PE EXAM Practice Question: Calculating Shear Force

In this article (and video above), we calculate the maximum shear force and moment resultant from a point load applied to a beam by drawing the shear force diagram and moment diagram associated with the condition. This question forms part of the Structural Mechanics section of the PE Exam and was created and solved by Engineer in training, Enrique Ivers

Question:

For the beam shown, what is the maximum shear force and the maximum moment?

Explanation:

For the beam shown, what is the maximum shear force and the maximum moment?

There are reaction forces resultant from the 40 K point load. We know the distance from pinned support A and point load P. We can easily calculate the distance from P to roller support B.

Shear at the left support, V_a, is equal to the Reaction force, R_a. This force is calculated by the equation

where P is the point load, b is the distance between the load and support b, and L is the total length of the beam:

As the forces must be in equilibrium, we know that  R_b must be equal to the difference between the Point load P, and R_a: R_b = P – R_a = 40K – 8K = 32K

Similar how V_a, V_b= R_b, so the shear force at support B is also 32K.

Although not required for this problem, you may be asked to select the appropriate shear diagram in similar problems.

Recall that shear diagrams are linear whereas moment diagrams are parabolic. In a uniform load, the linear shear diagram will exhibit a non-zero slope.

However, in a case like this one, the shear diagram has a slope of zero, and since we’re “tracing the impact of the load”, the impact of the point load is apparent in the diagram at that specific point.

Let’s take a look!

The first load on the structure is V_a = R_a = 8 K.

This force is acting upwards, so it raises the shear force diagram from zero to +8 K at point A

The shear force them remains constant as we move from left to right until we hit the point load of 40 K acting downward.

The first load on the structure is V_a = R_a = 8 K.

This force is acting upwards, so it raises the shear force diagram from zero to +8 K at point A.

The shear force them remains constant as we
move from left to right until we hit the point load of 40 K acting downward.

This will cause the shear force diagram to “drop” down by 40 K to a value of + 8 K – 40 K = -32 K

We continue until we hit support B, where
V_b = R_b = 32 K, also pointing upwards. This results in the shear diagram ending up back at zero.

We’ll now calculate the maximum moment. And, as we did an exercise with the shear diagram, we’ll also draw the moment diagram.

We can use the Maximum Moment Equation:

where P is the point load, a is the distance between support A and the point load, b is the distance between support B and the point load, and L is the length of the beam.

Knowing our maximum moment is 160K·ft, we can draw our moment diagram.

Recall that the magnitude of the moment at the supports will be zero, and the maximum moment coincides with the location of the point load.

 

Answer:

The correct answer is B.

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Anthony Fasano, P.E.
Engineering Management Institute
Author of Engineer Your Own Success