# Simple Statics - Determinate Beams

Some simple beams can be analysed using simple statics, this is a simple method that uses vertical equilibrium and moment equilibrium to derive equations for bending moment and shear force along these simple beams.

• Vertical equilibrium – vertical equilibrium or more generally force equilibrium is when the vector sum of all forces at every point on a structure is zero, i.e. there are no out of balance forces that would otherwise give rise to an acceleration (for more information please refer to this article)
• Moment equilibrium – moment equilibrium is when the sum of all moments at every point on a structure is zero, i.e. there are no out of balance moments that would otherwise give rise to a rotation (for more information please refer to this article)

Only determinate beams can be analysed in this way, a determinate beam (or determinate structure) is a beam with the same number of equilibrium equations as unknowns in the system, determinate can therefore be solved simply by rearranging equations.

Where there are more unknowns in the system than there are equilibrium equations then the structure is statically indeterminate. Statically indeterminate structures cannot be solved using simple statics and must be solved by taking into account stiffness and displacements in the structure.

Prior to diving into simple static calculations it is useful to discuss the support systems that typically restrain a beam, the typical loads on a beam and a consistent sign convention.

#### Support Systems

Typically two different support systems are used for beams:

• Simple supports – Simple supports provide vertical restraint, but no moment restraint, they are typically denoted using a triangular support or an upwards arrow. Simple supports can be laterally fixed or "roller" supports that allow lateral movement. When undertaking calculations with simple supports it is often implied that enough roller supports are present to allow a beam to bend freely.
• Fully fixed supports – Fully fixed supports, sometimes called Encastré supports, provide both vertical restraint and moment restraint. Fully fixed supports are denoted as a vertical line with hatching at the beam end, as if the beam was fixed into a wall.

• Point load – As the name suggests this is a load applied at a single point on a beam. Point loads are denoted using a downwards arrow. The metric units for point loads are newtons (N) or kilonewtons (kN = 1000N)
• Uniformly distributed load – A uniformly distributed load (often abbreviated to a UDL) is a load that is spread over a length of beam, with the same intensity of loading through the length. UDLs are typically denoted with a vertically hatched area or wavy line over the length of load. As a UDL is a load over length of beam then the metric units are newtons per metre (N/m) or kilonewtons per metre (N/m)
• Point moment - As the name suggests this is a moment applied at a single point on a beam. Point moments are denoted using a rotating arrow. The metric units for point moments are newton-metres (Nm) or kilonewton-metres (kNm = 1000Nm)

#### Sign Convention

Prior to undertaking a simple static analysis it is important to establish a consistent sign convention. This is very important as some moments and forces will act in an opposite sense to other moments and forces, we need to be clear with where we need to add or subtract forces and moments.

The direction of a moment is usually defined by whether the moment has either a hogging or a sagging effect on a beam.

• A sagging moment will cause a beam to have a tensile stress at the bottom of the beam and a compressive stress on the top of the beam. For more information on how bending causes stresses in a beam, please refer to this article. If a beam was positioned between two simply supports and a downward force applied to the beam, the beam would bend downwards, this is sagging.
• A hogging moment is the bending in the opposite direction to sagging and will cause a beam to have a tensile stress at the top of the beam and a compressive stress on the bottom of the beam. If a beam was positioned between two simply supports and an upwards force applied to the beam, the beam would bend downwards, this is hogging.

The following sign convention is used in this website, but it is worth noting that sign conventions do vary:

• Hogging is caused by a positive moment
• Sagging is caused by a negative moment

The direction of shear force also needs to be taken into account, shear force sign convention is established by cutting a small element out of the beam and looking at the shear force on either side of the element.

• If the shear force on the left hand side of the element is upwards and the shear force on the right hand side of the element is downwards then this is a negative shear force.
• If the shear force on the left hand side of the element is downwards and the shear force on the right hand side of the element is upwards then this is a positive shear force.

