108k views
5 votes
The 80mm diameter shaft is made A-36 steel. if its subjected to the triangular distributed load, determine the angle of twist of end A with respect C.



User Aseagram
by
8.5k points

1 Answer

0 votes
To determine the angle of twist of end A with respect to C, we can use the formula:

theta = (T L) / (G J)

where theta is the angle of twist, T is the torque, L is the length of the shaft, G is the shear modulus of elasticity, and J is the polar moment of inertia of the shaft.

The torque can be found from the distributed load. The triangular load can be divided into two parts: a rectangle of width 80 mm and height 1 kN/m, and a triangle of base 80 mm and height 2 kN/m. The area of the rectangle is:

A1 = 80 x 1000 = 80000 N.mm

The area of the triangle is:

A2 = 1/2 x 80 x 80 x 2000 = 640000 N.mm

The total torque is:

T = (80000 + 640000) / 2 = 360000 N.mm

The polar moment of inertia of a solid circular shaft is:

J = pi/32 x D^4

where D is the diameter of the shaft. Substituting the given values, we get:

J = pi/32 x 80^4 = 2.015 x 10^8 mm^4

The shear modulus of elasticity for A-36 steel is 79.3 GPa = 79.3 x 10^3 MPa = 79.3 x 10^3 N/mm^2.

Substituting the given values, we get:

theta = (360000 x 1000) / (79.3 x 10^3 x 2.015 x 10^8) = 0.002 radians

Therefore, the angle of twist of end A with respect to C is 0.002 radians.
User Blitzkriegz
by
7.8k points