Question

The following is the torque versus angle of twist diagram for a 0.75 inch diameter steel bar tested in torsion. The gage length of the bar is 3 inches. The angle of rotation and torque at points Q and P are (0.8 radian, 1500 in-lb) and (11.3 radian, 4000 in-lb) respectively. The power-law curve fitting relation for the data above the yield point is given by T = 329880.21, where T is in in-lb and g is in radian. Torque T-in. ib. TWIST-O (rad) The engineering shear strain at point P should be closest to: o 2 0.1 O 1.5 0.5

Answer 1

Criven Gange length 1- 3 in d = 0.75 in Gauge diameter Gauge Padius = n = 0.375 in . At point a Op = 0.8 rad To 1500 in-16 Tu J Gy® d where To = torqu for thal- seebon J = Polas moment of herbia G = Shear modulus G (0.8) (075) 1500 ཁེད Gy = 181082-9572 lb in2 At point P. torque = Tp2-4000 tb-in Shear sy, Engeweng a shroun will compopond to this torque only not the torque obtained by powry law.

to this un asels brosus bouz sponding 191 - 2 Td3 16 x 4000 2 1 (0.75)3 T: 48288.78 lb in Enggi gering Y shear strain shrain = E enegg- ". 2G Engj 48288.78 2 (181082-9572) E ongg = 0.133 ophon. Ans.

Please forgive me for any calculation mistake and do comment for any short of queries.