Question

(Torsion: Angle of Twist Problem 5.49 < 3 of 3 > The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown where T = 200 N.m. 1200 Nm 400 Nm B 500 mm The shaft has a diameter of 30 mm. Part A Determine the angle of twist of end B with respect to end A. Express your answer to three significant figures and include appropriate units. HA O 2 ? $B/A = Value Units Submit Request Answer

Answer 1

Properties of A 992 steel :-

400Nm 800Nm 400NM D D K 6oomm OOONM Free body diagrams of segments of shaft A 500mm T-200Nm 200NM A 400mm Figl

Shear Modulus of Elasticity = G = C = 75 GPa = 75 x 103 N/mm2

Diameter of shaft = d = 30 mm

Polar moment of inertia = J = IP = (pi) (d4)/ (32)

J = IP = 79521.564 mm4

From theory of Torsion, we have

(T / IP) = ( q / r) = (G$\Theta$/L)

where T is Torque moment

IP is Polar moment of inertia

q is shear stress

G is Shear modulus of elasticity

$\Theta$ is angle of twist in radians

L is length of shaft

Figure 1 shows that free body diagrams of each segments of shaft.

Segments AC and CD will have angle of twist in same direction but segment DB will have angle of twist in direction opposite to that of segments AC and CD.

consider first and third term of above equation, we get

(( T Nmm) /(79521.567 mm4)) = (75 x 103 N/mm2) x $\Theta$/L

This implies, $\Theta$ = (TL) / (5964117303 )

For segment, AC,

We have, TAB = 200 x 103 Nmm

LAB = 400 mm

$\Theta$AB = (200 x 103 x 400)/(5964117303)=0.013414 rad

For segment CD, T = 800 x 103Nmm, L = 500 mm

$\Theta$CD = (800 x 103 x 500)/ (5964117303)=0.067068 rad

For segment DB, T = 400 x 103 Nmm, L = 600 mm

$\Theta$DB =(400 x 103 x 600)/(5964117303)= 0.040241 rad

There for, angle of twist of end B with respect to end A = $\Theta$B / A = $\Theta$AC + $\Theta$CD - $\Theta$DB = 0.013414 + 0.067068 - 0.040241 = 0.040241 radians