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Beam Theory - EngineeringNotes.net

Mechanics & Structures University Engineering

Beam Theory

Chief Noter

July 27, 2021
8 min read
Beams Bending Stress Stress Analysis Supports

Regardless of their shape, length, supports or loads, all beams bend slightly. However, the extent of this
bending depends hugely on these factors. This is where beam theory comes in, to calculate the bending & deflection of beams.

Supports
Shear Forces & Bending Moments
Stress & Second Moment of Area
Deflection of Beams
- Macaulay’s Method
- Superposition
Summary

Supports

Simply Supported Beams

Simply supported beam, bending & deflection of beams

There is a vertical reaction at both ends
Deflection at each end is zero
Angle at each end is not necessarily zero
Maximum deflection is in the centre (when no loads are applied)

Bi-Support Cantilever

One end has two vertical reaction forces next to one another, pointing in opposite directions
Deflection at fixed end is zero
Deflection at cantilevered end is always maximum
Angle at fixed end is not necessarily zero

Built-in Cantilever

Deflection at built-in end is zero
Angle at built in end is zero
Maximum deflection is always at cantilevered end
The built-in end provided both a moment and a reaction force

Built-in at Both Ends

Deflection at both ends is zero
Angle at both ends is zero
Maximum deflection is in the middle (when no loads are applied)
There is a moment and reaction force at both ends

Remember: a built in end provides a moment!

Finding Shear Force & Bending Moments

Often, shear force and bending moment diagrams need to be drawn in order to find the maximum
bending moment in a beam. To do this, the beam is split into regions – areas between two changes
in load, and each region is sectioned in the middle

Finding shear force and bending moments in beams, beam theory, bending & deflection of beams

Asimply supported beam with no loads on it has only one region. Add a point load in the middle,
and there are two regions, one on either side, etc.

The example above has four regions:

From the left end to $F_1$ , with the distributed load $W$ acting throughout
From $F_1$ to the end of the distributed load
From the end of the distributed load to $F_2$
From $F_2$ to the right end

Note that the arrow labelled $W$ only represents the total weight of the distributed load, nothing
changed in terms of load at this point, so there is no new region.

Sectioning Example & Shear Force/Bending Moment Diagrams

It is important to stick to the sign convention whenever solving problems that involve sectioning
beams:

sign convention for beam theory, beam theory sign convention

Now let’s add some numbers to the example above and go through the calculations, step by step:

1. Taking moments about the left side to find the reaction forces

$12(3)+5(5)=5(9)=10R_2$

$R_2=10.6 N$

$\sumF_y=0: \quad R_1+R_2=12+5+5$

$R_1=11.4 N$

2. Section in the middle of region 1

$\sum F_y=0: \quad 11.4 + F_1 = 2x$

$F_1 = 2x - 11.4$

$\sum M = 0: \quad 11.4x + M_1=2x(\frac{x}{2})$

$M_1=x^2-11.4x$

3. Section in the middle of region 2

$\sum F_y = 0: \quad 11.4 + F_2 = 2x+5$

$F_2=2x-6.4$

$\sum M = 0: \quad 11.4x+M_2=2x(\frac{x}{2}) +5(x-5)$

$M_2=x^2-6.4x-25$

4. Section in the middle of region 3

$\sum F_y=0: \quad 11.4+F_3=12+5$

$F_3=5.6 N$

$\sum M=0: \quad 11.4 x + M_3=12(x-3)+5(x-5)$

$M_3=5.6x-61$

5. Substitute in the region boundaries

Shear Forces

	Left Limit	Right Limit
$F_1=2x-11.4$	$x=0: \quad F=-11.4$	$x=5: \quad F=-1.4$
$F_2=2x-6.4$	$x=5:\quad F=3.6$	$x=6:\quad F=5.6$
$F_3=5.6$	$x=6:\quad F=5.6$	$x=9:\quad F=5.6$
$F_4=R_2$		$x=9:\quad F=10.6$

Bending Moments

	Left Limit	Right Limit
$M_1=x^2-11.4x$	$x=0:\quad M=0$	$x=5:\quad M=-32$
$M_2=x^2-6.4x-25$	$x=5:\quad M=-32$	$x=6:\quad M=-27.4$
$M_3 = 5.6x-61$	$x=6: \quad M=-27.4$	$x=9:\quad M=-10.6$

The moment at the right end must also equal zero.

Now we can draw the shear force and bending moment diagrams:

We can clearly see the maximum bending moment is -32 Nm. This is needed to calculate the
maximum stress in the beam, and the deflections.

Stress & Second Moment of Area

Often, we need to know the maximum stress in a beam, to predict where it will break. This is given
by the equation:

$\sigma=\frac{My}{I}$

$M$ is the maximum bending moment
$y$ is the distance from the neutral axis of the stress
$I$ is the second moment of area

The units of second moment of area are m⁴.

