Shear force and bending moment diagrams are powerful graphical methods that every mechanical
and civil engineer should know how to use to analyse a beam under loading.
In this video I’ll explain exactly how to master these diagrams, and we will see how
they can be used to understand how a beam is loaded.
I want to start by explaining what shear forces and bending moments actually are.
When a beam is loaded, internal forces develop within it to maintain equilibrium.
These internal forces have two components.
We have shear forces, oriented in the vertical direction.
And we also have normal forces, which are oriented along the axis of the beam.
If the beam is sagging, the top of the beam will get shorter, and so the normal forces
in the top of the beam will be compressive.
The bottom of the beam will get longer, and so the normal forces in the bottom of the
beam will be tensile.
Each of the tensile normal forces has a corresponding compressive force which is equal in magnitude
but opposite in direction.
As such these forces don’t produce a net normal force, but they do produce a moment.
This means that we can conveniently represent the internal forces acting on the beam cross-section
using just two resultants - one shear force, which is a resultant of the vertical internal
forces, and one bending moment, which is a resultant of the normal internal forces.
This is a very common way of representing the internal forces within a beam.
Drawing the shear force and bending moment diagrams is just figuring out what these internal
forces are at each location along the beam.
These resultant shear forces and bending moments will depend on the loads acting on the beam,
and the way in which the beam is supported.
Beams can be loaded in a number of ways, the most common being concentrated forces,
distributed forces, and concentrated moments.
Beams can also be supported in a number of different ways.
They can have pinned supports, roller supports, or be fully fixed, which each restrain the
beam in different ways.
Pinned supports prevent vertical and horizontal displacements but allow rotation.
Roller supports prevent vertical displacement but allow horizontal displacement and rotation.
Fixed supports prevent all displacements and rotation.
If a certain degree of freedom is restrained at a support, we will have a corresponding
reaction force or reaction moment at that location.
For example rotations are permitted for a pinned support, so there is no reaction moment,
but displacements in the vertical and horizontal directions are prevented, so we will have
horizontal and vertical reaction forces.
So how do you determine the shear forces and bending moments within a beam?
There are three main steps we need to follow.
First we draw a free body diagram of the beam.
This shows all of the applied and reaction loads acting on the beam.
The next step is to calculate the magnitude of the reaction forces and reaction moments
at all of the beam supports.
We do this using the concept of equilibrium.
To maintain equilibrium, all of the forces in the vertical and horizontal directions
should cancel each other out.
Similarly, all of the moments acting at every point along the beam should cancel each other out.
This gives us a set of simple equations we can solve to calculate the reaction forces
and moments.
If we can calculate all of the reaction loads using the three equilibrium equations, the
beam is said to be statically determinate.
For some beam configurations, like this one shown here, we won’t be able to calculate
all of the reaction loads because we have too many unknowns and not enough equilibrium
equations.
In this case the beam is said to be statically indeterminate.
This beam has 4 reaction forces, but we only have 3 equilibrium equations.
To solve this beam we would need to use slightly more complicated methods and consider boundary
conditions.
In this video I will only cover statically determinate cases, where we can use the equilibrium
equations to calculate all of the reaction loads.
Once we have calculated all of the reaction loads, the third and final step is to figure
out the internal shear forces and bending moments at every location along the beam.
To do this we will use the concept of equilibrium again.
If we cut our beam at any location, the internal forces and moments need to cancel out the
external forces and moments so that equilibrium is maintained.
This allows us to easily calculate the shear force and bending moment at each location
along the beam.
All we need to do is start from one side of the beam, and move the location of the cut
along the beam, calculating the shear forces and bending moments as we go.
Now is a good time to define the sign convention we will be using.
Applied forces will be positive if they are acting in the downwards direction.
For shear forces and bending moments, the positive sign convention will be as shown here.
If the beam is on the left side of our cut, shear forces pointing downwards will be positive.
If the beam is on the right side of our cut, shear forces pointing upwards will be positive.
Positive bending moments will be those that put the lower section of the beam into tension.
Another way to think about it is that bending moments which cause sagging of the beam are
positive, and those that cause hogging of the beam are negative.
Let's take a look at an example of a beam with pinned and roller supports, loaded by
two concentrated forces.
First we draw the free body diagram.
We can then use the equilibrium equations to determine the unknown reaction forces at
Point A and Point B.
The sum of the forces in the vertical direction is equal to zero, so R-A plus R-B is equal
to 15 plus 6.
Because H-A is the only horizontal force, it must be equal to zero.
We also know that the sum of the moments about any point along our beam must be zero.
Let's consider the moments about Point B. That gives us this equation, which we can
solve to determine that R-A is equal to 12.
By substituting R-A into the previous equation we can deduce that R-B is equal to 9.
Now that all of the external loads acting on the beam are defined, we can draw the shear
force and bending moment diagrams.
We will start from the left hand side of the beam.
Let's draw the free body diagram for a location immediately to the right of the 12 kN
reaction force.
To maintain equilibrium, the shear force must be equal to the reaction force.
We can draw this on our shear force diagram.
The shear force will be constant until we reach the next applied force.
The bending moment must be equal to the 12 kN reaction force multiplied by the
distance X to the reaction force.
