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Shear and Moment Diagrams – An Ultimate Guide

by Dr Seán Carroll
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Updated: 12th April 2021
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Tutorial

In this post we’re going to take a look at shear and moment diagrams in detail. Determining shear and moment diagrams is an essential skill for any engineer. Unfortunately it’s probably the one structural analysis skill most students struggle with most.

This is a problem. Without understanding the shear forces and bending moments developed in a structure you can’t complete a design. Shear force and bending moment diagrams tell us about the underlying state of stress in the structure. So naturally they’re the starting point in any design process.

Another reason every graduating engineer needs to have a solid grasp of shear forces and bending moments is because they’re absolutely going to be tested in almost every graduate interview. The quickest way to tell a great CV writer from a great graduate engineer is to ask them to sketch a qualitative bending moment diagram for a given structure and load combination!

So in this post we’ll give you a thorough introduction to shear forces, bending moments and how to draw shear and moment diagrams. We won’t be able to cover everything in this one post but hopefully you’ll reach the end knowing more than when you started! If you want to do a deep dive to really nail down this skill, you should take a look at my course, Mastering Shear Force and Bending Moment Diagrams [🎓 NOW FREE FOR STUDENTS]. In it, we’ll cover the fundamental theory and put it into practice with plenty of worked examples.

Mastering Shear Force and Bending Moment Diagrams | DegreeTutors.com
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Mastering Shear Force and Bending Moment Diagrams

Your complete roadmap to mastering these essential structural analysis skills.

After completing this course…
  • You will be fully competent in drawing shear force and bending moment diagrams for statically determinate beams and frames.
  • You will have a robust system of analysis that allows you to confidently tackle the analysis of any statically determinate structure.
  • You will understand the relationship between external loading and the shear forces and bending moments they induce.

In this post we’ll cover…

1.0 What is a Bending Moment?

Let’s start with a basic question; what is a bending moment? To answer this we need to consider what’s happening internally in a structure under load. Consider a simply supported beam subject to a uniformly distorted load.

Shear-and-moment-diagrams-beam | DegreeTutors.com

The beam will deflect under the load. In order for the beam to deflect as shown, the fibres in the top of the beam must contract or get shorter. The fibres in the bottom of the beam must get longer.

We can say the top of the beam is in compression while the bottom is in tension (notice the direction of the arrows on the fibres in the deflected beam). Now, at some position in the depth of the beam, compression must turn into tension. There is a plane in the beam where this transition between tension and compression occurs. This plane is called the neutral plane or sometimes the neutral axis.

Imagine taking a vertical cut through the beam at some distance x along the beam. We can represent the strain and stress variation throughout the depth of the beam with strain and stress distribution diagrams.

Shear-and-moment-diagrams-stresses | DegreeTutors.com

Remember, strain is just the change in length divided by the original length. In this case we’re considering the longitudinal strain or strain perpendicular (normal) to the cut face. 

Compression strains above the neutral axis exist because the longitudinal fibres in the beam are getting shorter. Tensile strains occur in the bottom because the fibres are extending or getting longer. 

We can assume this beam is made of a linearly elastic material and as such the stresses are linearly proportional to the strains. This simply means we need to multiple the strain at some point in the beam by the Young’s modulus (modulus of elasticity) to get the corresponding stress at that point in the beam. 

We know that if we multiply a stress by the area over which it acts, we get the resultant force on that area. The same is true for the stress acting on the cut face of the beam. The compression stresses can be represented by a compression force (stress resultant) while the tensile stresses can be replaced by an equivalent tensile force. So for example the compression force is given by,

(1)   \begin{equation*} F_c = \underbrace{\left[\sigma_t \times \frac{1}{2} \right]}_{\text{average stress}}\underbrace{\left[ b \times\frac{h}{2}\right]}_{\text{area}} \end{equation*}

Shear-and-moment-diagrams-force-couple | DegreeTutors.com

As a result of the external loading on the structure and the deflection that this induces, we end up with two forces acting on the cut cross-section. These forces are:

  • equal in magnitude (must be to maintain force equilibrium)
  • parallel to each other (and perpendicular to the cut face)
  • acting in opposite directions
  • separated by a distance or lever arm, z

You might recognise this pair of forces as forming a couple or moment M

(2)   \begin{equation*} M = F_c\times z = F_t\times z \end{equation*}

💡 The internal bending moment M, is the bending moment we represent in a bending moment diagram. The bending moment diagram shows how M (and therefore normal stress) varies across a structure.

