4.0 Shear Force and Bending Moment

Shear force and bending moment are the internal force and internal couple that arise when a beam is subjected to external loads. A beam is a structural member whose longitudinal dimension is much larger than its x-sectional dimensions and is loaded such that it bends transversely to its longitudinal axis. The distribution of the shear force and bending moment along the axis of the beam is dependent on the type of loading and support on the beam.

Depending on the type of support, a beam can be described as simply-supported (or simple beam) if it is pinned at one end and has a roller at the other end; cantilevered if it is fixed at one end only; continuous if it spans over one or more internal supports in addition to the end supports, or an overhang beam if some part of it extends beyond a pin or roller support.

Simple Beam                                                               Continuous Beam

Cantilever Beam                                                         Overhang Beam

The effect of the external force P is to create internal stresses and strains within the beam. The external force has to be resisted by internal forces and couples that act at every x-section of the beam. The intensities of these internal forces give rise to the stresses and strains within the beam. The external force is finally transferred to the beam supports and is resisted by the reactions at the supports.

Consider the beam below loaded by the force P. Imagine that the beam is cut at a cross-section m-n located at a distance x from the left support and the left-hand part is isolated as a free-body.

The free-body is held in equilibrium by the reaction RAv and RAH at the support and the internal forces Va and Ha and internal couple of moment Ma. The internal force Va, acting parallel to the cross-section is called the shear force while the internal force, Ha acting normal to the cross-section and in the longitudinal direction of the beam is called axial force. The internal couple of moment Ma is called the bending moment.

From equilibrium consideration,

Ha = RAH
Va = RAv
Ma = RAv. x

Note that for most beams, the force P will act in a direction perpendicular to the longitudinal axis of the beam(i.e. tranverse to the beam axis) hence there will be no internal axial force for most beams.

For design purposes, it is important to know the maximum value of Va and Ma and the cross-section where that occurs. This is generally achieved through the evaluation and plotting of the shear force and bending moment at every cross-section of the beam. This plot is known as the shear force and bending moment diagrams. Design of the beam is then based on the maximum values of the shear force and bending moment.

For statically determinate beams, the support reactions can be determined from only equilibrium equations. Having calculated the support reactions, the expressions for the shear force and bending moment are written for each segment or region of the beam. The shear force and bending moment diagrams are then drawn using the expressions determined for the various regions of the beam.

The procedure involved in solving problems concerned with determining the shear force and bending moment distribution on beams can be summarized in the following steps:

A Calculate the support reactions. Remove the supports and draw the freebody diagram of the
beam. Use equilibrium conditions to determine the support reactions.

B Determine the shear force and bending moment equations for each region of beam. Note that each expression is valid only within the region for which it was derived and cannot be applied outside that region.

C Draw shear force (S.F.) and bending moment (B.M) diagrams. Use the equations determined
from B and ensure that each equation is applied only within the region in which it is valid.

1. Establish how many regions you have on the beam.

2. Calculate the support reactions for the entire beam
- Consider equilibrium of the free body diangram for the entire beam

3. Write the shear force and bending moment equations for each region of the beam.
- Consider equilibrium of the left or right hand portions of the beam, obtained by passing a section at any point in the
origin.
- Keep track of your choice of co-ordinate origin and the range of validity of x for each region.

4. Plot the shear force and bending moment diagrams.
- Use the expressions / equations obtained in Step 3 for each region to plot the SF and BM diagrams.
- If the shear force or BM is a contant in a region it means that the SF and BM does not vary within the region but has
the same value at all points in that region.
- If the SF or BM equation in a region is linear or first order function, then SF or BM will vary linearly in the region.
You need to determine the value of SF or BM at two points in the region and connect these with a straight line.
- If the SF or BM equation is a higher order function (quadratic, cubic, etc.), determine if the function has a maximum
or minimum  value in the origin and obtain locations and values for the maximum and minimum SF or BM.  Compute
values of the function of 5 or more of the locations and join with a smooth curve.

SIGN CONVENTION

Example 4.1:

Determine the maximum shear force and bending moment in a simple beam of span L that is loaded by a concentrated force P at midspan. Draw the shear force and bending moment diagrams.

Example 4.2:

Determine the shear force and bending moment distributions in a simply-supported beam of span L that is loaded by two equal concentrated load P as shown.

Example 4.3:

For the cantilever beam of length L subjected to a uniformly distributed load (udl) of intensity q, draw the shear force and bending moment diagrams.

Example 4.4:

For the simply-supported beam of length, L loaded by a udl of intensity q along its entire length, draw the shear force and BM diagram.

Example 4.5: (Problem 4.2-1)

Determine the shear force and bending moment at the middle of the simple beam AB shown below.

