Mechanics & Structures University Engineering

Advanced Beam Theory

Chief Noter

July 23, 2021
6 min read
Elastic Failure Plastic Strain Stress

Stress Gradients
Shear Stresses in Beams
Plastic Bending in Beams
Plastic Hinges
Residual Stresses
Buckling of Struts
Summary

Stress Gradients

For small elements, we can assume the pane stress equilibrium applies: $\tau_{xy}=\tau_{yz}$

We know that the stresses along a beam can vary, and so if we look at a larger element, there must be a stress gradient. Performing an equilibrium of forces on these stresses, and the body force $X$ leads us to the stress equilibrium equations:

$\frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{yx}}{\partial y}+X=0$

$\frac{\partial\sigma_y}{\partial y}+\frac{\partial\tau_{xy}}{\partial x}+Y=0$

Body forces X and Y are often negligible.

Shear Stresses in Beams

We know that from the notes sheet on beam theory that at any section throughout the beam:

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

Substituting this into the stress equilibrium equations above gives:

$\partial\tau_{xy}=-\frac{\partial M}{\partial x}\frac{y}{I}\partial y$

$\partial \tau_{xy}=-\frac{S}{I}y \partial y$

Where $S$ is the shear force ating on the cross-section:

$S=\frac{\partial M}{\partial x}$

Integrating and applying the boundary conditions for zero shear stress at the top and bottom ( $\tau_{xy}=0 @ \pm \frac{d}{2}$ ) gives us the shear stress distribution throughout the section:

$\tau_{xy}=\frac{Sy^2}{2I}+\frac{Sd^2}{8I}$

The maximum shear stress occurs at the neutral axis, $y=0$ :

$\hat\tau_{xy}=\frac{Sd^2}{8I}$

Shear stresses are much smaller than axial stresses, and are only really important in short, stubby beams. For a cantilever, this is when L = ¼D.

I-Beams

Also known as rolled steel joists (RSJs), these are by far the most common form of steel beams because of their immense rigidity – weight ratio. The shear force distribution is as follows:

The majority of the tensile stresses are in the flanges, due to bending
The majority of the shear stresses are in the web, due to shear forces

The shear stress distribution is discontinuous, due to the varying cross-section. The maximum still occurs at the neutral axis:

$\hat\tau_{xy}=\frac{S}{8I}[d^2+\frac{B}{t}(D^2-d^2)]$

Meanwhie the maximum direct stress is:

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

The second moment of area is given as:

$I_{xx}=\frac{BD^3}{12}-{bd^3}{12}$

Lightening a Beam

Often, you see I-beams with cut outs in the flange. This reduces the weight significantly without affecting its strength (because shear stresses are so much lower than direct stresses):

The beam on the right has a segnificantly higeher second moment of area but weights the same (if not less) than the beam on the left.

Plastic Bending of Beams

When a beam is bent beyond its yield point, plastic deformation occurs. In real life, the stress continues to increase beyond this point through work hardening, but for this analysis we will model materials as elastic-perfectly plastic:

Elastic-perfctly plastic materials experience a constant stress after yielding.

Elastic Bending

This folllows the standard stress law that we know and love:

$\sigma_{x}=\frac{My}{I}$

If there is also an axial stress applied, the line is simply shifted o the left or right by its magnitude. For a rectangular beam width $b$ and height $d$ , the stress distribution then is:

$\sigma_x=\frac{6M_e}{bd^2}=\frac{F}{bd}$

$M_e$ is the elastic bending moment
$F$ is the applied axial force

Plastic Yield

If the stress in the beam reaches the yield stress, $\sigma_Y$ , a plastic region forms around an elastic core:

As the stress increases, the elastic core shrinks until the entire cross-section has deformed plastically and permanently.

The total plastic bending moment, $M_p$ , at the top and bottom is given as:

$M_p=\frac{\sigma_Y bd^2}{4}$

The shape factor, $f$ , represents the safety margin between initial yielding (when plastic deformation begins at the top and bottom) and total failure (when the deformation is fully plastic):

$f=\frac{M_p}{M_e}$

For a rectangular cross-section, $f=1.5$
For an I-Beam, $1.12 < f < 1.15$
For a diamond cross-section, $f=2.0$
For a circular cross-section, $f=1.698$

Asymmetric Cross-Sections

If the cross-section is not symmetric about the neutral axis, the neutral axis itself will change place during fully plastic bending:

During elastic bending, the neutral axis passes through the centroid of the cross-section
For compound cross-sections, use the parallel axis theorem
During plastic bending, the neutral axis splits the cross-section into two equal areas

Plastic Hinges

When the load on a beam is concentrated at a singular point, the greatest plastic bending moment occurs here. Often, the beam will not just snap immediately, however, and there will be a region of decreasing plastic deformation on either side of it.

This is called a plastic hinge.

Residual Stresses

When a beam has been loaded beyond yield, the permanent plastic stresses remain in the beam even if it is fully unloaded. These stresses are called residual stresses.

In a beam that deforms plastically but does not bend, the plastic stress is locked in as a constant throughout the cross-section.

If the beams bends plastically and is then unloaded elastically, a more complicated residual stress
distribution is left behind:

Buckling of Struts

Struts are compressive beams that support a greater structure. Buckling can cause instability and failure at loads significantly lower than the yield stress in the material.

The principle of buckling can be thought of in terms of elastic behaviour, where the struts bend under compression:

$kxL>Px$ Strut returns upright
$kxL<Px$ Strut is unstable and collapses

The ccritical load for buckling, then, is:

$P_c=kL$

Pin Jointed Struts

Euler’s critical buckling load:

$P_c]\frac{n^2\pi^2EI}{L^2}$

Where $n$ is the number of constraints. The central deflection, $u$ , is given as a sinusoidal function. $A$ is a constant to be determined:

$u=A\sin{\frac{\pi y}{L}}$

Built-In Struts

The example on the left experiences convex bending at the top and bottom, and concave bending in the central section (of length $\frac{L}}{2}$ . The critical load is:

$P_c=\frac{4\pi^2 EI}{L^2}$

The example on the right can be treated as a pin joint of twice its length:

$P_c=\frac{\pi^2 EI}{4L^2}$

Stresses in a Loaded Strut

The total axial stresses in a loaded strut are the sum of the applied load and those due to bending:

$\hat\sigma=-\frac{P}{A}+\frac{My}{I}$

The shear force acting on the cross section of a beam is:
$S=\frac{\partial M}{\partial x}$
The shear stress distribution is given as:
$\tau_{xy}=-\frac{Sy^2}{2I}+\frac{Sd^2}{8I}$
The plastic bending moment for a rectangular beam is:
$M_p=\frac{\sigma_Y bd^2}{4}$
With the form factor relating the elastic and plastic bending moments:
$f=\frac{M_p}{M_e}$
Residual stresses remain in a member even when all loads are removed, due to permanent plastic deformation.
The critical buckling load for a pin jointed strut is:
$P_c=\frac{n^2\pi^2EI}{L^2}$
The critical buckling load for a built-in strut depends on the form of buckling (see above)

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Stress Gradients