Question

The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft.

Determine its smallest diameter d if the allowable bending stress is ? allow = 130 MPa.
(Figure 1)

Answer 1

Concepts and reason

Support reactions:

If there is a force acting by a support due to which, the translational movement of the object is restricted or prevented, then the force developed is called as a support reaction.

Force Moment:

The magnitude of the moment can be determined as the product of force and the perpendicular distance to the force from the reference point.

Apply equilibrium condition for moments about A and calculate the reaction at B. Then calculate the reaction at A by applying the force law of equilibrium along y-direction. Then calculate the shear force and draw the shear force diagram. Then calculate the maximum bending moment from the bending moment diagram. Finally calculate the smallest diameter of the shaft using the values of the maximum bending moment, moment of inertia, and the allowable stress in the formula of allowable stress.

Fundamentals

Distributed load

A load on a member which is evenly distributed over the entire length of the member is distributed load. It is expressed in the units of weight per length.

Distributed loads acts on a beam or member as shown in Figure (1):

The resultant force of a distributed load is the area of the loading diagram. The equivalent force for the rectangular loading in Figure 1(a) is,

$R=WL $

Here, the distributed load is and the length of the beam is L.

Write the equivalent force for triangular loading in Figure 1(b).

The formula to calculate the moment is as follows:

Here, force is and perpendicular distance is

Internal forces:

Consider the beam as shown in Figure (2). To calculate the internal forces at point B in Figure 2(a), the section a-a have to be considered and solving any segment in Figure 2(b) gives the internal forces.

In the Figure 2(b), the force acting perpendicular to the cross section is normal force and the force acting parallel to the cross section is shear force . Also the couple moment at B is the bending moment at B.

Equilibrium conditions:

Write the equilibrium force equation in x direction.

Here, the total force acting in x direction is .

Write the equilibrium force equation in y direction.

Here, the total force acting in y direction is .

Write the equilibrium moment equation.

Here, the total moment acting is .

The relation for Moment of inertia of circular section is given follows:

Here, diameter is .

The formula to calculate the allowable stress of the shaft is given as follows:

Here, maximum bending moment acting on the shaft is and vertical distance from the upper part of section to its centroid is c.

General sign conventions of moment:

Moment is considered as positive in counter clockwise direction and moment is considered as negative in clockwise direction.

The free body diagram of the beam is drawn as shown in Figure (3).

Here, Reaction at A is and reaction at B is .

Convert the uniformly distributed load into a point load as shown in Figure (3).

Convert the uniformly varying load into a point load as shown in Figure (3).

Take the moment about point A.

$0=W3 $

Apply the force law of equilibrium along y-direction.

Substitute 28.5 kN for .

Draw the forces acting at point A as shown in Figure (4).

Here, normal force at point is , shear force at point is , and moment at point is .

Calculate the shear force at point A.

$V=RA $

Substitute for .

Draw the forces acting between the point A and point B as shown in Figure (5).

Here, shear force at just to the left of point B is .

Calculate the shear force between point A and point.

$V-B=R-W $

Substitute for and for Wl.

Draw the forces acting in the section AB as shown in Figure (6).

Here, normal force at point is , shear force just to the right of point B is and moment at point is .

Calculate the shear force between point B and point C

Substitute for , for , and for Wl

Draw the forces acting at point C as shown in Figure (7).

Here, normal force at point is , shear force at point C is and moment at point is .

Calculate the shear force at point A from Figure (7).

The shear force diagram for the beam is shown as in Figure (8).

Shear force is zero at a distance of x from A from Figure (8).

Substitute for V.

Form the bending moment relation from the diagram in terms of x in order to calculate the maximum bending moment at point A and point B.

…… (1)

Calculate the bending moment from Equation (1) when x is zero i.e point A.

Substitute 0 for x

Calculate the bending moment from Equation (1) when x is 1.375 m i.e .

Substitute for x.

Calculate the bending moment from Equation (1) when x is 3 m i.e point B.

Substitute for x.

From Figure (7), calculate the bending moment at point C.

$Mc=0 $.

The bending moment diagram for the beam is shown as in Figure (9).

From Figure (10), the maximum bending moment occurring on the beam is .

Calculate the smallest diameter of the shaft.

Substitute for , for , and for c.

Ans:

The smallest diameter of the shaft is .