Example 4.6: (Problem 4.2-10)

The beam ABC shown in the figure is attached to a pin support at C and a roller support at A. A uniform load of intensity q acts on part AB and a triangular load of maximum intensity 2q acts on part BC. (a) Obtain an expression for the bending moment M in part AB at distance x from support A. (b) From this expression, determine the maximum bending moment Mmax in part AB.

Example 4.7:

Given: Load Intensity Curve, show that:

1. Resultant load is equal to area under the curve.
2. Resultant load or total pressure acts through centroid of area under the curve.
Solution

Example

A simple supported beam carries a vertical load that increases uniformly from zero at the left support to a maximum value of q kN/m at the right support.  Draw the shear force and bending moment diagrams.  Assume the q = 9 kN/m and L=6m

Note:  For purposes of writing equilibrium equations, the entire distributed load associated with the relevant free body diagram is replaced by it's resultant (which is equal to the area occupied by the distributed load) which acts through the centroid of the distributed load.

Example

Given:  Load intensity curve or pressure intensity curve.

Show that:
1.    Resultant load or total pressure is equal to area under the curve.

2.    Resultant load or total pressure acts through centroid of area under the curve.

Relationship Between Load, Shear Force and Bending Moment

Consider an element of a beam that is cut out between two x-sections that are distance dx apart (fig. 4.8a).  It is assumed that a distributional load of intensity q acts on the top surface of an element.  The internal forces on both faces of the element are as shown below:

Note: The distance dx  is infinitesimally small.

From equilibrium forces in the vertical direction, we have that

V - (V + dV) - qdx = 0
dV/dx  = -q                     ---- (1)

This expression shows that the rate of change of V with respect to x is equal to -q, where q is the udl and is positive when acting downward.  From equation 1 above, we have that

This expression shows that the difference VB - VA of the shear forces at two sectors B and A, is equal in magnitude to the resultant of the distributed load between the two sections.  The area of the load-intensity diagram may be treated as either positive or negative, depending on whether q acts downward or upward, respectively.

Note that these equations were derived for the case of uniformly distributed loading and will not apply where a point load is acting on the beam.  Equation 2 cannot be used to find the difference in shear forces between the two points if a ponit load acts on the beam between the two points, since the intensity of load q is undefined for a point load.
Next, let us consider equilibrium by summing moments about an axis through the left hand face of the element and perpendicular to the plane of the figure.

m + qdx (dx/2) + (V + dV)dx - (m+ dm) = 0

Neglecting products oof differentials because they are very small in comparison to the other terms, we obtain that

dm/dx = V                  ----- (3)

This experssison shows that the rate of change of the bending moment m with with respect to x is equal to the shear force.  Note again that this equation applies only in regions where distributed loads or zero load (ie. q=0) act on the beam.  At a point where a concentrated load acts, a sudden change (or discontinuity) in the shear force occurs and the derivative dm/dx is undefined.

Observe that from equation (3)

This implies that the difference in bending moment between any two sections A and B is equal to the area of the shear force diagram between two points.

Note that equation 3 shows that the bending moment is a maximum where the shear force is zero.

Considering a concentrated load P acting on top of the elemental area,

Considering equilibrium of forces in the vertical direction:

S fq = 0;              V-P-(V + Vi) = 0
==> Vi = -P    ---- (5)

This implies that as we move from left to right of the point of application of a concentrated load,  the shear force changes abruptly and decreases by an amount equal to the maginitude of the downward point load.

Considering equiblibrium of moments about the left-face:

-M - P(dx/2) - (V + V1)dx + M + M1 = 0
==>  M1 =  P(dx/2) + Vdx + V1dx = 0

since dx is infinitesimally small.  Hence, we see that the BM does not change as we pass through the point of applications of a concentrated load.  Note that even though the BM does not change dM/dx changes abruptly since dM/dx = V = -P for the elemental area shown above, dM/dx = V on the left face and dM/dx = V + V1 on the right face.  Thus at the point of application of the concentrated load dM/dx changes abruptly and will decrease by an amount equal to the concentrated load.  ie.  V1 = -P

1).    Observe that the shear force to the left of the concentrated load id constant and equal to P/2 while to the right of the concentrated load, the shear force is -P/2.

2).    Observe that the slope of the shear force diagram, dV/dx, on either side if the concentrated load is zero, indicating zero load density, q.

3).    The slope of the BM diagram dM/dx to the left of the concentrated load is PL/4/(L/2) = P/2  while that to the right of the concentrated load is PL/4/(-L/2) = -P/2.  These values are equal to the shear force on these segments of the beam.

4).    At the point of application of the concentrated load, there is an abrupt change in the shear force diagram (equal to p) and a corresponding change in the slope of the BM diagram.

5).    Observe that the difference in the BM between any two points A and B is equal to the area of the shear force diagram between the two points.  Note that this would not be the case if the beam were subjected to a load in the form of a couple.

Example:

Using the relationship between distributed load, shear force and bending moment, obtain the expressions for the shear force and bending moment for the two beams shown below.

a).

b).

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