The aim of this lecture is to help students learn how to come up with
shear and moment equations for beams subjected to various loads.
Representing shear and moment algebraically is very useful,
as it allows the structural engineer quickly identify critical stress areas in the beam.
In most cases, it may not be possible
to write down only one equation for shear or moment for the entire beam.
Rather it would be necessary to write multiple equations,
one for each beam's segment.
This happens when the beam is subjected to one or more concentrated loads.
For example, here 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.
A 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 support reactions.
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 always 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:
Here the beam 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 one equation only.
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 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,
and solve them for the unknowns.
Here, the unknowns are the shear and moment at the cut point.
Let's go through this process using a simple example.
Consider the simply supported beam 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 and moment equation for the left segment of the beam,
and 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.
Here is the free-body diagram when the beam is cut in segment AB.
Here is the free-body diagram when the beam is cut in segment BC.
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:
Sum of the forces in the Y direction equals: 6 - V = 0
Sum of the moments about point O, the cut point, equals: 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 have:
Sum of the forces in the Y direction equals: 6 - 10 - V = 0
Sum of the moments about the cut point, point O, equals: 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 work 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.
Moment at X equals 4 can be calculated using either of the moment equations.
Before we end this session let's examine another case.
Here, half of the beam is subjected to a distributed load,
the other half is not subjected to any loads.
We wish to write the shear 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.
A 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:
Sum of the forces in the Y direction equals: 375 - 100X - V = 0
Sum of the moments about the cut point equals: 375X - 100X(X/2) - M = 0
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:
Sum of the forces in the Y direction equals: 375 - 100(5) - V = 0
Sum of the moments about the cut point equals: 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: