Question

help

The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.

Answer 1

Concepts and reason

The yoke and rod connection is held using a pin at A. the load acting on the two rods are equal and the shear force acting on the pin is equally distributed at both sides of the pin. Hence, the average load acting on the pin is half of the load applied at any one of the rod.

First, the average normal stress in 40 mm diameter is obtained by using the applying load and diameter of the 40 mm rod. Then, the average normal stress in 30 mm diameter is obtained by using the load and diameter of the 30 mm rod. Finally, the average shear stress for pin A is obtained by using the shear force and diameter of the 25 mm rod.

Fundamentals

The expression to calculate average normal stress is given as follows:

Here, load acting is , and area of the cross section is .

The expression to calculate shear stress is given as follows:

Here, shear force is V.

The expression to calculate the area of the circular cross section is as follows:

Here, diameter of the circular cross section is d.

Find the area of the 40 mm diameter rod .

Substitute 40 mm for .

Find the average normal stress in 40 mm diameter rod.

Substitute 5 kN for P and for .

Find the area of the 30 mm diameter rod .

Substitute 30 mm for .

Find the average normal stress in 30 mm diameter rod.

Substitute 5 kN for P and for .

Find the area of the 25 mm diameter rod .

Substitute 25 mm for .

Find the average shear stress for pin A.

Substitute 2.5 kN for V and for .

Ans:

The average normal stress in 40 mm diameter rod is 3.9808 MPa.