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Calculating Beam Deflection from First Principles

In this tutorial we’re going to explore beam deflection and see how we can calculate the deflection of any beam from first principles using the differential equation of the deflection curve. We’ll work our way through a numerical example before discussing how we can use superposition along with tabulated formulae to speed up the process. 

The table of contents below will give you an idea of what we’ll be discussing.

1.0 Differential Equation of the Deflection Curve

The differential equation of the deflection curve is used to describe bending behaviour and so it crops up when examining beam bending and column buckling behaviour. The equation simply describes the shape of the deflection curve of a structural member undergoing bending. So, if x measures the distance along a beam and v represents the deflection of the beam, the equation says,

(1)   \begin{equation*}EI \frac{\mathrm{d}^2v}{\mathrm{d}x^2} = M(x)\end{equation*}

where, EI is the flexural rigidity of the beam and M(x) describes the bending moment in the beam as a function of x. We won’t go into the derivation of the equation in this tutorial, rather we’ll focus on its application. 

Our objective is to use this equation to calculate beam deflection, v, so we need to integrate the equation twice to get an expression for v. The best way to get to grips with this is to work through an example. 

1.1 The ‘Small Deflection’ Assumption

Before we work our way through the example below we need to state the assumptions on which our analysis is based. The first is the so-called ‘small deflection’ assumption. In order to obtain equation 1, we made the assumption that the deflection of our beam (or any deflecting structure we apply this equation to) is small. In other words, if we consider a short curved length of our beam undergoing deflection, the curved length, ds, should be approximately equal to its projection onto a horizontal plane, dx.

We also must assume that at any point along our beam, the rotation of the beam, \theta is small enough that we can say \theta \approx \tan\theta, i.e. the angle of rotation at a point is approximately equal to the slope of the deflection curve. For most practical cases, deflection is a serviceability issue and we expect it to be relatively small and largely imperceptible to the naked eye. As such, this small deflection assumption is satisfied in most cases, but you need to be aware of its existence. 

1.2 Linearly Elastic Assumption

In order to derive equation 1 it was also assumed that the material the beam was made of was linearly elastic and therefore followed Hooke’s law. This must be the case because we are relying on the fact that the curvature of the beam, is proportional to the associated bending moment. This is important to remember, because our deflection equations will become inaccurate for plastic deformations, which would likely also invalidate our small deflection assumption. Now that we know the boundaries we’re working within, we can crack on with the example.

2.0 Determining the Bending Moment Equations

Consider the simply supported beam in Fig. 1 below. The beam is subject to two point loads and a uniformly distributed load. Our task is to determine the mid-span deflection and the maximum deflection. Note that because the beam isn’t symmetrically loaded, the maximum deflection need not occur at the mid-span location. Static analysis of the beam reveals the support reactions at A and D,

    \begin{align*}V_A &= 139.375 \:kN\\V_D &= 145.625 \:kN\end{align*}

Simply-supported-beam | DegreeTutors.com
Fig. 1. Simply supported beam.

Looking again at the differential equation of the deflection curve, we see that we need expressions that describe the bending moment as a function of x. Looking at the loading here, we note that the bending moment diagram will not be described by one continuous function. The presence of two point loads means we’ll actually need three equations to fully describe how the bending moment varies along the beam; to this end we’ll consider the beam as three different regions:

  • Region 1: 0\leq x \leq 3
  • Region 2: 3\leq x \leq 6
  • Region 3: 6\leq x \leq 8

where x is measured from left to right with the origin at position A. The equations for M(x) are obtained by making cuts in the structure to reveal the internal bending moment and then evaluating the internal bending moment as a function of x by considering moment equilibrium of the sub-structure. If you’re unsure about any of that, take a detour over to this article on shear and moment diagrams for a refresher.

2.1 Internal Bending Moment in Region 1

To evaluate the internal bending moment in region 1, we cut the structure in this region to reveal the bending moment M(x). Our cut is made at a distance x to the right of support A, Fig. 2.

Simply-supported-beam-cut1 | DegreeTutors.com
Fig 2. Sub-structure created by the imaginary cut made in region 1. The cut reveals the internal bending moment in this region, M(x).

Now, we know the sub-structure is in equilibrium under the action of the internal shear (not shown) and the internal bending moment. So we can evaluate moment equilibrium in order to determine an expression for M(x).