If we know the state of longitudinal or normal stress due to bending at a given section in a structure we can work out the corresponding bending moment. 

However, more often it’s the case that we know the value of the bending moment at a point and use this to work out the maximum values of normal stress at that location. 

We do this using the Moment-Curvature equation a.k.a. the Engineer’s Bending Equation… 

(3)   \begin{equation*} \boxed{\sigma = \frac{-M\:y}{I}} \end{equation*}

…which relates the stress, \sigma at a distance y from the neutral axis, to the moment, M. Where I is the second moment of area for the cross-section. 

Hopefully now you can clearly see how bending moments arise;

  • external forces induce deflections
  • strains develop (which we see at a larger scale as structural deflections)
  • where we have strains, we must have stresses (remember Young’s modulus)
  • these stresses, can be represented with their force resultants that ultimately form a couple or internal bending moment, M

2.0 What is a Shear Force?

We can now turn our attention to shear forces and start with a simple definition; 

💡 A shear force is any force acting perpendicularly to the longitudinal axis of the structure. We’re typically interesting in internal shear forces that are the resultant of internal shear stresses developed in the structure.

Building on our discussion of bending moments, the shear force represented in the shear force diagram is also the resultant of shear stresses acting at a given point in the structure. Consider the cut face of the beam discussed above.

Shear-and-moment-diagrams-force-shear-stress | DegreeTutors.com

The shear stress, \tau acting on this cut face is evenly distributed across the width of the face and acts parallel to the cut face. The average value of the shear stress, \tau_{avg} is simply the shear force at this point in the structure V divided by the cross-sectional area over which it acts, A

(4)   \begin{equation*} \tau_{avg} = \frac{V}{A} \end{equation*}

However, this is just the average value of the shear stress acting on the face. The shear stress actually varies parabolically through the depth of the section according to the following equation,

(5)   \begin{equation*} \tau = \frac{VQ}{Ib} \end{equation*}

where, Q is the first moment of area of the area above the level at which the shear stress is being determined, I is the second moment of area of the cross-section and b is the width of the section.

Shear-and-moment-diagrams-force-parabolic-shear-stress | DegreeTutors.com

We don’t want to go too far down the rabbit hole with shear stresses. For the purposes of this tutorial, all we want to do is establish the link between the shear force we observe in the shear force diagram and the corresponding shear stress within the structure. Equations (4) and (5) do that for us.  

3.0 Calculating Internal Shear Forces and Bending Moments

Up to this point we’ve considered the link between the normal (bending) stress and associated bending moment and the shear stress and associated shear force. Based on this you should be comfortable with the idea that knowing the value of bending moment and shear force at a point are important for understanding the stresses in the structure at that point.

Now we’re going to consider the problem of calculating shear forces and bending moments not from the point of view of internal stresses but by considering equilibrium of the structure. 

In reality, this is practically how we determine the shear force and bending moment at a point in the structure. Again, let’s consider the simply supported beam from above, subject to a uniformly distributed load, w kN/m. 

Simple statics tell us that if the beam is in a state of static equilibrium, the left and right hand support reactions are,

(6)   \begin{equation*}  V_a = V_b = \frac{wL}{2} \end{equation*}

Shear-and-moment-diagrams-beam-udl | DegreeTutors.com

If the structure is in a state of static equilibrium (which it is), then any sub-structure or part of the structure must also be in a state of static equilibrium under the stabilising action of the internal stress resultants. 

