Mechanical advantage refers to the amplification of force and can be quantified as the ratio of the output force generated by a system to the input force applied to it. It is useful for a wide range of applications, like lifting heavy loads.
MECHANICAL ADVANTAGE
$$MA = \frac{F_{\mathrm{out}}}{F_{\mathrm{in}}}$$
The video below explores different methods that can be use to amplify a force, with a focus on levers, pulleys and gears, three types of machine that generate mechanical advantage.
Trusses are important structures in engineering that are used in all sorts of applications, from bridges and roofs to parts of the International Space Station. The reason they are so useful is that they allow us to create strong and rigid structures while using materials in a very efficient way.
This page will cover everything you need to know to understand and analyse trusses, but you can also watch the video below for the animated version!
A truss in everyday language is a rigid structure that is made up of a collection of straight members. But in an engineering and strength of materials context it has a more specific meaning – in these contexts a truss is a structure made up of members that only carry axial loads.
The members of a structure can be considered to only carry axial loads if the following assumptions are applicable:
Assumption 1
The joints of the structure behave like pinned connections
Assumption 2
Loads are only applied at the joints of the structure
If these assumptions can be applied, the truss members are not acted on by any bending moments – they only carry axial loads. This means that each member of a truss must be either in tension or in compression.
Truss members can only carry axial loads so they are either in tension or in compression
The fact that truss members only carry axial loads makes the analysis of trusses much easier than the analysis of frames, for example, where the members can carry both bending moments and axial loads.
Truss on the left (pinned connections and loads acting at joints only) and frame on the right (rigid connections and loads acting in the middle of a member)
When can a structure be analysed as a truss?
Few real structures have actual pinned connections, so how do you decide if you can analyse a real structure as a truss?
The members of real structures like bridges are often rigidly connected using what is known as a gusset plate, a plate that is bolted or riveted to several structural members, as shown in the photograph below. If the centre-lines of all of the members at a joint intersect at the same point, then it is reasonable to assume that the joint behaves in a similar way to a pinned connection.
Gusset plates on a bridge
For analysis purposes it is often reasonable to assume that any loads carried by the structure can be modelled as acting at the joints, instead of being distributed over the length of the members. This is usually valid if the members are relatively short.
This means that structures like the bridge shown above can often be analysed as trusses.
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The base shape of a truss is three members connected to form a triangle. If a load is applied to the structure the angles of the triangle won’t be able to change if the length of each of the members stays the same. This means that the triangle is a very stable shape that won’t deform when loads are applied to it, and so it is a great base from which to build a larger structure.
Joining four members together does not form a stable structure. The angles between members can change without any change in the length of the members, so using a four-sided shape as the base for building a truss structure would be a terrible choice. An easy way to stabilize this configuration is to add a diagonal bracing member to split it into triangles.
Unstable configuration made stable by adding a diagonal member
The figure below shows some common truss designs, which as you can see are made up of triangular elements.
Common truss configurations
Analysing Trusses
When analysing a truss the main objective is to determine the force in each of its members for a certain set of applied loads. This allows us to check that the members can carry the loads without failing, or gives us the information we need to select the best cross-section for each of the members. There are two main methods we can use to do this – the Method of Joints, and the Method of Sections.
The Method of Joints
The method of joints involves applying the static equilibrium equations to the joints in the truss to determine the forces acting on each joint. From there the tensile or compressive axial force carried by each member of the truss can easily be determined.
Here’s how it works:
Step 1
Draw a free body diagram of the truss and solve reaction forces
The first step is to draw a free body diagram showing all of the external loads acting on the truss.
The reaction forces at the truss supports can then be calculated using the three static equilibrium equations.
Draw a free body diagram of each joint and solve internal forces
The next step is to draw a free body diagram for every single joint, and work through them one by one to solve the unknown forces acting at the joint.
