Structural analysis: Shear and Moment in Beams
A beam is a properly supported structural member,
capable of withstanding loads that are applied in a direction perpendicular to its central axis.
In this session, we are going to discuss the type and nature of the forces internal to a beam.
Consider a cantilever beam subjected to a concentrated load at its free end.
Suppose we conceptually cut the beam at its midpoint and choose to examine its right segment.
Since the conditions of equilibrium must be satisfied for the segment, then we can write:
Sum of the forces in the X direction must be zero.
Sum of the forces in the Y direction must be zero.
And sum of the moments about, say point C, must be zero.
The first equation here is automatically satisfied since there are no applied loads in the X direction.
But how do we satisfy the second and third equations?
A downward force P is applied to the right end of the segment.
To satisfy the second equation, therefore, there must be an upward force at point C,
equal in magnitude to P, in order for the sum of the forces in the Y direction to be zero.
This upward force labeled V on the diagram is called the shear force.
It is one of the internal forces in the beam.
To satisfy the third equilibrium equation,
it is important to recognize that P creates a bending moment about point C.
The moment has a magnitude of P(L)/2, and a clockwise direction.
For the sum of the bending moments to be zero,
there must be a bending moment present at C.
The magnitude of this moment must be equal to the magnitude of the moment that P creates about C.
The two moments however must act in opposite directions in order for their sum to be zero.
Let's denote the bending moment at point C as M.
Here are the two equilibrium equations in terms of V and M:
Solving for the two internal forces, we get:
V = P, and M = -P(L)/2
Here, we can conclude that when a beam is subjected to a transversal load,
a load that is perpendicular to the beam's central axis,
then two internal forces develop in the beam: A shear force and a bending moment.
Generally speaking, these forces do not remain constant in the beam.
Therefore, their values need to be determined at critical locations throughout the beam.
In order to calculate the shear force and bending moment at a point in the beam,
it is necessary to conceptually cut the beam at that point,
and draw the free-body diagram for each beam's segment.
Here, it is important to recognize that any time we cut a beam,
we need to show the beam's internal forces at the cut point.
So if we cut this beam at its midpoint, point C,
we end up with two beam segments: the left segment and the right segment.
The internal forces at point C in the left segment of the beam are shown like this:
In the right segment, these forces are shown like this:
Note that at C, the shear force in the left segment
has the same magnitude as the shear force in the right segment.
We are using the same variable V, to denote the magnitude of both force vectors.
However, they do have opposing directions.
One is drawn upward, the other one is drawn downward.
This is necessary as the algebraic sum of the two vectors must be zero.
The same is true for the bending moment at C.
The moment in the left segment is drawn in the counterclockwise direction.
Whereas for the right segment, the moment is drawn in the clockwise direction.
Since they both have the same magnitude, their algebraic sum, therefore, is zero.
Now, let's see how we can actually calculate shear and moment at a point in a beam.
The procedure is rather straightforward:
1. Calculate the beam's support reactions.
2. Conceptually cut the beam at the point of interest.
3. Show the internal forces at the cut point and draw each segment's free-body diagram.
4. Formulate the equilibrium equations for either the left or the right segment of the beam
5. Solve the equations for the unknown shear force and bending moment.
Let's illustrate this process using an example.
Suppose we have a simply supported beam subjected to two concentrated loads.
We wish to determine shear and moment at points C and D.
Let's start by calculating the support reactions.
Here is the beam's free-body diagram:
Here are the equilibrium equations:
From the last equation, we can calculate By.
By = 3.5
From the second equation we get: Ay = 2.5
And the first equation gives us: Ax = 0
Here is the results shown graphically on the beam:
Now, to determine the shear and moment at point C,
We conceptually cut the beam at C,
and draw the free body-diagram for either the left segment or the right segment,
whichever is more convenient.
Let's pick the left segment. Here is the free body diagram:
Note that I did not draw the internal axial force at C, since I already know there is no axial force in play here.
How do I know that?
Because, there are no applied forces in the X direction.
The relevant equilibrium equations, therefore, are:
From the first equation we get: Vc = 2.5
From the second equation we get: Mc = 10
Here is the results shown on the beam:
To determine shear and moment at point D, let's redraw the beam.
Now, cut the beam at point D and draw the free-body diagram for either the left or the right segment.
Let's pick the right segment this time.
Here is the free-body diagram:
Again, since there's no axial force present on the beam,
I'm going to omit drawing the internal axial force at point D,
then, the relevant equilibrium equations are:
From the first equation, we get: Vd = -1.5
From the second equation, we get: Md = 17
Here are the results shown graphically on the beam:
Let's consider a more interesting problem.
Suppose we have a cantilever beam subjected to a uniformly distributed load.
Here we wish to come up with a couple of algebraic equations for calculating shear and moment,
at any point in the beam.
That is, I want to write an equation in the form V of X,
where X is the position of an arbitrary point in the beam.
When I substitute a value for X in the equation, it gives me the value of shear at that point.
I also want to write a similar equation for bending moment in the beam,
which can be used to calculate internal bending moment at a desired point.
In other words, I'm interested in representing shear and moment in the beam algebraically.
To do this, let's find the support reactions.
Here is the free-body diagram of the beam:
Here are the equilibrium equations:
Solving them for the unknowns, we get:
Ax = 0, Ay equals 5000.
M equals -25000.
Now let's cut the beam at an arbitrary point, say point P.
Draw the free-body diagram of the left segment of the beam.
Since we do not know the length of the segment, I'm going to label it as X.
That is, the distance from the left end of the beam to the point is X.
Now write the equilibrium equations in terms of X.
We have: sum of the forces in the Y direction must be zero.
That is, 5000 - 500X - V = 0
Here, 500 X is the area of the distributed load applied to the segment.
The rectangular load has a height of 500, and a base of X.
Therefore, its area is 500 times X.
Sum of the moments about point A must be 0.
Therefore, V X - M + 500 X (X/2) - 25000 = 0
Just like the first equation, here,
500 times X is the area of the rectangle representing the distributed load.
X over 2 is the distance from the center of the area to point A.
Hence, the product gives the moment of the distributed load about point A.
Solving these equations for V and M in terms of X, we get:
V = 5000 - 500 X
M = 5000 X - 250 X^2 - 25000
So, these are our shear and moment equations for the beam.
We can use them to calculate shear and moment at any point in the beam.
For example, what is shear and moment at the left end of the beam, where X equals zero?
V of 0 = 5000 - 0, or 5000.
M of 0 = 0 - 0 - 25000, or -25000.
Notice that these are the support reactions.
The equations correctly give us the support reactions for shear and moment when X equals zero.
What is the shear and moment at the midpoint of the beam where X equals five?
V of 5 = 5000 - 500 (5), or 2500.
M of 5 = 5000 (5) - 250 (5)(5) - 25000, or -6250.
Let's do one more point.
What is shear and moment at the free end of the beam, where X equals ten?
V of 10 = 5000 - 500 (10), or 0.
M of 10 = 5000 (10) - 250 (10)(10) - 25000, or zero.
The equations correctly indicate that the shear and moment at the free end of the beam are zero.
For design purposes, it is generally useful to graph these equations.
We will discuss the graphical representation of shear and moment in beams in the next session.