Pin jointed frames, generally, transfer the applied loads by inducing axial tensile or compressive forces in the individual members. The magnitude and sense of these forces can be determined by using standard methods of analysis.

The following tutorial questions require that the forces in the individual elements be solved by either the * "method of sections"* or the

The * "method of sections"* involves the application of the three equations of static equilibrium to a two dimensional plane frame. An imaginary section line cuts the frame in two. Since there are only three equations of equilibrium, the section through the frame must not include any more than three members for which the internal forces are unknown.

The * "method of joints"* considers the isolation of each individual joint. For each joint, as the forces are coincident the moment equation is of no value leaving only two equations of equilibrium available to resolve the forces in the members. The equilibrium of each joint must be considered in a sequence that ensures that there are no more than two unknown member forces in each joint under consideration.

**Notes:**

- The analysis associated with these tutorial questions will round all your input values as follows:
- Forces, e.g. vertical and horizontal reactions (kN) - 1 decimal place.
- Dimensions (m) - 3 decimal places.
- Angles (°) - 1 decimal place.

- Numeric answers are required in all the fields. If you calculate a force as being zero, enter zero (0) in the appropriate field.
- You are only able to print the current question as a tutorial question if you have not submitted your answers.
- Once a tutorial question has been printed, you will not be able to submit your answers, or see the solution to that particular question.

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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 will be used to identify the forces present in each member of the frame.

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:

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.

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.

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.

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 will be used to identify the forces present in each member of the frame.

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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 sections will be used to identify the forces present in each member of the frame.

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:

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.

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.

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 sections will be used to identify the forces present in each member of the frame.

The reactive forces should be identified and marked onto the free body diagram of the frame as shown on the diagram above. At joint E there is only a vertical reaction because the connection is a roller. At joint A there is a vertical and a horizontal reaction because the connection is pinned. 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.

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 will be used to identify the forces present in each member of the frame.

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:

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.

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.

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.

The reactive forces should be identified and marked onto the free body diagram of the frame as shown on the diagram above. At joint A there is only a vertical reaction because the connection is a roller. At joint A there is a vertical and a horizontal reaction because the connection is pinned. 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.

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.

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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.

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

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

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

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

Note: It is possible to in this example to start resolving at joint E without first calculating the value of the reaction forces as there are only two unknown member forces.

Note: At joint D there are two unknown forces which are the reaction force from the roller support and the force in M_{BD}. The reaction acts at a right angle and M_{BD} parallel to the members M_{BDF} . It is more conveniant to resolve the forces in these two directions as opposed to the vertical and horizontal directions.

Note: It is possible to in this example to start resolving at joint E without first calculating the value of the reaction forces as there are only two unknown member forces.

Note: At joint D there are two unknown forces which are the reaction force from the roller support and the force in M_{BD}. The reaction acts at a right angle and M_{BD} parallel to the members M_{BDF} . It is more conveniant to resolve the forces in these two directions as opposed to the vertical and horizontal directions.

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

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

The reactive forces should be identified and marked onto the free body diagram of the frame as shown on the diagram above. At joint A there is only a vertical reaction because the connection is a roller. At joint A there is a vertical and a horizontal reaction because the connection is pinned. 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.

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

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

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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.

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

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

Note: It is possible to in this example to start resolving at joint E without first calculating the value of the reaction forces as there are only two unknown member forces.

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.

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

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

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

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 procced 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.

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.

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.)

_{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.)

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

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

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

The reactive forces should be identified and marked onto the free body diagram of the frame as shown on the diagram above. At joint A there is only a vertical reaction because the connection is a roller. At joint A there is a vertical and a horizontal reaction because the connection is pinned. 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.

_{EG} the resolution of forces at the joint can be carried out as a check. ( A force may be present in M_{EG} due to rounding procedure undertaken but this should be assumed to be zero.)