The unknown forces can be determined by once again applying the equilibrium equations. All of the joints are pinned connections so there are no moments – you only need to consider equilibrium of the horizontal and vertical forces.
$$\sum F_\leftrightarrow = 0$$
$$\sum F_\updownarrow = 0$$
This calculation gives the forces imparted on each of the joints by the members of the truss. The force each member imparts on a joint is equal and opposite to the axial force within the member. If the force acting on a joint because of a specific member is acting away from the joint, it means that the member is in tension, and if the force is acting towards the joint it means that the member is in compression.
Remember that each member is carrying a uniform axial load. This means that if you know the force acting on the joint at one end of the member, the force acting on the joint at the other end of the member will be the same.
The method of joints is useful when you need to calculate the forces in all of the members in the truss, but sometimes you only need to determine the force in one or two members. In this case an alternative method, called the Method of Sections is quicker.
The Method of Sections
When using the Method of Sections, an imaginary cut is made through the truss. Beause the internal forces within the truss members develop in such a way that they balance the external forces (i.e. the reaction forces and the applied loads), the static equilibrium equations can be applied to determine the forces in the members that have been cut through.
Here’s a more detailed summary of the process:
Step 1
Draw a free body diagram of the truss and solve reaction forces
The first step is to draw a free body diagram showing all of the external loads acting on the truss. The three static equilibrium equations can be applied to calculate the reaction forces.
This is the exact same first step as for the Method of Joints.
Step 2
Cut the truss through the members of interest and draw the internal forces
The next step involves making an imaginary cut through the members of interest in the truss, and drawing the internal forces in the cut members.
Step 3
Apply the equilibrium equations to solve the internal forces
The internal forces and external forces will always be in equilibrium, no matter where you have decided to cut the truss. This means that the static equilibrium equations can be applied to solve the internal forces in the cut members.
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:
Shear forces, that are oriented in the vertical direction, parallel to the beam cross-section
Normal forces, that are oriented along the axis of the beam, perpendicular to the beam cross-section
Shear (left) and normal (right) internal forces
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.
Normal internal forces are either tensile or compressive
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.
The effect of the internal forces on the beam cross-section can be represented by two resultants – a shear force and a bending moment
Why Are Shear Force and Bending Moment Diagrams Useful?
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:
How the beam is loaded
How the beam is supported
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.
The three most common load types acting on beams are concentrated forces, distributed forces and concentrated moments
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.
Pinned Support
vertical displacement
horizontal displacement
rotation
Roller Support
vertical displacement
horizontal displacement
rotation
Fixed Support
vertical displacement
horizontal displacement
rotation
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.
Example of a beam configuration (above) and the corresponding free body diagram (below)
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.
Test Your Understanding
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
Explanation
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.
How To Draw the Shear Force and Bending Moment Diagram
There are three main steps that need to be followed to determine the shear force and bending moment diagrams:
Step 1| Draw a free body diagram
To correctly determine the shear forces and bending moments along a beam we need to know all of the loads acting on it, which includes external loads and reaction loads at supports. By drawing the free body diagram you identify all of these loads and show then on a sketch.
Step 2| Calculate the reaction forces and moments at the beam supports
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.
Step 3| Determine the internal shear forces and bending moments at every location along the beam
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.
Sign Conventions
A consistent sign convention needs to be used when calculating shear forces and bending moments. The most common sign convention is as follows:
Applied forces are positive if they are acting in the downwards direction.
If the beam is on the left side of the imaginary cutting plane shear forces pointing downwards are positive. If the beam is on the right side of the cutting plane, shear forces pointing upwards are positive.
If a bending moment causes sagging then it is positive, and if it causes hogging then it is negative.
These sign conventions are summarised in the figure below.
A common sign convention for shear forces and bending moments
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 Beam
In this example we’ll determine the shear force and bending moment diagram for a simply supported beam that is carrying two loads.
The simply supported beam configuration for this example
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).
The free body diagram for Example 1
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:
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.
Updated free body diagram showing the magnitudes of all external loads
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.
Free body diagram for an imaginary cut directly to the right of the pinned support
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:
$$M(x) = 12x$$
Calculating the shear force and bending moment from the free body diagram
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.
Related Topics
Trusses
Trusses are structures (e.g bridges or roofs) made up of members that can only carry tensile or compressive axial loads.