This is a key point! Imagine taking a cut through the structure and separating it into 2 sub-structures. When we cut the structure, we ‘reveal’ the internal stress resultants (bending moment and shear force). 

Shear-and-moment-diagrams-sub-structure | DegreeTutors.com

M_L and M_R are the internal bending moments on either side of the imaginary cut while V_L and V_R are the internal shear forces on either side of the imaginary cut. 

💡 M_R and V_R represent the influence of the left hand side of the structure (sub-structure 1) on the right hand side of the structure (sub-structure 2) and vice versa.

We’ve just said that each one of these sub-structures is stabilised by the influence of the internal bending moment and shear force revealed by the imaginary cuts. 

This means, if we want to find the value of internal bending moment or shear force at any point in a structure, we simply cut the structure at that point to expose the internal stress resultants (M and V). Then calculate what values they must have to ensure the sub-structure remains in equilibrium! For example the sub-structure below must remain in equilibrium under the combined influence of:

  • the external distributed load acting on the sub-structure
  • the left hand reaction, V_A (note it has not been reduced just because we’re considering the sub-structure. It retains at its original value)
  • the internal bending moment at the cut, M_L
  • the internal shear force at the cut, V_L
Shear-and-moment-diagrams-sub-structure-1 | DegreeTutors.com

This starts to make more sense when we plug some numbers into an example. For the beam above, let’s imagine it has a span L=6 m, applied loading of w=10 kN/m and imagine we cut the beam at a=2 m from the left hand support.

The left hand reaction, V_A is,

(7)   \begin{equation*} V_A = \frac{w L}{2} = \frac{10\times 6}{2} = 30 \:\text{kN} \end{equation*}

Now taking the sum of the moments about the cut and assuming clockwise moments are positive,

(8)   \begin{align*} &\sum M_{\text{cut}}=0\\ &(30\:\text{kN}\times 2\:\text{m}) - (10\:\text{kN/m}\times 2\text{m}\times 1\text{m}) -M_L=0\\ & \therefore \: M_L = 40 \:\text{kNm} \end{align*}

So, the internal bending moment required to maintain moment equilibrium of the sub-structure is 40 kNm. Similarly, if we take the sum of the vertical forces acting on the sub-structure, this would yield V_L = 10 kN. 

4.0 Building Shear and Moment Diagrams

In the last section we worked out how to evaluate the internal shear force and bending moment at a discrete location using imaginary cuts. But to draw a shear force and bending moment diagram, we need to know how these values change across the structure.

What we really want is an equation that tells us the value of the shear force and bending moment as a function of x. Where x is the position along the beam. Consider making an imaginary cut, just like above, except now we can make the cut at a distance x along the beam. 

Now the internal shear force and bending moment revealed by the cut are functions of x, the cut position. Here, we’ll determine an expression for M(x). But the procedure is exactly the same to determine V(x)

Shear-and-moment-diagrams-BMD | DegreeTutors.com

Taking the sum of the moments about the cut and again assuming clockwise moments are positive,

(9)   \begin{equation*} \sum M_{\text{cut}}=0 \end{equation*}

(10)   \begin{equation*} \frac{wL}{2}x - w x \frac{x}{2}-M(x) &= 0\\ \end{equation*}

(11)   \begin{equation*} M(x) &= \frac{wLx}{2} - \frac{wx^2}{2}\\ \end{equation*}

(12)   \begin{equation*} M(x) &= \frac{wx}{2}(L-x) \end{equation*}

Now we can use equation (12) to determine the value of the internal bending moment for any value of x along the beam. Plotting the bending moment diagram is simply a matter of plotting the equation. 

Shear-and-moment-diagrams-BMD | DegreeTutors.com

4.1 Finding the location of the maximum bending moment

In the example above, the structure and loading is symmetrical so it’s pretty easy to recognise the location of the maximum moment and then subsequently to evaluate it.

However this may not always be the case. So it’s helpful to have a technique to identify the location of the maximum moment without needing to plot the full bending moment diagram.