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

    \begin{equation*}M(x)= 139.375x - 20\frac{x^2}{2}\end{equation*}

(2)   \begin{equation*}\boxed{M(x) &= -10x^2 + 139.375x}\end{equation*}

Remember that this equation is valid for values for 0\leq x \leq 3

2.2 Internal Bending Moment in Region 2

We can repeat the process now to determine the relevant equation for region 2. Fig. 3. shows the sub-structure created by the cut to reveal the internal bending moment.

Simply-supported-beam-cut2 | DegreeTutors.com
Fig 3. Sub-structure created by the imaginary cut made in region 2. The cut reveals the internal bending moment in this region, M(x).

Evaluating the sum of the moments about the cut as above,

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

    \begin{equation*}M(x) = 139.375x -75(x-3)-20\frac{x^2}{2}\end{equation*}

(3)   \begin{equation*}\boxed{M(x) &= -10x^2 + 64.375x+225}\end{equation*}

Again we note that this equation is valid for 3\leq x \leq 6

2.3 Internal Bending Moment in Region 3

Finally, we can establish the relevant equation for region 3, Fig. 4 below.

Simply-supported-beam-cut3 | DegreeTutors.com
Fig 4. Sub-structure created by the imaginary cut made in region 3. The cut reveals the internal bending moment in this region, M(x).

Evaluating moment equilibrium of the sub-structure yields, 

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

    \begin{equation*}M(x) = 139.375x -75(x-3)-20\frac{x^2}{2}-50(x-6)\end{equation*}

(4)   \begin{equation*}\boxed{M(x) &= -10x^2 + 14.375x+525}\end{equation*}

And again, for completeness we’ll note that this equation is valid only for 6\leq x \leq 8.

3.0 Integrating the Differential Equation of the Deflection Curve

Now that we have established how the bending moment varies, we can substitute our relevant expressions for M(x) into the differential equation and perform our integrations. After substituting our expressions for M(x) into equation 1 we have,

(5)   \begin{equation*}EI \frac{\mathrm{d}^2v}{\mathrm{d}x^2} = -10x^2 + 139.375x\:\:(\text{Region } 1)\end{equation*}


(6)   \begin{equation*}EI \frac{\mathrm{d}^2v}{\mathrm{d}x^2} = -10x^2 + 64.375x+225\:\:(\text{Region } 2)\end{equation*}


(7)   \begin{equation*}EI \frac{\mathrm{d}^2v}{\mathrm{d}x^2} = -10x^2 + 14.375x+525\:\:(\text{Region } 3)\end{equation*}

Integrating each expression yields, 

(8)   \begin{equation*}EI \frac{\mathrm{d}v}{\mathrm{d}x} = -10\frac{x^3}{3} + 139.375\frac{x^2}{2} + C_1 \:\:(\text{Region } 1)\end{equation*}


(9)   \begin{equation*}EI \frac{\mathrm{d}v}{\mathrm{d}x} = -10\frac{x^3}{3} + 64.375\frac{x^2}{2} + 225x + C_2  \:\:(\text{Region } 2)\end{equation*}


(10)   \begin{equation*}EI \frac{\mathrm{d}v}{\mathrm{d}x} = -10\frac{x^3}{3} + 14.375\frac{x^2}{2} + 525x + C_3  \:\:(\text{Region } 3)\end{equation*}

Here we note that our integration has generated three unknown constants of integration, C_1, C_2 and C_3. We also note that we now have the term \frac{\mathrm{d}v}{\mathrm{d}x} in our equations which corresponds to the slope of the deflection curve. We need to perform one further integration to reduce this back to the displacement itself, v. This integration yields,

(11)   \begin{equation*}EI v = -10\frac{x^4}{12} + 139.375\frac{x^3}{6} + C_1x + C_4 \:\:(\text{Region } 1)\end{equation*}


(12)   \begin{equation*}EI v = -10\frac{x^4}{12} + 64.375\frac{x^3}{6} + 225\frac{x^2}{2} + C_2 x + C_5  \:\:(\text{Region } 2)\end{equation*}


(13)   \begin{equation*}EI v = -10\frac{x^4}{12} + 14.375\frac{x^3}{6} + 525\frac{x^2}{2} + C_3x + C_6  \:\:(\text{Region } 3)\end{equation*}

Again we can see that this integration has given us a further 3 constants of integration, C_4, C_5 and C_6. In total we have six unknown constants that we need to identify. The good news is that we now finally have an equation for the deflection v in each region.

