-(2*3/8)=1.25in

Polar moment of inertia of steel shaft=J1 =pi*(24-1.254)/32 =1.331 in4

Diameter of titanium solid shaft = 1.25in

Polar moment of inertia of titanium shaft =J2= pi*1.254/32=0.240 in4

G steel = G1=11000ksi

G titanium = G2 = 6400 ksi

c)Angle of twist of steel shaft,with respect to fixed end, at point of application of 26kip-in tosion = (14.8*3*12)/(11000*1.331) = 0.0364 rad(anticlockwise when looked from right towards left)

Angle of twist of junction of the two shafts with respect to the point of application of 26 kip-in torsion = (11.2*2.5*12)/(11000*1.331) = 0.0229 rad(clockwise when looked from right towards left)

Angle of twist of free end with respect to the junction of the shafts = (2.2*2*12)/(6400*0.24) = 0.0344rad(clockwise when looked from right towards left)

Therefore, total rotation of free end with respect to fixed end = 0.0364-0.0229-0.0344 = -0.0209rad=0.0209(clockwise when looked from right towards left)

d) Maximum shear strain is given by Tr/GJ = (14.8*1)/(11000*1.331)=0.001

e)Maxiumu shear strain in titanium shaft = Tr/GJ = (2.2*0.625)/(6400*0.24) = 8.95*10-4

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