In this example, the bending moment for the whole structure is described by a single equation…equation (12). You might remember from basic calculus that to identify the location of the maximum point in a function we simply differentiate the function to get the equation for the slope. Then it’s just a matter of setting this function equal to zero and solving for x. 

In other words, at the location of the maximum bending moment, the slope of the bending moment diagram is zero. So we just need to solve for this location. Once we have the location we can evaluate the bending moment using equation (12).

So, to demonstrate let’s first evaluate the differential of equation (12),

(13)   \begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} M(x) = \frac{wL}{2} - wx \end{equation*}

Remember, equation (13) represents the slope of the bending moment diagram. So we now let it equal to zero and solve for x.

(14)   \begin{equation*} \frac{wL}{2} - wx = 0 \end{equation*}

(15)   \begin{equation*} \therefore x=\frac{L}{2} \end{equation*}

Surprise surprise, the bending moment is a maximum at the mid-span, x=L/2. Now we can evaluate equation (12) at x=L/2 = 3 m. 

(16)   \begin{equation*} M(x=L/2) = \frac{wL^2}{4} - \frac{WL^2}{8} \end{equation*}

(17)   \begin{equation*} \therefore M_{\text{max}} = \frac{wL^2}{8} = \frac{10\times 6^2}{8} = 45 \:\text{kNm} \end{equation*}

There we have it; the location and magnitude of the maximum bending moment in this simply supported beam, all with some basic calculus. 

5.0 Drawing Shear Force and Bending Moment Diagrams – An Example

Now that we have a grasp of the fundamentals, let’s see how it all ties together with a bigger more complex worked example. This example is an extract from this course. Just a quick heads up, if you’re new to shear force and bending moment diagrams, this question might be a bit of a challenge. If you get a bit lost with this example, it might be worth your time taking a look at this DegreeTutors course. It’s aimed at bringing you from scratch all the way up to being comfortable analysing complex shear and moment diagrams.  

Ok, let’s get on with it. We want to determine the shear force and bending moment diagrams for the following simply supported beam. 

Shear-and-moment-diagrams-example-question| DegreeTutors.com

You can continue reading through the solution below…or if you prefer video, you can watch me walk through the solution here.

5.1 Video Tutorial

5.2 Calculating the support reactions

The first step in analysing any statically determinate structure is working out the support reactions. We can kick-off by taking the sum of the moments about point A, to determine the unknown vertical reaction at B, V_B,

(18)   \begin{equation*} \sum M_{\text{A}}=0\:\:\text{(assume clockwise positive)} \end{equation*}

(19)   \begin{equation*} (90\text{kN}\times 3\text{m}) + 50\text{kNm} +(10\text{kN/m}\times 9\text{m}\times 8\text{m}\times 0.5) - 10\text{m}\times V_B = 0 \end{equation*}

(20)   \begin{equation*} \therefore V_B = 68\text{kN} \end{equation*}

Now with only one unknown force, we can consider the sum of the forces in the vertical direction to calculate the unknown reaction at A, V_A,

(21)   \begin{equation*} V_A + 68\text{kN} - 90\text{kN} - 10\text{kN/m}\times 9\text{m}\times 0.5 = 0 \end{equation*}

(22)   \begin{equation*} \therefore V_A = 67\text{kN} \end{equation*}

5.3 Drawing the shear force diagram

Our approach to drawing the shear force diagram is actually very straightforward. We’re going to ‘trace the impact of the loads’ across the beam from left to right. 

The first load on the structure is V_A = 67\text{kN} acting upwards, this raises the shear force diagram from zero to +67\text{kN} at point A. The shear force then remains constant as we move from left to right until we hit the external load of 90\text{kN} acting down at D. This will cause the shear force diagram to ‘drop’ down by 90\text{kN} at D to a value of +67 - 90 = -23\text{kN}

This process of following or tracing the loads across the structure continues across the full beam until you’ve completely traced out the shear force diagram.

