Torsion is the twisting of an object caused by a moment acting about the longitudinal axis of the object. A moment that tends to cause twisting is called torque.
The video below takes an in-depth look at torsion, including how to calculate the angle of twist, and the shear stresses and shear strains associated with torsional deformation. It also covers how ductile and brittle materials fail under torsional loads.
Stress and strain are fundamental concepts in engineering, and in strength of materials in particular, that describe how an object responds to applied loads. This page covers the basics, which are also summarised in the following video:
When a body like the bar shown below is loaded by external forces, internal forces develop within it to resist the applied forces. We can visualise these forces by making an imaginary cut through the object. The internal forces develop in such a way that they balance the external forces, to maintain equilibrium.
Instead of discussing the magnitude of these imaginary internal forces, it’s more useful to talk about how the internal forces are distributed over a surface, using a parameter called stress, that quantifies the internal force per unit area. Stress describes the distribution of internal forces within a body.
The concept of stress gives us a way of describing the internal state that develops within a body as it responds to externally applied loads. This is important because it allows us to predict when an object will fail. By comparing the stress in a body with the yield or ultimate strengths of the material, obtained from tests, we can estimate how close an object is to deforming permanently or to fracturing. Using stress to predict failure is covered in more detail in the page on failure theories.
Stress is split into two different types, normal stress and shear stress, that depends on whether the internal forces are perpendicular or parallel to the surface of interest.
The type of stress where the internal forces act perpendicular to a surface, like for the bar shown above, is called normal stress. This type of stress is denoted using the symbol $\sigma$ (the greek letter sigma), and is calculated as the force divided by the area over which it acts.
Stress is a measure of the internal force per unit area, so it has units of Newtons per square meter ($\mathrm{N/m^2}$) in SI units and pounds per square inch in US units. Units of $N/m^2$ are also called Pascals ($\mathrm{Pa}$).
Normal stresses can be either tensile, when the object is getting stretched, or compressive, when it is getting compressed.
The type of stress where the internal forces act parallel to a surface is called shear stress.
It can be calculated in a similar way to normal stress, as the applied force $F$ divided by the cross-sectional area $A$. Shear stress is denoted using the symbol $\tau$ (the greek letter tau).
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You might be wondering why the stress at a single point should change from a shear stress to a normal stress or the other way around depending on how we choose to orient our imaginary cut through the bar.
The truth is that it doesn’t. The stress state at a single point within a body will actually have components in the normal and shear directions. The magnitude of the normal and shear components depends on the angle of the plane being used to observe the stresses, as illustrated below.
This is because stress is actually a tensor quantity – at each point within a body the stress state can be represented by three normal stresses and six shear stresses. These stress components are shown on the stress element, a small cube that is used to represent the stress state at a single point within a larger body. You can learn more about the stress tensor and stress element by having a read of our page on stress transformation and Mohr’s circle.
Strain is a quantity that describes the deformations that occur within a body. Like with stress, we can differentiate between normal strain and shear strain.
Normal strain is defined as a change in length $\Delta L$ divided by the original length $L_0$. This makes it a dimensionless quantity, that is often expressed as a percentage.
Shear strain is a measure of the deformation caused by a shear stress. It corresponds to the change in angle between two lines that are initially perperpendicular to one another, as shown in the image below. Like normal strain, shear strain is a dimensionless quantity that is usually expressed as a percentage.
Stress and strain are two closely linked parameters – it makes sense that the internal forces that develop within a body (stress) depend on how much the body is being deformed (strain).
The relationship between these two parameters can be described using a stress-strain diagram. Stress-strain diagrams are different for different materials. The image below shows typical diagrams for glass, steel and rubber.
The diagram for a specific material can be obtained by performing a tensile test, which involves applying a known force to a test piece and measuring the normal stress and normal strain in the test piece as the applied force is increased.
For many materials the relationship between stress and strain is linear for relatively low values of stress and strain. In this region the linear relationship between normal stress and normal strain is defined by Hooke’s law, where $E$ is Young’s modulus, a material property that defines the stiffness of a material.
Other important material properties that can be determined from the stress-strain curve include strength, ductility and toughness.
