The aim of this lecture is to help students learn how to write
shear and moment equations for beams under various loads.
Representing shear and moment algebraically is useful,
as it allows the engineer to quickly identify critical stress areas in the beam.
In most cases, it may not be possible
to write only one equation for shear or moment for the entire beam.
Rather it would be necessary to write multiple equations,
one for each beam segment.
This happens when the beam is subjected to one or more concentrated loads.
For example, in this beam, there is a concentrated load at B.
Since the load divides the beam into two segments,
we need to write two equations for shear,
and two equations for moment,
a pair of equations for each segment.
Here, three pairs of equations are needed for representing shear and moment,
since the concentrated loads divide the beam into three segments.
The need for having multiple shear and moment equations also arises,
when the beam is partially subjected to distributed loads.
For example, here, the uniformly distributed load
divides the beam into two segments:
segment AB, which is directly under the load,
and segment BC which is load free.
Therefore, we need one set of shear and moment equations for the left segment of the beam,
and another set of equations for the right segment of the beam.
In this example, the triangular load divides the beam into three segments:
AB, BC, and CD.
Therefore, we need three pairs of shear and moment equations,
a pair for each segment.
Now that we know how it may be necessary to have multiple equations
for representing shear and moment in beams,
let's talk about how we should go about actually formulating such equations.
Given a beam, we always start by calculating its support reactions.
So, step one is to calculate the reaction forces.
Step two is to decide how many shear and moment equations we need,
that is, how many segments the beam is divided into by the loads.
Let's spend a few minutes examining several cases here.
How many pairs of shear and moment equations do we need for this beam?
The load divides the beam into two segments, therefore we need two sets of equations,
one set of equations for the left segment, another set of equations for the right segment.
Here, the load does not divide the beam into multiple segments,
as it is applied at the tip of the beam.
Therefore, we only need one shear equation and one moment equation for the entire beam.
Here, the loads divide the beam into three segments.
Therefore, three pairs of shear and moment equations are needed.
From these examples, it may appear that the required number of shear equations
is equal to the required number of moment equations,
but this is not necessarily true.
We could have loading patterns
that make the required number of moment equations,
more than that of shear equations.
Consider this beam:
It is subjected to a concentrated moment at its midpoint.
This causes a sudden drop, a discontinuity, in the beam's internal bending moment at the point,
consequently, two equations are needed for representing moment in the beam.
We need one equation for the left segment up to the point of discontinuity.
We need another equation for representing moment past the point of discontinuity.
However, the concentrated moment does not cause any abrupt change in shear,
therefore we can represent shear in the beam using a single equation.
So let's summarize what we have so far.
To write shear and moment equations, first we calculate the beam's support reactions.
Second, we determine the number of segments that the beam needs to be divided into
for the purpose of formulating shear and moment equations.
Now let's talk about the third step.
In this step, we are going to cut the beam in each of the identified segments
at some distance, say X, from the left end of the beam.
For example, if this is our beam, we are going to cut it twice:
Once in segment AB,
and once in segment BC.
Note that the distance from the left end of the beam to the cut point is labeled X in both cases.
Then, we draw the free-body diagram for each segment.
The fourth step is to formulate the equilibrium equations for each segment,
then solve them for the unknowns.
Here, the unknowns are the shear and moment values at the cut point.
Let's go through this process using a simple example.
Consider the simply supported beam which is subjected to a concentrated load of 10 kN.
The load is applied four meters away from the left end of the beam.
We wish to formulate the shear and moment equations for the beam.
Step 1: find the support reactions.
This involves drawing the beam's free-body diagram,
formulating the equilibrium equations for the entire beam,
then solving them for the unknowns.
Step 2: determine the required number of shear and moment equations.
Here, the load divides the beam into two segments.
Therefore we need two sets of equations:
We need a shear equation and a moment equation for the left segment of the beam,
and we need another set of equations for the right segment of the beam.
Step three: cut the beam in each segment,
then draw the free-body diagram of the left part of the beam.
The free-body diagram for the beam, when cut in segment AB, looks like this:
When the beam is cut in segment BC, its free-body diagram looks like this:
Step four: Now formulate the equilibrium equations for each free-body diagram,
then solve them for the unknown shear and moment.
For the first free-body diagram, we get:
For the sum of the forces in the Y direction, we get: 6 - V = 0
For the sum of the moments about point O, the cut-point, we get: 6X - M = 0
Solving the first equation for V, we get: V = 6
Solving the second equation for M, we get: M = 6X
Since these equations are written for segment AB, they are valid only for X between 0 and 4.
For the second free-body diagram we can write:
6-10-V=0
6x-10(x-4)-M=0
The first equilibrium equation gives us: V = -4
The second equation gives us: M = 40 - 4X
This pair of equations works only for X between 4 and 10, where segment BC is located.
Let's summarize the results.
Shear in the beam is expressed algebraically using two equations:
The first equation gives us the shear value in the beam when X is between 0 and 4.
The second equation gives us the shear value when X is between 4 and 10.
Similarly, moment in the beam is expressed algebraically using two equations.
The first equation gives us the moment value in the beam when X is between 0 and 4.
The second equation gives us the moment value when X is between 4 and 10.
Here also, moment at X=4 can be calculated using either of the two equations.
Before we end this session, let's examine another case.
Here, half of the beam is subjected to a distributed load,
but the other half is not subjected to any loads.
We wish to write the shear and moment equations for the beam.
Step 1: find the support reactions.
Step two: determine the required number of shear and moment equations.
Just like the previous example, here the load divides the beam into two segments: AB and BC.
Therefore we need two sets of equations.
We need one pair of shear and moment equations for segment AB,
and another pair of equations for segment BC.
Step three: cut the beam in each segment and draw the free-body diagram of the left part of the beam.
Step four: write the equilibrium equations for each free-body diagram and solve them for the unknowns.
For the first free-body diagram, we have:
For the sum of the forces in the Y direction: 375 - 100X - V = 0
For the sum of the moments about the cut-point : 375X - 100X(X/2) - M = 0
Now solve the first equation for V and the second equation for M.
So V equals: 375 - 100X
And M equals: 375X - 50X^2
Both equations are valid for X between 0 and 5.
For the second free-body diagram, we have:
For the sum of the forces in the Y direction: 375 - 100(5) - V = 0
For the sum of the moments about the cut-point: 375X - 100(5)(X - 2.5) - M = 0
Solving the equations for V and M, we get:
V = -125
M = 1250 - 125X
In summary, shear in the beam is represented using these equations:
And moment in the beam is represented using these equations:
We will continue our discussion on shear and moment in beams in the next lecture.