When we reach the linearly varying load at E, we make use of the relationship between load intensity, q and shear force V that tells us that the slope of the shear force diagram is equal to the negative of the load intensity at a point,

(23)   \begin{equation*} \frac{\mathrm{d}V}{\mathrm{d}x} = -q \end{equation*}

This is telling us that the linearly varying distributed load between E and F will produce a curved shear force diagram described by a polynomial equation. In other words, the shear force diagram starts curving at E with a linearly reducing slope as we move towards F, ultimately finishing at F with a slope of zero (horizontal). When the full loading for the beam is traced out, we end up with the following, 

Shear-and-moment-diagrams-example-SFD| DegreeTutors.com

It’s worth pausing for a moment to explain how the shear force to the left of B, V_{B,L} = -59.1\text{kN} was calculated. This is obtained by subtracting the total vertical load between E and B from the shear force of -23 \text{kN} at E.

(24)   \begin{equation*} V_{B,L} = -23\text{kN}-(4.44\text{kN/m}\times 5\text{m})  - (10-4.44)\text{kN/m}\times 5\text{m}\times 0.5 \end{equation*}

(25)   \begin{equation*} V_{B,L} = -59.1\text{kN} \end{equation*}

5.4 Drawing the bending moment diagram

Once we’ve completed the shear force diagram, the bending moment diagram becomes much easier to determine. This is because we can make use of the following relationship between the shear force V and the slope of the bending moment diagram,

(26)   \begin{equation*} \frac{\mathrm{d}M}{\mathrm{d}x} = V \end{equation*}

Similarly to equation (23), this expressions allows us to infer a qualitative shape for the bending moment diagram, based on the shear force diagram we’ve already calculated. 

Consider the shear force between A and D for example; it’s constant, which means the slope of the bending moment diagram is also constant (an inclined straight line). Between D and E, the shear force is still constant but has changed sign. This tells us the slope of the bending moment diagram has also changed sign, i.e. the bending moment diagram has a local peak at D. 

The fact that the shear force is a polynomial (curve) between E and F also tells us the bending moment’s slope is continuously changing, i.e. it’s also a curve. But the fact that the shear force changes sign at B, means the bending moment diagram has a peak at that point.

Finally, the externally applied moment at F tells us that the bending moment diagram at this location has a value of 50\text{kNm}. We can combine all this information together to sketch out a qualitative bending moment diagram, based purely on the information encoded in the shear force diagram.

Shear-and-moment-diagrams-example-BMD-qualitative| DegreeTutors.com

Now we simply have to cut the structure at discrete locations (indicated with red dashed lines above) to establish the various key values required to quantitatively define the bending moment diagram. In this case three cuts are sufficient:

  • at D to determine the local peak – Cut 1-1
  • at E to determine the value on the boundary between the straight and curved sections of the bending moment diagram – Cut 2-2
  • at B to determine the local peak – Cut 3-3 

Cut 1-1

As we’ve seen above, to determine the internal bending moment at D, M_D, we cut the structure to reveal the internal bending moment at this point. Then by considering moment equilibrium of the sub-structure we can solve for the value of M_D.

Shear-and-moment-diagrams-example-Cut-1| DegreeTutors.com

Taking the sum of the moments about the cut,

(27)   \begin{equation*} \sum M_{\text{cut 1}}=0\:\:\text{(assume clockwise positive)} \end{equation*}

(28)   \begin{equation*} -M_1 + (67\text{kN}\times 3\text{m}) = 0 \end{equation*}

(29)   \begin{equation*} M_1 = 201\text{kNm} \end{equation*}

Cut 2-2

Repeating this process for cut 2-2,

Shear-and-moment-diagrams-example-Cut-2| DegreeTutors.com

(30)   \begin{equation*} \sum M_{\text{cut 2}}=0\:\:\text{(assume clockwise positive)} \end{equation*}

(31)   \begin{equation*} -M_2 + (67\text{kN}\times 5\text{m}) - (90\text{kN/m}\times 2\text{m}) = 0 \end{equation*}