A steel bar is being designed to support a suspended 2000 kg mass, as illustrated below. The bar is 1 m in length and has a circular cross-section. The maximum allowable stress in the bar is 80% of the material yield strength and the selected steel has a yield strength of 300 MPa. What is the minimum allowable radius of the bar, to the nearest 0.1 mm?
0.4 mm
1.1 mm
5.1 mm
6.4 mm
The tensile stress in the bar is:
$$\sigma = \frac{F}{A} = \frac{9.81 \cdot 2000}{\pi r^2}$$
The maximum allowable stress is:
$$\sigma_{max} = 0.8 \sigma_y$$
The limiting radius can be calculated as follows by equating the tensile stress in the bar with the allowable stress:
$$\frac{9.81 \cdot 2000}{\pi r^2} = 0.8\sigma_y$$
$$r = \sqrt{\frac{9.81 \cdot 2000}{\pi 0.8 \sigma_y}} = 5.1 \mathrm{mm}$$
Young's Modulus
Young's modulus is a measure of the stiffness of a material – it describes the relationship between stress and strain.
Strength, Ductility & Toughness
Strength, ductility and toughness are three important material properties that can be determined from a stress-strain curve.
Shear force and bending moment diagrams are powerful graphical methods that are used to analyze a beam under loading. This page will walk you through what shear forces and bending moments are, why they are useful, the procedure for drawing the diagrams and some other keys aspects as well.
If you’re not in the mood for reading, just watch the video!
When loads are applied to a beam, internal forces develop within the beam in response to the loads. We can visualise these forces by making an imaginary cut through the beam and considering the internal forces acting on the cross-section. These internal forces have two components:
The internal forces develop in such a way as to maintain equilibrium. No matter where the imaginary cut is made along the length of the beam, the effect of the internal forces will always balance the effect of the external forces.
The normal stresses will be tensile on one side of the cross-section, and compressive on the other. If the beam is sagging the top of the beam will get shorter, and so the normal forces acting at the top will be compressive. The bottom of the beam will get longer, and so the normal forces acting at the bottom of the beam will be tensile.
These forces cancel each other out so they don’t produce a net force perpendicular to the beam cross-section, but they do produce a moment.
This means that the internal forces acting on the cross-section of the beam can be represented by one resultant force, called a shear force, that is the resultant of the internal shear forces, and by one resultant moment, called a bending moment, that is the resultant of the internal normal forces.
Shear force and bending moment diagrams are used to analyse and design beams. By showing how the shear force and bending moment vary along the length of a beam, they allow the loading on the beam to be quantified.
They are often used as a starting point for performing more detailed analysis, which might include calculating stresses in beams or determining how beams will deflect.
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The shear forces and bending moments along a beam do not depend on the geometry of the beam cross-section or the material the beam is made of. They depend on two factors only:
Let’s look at loads first. The most common ways of applying loads to a beam are concentrated forces, distributed forces, and concentrated moments. Distributed forces can be uniform, or non-uniform.
In the real world beams can be supported in many different ways, but for analysis purposes real supports can be modelled as one of three idealised supports – pinned supports, roller supports and fixed supports. Each of these supports restrains the beam in different ways.
When a certain degree of freedom (rotation or translation) is restrained at a support, there will be a corresponding reaction force or reaction moment at that location. At a pinned support for example a beam will be experiencing horizontal and vertical reaction forces, because horizontal and vertical displacements are restrained, but there will be no reaction moment because the beam can rotate at the pinned support.
A free body diagram is a simple sketch that shows all of the external loads acting on a beam and any reaction forces from supports. Figuring out the free body diagram is an important first step in determining the shear force and bending moment diagrams.
Consider the pinned support in the beam configuration shown above. The beam can rotate at this support, so there is no reaction moment, but displacements in the vertical and horizontal directions are prevented, so there will be horizontal and vertical reaction forces. At the roller support there is just a vertical reaction force. These three reaction forces and the applied distributed and concentrated forces are shown on the free body diagram.
What is the reaction moment at the fixed support for the cantilever beam configuration shown in the image below?