3.1 Finding the constants of integration

In order to find the constants of integration, we need some conditions or constraints that we can represent in equation form. As we have six unknowns to solve for, we’re going to need 6 constraint equations. These are as follows:

  1. at x=3, the slopes \frac{\mathrm{d}v}{\mathrm{d}x}, in regions 1 and 2 are the same.
  2. at x=3, the deflections v, in regions 1 and 2 are the same.
  3. at x=6, the slopes \frac{\mathrm{d}v}{\mathrm{d}x}, in regions 2 and 3 are the same.
  4. at x=6, the deflections v, in regions 2 and 3 are the same.
  5. at x=0 (support A), the deflection v is zero.
  6. at x=8 (support D), the deflection v is zero.

The first four conditions are referred to as continuity conditions and result directly from the fact that the beam and therefore deflections and slopes are continuous. The final two are classic boundary conditions. We can now use these statements to build six equations from which the constants of integration can be identified.

Condition (1)

At x=3, the slopes \frac{\mathrm{d}v}{\mathrm{d}x}, in regions 1 and 2 are the same. Therefore we can equate equations 8 and 9 and substitute in x=3.

    \begin{equation*}-10\frac{3^3}{3} + 139.375\frac{3^2}{2} + C_1 = -10\frac{3^3}{3} + 64.375\frac{3^2}{2}+225(3)+C_2\end{equation*}


(14)   \begin{equation*}\boxed{C_1 - C_2 = 337.5}\end{equation*}

Condition (2)

At x=3, the deflections v, in regions 1 and 2 are the same. So, equating equations 11 and 12 with x=3 gives us,

    \begin{equation*}-10\frac{3^4}{12}+139.375\frac{3^3}{6}+C_1(3)+C_4 = -10\frac{3^4}{12}+64.375\frac{3^3}{6}+225\frac{3^2}{2} + C_2(3) + C_5 \end{equation*}


(15)   \begin{equation*}\boxed{3C_1 + C_4-3C_2-C_5=675}\end{equation*}

Condition (3)

At x=6, the slopes \frac{\mathrm{d}v}{\mathrm{d}x}, in regions 2 and 3 are the same, so using equations 9 and 10 with x=6,

    \begin{equation*}-10\frac{6^3}{3}+64.375\frac{6^2}{2}+225(6) + C_2 = -10\frac{6^3}{3} + 14.375\frac{6^2}{2}+525(6) + C_3\end{equation*}


(16)   \begin{equation*}\boxed{C_2-C_3 = 900}\end{equation*}

Condition (4)

At x=6, the deflections v, in regions 2 and 3 are the same. Equating equations 12 and 13 with x=6

    \begin{equation*}-10\frac{6^4}{12}+64.375\frac{6^3}{6}+225\frac{6^2}{2}+C_2(6) + C_5=-10\frac{6^4}{12}+14.375\frac{6^3}{6}+525\frac{6^2}{2}+C_3(6)+C_6\end{equation*}


(17)   \begin{equation*}\boxed{6C_2 + C_5-6C_3-C_6=3600}\end{equation*}

Condition (5)

The fifth condition is a standard boundary condition; at x=0 the deflection v is zero. So we can let equation 11 equal zero with x=0

    \begin{equation*}0=-10\frac{0^4}{12}+139.375\frac{0^3}{6}+C_1(0)+C_4\end{equation*}


(18)   \begin{equation*}\boxed{C_4=0}\end{equation*}

Condition (6)

The final condition relates to the other boundary; at x=8 the deflection v is also zero. So applying this to equation 13 with x=8 gives,

    \begin{equation*}0=-10\frac{8^4}{12}+14.375\frac{8^3}{6}+525\frac{8^2}{2}+C_3(8) + C_6\end{equation*}


(19)   \begin{equation*}\boxed{C_6 + 8C_3 = -14613.334}\end{equation*}

Now that we have our six equations, we need to use them to solve for the unknown constants. By far the easiest way to do this is to arrange them in matrix form and solve the system by inverting the matrix of coefficients. The matrix representation of the system is,