(32)   \begin{equation*} M_2 = 155\text{kNm} \end{equation*}

Cut 3-3

And finally for cut 3-3, this time considering equilibrium of the sub-structure to the right-hand side of the cut

Shear-and-moment-diagrams-example-Cut-3| DegreeTutors.com

(33)   \begin{equation*} \sum M_{\text{cut 3}}=0\:\:\text{(assume clockwise positive)} \end{equation*}

(34)   \begin{equation*} -M_3 + 50\text{kNm} + (4.44\text{kN/m}\times 4\text{m}\times 0.5 \times (4/3)\text{m}) = 0 \end{equation*}

(35)   \begin{equation*} M_3 = 61.84\text{kNm} \end{equation*}

We can now sketch the complete quantitative bending moment diagram for the structure. In fact at this point we can summarise the output of our complete structural analysis.

Shear-and-moment-diagrams-example-Summary-sketch| DegreeTutors.com

6.0 Relating Loading, Shear Force and Bending Moment

In the previous example, we made use of two very helpful differential relationships that related loading with shear force and shear force with bending moment. However we didn’t properly introduce them. Now that we have a good idea of the general workflow for generating shear and moment diagrams, we can dig a bit deeper into these differential relationships. Understanding these, is the key to being able to build shear force and bending moment diagrams quickly and reliably. 

Fully understanding the relationships we derive next will allow you to more ‘intuitively extract’ qualitative shear and moment diagrams ‘by eye’, with cuts used to confirm numerical values at salient points. We’re going to explore 3 cases:

  • Case 1: Uniformly distributed loading
  • Case 2: Point force loading
  • Case 3: Point moment loading

In each case, our objective is to determine the relationship between the applied loading and the shear force and bending moment it induces.

6.1 Case 1: Uniformly distributed loading

Consider a short segment of length \mathrm{d}x cut from a beam and subject to a uniformly distributed load with intensity w kN/m. As we saw above, these cuts have revealed the internal moment and shear on either side of the segment. Note the infinitesimal increase in moment (\mathrm{d}M) and shear (\mathrm{d}V) on the right side of the cut.

Beam segment with UDL | DegreeTutors.com

Shear Force

We can start by considering vertical force equilibrium for the segment. Since it must be in a state of static equilibrium, the sum of the vertical forces must equal zero.

    \begin{equation*}\sum F_y = 0\:\:(\uparrow+ve)\end{equation*}

    \begin{equation*}V-(v+\mathrm{d}V)-w\mathrm{d}x=0\end{equation*}

(36)   \begin{equation*}\therefore \boxed{ \frac{\mathrm{d}V}{\mathrm{d}x} = -w}\end{equation*}

In other words, the slope of the shear force diagram (\mathrm{d}V/\mathrm{d}x) at a point is equal to the negative of the load intensity at that point. We can demonstrate this with a simple example. Consider the beam below subject to a distributed load with linearly increasing intensity. By making a cut at a distance x from the left support with reveal the internal shear force V(x).

Beam with increasing load | DegreeTutors.com

If the load intensity increases linearly from zero to w, then at the cut the load intensity is wx/L. We can now evaluate vertical force equilibrium for the sub-structure,

    \begin{equation*}\sum F_y = 0\:\:(\uparrow+ve)\end{equation*}

    \begin{equation*}V(x) = \frac{wl}{3} - \frac{wx^2}{2L}\end{equation*}

We can now differentiate the expression for V(x) yielding,

    \begin{equation*}\frac{\mathrm{d}V}{\mathrm{d}x} = -\frac{wx}{L}\end{equation*}

So we can see that the differential of the shear force is equal to the negative of the load intensity. It’s also worth noting the shape of the SFD, pictured below. At the left hand support when the load intensity is zero, the SFD has a value of wL/3 (the value of the left reaction) but it is horizontal, i.e. has a slope of zero. As the load intensity increases as we move from left to right, the SFD gets steeper. i.e. the slope increases.