This gives us the equation for a straight line, which we can draw on our
bending moment diagram.
We then repeat the process by moving the location of our cut further to the right.
This time we place the cut immediately after the 15 kN force, and we draw the free
body diagram again to determine the shear force and the bending moment.
We repeat this process until we have covered the full length of the beam.
We end up with the complete shear force and bending moment diagrams for the beam.
That example was a fairly simple one.
For cases with more complex loading, drawing the shear force and bending moment diagrams
can be more difficult.
There are relationships between the applied loads, shear forces and bending moments which
will help us better understand what our diagrams should look like.
Let's consider a beam loaded by an arbitrary distributed force.
We can zoom in to look at an infinitesimally small segment of the beam with a width equal
to D-X, and draw the free body diagram.
Over such a short section of the beam the distributed force can be assumed to be uniform,
and we can replace it with an equivalent concentrated force.
By applying the equilibrium equations to this free body diagram, it is possible to demonstrate
that the following relationships exist between the applied distributed force, the shear force
graph and the bending moment graph.
The quantity D-V over D-X is the slope of the shear force curve, and at a given point
along the beam it is equal to minus the distributed force.
Similarly D-M over D-X is the slope of the bending moment curve, and at a given point
it is equal to the shear force.
If we integrate the first equation, we can show that the change in shear force between
two points is equal to the area under the loading diagram between those two points.
And if we integrate the second equation we can show that the change in bending moment
between two points is equal to the area under the shear force curve.
This is really useful information we can use to help construct or sense check our shear
force and bending moment diagrams.
Let's take a look at an example.
This beam has an applied distributed force and a concentrated force.
Let’s quickly draw the shear force and bending moment diagrams.
By using the free body diagram method, we can show that the bending moment curve for
the section of the beam under the distributed force is defined by the quadratic equation
-4 X^2 + 34 X + 68.
If we differentiate this equation we get -8 X + 34, which based on the D-M over D-X
equation above we now know is the equation for the shear force curve in this section
of the beam.
If we differentiate again we get -8, which is the equation for the distributed force.
This is a great way to sense check your shear force and bending moment diagrams.
Another way of checking your diagrams is using the area equations I mentioned earlier.
The area under the shear force curve highlighted here is equal to 34 times 2, which is 68.
This is equal to the change in bending moment over this section of the beam.
We can also calculate the area under the shear force diagram for the beam section under the
distributed force.
The total area of this section is equal to 72.3 minus 12.3, which is 60.
This is equal to the change in bending moment of 60 kNm over this section of the beam.
Where concentrated forces are applied there is a sudden jump in the shear force diagram,
and where concentrated moments are applied there is a sudden jump in the bending moment diagram.
These equations will not be applicable across discontinuities in the diagrams.
One final observation we can make based on these equations is that when the shear force
is equal to zero, the bending moment curve will be at a local minimum or maximum.
Let’s look at one last example.
Here we have a cantilever with an applied concentrated
moment of 120 kNm and a distributed force of 6 kN/m.
Again we start by drawing the free body diagram.
Because the support is fully fixed, we have vertical and horizontal reaction forces R-A
and H-A, and a reaction moment M-A.
Let's look at our first equilibrium equation.
The sum of forces in the vertical direction is equal to zero.
In this case the only forces acting in the vertical direction are the reaction force
R-A and the distributed force, so R-A is equal to 6 times 3, which is 18.
H-A is the only force in the horizontal direction, so it must be equal to zero.
Next we can take the sum of the moments acting at point A. In calculating the moment caused
by a uniformly distributed force, you can remember that it is equal to a concentrated
force located in the middle point of the load.
This gives us M-A equals 21.
To calculate our shear forces and bending moments we will start on the left side of
the beam and move towards the right.
This is our first free body diagram.
The shear force calculation is easy, as we only need to consider the reaction force of 18 kN.
The bending moment needs to take into account the reaction moment and the reaction force.
At X equals zero the bending moment is equal to the reaction moment of 21 kNm.
As we move to the right we also need to consider the moment caused by the 18 kN reaction force.
This gives us the equation for a straight line.
We can then move our cut to the right of the concentrated moment.
The moment won't affect the shear force, which will remain constant at 18 kN until
we reach the distributed force.
But it does cause the bending moment to suddenly drop by 120 kNm.
After the drop, the bending moment is again defined by a straight line.
Things get a little more tricky when we reach the distributed force.
We can replace the uniformly distributed force by an equivalent concentrated force with a
magnitude of 6 multiplied by the length X over which the force is applied.
This force is located at a distance of X/2 from our cut.
We can then calculate the shear force and bending moment equations using the normal
approach.
The bending moment in this section of the beam is defined by a quadratic equation.
No loads are acting on the small one metre section to the right of the distributed force,
so shear forces and bending moments in that section will be equal to zero.
Although we can't calculate displacements from these diagrams, we can use the bending
moment information to predict the deformed shape of the beam.
Where the bending moment is positive the beam will be sagging, and where it is negative
it will be hogging.
Where the bending moment is zero the beam will be straight.
That will give us a deformed shape that looks something like this.
That's it for this quick look at shear forces and bending moments in beams.
I hope you learned something new, and if you enjoyed the video please don’t forget to subscribe!