(20)   \begin{equation*}\underbrace{\begin{bmatrix}1 & -1 & 0 & 0 & 0 & 0\\3 & -3 & 0 & 1 & -1 & 0\\0 & 1 & -1 & 0 & 0 & 0\\0 & 6 & -6 & 0 & 1 & -1\\0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 8 & 0 & 0 & 1\\\end{bmatrix}}_{A} = \underbrace{\begin{Bmatrix}C_1\\C_2\\C_3\\C_4\\C_5\\C_6\end{Bmatrix}}_{C} = \underbrace{\begin{Bmatrix}337.5\\675\\900\\3600\\0\\-14613.3\end{Bmatrix}}_{B}\end{equation*}

The vector of unknown constants are obtained as,

(21)   \begin{equation*}\begin{Bmatrix}C\end{Bmatrix} = \begin{bmatrix}A\end{bmatrix}^{-1}\begin{Bmatrix}B\end{Bmatrix}\end{equation*}

At this point we need a way to invert a matrix, and as it’s a 6\times 6 matrix, we won’t be doing this by hand! I’ll use the following Python code to perform the operation in equation 21.

import numpy as np #Numpy for working with arrays

#Define each row of the coefficient matrix
row1 = [1, -1, 0, 0, 0, 0]
row2 = [3, -3, 0, 1, -1, 0]
row3 = [0, 1, -1, 0, 0, 0,]
row4 = [0, 6, -6, 0, 1, -1]
row5 = [0, 0, 0, 1, 0, 0]
row6 = [0, 0, 8, 0, 0, 1]

A = np.mat([row1,row2,row3,row4,row5,row6]) #Define matrix of coefficients
B = np.array([[337.5],[675],[900],[3600],[0],[-14613.334]]) 

C = A.I*B #Unknown constants

If you want to get Python set up on your machine, you can follow this lecture. This will get you set up with a handy Python coding environment. Assuming you’ve done that, or have your own way of inverting matrices, the constants evaluate to,

    \begin{equation*}\begin{Bmatrix}C\end{Bmatrix}=\begin{Bmatrix}-856.354\\-1193.854\\-2093.854\\0\\337.5\\2137.5\end{Bmatrix}\end{equation*}

4.0 Calculating Beam Deflection

At this point we can summarise the three equations that describe the deflection in the three regions of our beam:

(22)   \begin{equation*}EI v = -10\frac{x^4}{12} + 139.375\frac{x^3}{6}  -856.354x \:\:(\text{Region } 1)\end{equation*}


(23)   \begin{equation*}EI v = -10\frac{x^4}{12} + 64.375\frac{x^3}{6} + 225\frac{x^2}{2} -1193.854x + 337.5  \:\:(\text{Region } 2)\end{equation*}


(24)   \begin{equation*}EI v = -10\frac{x^4}{12} + 14.375\frac{x^3}{6} + 525\frac{x^2}{2} -2093.854x + 2137.5  \:\:(\text{Region } 3)\end{equation*}

4.1 Mid-span deflection

To calculate the mid-span deflection, we substitute x=4 into equation 23 giving us,

    \begin{equation*}EI v = -10\frac{4^4}{12} + 64.375\frac{4^3}{6} + 225\frac{4^2}{2} -(1193.854)(4) + 337.5\end{equation*}


    \begin{equation*}\boxed{v(4) = \frac{-2164.6}{EI}}\end{equation*}

Now that we have a complete definition of the deflection in the beam, we can plot it to get a better sense of the deflected shape. Fig. 5 below shows a plot of the internal bending moment and the deflected shape. Note the y-axis gives the deflection is a function of EI

Simply supported beam moment and deflection | DegreeTutors.com
Fig. 5. Simply supported beam, plot of bending moment and plot of deflected shape.

4.2 Location of Maximum Deflection

We can see from Fig. 5 above that despite the unsymmetrical loading, the maximum deflection occurs very close to the mid-span location. We can confirm the exact location of the maximum deflection by recognising that at this location, the slope of the deflection curve is zero. In other words, a tangent to the deflection curve at the point of max deflection would be horizontal and therefore have a slope of zero. 

