Mechanics of Materials(ME-294)

Lecture 9:

Shear Force Diagrams(SF Diagrams)and

Bending Moment Diagrams (BM Diagrams)

Introduction

• Loaded beam are subjected to shear forces and bending moments

• and it is usual to draw the loaded beam followed by diagrams that depict the way the shear force and bending moment vary along the length of the beam

• With a shear diagram, we can identify the location and size of the largest shear load in a beam. Therefore, we know the location of the largest shear stress, and we can calculate the value of this stress.▫ Once we know the actual stress in the material, we can compare this

values with the shear strength of the material, and we can know whether the beam will fail in shear.

• Shear diagrams are necessary for drawing bending moment diagrams (“moment diagrams”, for short), which we can use to identify the location and size of bending stresses that develop within beams.▫ We can compare the actual bending stresses with the yield strength of

the material, and we can know whether the beam will fail in bending.

Loads on Beams

You can load a beam with• Point loads: A swimmer standing on

the end of a diving board is an example of a point load: a force applied at a single point on the beam.

• uniformly distributed loads (UDL)

• Non-uniformly distributed loads

If loading is perpendicular (transverse) to its axis so that the rod bends, then the rod called a beam.

Imagine a simply-supported beam with a point load at the mid-span. Cut the beam to the left of the point load, and draw a free-body diagram of the beam segment. In a free body diagram, forces must balance. Therefore, a downward force at the cut edge balances the support reaction RA. We call this shear force V. It is a shear force because the force acts parallel to a surface (the cut edge of the beam)

The forces RA and V are in balance (equal in value; opposite in sign), but our segment wants to spin clockwise about point A. To counteract this tendency to spin, a moment M develops within the beam to prevent this rotation. The moment equals the shear force times its distance from point A. Cut the beam to the right of the point load, and draw the free-body diagram. Since P is larger than RA, force V points upwards.

Beams ReactionsThis point load P could be the weight of an object on the beam, or it could be a load applied by a cable or rod attached to the beam at that point. Typically, the left end of the beam is marked “A” and the right end is marked “B”.

For each loaded beam we draw beam reactions: reaction forces RA and RB for a simply supported beam; reaction force RB and reaction moment MB for a cantilever beam. The symbols for supports indicate the kinds reactions that develop at the support.

Beams ReactionsFor example, support “A” is pinned, like a hinge, so the symbol for the support is a triangle. A pinned support may have vertical and horizontal reaction forces, but the beam at the right has no applied horizontal loads, therefore RAx = 0. In the beam problems that follow, there are no applied horizontal forces, so the horizontal reaction force is zero, and the vertical reaction forces RAy and RBy are abbreviated RA and RB. A roller support allows the beam to move freely horizontally; the symbol is a circle. A roller support has only a vertical reaction force. A beam supported by a pin at one end and a roller at the other end is called a simply-supported beam

A cantilever beam is embedded in a wall, therefore the beam has vertical and horizontal reaction forces as well as a reaction moment. The horizontal reaction force RBx=0 as long as there are no horizontal applied forces, so the vertical reaction force RBy is usually abbreviated RB.

Shear Force :

Beam Rigidly Supported at One End with Concentrated Load at Other This is usually referred to as a cantilever.

Consider any section XX of the cantilever beam shown below, the concentrated load W tends to force the piece of length x downwards relative to the remainder of the beam. i.e. to shear it off as shown in the second part of the diagram.

The net transverse force on XX is called the Shear Force and denoted F.

From one end of the beam to the other there is no change in the load therefore the shear force F is constant over the entire length of the beam and is equal to –W according to the sign convention. This is shown in the diagram which follows:

Bending Moment The moment of a force about a point is the product of the force and the perpendicular distance of the point from its line of action. Such a moment, when applied to a component (shaft, beam or strut) in such a way as to result in bending, is referred to as a Bending Moment and is usually denoted by B.M. The units are Nm ( kNm or MNm).

If you study Mechanics beyond this level you will see that knowledge of bending moments are important when calculating bending stresses. Sign convention We need to agree a sign convention for our bending moments and we shall adopt the one shown below:

for a non-uniformly distributed load, the location of the equivalent load is at the centroid of the distributed load.

The centroid of a triangle is one third of the distance from the wide end of the triangle, so the location of the equivalent load is one third of the distance from the right end of this beam, or two thirds of the distance from the left end.

If a uniformly distributed load is not symmetrical, then we need to convert the distributed load into a point load equivalent to the total load W=wL1 where L1 is the length of the distributed load. The equivalent point load is located at the centroid of the distributed load…the center of the rectangle. Use the equivalent load diagram for calculating the reaction forces.

Reaction Forces

For beams with nonsymmetrical loading, we need two equations from Statics: the sum of the vertical forces equals zero, and the sum of the moments about a point equals zero. The moment about a point is the force acting on an object times the perpendicular distance from the force to the pivot point. Whether the object is a blob or a beam, the moment about point A is M A=P⋅x . You can pick a pivot point at either end of a simply-supported beam.

Most students find it easier to select the left end of the beam, point A. Since moment has a magnitude and a direction (clockwise or counterclockwise), we need to establish a convention for positive and negative moments. We’ll select counterclockwise as positive, symbolized as , and start adding up the moments about point A.

Reaction ForcesThe load acts at a distance x from point A. Think of point A as a hinge point…the load causes the beam to rotate clockwise about point A, so the moment is negative. The reaction force RB acts at a distance L from point A, and causes the beam to rotate counterclockwise about point A, so the moment is positive. The moment about point A is ∑M A=0=−Px+RB L. Now solve for the reaction force RB=Px L . Use the sum of the forces in the vertical direction to calculate the other reaction force. Forces have magnitude and direction; pick upwards as positive, so ∑F y=0=RA−P+RB . Now solve for the reaction force RA=P−RB

Reaction Forces