Calculate and Draw Shear Force Diagrams

In Fundamentals, Structural Analysis by Tom OrmondeLeave a Comment

Shear force is an extremely important phenomena in structural engineering and also useful for mechanical, civil, aerospace and others (so don’t worry if you’re not the best). 

We will be looking at what shear force actually is and how to calculate it. On top of this and most importantly, going through some step by step examples. Be sure to be aware of the various different types of reactions in beams as this will add to your understanding of beam theory.

By the end of this post, you will be able to find the shear force and plot shear force diagrams, including the reactions in the beam. Subsequent posts will cover how to calculate the shear resistance of members, plates, connections and more.

So, What Is Shear Force?

Lets get cracking. What exactly is shear force? It is a force that is transferred perpendicular to the member, or through the member. The support conditions affect the magnitude and location of the shear force because the shear is transferred through the member into the supports.

Take a look at the example below. Can you figure out which force is the shear force from the description above?

Reactions

The shear is the forces in the X direction, Fx, A and B. Well done if you got it! The force can also be angular. If this is the case, we have to resolve the forces into the correct direction using trigonometry.

What Shear Failure Looks Like..

Before going into the worked examples, it is interesting to see what happens when the shear force being applied is greater than the resistance provided. Thus the member fails, hopefully providing you all motivation not to let this happen!

Warning, disturbing image below, it shows what would happen if a concrete beam was to endure shear force. Or at least a typical graphical representation of such an event.

Shear Failure in a Concrete Slab / Beam

From the image above, the force applied to the left (arrow) is causing a shear failure in the slab or beam. The location of the failure occurs under the applied force and will then spread to the support, the closest support available, as this is the load path of least resistance, or the closest. As we know, force is lazy and never wants to travel far. The support provides the reaction and due to the reaction, an actual section is displaced, both vertically and horizontally. The displacements can be calculated and are dependent upon geometric and material properties.

Concrete is shown in the example, but the failure pattern will be very similar for other materials. Normally we see diagonal cracking to a member which is stiffer (i.e the support).

When Should We Worry About Shear Force?

Well to put it bluntly, all the time. The major issue of shear failure is the type of failure that occurs, namely a brittle failure. This type of failure can occur without any prior warning and therefore is the most dangerous type.

Imagine a scenario with a hurricane and an earthquake. From a distance, we can see the hurricane approaching and at least prepare; take shelter, board up windows etc. With an earthquake, we don’t have that luxury and have to hope for the best (unless ground movements are detected in advanced).

With structures, deflections can occur ahead of a full failure. This allows time for evacuations at the very worst, or remedial works to strengthen as required. Brittle failures such as a shear failure of a beam, column or connection can happen instantly.

Does The Type of Load and Direction Matter?

Yes and no is the best answer. Forces have their own unique set of characteristics. For example, a uniformly distributed load (UDL) has the force spread out across the whole of the beam. A point load has a concentration of load at one point (the name says it all). The shear force  is the summation of the forces in the vertical direction (of a horizontal beam) and therefore the load does have an effect. However, the location of the maximum shear force will not be affected since these occur at supports. 

But, and this is important to understand, it affects the shear force diagram that we produce to show the shear force in the beam! We will jump into this in the next chapter and the examples will make this crystal clear.

Shear Force Formula

The equation for shear force is easy to remember; it is the summation of all the forces in the direction of the shear force, perpendicular to the member. From the diagram used to show the reactions, we will compute the formula for the shear force we would have in the reactions.

Shear Force Formula

We already know the shear forces are Fxa and Fxb. So what force would we expect to see in each? Well, the sum of the forces has to be equal for it to be in equilibrium according to Isaac Newton’s first law.  

 ∑Fv = 0

External Force = FxA + FxB

If the force is in the center, from taking moments at each, we can conclude that they are equal. 

FxA  = FxB = External Force / 2

What Is A Shear Force Diagram?

The best way to show the analysis of shear force is graphically.  By producing a sketch of the member (free body diagram) as we have above and including the force along the beam. This allows us to look at the shear force at any point in the beam and design accordingly.

Typically, we would use a linear member (linear means to have the same sectional and material properties along the whole beam length) and therefore we would only need to check the shear force at the supports, the point of maximum shear force. However, with design constraints and the evolution of machinery, architectural requirements etc., this is not always the case. I’m personally seeing it more and more going forward in my career. Therefore, it is important to be able to access the beam at different locations and we can do this by hand. Once this is fully understood, there is a lot of commercial software available to use for the analysis.

Recently I designed a tapered beam. To give reasoning for how and why this occurred, I will put some background information about the job (everyone loves a good story right?). Due to complications and revised client specifications, the beam sizes needed to be altered due to an increase in the span, but most of the steel frame was already fabricated. It was agreed that we would taper the ends of the beams to allow the columns to remain at the same height, therefore not altering the headroom or creating an uneven surface. 

One problem with tapering the end of a beam is that we have reduced sectional properties of the beam. Coincidentally, this is the location of the maximum shear.

Tapered Beam. Maximum Shear at Tapered End.

For  the examples below, we are going to calculate the maximum shear force at the supports and draw the shear force diagrams. Go through it step by step. The process will provide discipline for more complex cases and it is also a great way to avoid mistakes! Note, in all the examples we are only going to consider the external forces. Self weight is excluded as is partial factors of safety. 

Ex.1 – Simply Supported Beam – 2 Point Loads

A nice easy one to get started. Calculate the shear force at both supports and show the shear force diagram.


Ex. 1 – Simply Supported Beam with 2 Point Loads

First of all, we need to calculate the reactions, or shear force, at the supports. Due to the forces being uneven, each support will take a different proportion of the load.

Taking moments about support A. RA. No moment can occur at either RA or RB as it is simply supported.

 ∑M @ RA = 0

 0 = 10*1 + 20*2 – RB*3

RB = 16.67kN

Now use the shear force equation to calculate the force at support RA.

 ∑Fv = 0

RA + RB = 10kN + 20kN

RA = 13.33kN

Ex. 1 – Shear Force at Supports

Now for the final part, plot the shear force diagram. We just follow each force in the above diagram. Where no force exists between the point loads and supports, the shear force does not change.


Ex. 1 – Shear Force Diagram

PRO TIP – THE MAXIMUM MOMENT IS AT THE LOCATION OF ZERO SHEAR FORCE.

Ex.2 – Simply Supported Beam – Uniformly Distributed Load

Now for another example, this time with a uniformly distributed load.



Ex.2 Simply Supported Beam with UDL

First step, calculate the force in the reactions. We can do this by either remembering the equation for a UDL, or manually. We will do it manually by taking moments. The UDL is converted to a point load at the center of the beam.



Ex. 2 Finding Reactions from a UDL

 ∑M @ RA = 0

 0 = 50*5 – 10*RB

RB = 25kN

Now use the shear force equation to calculate the force at support RA.

 ∑Fv = 0

RA + RB = 50kN

RA = 50kN

Ex. 2 – Shear Force Diagram

I hope you found the two examples useful. Shear force is extremely important to be able to calculate and knowledge of how the shear force diagrams look will help validate structural analysis models.

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