Question

How many N*m of torsional strain energy accumulates at 1kN*m as shown in the figure?

(However, the vertical modulus of elasticity is 100 GPa, and the Poisson's ratio is 0.25.)

„25old.) ET= те L=100cm T=1 kN-m d=4cm 1 0.5 25 (3) 50 4 500

you gotta pick one answer.

Answer 1

Given T = 1 kNm = 1000 Nm L = 100 cm = 1m d = 4 cm = 0.04 m E = 100 GPa = 100 x 10°N/m2 v = 0.25 Modulus of rigidity for the bar can be computed as G= E 2(1 + v) = 100 x 109 2(1 +0.25) 40 x 109N/m2 Torsional strain energy accumulated on the bar can be calculated as 1 Te II ET But we know T GO L J Therefore II TL G] Substituting this ET T2L 2G) Or Ey = T2L 16T2L 2G (Ted4/32) G(T1d4)

Substituting the known values ET 16 x 10002 x 1 40 x 10' X 7 X (0.04) = 49.74 N.m - 50 N.m Correct option is option (3) 50 N.m