0 kNm
-50 kNm
-100 kNm
25 kNm
To ensure static equilibrium, the reaction moment at the fixed support must be equal and opposite to the sum of the external moments acting about that point.
$$M = 5 \cdot 10 - 10 \cdot 10 = -50\,\,\mathrm{kNm}$$
There are three main steps that need to be followed to determine the shear force and bending moment diagrams:
Although the external loads are known, the reaction forces and moments at the supports are not – they need to be calculated. This can be done by considering the fact that all of the external loads, both the applied loads and the loads at the supports, must balance each other. If they didn’t the beam would not be in static equilibrium.
This can be represented by three equilibrium equations. These equations state that the sum of the forces in the horizontal direction, the sum of the forces in the vertical direction, and the sum of the moments taken about any point must all be equal to zero. $$\sum{F_x} = 0$$ $$\sum{F_y} = 0$$ $$\sum{M} = 0$$
The reaction forces and moments at the supports can be calculated using these equations.
We now have all the information needed to determine the shear force and bending moment diagrams.
As mentioned in an earlier part of this article, if we make an imaginary cut through the beam at any location, the internal forces and moments acting on the cut cross-section must balance the external forces and moments. This means that we can once again apply the equilibrium equations to calculate the shear forces and bending moments along the beam.
Start at one end of the beam and move the location of the imaginary cut to the other end, applying the equilibrium equations and calculating the shear forces and bending moments as you move along the beam.
Doing this along the full length of the beam will give you the shear force and bending moment diagrams.
A consistent sign convention needs to be used when calculating shear forces and bending moments. The most common sign convention is as follows:
These sign conventions are summarised in the figure below.
Let’s work through an example to show the process of determining the shear force and bending moment diagrams from start to finish.
Example - Simply Supported BeamIn this example we’ll determine the shear force and bending moment diagram for a simply supported beam that is carrying two loads.
The first step is drawing the free body diagram. There are horizontal and vertical reaction forces at the pinned support (Point A) and a there is one vertical reaction force at the roller support (Point B).
Next we need to apply the equilibrium equations to determine the unknown reaction forces at Point A and Point B. The sum of the forces in the horizontal direction must be zero. $H_A$ is the only force in the horizontal direction, so it must be equal to zero:
$$\sum{F_x} = 0$$
$$H_A = 0 \mathrm{kN}$$
The sum of the forces in the vertical direction must be equal to zero:
$$\sum{F_y} = 0$$
$$R_B + R_A = 15 + 6$$
And the sum of the moments about any point must be equal to zero. If we choose to take the moments about Point B we get the following:
$$\sum{M_B} = 0$$
$$(-R_A \cdot 6) + (15 \cdot 5) + (6 \cdot 2) = 0$$
We can solve this equation to determine that:
$$R_B = 9 \mathrm{kN}$$
Substituting $R_B$ into the vertical equilibrium equation gives:
$$9 + R_A = 15 + 6$$
$$R_A = 12 \mathrm{kN}$$
All of the external loads acting on the beam have now been determined, and the free body diagram can be updated.
Now we can start to draw the shear force and bending moment diagrams, starting from the left side of the beam. We can draw the free body diagram showing a beam segment which has been cut immediately the right of the 12 kN reaction force at Point A.
To maintain equilibrium, the shear force V(x) must balance the 12 kN reaction force:
$$V(x) = 12 \mathrm{kN}$$
Similarly, the bending moment must balance the moment generated by the 12 kN reaction load:
$$M(x) = 12x$$
The shear force will be constant until we reach the next applied force, so we can draw this on shear force diagram. And the equation obtained for M(x) is the equation for a straight line, which can be drawn on the bending moment diagram.
The process is then repeated, moving the location of the imaginary cut further to the right. This time the cut is placed immediately after the 15 kN force, and the free body diagram is drawn. The shear force and bending moment can be calculated by applying the equilibrium equations.
The shear force and bending moment diagrams can be updated.
This process is repeated until the full length of the beam has been covered. The completed diagrams are shown below.
Trusses
Trusses are structures (e.g bridges or roofs) made up of members that can only carry tensile or compressive axial loads.
Area Moment of Inertia
The area moment of inertia describes how the material of a cross-section is distributed relative to a specific bending axis.
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