Question

The overhang beam is subjected to the uniform distributed load having intensity of w = 55 kN/m. Determine shear stress developed in the beam.

Answer 1

Concepts and reason

A beam may be defined as a structural element that is used to bear load in transverse direction or horizontal direction.

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

A roller support restricts only the perpendicular force acting on it and allow both rotation and translation in horizontal direction. Therefore, it will give only single reaction in the vertical direction.

A fixed support restricts horizontal force, vertical forces and the moment too. They will not allow the structure to translate or rotate. Therefore, it produces two reaction forces and a reaction bending moment.

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. When their resultant force and couples becomes equal to zero then the body is said to be in equilibrium.

To apply these equilibrium equations, we need to know the known and unknown forces that act on the body.

The load acting on the beams may be a point load or a distributed load.

A distributed loading over a surface may be defined as the load acting over a length or area of the surface instead of a point.

From analysis point of view, to simplify their analysis these can be replaced by a equivalent concentrated point load which can be applied at a specific point. The magnitude of this equivalent load can be determined by geometry method or by integral method.

If the beam is cut at some location located at and the portion is removed as free body, an internal shear force and bending moment must act on the cut surface for equilibrium.

Shear force and bending moment diagrams represent the shear force at each section and bending moment at each point of the beam.

The shear force and moment acting on a surface will manifest themselves as force distribution across the entire area. These force distributions have two components in normal and tangential directions called normal stress and tangential shear stress.

Fundamentals

‎Write the equilibrium equations.

Here, the resultant force is and the resultant moment about any arbitrary point is .

To determine the equivalent force of distributed loads, two methods are there:

1.Geometry method

2.Integral method

In Geometry method, the distribution curve is required for analysis. For this we have to draw the distribution curve. The area under the curve gives the magnitude of the equivalent load and the location of this point is the centroid of this distribution curve. This method is generally preferred when the curve is simple, that is, when the area and the centroid of the curve can be easily calculated.

In integral method, the distributed pressure is required as a function of length variable for example . It is integrated over the area to get the resultant pressure.

The resultant (equivalent) load is given as:

And the point of application of this is given as:

The internal forces at any section in the beam are represented in two different ways:

First representation: When the internal forces at any section in the beam are considered from the left end of the beam, the representation of the forces will be as follows:

Second representation: When the internal forces at any section in the beam are considered from the right end of the beam, the representation of the forces will be as follows:

Here, V is the internal shear force, N is the internal normal force, and M is the internal bending moment.

Basic equations for the calculation of internal forces are as follows:

Moment equilibrium condition,

Force equilibrium condition along vertical direction,

Force equilibrium condition along horizontal direction,

The maximum shear stress in the beams due to vertical loads are given as:

Here, K is a factor that depends upon the section, V is the vertical load and A is the cross-sectional area.

For circular sections:

For rectangular and square sections:

General sign convention for bending moment: The bending moment is considered positive in counter-clockwise direction (also called as sagging bending) and negative in clockwise direction (also called as hogging bending).

General sign convention for vertical force: Forces in the upward direction are considered positive and in the downward direction are considered negative.

General sign convention for horizontal force: Forces towards the right direction are considered positive and towards the left direction are considered negative.

Draw Free body diagram representing forces in all directions.

Apply the equilibrium of forces in the horizontal direction.

Here, is the horizontal reaction force at point A.

Take moment about point A.

Here, is the vertical reaction force at point B.

Substitute for w.

Apply the equilibrium of forces in the vertical direction.

Here, is the vertical reaction force at point A.

Substitute for w and for .

Consider a section at a distance of x from point A as shown below.

Calculate the internal forces in this section.

Apply the equilibrium of forces in the vertical direction.

Here, V is the shear force in the section.

Substitute for .

Calculate the shear force at and .

Substitute for x.

Substitute for x.

Consider a section at a distance of x from point A as shown below.

Calculate the internal forces in this section.

Apply the equilibrium of forces in the vertical direction.

Here, V is the shear force in the section.

Substitute for and for .

Calculate the shear force at and .

Use the values of shear force calculated at various points to plot shear force diagram as shown below.

From the shear force diagram the maximum shear force in the beam is at B and equals to .

Calculate the cross-sectional area of the beam.

The maximum shear stress in the beam of rectangular cross-section is:

Substitute for A and for .

Ans:

The maximum shear stress developed in the beam is .