Question

Saved Question 20 1 points 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 and Pare (0.8 radian, 1500 in-lb) and (11.3 radian, 4000 in-Ib) respectively. The power-law curve fitting relation for the data above the yield point is given by T-32980021, where T is in in-lb and is in radian Torque T-in. Ib. Q TWIST-6 (rad) The shear stress at point P should be closest to: O 18000 psi 40000 psi 9000 psi 54000 psi Question 20 of 22 Moving to another question will save this response.

Answer 1

Solution Q-20 To find use torsion С ४ Shear stress at pointa p equation T a shear stress Lor in radius of bar T Torque it of twist Gro L o Angle G7 shear modulus L> Length of bar = Tp = Toxa J J> poler moment of incotia d x3 Tpx 2x ax at ग 32 x dt uz 2 G Cp 16 TP T a3 a diameter of i bar d= 0.75 in Tp= 16x4000 Tx(0.753 Tp - Torque at pointp Tp= 4000 in- eb Cp - 48288.788 Psi Shear stress at point at point P is 488 48288.788 Psi CO From option's is correct answer option 'i 54000 Psi

0-12 5000 Psi na given 02= 5000 Psi oy and Cay not present Principle stress 5... ontoy xay) + tay 01, 2 = σα 2 뜰 뜰 + o 5 = 5000 Psi 02 = 5 ox oll 2 2 = = 0 The value value of principle stress or is o i the correct option above nonc of the