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
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