Beam and shear force diagram | DegreeTutors.com

Another implication of this differential relationship between shear force and load intensity can be seen if we integrate both sides of the equation,

    \begin{equation*}\frac{\mathrm{d}V}{\mathrm{d}x} = -w\end{equation*}

    \begin{equation*}\int_A^B \mathrm{d}V = -\int_A^B w\: \mathrm{d}x\end{equation*}

(37)   \begin{align*}V_B - V_A &= -\int_A^B w\: \mathrm{d}x\\&= - \:\text{area under loading diagram between A and B}\end{align*}

We can see this represented graphically in the image below.

Beam SFD and Area | DegreeTutors.com

Bending Moment

Having established the key relationship for shear, now we can turn our attention to bending moments. Referring back to our beam segment of length \mathrm{d}x and considering moment equilibrium of the segment by taking moments about the left hand side of the segment,

    \begin{equation*}\sum M_{\text{LHS}} = 0 \:\:(\curvearrowright +ve)\end{equation*}

    \begin{equation*}M-(M+\mathrm{d}M)+(V+\mathrm{d}V)(\mathrm{d}x)+w. \mathrm{d}x. \frac{\mathrm{d}x}{2} = 0\end{equation*}

(38)   \begin{equation*}\therefore \boxed{ \frac{\mathrm{d}M}{\mathrm{d}x} = V}\end{equation*}

So, the slope of the BMD at a point equals the shear force at that point. Combined with the previous differential equation we derived, this is a very helpful equation. Whenever we have a beam subject to a distributed load, we can use these equations to infer the shape of the SFD and BMD. Consider the SFD and BMD for our beam below.

Beam, SFD and BMD | DegreeTutors.com

We note that when the shear force is zero, the slope of the BMD is also zero indicating a local maximum in the BMD. We also note the change in sign of the slope of the BMD as the shear force goes from positive to negative. Remember that the shape of the SFD was itself deduced from the shape of the loading diagram. By making use of these relationships between loading, SFD and BMD, we can build up a qualitative picture of structural behaviour.

6.2 Case 2: Point force loading

Now we repeat the same process as above but this time our beam segment is subject to a point load P located at \mathrm{d}x/2. Note that on the right hand side of the element, the internal shear force and bending moment have increased by a finite amount rather than an infinitesimal amount as was the case previously.

Beam segment with point loading | DegreeTutors.com

Shear Force

Evaluating the sum of the vertical forces yields,

    \begin{equation*}\sum F_y = 0\:\:(\uparrow+ve)\end{equation*}

    \begin{equation*}-P +V - (V+V_1)=0\end{equation*}

(39)   \begin{equation*}\therefore \boxed{ V_1 = -P_1 }\end{equation*}

From this we see that a point load induces a step change in the SFD. We’ve already seen this when we followed the loads across the structure to build the shear force diagram above. This equation is simply the mathematical representation of this. Consider for example the simple case below of a beam subject to two point loads.

Beam with two point loads | DegreeTutors.com

We can readily see the step changes in the shear force diagram being equal to the magnitude of the point loads at that location.

Bending Moment

If we now consider moment equilibrium of our segment,

    \begin{equation*}\sum M_{\text{LHS}} = 0 \:\:(\curvearrowright +ve)\end{equation*}

    \begin{equation*}M-(M+M_1) + P\frac{\mathrm{d}x}{2} + (V+V_1)\mathrm{d}x\end{equation*}

    \begin{equation*}M_1 = \underbrace{P\frac{\mathrm{d}x}{2} + (V+V_1)\mathrm{d}x}_{\text{infinitesimally small}}\end{equation*}

    \begin{equation*}\therefore \boxed{M_1 \approx 0}\end{equation*}

The presence of infinitesimally small segment lengths on the right hand side of the equal sign means that M_1 is infinitesimally small. From this we conclude that the presence of a point load does not change the value of the bending moment diagram at a point.