Using the above sign convention the relationship that the shear force diagram is the same as the gradient of the bending moment diagram remains true:

$$V = \frac{dM}{dx}$$

#### Establishing Beam Bending Equations Using Simple Statics

Typical beam bending problems involve a known support system and a known applied load, with the unknowns being the support reactions, support moments (if applicable) and the beam bending and shear diagrams. For problems of this nature a typical solution process is:

1. Establish an equation for vertical equilibrium, with applied loading on one side of the equation and support reactions on the other.
2. Establish equations of moment equilibrium, with applied moments on one side of the equation and support reactions and moments (if applicable) on the other.
3. Solve using simultaneous equations to find the support reactions and moments (if applicable).
4. Using the expressions for support reactions, support moments (if applicable) and applied loading to develop equations for bending moment and shear force along the beam. This is done by taking cuts through key parts of the beam in turn and analysing the bending and shear effects to the left of the cut.
5. Use the beam bending and shear force equations to plot a bending moment and shear force diagram.

#### Example – Cantilever Beam with Applied Point Load.

A cantilever beam with an applied point load, P, is shown in the diagram below:

Step 1: From vertical equilibrium:

$$R_{A} = P$$

Step 2: From moment equilibrium:

$$M_{A} = Pa$$

Step 3: As there is only a single vertical support and a single vertical loading and a single moment support and a single applied moment the above equations are already solved, so solving using simultaneous equations are not needed.

Step 4: To build the bending moment and shear force equations we take cuts at key points along the beam and analyse the loading applied to the left of the cut. We need to take cuts after every new load position, i.e. we will take one cut after the support reaction and support moment and another cut after the point load.

Taking a cut through the beam after the support reaction and support moment, but before the point load, gives the following loads to the left of the beam:

At our position, x, along the beam the following bending moments and shear forces are applied to the beam:

This is expressed in the equations below:

$$M_{x} = M_{a} – R_{a}x$$ $$V_{x} = – R_{a}$$

Taking a cut through the beam after the point load, gives the following loads to the left of the beam:

At our position, x, along the beam the following bending moments and shear forces are applied to the beam:

This is expressed in the equations below:

$$M_{x} = M_{a} – R_{a}x + P(x-a)$$ $$V_{x} = – R_{a} + P$$

We therefore end up with two equations for moment and shear force distribution, one for before the point load and one for after the point load. This can then be plotted on a bending moment and shear force diagram.

Step 5: The bending moment and shear force diagram for the above bending and shear force equations is shown below.

It is worth noting that for when the applied loading consists of point loads the bending moment diagram is linear, with the angle of the bending moment diagram only changing at the position of a load or a reaction. Because $$V = \frac{dM}{dx}$$ the shear force diagram is horizontal between the position of a load or a reaction.

#### Example – Simply Support Beam with Applied Uniformly Distributed Load

A simply support beam with an applied uniformly distributed load, ω, is shown in the diagram below:

Step 1: From vertical equilibrium:

$$R_{A} + R_{B}= \omega L$$

Step 2: From moment equilibrium, taking the moments acting about $$R_{B}$$:

$$R_{A}L = \omega L \times \frac{L}{2}$$

Step 3: By rearranging the equation for moment equilibrium, the values for $$R_{A}$$ can be found:

$$R_{A} = \frac{\omega L}{2}$$ $$\text{Substituting } R_{A} \text{ into the vertical equilibrium equation}$$ $$\frac{\omega L}{2}+ R_{B}= \omega L$$ $$Therefore,$$ $$R_{B} = \frac{\omega L}{2}$$

This solution can also been established by looking at the symmetry of loading, as the loading is symmetrical on the beam it can be seen by inspection that the support reactions will be equal.

Step 4: To build the bending moment and shear force equations we take cuts at key points along the beam and analyse the loading applied to the left of the cut. For the UDL case the load pattern is the same along the span, so we only need to take a single cut in the beam.

Taking a cut through the beam at a point, x, gives the following loads to the left of the beam:

At our position, x, along the beam the following bending moments and shear forces are applied to the beam:

This is expressed in the equations below:

$$M_{x} = – R_{a}x + \frac{\omega x^{2}}{2}$$ $$V_{x} = – R_{a} + \omega x$$

This is the only equation we need for this span, this can then be plotted on a bending moment and shear force diagram.

Step 5: The bending moment and shear force diagram for the above bending and shear force equations is shown below.

It is worth noting that for when the applied loading consists of a UDL the bending moment diagram is parabolic. Because of the relationship $$V = \frac{dM}{dx}$$, for a parabolic benging moment diagram, the matching shear force diagram is linear.