The Pin jointed frame displayed above can be solved either using the method of joints or method of sections. In this worked example the method of joints and sections will both be utilised used to identify the forces present in each member of the frame.

#### Frame dimensions and loading

#### Static Determinacy

Before analysis of the frame can be undertaken it must be confirmed that it is statically determinate and stable. To achieve this, we must establish that the total number of unknowns is equal to the number of equations that are available. At each node (joint) there is two degrees of freedom present which require a restraint which is provided by the members and supports of the frame. Therefore a pin jointed frame can be said to be statically determinate when:

*(where m = members, r = reactions, j = joints)*

This pin jointed frame can be shown to be statically determinate as:

#### Support reactions

The reactive forces should be identified and marked onto the free body diagram of the frame as shown on the diagram above. The reactive forces have been assumed as positive which means if a negative value is obtained then the assumed direction is incorrect. The three equations of equilibrium should be applied to determine the unknown reaction forces located at the supports.

Note: The loading that is present on this system is symmetrical therefore the two vertical reactions will be of an equal magnitude.

#### Frame Symmetry

From inspection, we can identify that the frame displays symmetry with respect to a vertical axis between Joints K and D as the loading and frame geometry are identical. This means that it is possible for us to simplify the structure as it is only necessary for us to calculate the force in the members on one side of the axis of symmetry as the forces will be the same on either side. This means we can split the structure at the axis of symmetry as displayed above and proceed to calculate the member forces in this new system. It is important to be able to identify symmetry present in structures as this can reduce the time required to perform analysis.

#### Frame Geometry

To carry out the resolution of forces at the individual joints we are required to know the angles at which the members are positioned related to the joint. This is calculated from the given data using the rules of trigonometry. The angles that are required to enable us to solve this tutorial are displayed above and calculated below.

#### Resolution of Forces

Each individual joint should now be isolated in sequence. When using the method of joints it is important to ensure that a sequence is chosen where there are no more than two individual forces that are unknown at the joint. This is because there are only two equations of equilibrium available to us; as the forces are concurrent the moment equation cannot be used.

A free body diagram should be drawn for each individual joint detailing the members present and direction of the force in the members. For any member where a force is unknown it should be assumed to be in tension and if a negative value is obtained we know the assumption was incorrect and the member is in fact in compression.

#### Joint A

#### Joint H

#### Joint B

#### Section Cut

The frame should be cut at a section that passes through the members M_{CD} M_{OK} and M_{JK} which allows to use the three equations of equilibrium to calculate the forces present in the members.

#### Joint C

#### Joint J

#### Joint I

#### Joint K

Note: The force in the members in M_{OK} and M_{IO} are identical due to symmetry.

From inspection there should be no force present in M_{KD} the resolution of forces at the joint can be carried out as a check. ( A force may be present in M_{KD} due to rounding procedure undertaken but this should be assumed to be zero.)