We know by inspection that the max deflection occurs in region 2. But let’s assume we didn’t know this. We can let each equation for the slope of the deflection curve, \frac{\mathrm{d}v}{\mathrm{d}x}, equations 8, 9 and 10, equal to zero and solve for the roots of each equation, i.e. the values of x at which the slope is zero. I’m going to let Python do the manual labour here…

import numpy as np 

#Define the polynomials
p1 = np.poly1d([-10/3, 139.375/2, 0, -856.354]) #Region 1
p2 = np.poly1d([-10/3, 64.375/2, 225, -1193.854]) #Region 2
p3 = np.poly1d([-10/3, 14.375/2, 525, -2093.854]) #Region 3

#Extract the roots
rootRegion1 = p1.r #Region 1 
rootRegion2 = p2.r #Region 2
rootRegion3 = p3.r #Region 3

The roots are,

    \begin{align*}&[20.282, 3.885, -3.26] \:\:\text{Region 1}\\&[12.747, -7.067, 3.976] \:\:\text{Region 2}\\&[-13.274, 11.208, 4.222] \:\:\text{Region 3}\\\end{align*}

Values that fall outside the boundaries of the relevant region can be immediately discarded. This leaves only x=3.976 in region 2. As we suspected, this is very close to the mid-span location. We could now substitute this value back into equation 12 to confirm the value of max deflection but since we’ve already worked out the deflection at x=4, we won’t bother doing this.

5.0 Using Superposition to Calculate Beam Deflection

We’ve seen above how to determine beam deflection from first principles. It has given us a complete picture of the deflection but it has been a relatively lengthy process to get here. We can make use of the principle of superposition to arrive at an answer for the mid-span deflection much faster by making use of tabulated formulae for beam deflection. These formulae have already been determined and tabulated for common load cases using the technique we’ve just demonstrated. By evaluating the mid-span deflection for each load individually and summing the deflections induced, we obtain the same result identified above.

5.1 Uniformly Distributed Load

Consider the formula for the deflection of the beam subject to a uniformly distributed load, Fig. 6,

(25)   \begin{equation*}v_1 = -\frac{wx}{24EI}(L^3-2Lx^2 + x^3)\end{equation*}

At x=4, this formula evaluates to v_1=-1066.67/EI

Simply-supported-beam-case-1 | DegreeTutors.com
Fig. 6. Simply supported beam subject to a uniformly distributed load.

5.2 Point Load #1

The formula for the deflection of a beam subject to a single point load, Fig. 7, where the distance a is less than the distance to the position at which the deflection is being evaluated, x is,

(26)   \begin{equation*}v_2 = -\frac{Pbx}{6LEI}(L^2-b^2-x^2) - \frac{P(x-a)^3}{6EI}\end{equation*}

This evaluates to v_2 = -731.25/EI

Simply-supported-beam-case-2 | DegreeTutors.com
Fig. 7. Simply supported beam subject to a point load P=75\:kN at a distance a=3\:m from the left support and b=5\:m from the right support.

5.3 Point Load #2

Finally, evaluating the formula for deflection where x>a, Fig. 8, 

(27)   \begin{equation*}v_3 = -\frac{Pbx}{6LEI}(L^2-b^2-x^2) \end{equation*}

This evaluates to v_3 = -366.67/EI

Simply-supported-beam-case-3 | DegreeTutors.com
Fig. 8. Simply supported beam subject to a point load P = 50\:kN at a distance a=6\:m from the left support and b=2\:m from the right support.

Summing up the three mid-span deflections yields,

    \begin{align*}v_{\text{total}} &= -\frac{1066.67}{EI} -\frac{731.25}{EI} - \frac{366.67}{EI}\\v_{\text{total}} &= -\frac{2164.6}{EI}\end{align*}

Sure enough, this is the same value we obtained above. By plotting these equations, we can further visualise the contribution of each load towards the overall deflected shape, Fig. 9. 

Superposition_deflection | DegreeTutors.com
Fig. 9. Total deflected shape obtained as the superposition of individual deflections from each load considered separately.

So that about wraps up our discussion of beam deflection. In the end, it’s up to you what approach you decide to take to calculate deflections. There are of course other methods that you may use to evaluate deflection, but in any event it’s good to have an understanding of how we can go about it from first principles. Remember, just like any derivation, this one has its limiting assumptions, discussed above. Everything we’ve discussed about is only valid provided we satisfy these limiting assumptions. 

That’s all for this one, I’ll see you in the next one.


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