Second Moment of Area, $I$

The second moment of area is a property of a cross-section. For rectangles and circles it is given
as:

Second moments of area can be added and subtracted if the cross sections lie on the same
neutral axis:

Addition and subtraction of second moments of area, compound second moments of area, adding second moments of area, second moment of area of hollow shapes

However, if they do not lie on the same neutral axis, the parallel axis theorem must be used. Take
the T-section as an example:

First, the overall neutral axis must be found:

$\bar y(A_{total})=y_1(A_1)+y_2(A_2)$

Then, the overall second moment of area can be found:

$I_{xx}=[I_1+A_1(y_1-\bar y)^2]+[I_2+A_2(\bar y - y_2)^2]$

This equation applies for a T-section, where y₁ is above the overall neutral axis and y₂ is
below the overall neutral axis. The squared brackets should be replaced by the distance
between the section’s and the overall neutral axes.

Deflection of Beams

A measure of the deflection of a beam is the radius of curvature, R. This brings together everything
important about beam theory, through the fundamental equation:

$\frac{\sigma}{y}=\frac{M}{I}=\frac{E}{R}$

$\sigma$ is the maximum stress in the beam
$y$ is the distance from the neutral axis
$M$ is the maximum bending moment
$I$ is the second moment of area
$E$ is the Young’s modulus
$R$ is the radius of curvature

Often, the second moment of area and Young’s Modulus are combined to give a constant the
structural rigidity, $EI$ .

Rearranging this equation gives $R$ , the second derivative of deflection:

$R=\frac{d^2\nu}{dx^2}=\frac{1}{EI}M$

Integrating this gives the angle of deflection, also known as slope:

$\frac{d\nu}{dx}=\frac{1}{EI}\int M dx$

Integrating again gives the deflection of the beam:

$\nu=\frac{1}{EI}\int \int M dx$

There are two ways of solving for deflection, Macaulay’s method and superposition:

Macaulay’s Method

This defines the bending moment as a step function in terms of Macaulay brackets (pointy
brackets). Each Macaulay bracket is turned on or off, depending on whether or not it applies to the
part of the beam you are looking at.

Macaulay brackets are integrated as single entities, not using reverse chain rule.

To find the Macaulay function, section the beam in the last region, using Macaulay brackets for
the distances.

Using this beam setup, sectioned at the end, we get:

$\frac{d^2\nu}{dx^2}=\frac{1}{EI}[wa\langle x-\frac{a}{2}\rangle + P\langle x-b\rangle - R \langle x\rangle +M]$

The first term in the brackets only applies when $x>a$
The second term in the brackets only applies when $x>b$
The third term in the bracket always applies

Integrating the above once gives the slope:

$\frac{d\nu}{dx}=\frac{1}{EI}[\frac{wa}{2}\langle x-\frac{a}{2}\rangle^2 +\frac{P}{2}\langle x-b\rangle^2 - \frac{R}{2}\langle x\rangle^2 +M\langle x \rangle +A]$

Integrating again gives the deflection when $x>a$ :

$\nu=\frac{1}{EI}[ \frac{wa}{6}\langle x-\frac{a}{2}\rangle^3 +\frac{P}{6}\langle x-b\rangle^3 - \frac{R}{6}\langle x\rangle^3 +\frac{M}{2}\langle x \rangle^2 +A\langle x\rangle+B]$

When integrating the function for the bending moment, the two constants of integration can be
found from the types of support, e.g. at a built-in support, both slope and deflection are zero,
whereas at a simple support, only deflection is zero (see Supports).

Therefore:

$@x=0:\quad \frac{d\nu}{dx}=0 \rightarrow A=0 \quad \nu=0 \rightarrow B=0$

Superposition

Alternatively, you can treat the different loads on a beam individually, find the deflection caused
by each one, and then add these all together. To do this, use standard results:

Simply Supported Beams:

Moment-Held Beams:

Built-in Cantilevered Beams:

deflection & slope of built in cantilever beams

Built-in at both ends:

Remember that if you want to find the end deflection of a load that does not go all the way to the
end, you need to extrapolate the deflection by multiplying the slope by the length to the end of the
beam:

To find the shear forces and bending moments along a beam, section it and resolve forces & moments
The second moment of area for a rectangle is given as $I=\frac{bd^3}{12}$
The second moment of area for a circle is given as $I=\frac{\pi r^4}{4}$
Second moments of area can be added and subtracted if they share a neutral axis. If they do not, the parallel axis theorem must be used.
Deflection of beams and maximum bending moments can be calculated with the fundamental beam equation: $\frac{\sigma}{y}=\frac{M}{I}=\frac{E}{R}$
Integrating the radius of curvature gives the slope
Integrating the slope gives the deflection

Beam Theory

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