This notes sheet looks at structures that are static: they do not move. The equilibrium of forces and
moments are analysed, and their effects on stresses, strains, and deformations.

To begin with, a recap of Newton’s Laws of Motion:

1. An object will stay at rest or constant velocity unless acted upon by a net resultant force.
2. The resultant force on an object is directly proportional to the rate of change of momentum
of the object, and acts in the same direction (F = ma)
3. For every action, there is an equal but opposite reaction.
Static Equilibrium
One condition for static equilibrium is that there is no net resultant force acting on the particle, but
another is that there is no net rotational moment:

### Static Equilibrium

One condition for static equilibrium is that there is no net resultant force acting on the particle, but
another is that there is no net rotational moment:

Free body diagrams are used to analyse systems of moments and forces.

Before any analysis is done, the coordinate system must be defined.

This is to ensure the directions are all dealt with correctly, to avoid sign errors, and helps to
separate systems into horizontal, vertical, and normal components.

## Joints & Connections

Different types of supports can withstand different moments and forces:

### Built-in Supports

• Horizontal and vertical reaction forces
• Support moments without rotation

### Pin Supports

• Horizontal and vertical reaction forces
• Cannot resist rotation

### Roller/Sliding Support

• Can only support a reaction force in one direction (normal to the surface)
• Free to move along the surface
• Cannot resist rotation

Pin joints are generally used to represent the connections between different beams in a ‘pin jointed structure’, for example a truss bridge. Therefore, when analysing such structures, we assume the beams are free to rotate at each connection.

## Statical Determinacy

If for any particular structure enough forces and moments are known to calculate the remaining unknown forces from equilibriums only, it is said to be statically determinate. If more information is required, the system is statically indeterminate.

To find out of a system is statically determinate, count the number of bars, b, reactions, n, and pins, j. There are three scenarios:

• The system may be statically determinate, though this depends on the positioning of the beams
• The system is statically indeterminate
• The structure is a mechanism

In the system below, there:

• 4 bars,
• 6 end reactions,
• 5 pins,

: the system is statically determinate.

## Sectioning Structures

To simplify issues with complex pin jointed structures, section the system across the beams that
you want to find the tension in. This allows you to treat the new, smaller system, as a body with
only external forces.

Often, the maximum allowable load on a frame needs to be calculated. This will depend on a
number of factors: tensile/compressive yield in the beams, failure of pins, buckling etc.

Generally, there will be one beam which experiences a greater tensile/compressive force than any
other. Provided that all beams have the same elastic modulus and cross-section (so the same
second moment of area, see beam theory), this beam will be the most likely to fail due to
tensile/compressive yielding.

The beam most likely to fail due to buckling, however, is the compressed beam with the greatest value of .

The beam is in tension if the force points away from the pin, and compression if the force
points towards the pin.

Tabulating the compressive, tensile and TL² values shows which beam will fail in which condition