Question

Member CB has a square cross section of 30 mm on each side. (Figure 1).

Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section b–b to exceed σ=18 MPa and τ=20 MPa, respectively.

Answer 1

Concepts and reason

Shear stress is defined as the stress component which acts in the plane of the sectional area.

Normal stress:

The normal stress is the normal force acting per unit normal area of the body.

Transverse shear stress:

Due to transverse loads, loads that act perpendicular to the beam, shear stress is induced in the cross section of the beam. This shear stress is called as transverse shear stress. It is along the cross section of the beam. It is parabolic in nature with maximum at neutral axis and zero at outer fibers.

 

For a body or system to be in mechanical equilibrium the net forces acting on the body must be zero and the net moment must also be zero.

Magnitude of moment can be calculated by multiplying the force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.

To apply these equilibrium equations, we need to know the known and unknown forces that act on the body. When all the supports are removed by replacing them with forces that prevents the translation of body in a given direction that diagram is called free body diagram.

A pin support restricts both horizontal and vertical forces but not a moment. They will allow the structure to rotate, but not to translate in any direction. Therefore, it produces two reaction forces, one in vertical and other on horizontal direction.

Fundamentals

The equilibrium equations are:

1.

Net force on the body must be zero:

2.

And net moment on the body must be zero:

For simplicity, take the components of forces in vertical and horizontal direction and apply the equilibrium of forces in vertical and horizontal direction.

And,

A normal force is defined as a force that acts perpendicular to the surface or a body and shear force is the force that acts parallel to the surface, that is; the force which tends to shear the surface.

The normal stress developed in a body under the action of a normal force is:

Here, P is the normal force and A is the cross-sectional area of the body.

The shear stress developed in a body under the action of a shear force is:

Here, V is the shear force acting on the body.

Draw the free-body diagram of the link AB as shown below:

Calculate the angle as shown in the figure.

Take moment about point A.

Here, is the force in member BC and w is the intensity of uniformly distributed load.

Substitute for .

Draw the free-body diagram of the link BC isolate by a section b-b as shown below:

Calculate the shear force and normal force at section b-b.

Apply the equilibrium of forces in horizontal direction.

Here, is the shear force at section b-b.

Substitute for and for .

Apply the equilibrium of forces in the vertical direction.

Here, is the normal force at section b-b.

Substitute for and for .

Calculate the cross-sectional area at section b-b.

Here, A is the normal square cross-sectional area of the bar BC.

Substitute for A and for .

Consider the formula for maximum normal stress at section b-b.

Substitute for , for and for .

…… (1)

Calculate the average maximum shear stress at section b-b.

$==(442) $

Substitute for , for and for .

…… (2)

Compare equations (1) and (2), for fulfilling maximum normal stress criteria the maximum value of w should be and for fulfilling maximum average shear stress criteria the maximum value of w should be . This mean that when the value of load w is , the member will fail in second condition; that is, in shear criteria.

Hence, the maximum value of load intensity w should be minimum of the above two values; that is, .

Ans:

The largest intensity of uniform load w is .