However, noting that the shear force changes from V to V-P, we can say, according to the expression,

(40)   \begin{equation*}\frac{\mathrm{d}M}{\mathrm{d}x} = V\end{equation*}

that the slope of the bending moment diagram changes by an amount P. Again, we can see how this maps onto our simple example below. Note that at the point of application of P_1, the slope of the bending moment diagram changes. Also, where the shear force is zero, the bending moment diagram is horizontal.

So we have added two more equations into our toolbox for establishing qualitative structural behaviour.

Beam with two point loads and diagrams | DegreeTutors.com

6.3 Case 3: Point moment loading

Finally we can repeat the analysis for the case of moment applied at a point.

Beam segment with point moment | DegreeTutors.com

Shear Force

Take the sum of the forces in the vertical direction,

    \begin{equation*}\sum F_y = 0\:\:(\uparrow+ve)\end{equation*}

    \begin{equation*}V-(V+V_1)\end{equation*}

    \begin{equation*}\therefore \boxed{V_1 = 0}\end{equation*}

So, the shear force diagram does not change with the application of a moment.

Bending Moment

Taking the sum of the moments about the left hand side of the cut,

    \begin{equation*}\sum M_{\text{LHS}} = 0 \:\:(\curvearrowright +ve)\end{equation*}

    \begin{equation*}M+M_o-(M+M_1)+\underbrace{(V+V_1)\mathrm{d}x}_{\text{infinitesimally small}\rightarrow 0} = 0\end{equation*}

    \begin{equation*}\therefore \boxed{M_0 = M_1}\end{equation*}

This means that at the point of application of a bending moment, there is a step change in the bending moment diagram, equal to the magnitude of the moment applied.

The 6 boxed equations in this section above can be used to infer a huge amount of information about the behaviour of a structure under load. Let’s put this into practice with another worked example. 

7.0 Another Example

Determine the shear force diagram and bending moment diagram for the following simply supported beam. Make sure to attempt this yourself before watching the solution videos.

Shear and moment example | DegreeTutors.com

7.1 Setup and shear force diagram

7.2 Building the bending moment diagram

7.3 Confirming maximum moment with calculus

So there you have it. We’ve linked together the internal normal and shear stresses with the bending moment and shear force diagrams. And we’ve derived a toolbox full of helpful differential equations to help us quickly and intuitively build shear force and bending moment diagrams. There is quite a lot more we could say about shear and moment diagrams. But that’s probably enough for one post.

The best way for you to get better at evaluating shear force and bending moment diagrams is through practice. There really are no shortcuts I’m afraid. The good news is, the more you practice, the quicker you get and the stronger your intuition for structural behaviour becomes. That’s all for now, I hope you got some value from reading this post and I’ll see you in the next one. 


Build your own shear force and bending moment calculator

Understanding how to build shear force and bending moment diagrams the way we’ve demonstrated above is an essential skill. However, the process is time consuming, particularly when you enter the iterative process of analysis and design. And that’s before we even start talking about how to handle indeterminate structures! For these reasons, we generally make use of structural analysis software to speed the process up. But this software is typically expensive and for the vast majority of cases, has way more functionality than we need. So – why not just build you own, for (almost) free! In my course below, we use the Direct Stiffness Method to build our own 2D beam and frame analysis programme using Python. You don’t need to be a programmer to take this course. When you’re finished it…you’ll have your own DIY structural analysis programme. 

Play Video

Beam & Frame Analysis using the Direct Stiffness Method in Python

Build a sophisticated structural analysis software tool that models beams and frames using Python.

After completing this course…
  • You’ll understand how to model beam elements that resist axial force, shear forces and bending moments within the Direct Stiffness Method.
  • You’ll have your own analysis software that can generate shear force diagrams, bending moment diagrams, deflected shapes and more.
  • You’ll understand how to model rotational pins within your structures, inter-nodal distributed loading and realistic flexible supports.

Tutorial by:

Dr Seán Carroll

Hi, I’m Seán, the founder and lead tutor at DegreeTutors.com. I hope your found whatever I wrote up there helpful! Feel free to get in touch or follow DegreeTutors on any of the social accounts.