# Determinate Truss Analysis

Determinate trusses are trusses that can be solved using simple statics and resolving of forces. Determinate trusses are formed of the following components:

• Elements – Truss elements carry either pure tension or pure compression and are connected together at nodes. Elements are usually modelled to act along the centroidal axis (the axis passing through the centroid of a section) of the structural member forming them. Elements are typically named after the nodes that they are connected to, i.e. element 1-2 will connect nodes 1 and 2.
• Nodes – Nodes are the points at which elements are connected, for determinate trusses the nodes are usually assumed to be “pin jointed”, i.e. no moment is transferred through the node. For a truss to be determinate, the centroidal axes of the elements connected at a node must all meet at a single point, or within a small distance of each other. If the centroidal axes of the elements didn’t meet at the same point then an eccentricity would be present at the connection and a force times by an eccentricity gives a secondary moment. Nodes are typically given numbers to help identify them.
• Loads – Loads are assumed to be applied to the truss at node locations.
• Supports – The truss supports are located at a node location.

#### Sign Convention

When analysing a determinate truss it is important to ensure the directions of compressive and tensile forces are consistent. A typical approach is to indicate the direction of a force on the truss elements.

• Compression – Compression is indicated with an arrow pointing away from the centre of an element towards its nodes.
• Tension – Tension is indicated with an arrow pointing towards the centre of an element away from its nodes.

#### Analysis of a Determinate Truss

Determinate trusses are analysed by ensuring force equilibrium applies at every node on a structure. The typical process is shown below:

1. Draw up the truss frame and add all known forces, initially this will only be forces or maybe reactions.
2. A node is selected with only two unknown element forces.
3. All the known loads from elements connected to a node are drawn as vectors, ensuring the direction of loads matches the sign convention detailed above.
4. Equations for vertical equilibrium and horizontal equilibrium are established.
5. The equations are solved using simultaneous equations if needed, establishing values for the two unknown forces.
6. The newly established element loads are added back into the truss model. These new loads will usually mean that another node now only has two unknowns and can be solved next.
7. Repeat steps 2. to 6. for a new node until all element forces are found.

Sometimes it may appear that there are no elements with only two unknowns, however it is often possible to use a couple of tricks to reduce the number of unknowns to two or less:

• As with beams, vertical and moment equilibrium can be used to find reactions for a given applied load. It is important to ensure moments are resolved about the correct axes, with the lever arm taken perpendicular to the direction of the applied load.
• Where an element is at 90 degrees to a load then it will not provide direct resistance to the load and can therefore be ignored when working out the distribution of load in the other truss elements. Figure 3: Example of using vertical equilibrium, moment equilibrium and symmetry to help solve a determinate truss

#### Determinate Truss Example – Simple Warren Truss

An example of how to solve a simple Warren Truss is shown below:

By inspection, as the load is in the centre of the span and the truss is symmetrical then the reactions will be equal.

$$\text{By vertical equilibrium}$$ $$R_{a} + R_{b} = P$$ $$\text{By inspection}$$ $$R_{a} = R_{b}$$ $$\text{Therefore}$$ $$R_{a} = R_{b} = 50kN$$

As we now know the reaction it is possible to fully analyse Node 1:

$$\text{By vertical equilibrium}$$ $$50 = F_{1,2} \sin60^{\circ}$$ $$F_{1,2} = \frac{50}{\sin60^{\circ}} = 57.74 (compressive)$$ $$\text{By horizontal equilibrium}$$ $$F_{1,3} = F_{1,2} \cos60^{\circ}$$ $$F_{1,3} = 28.79 (tensile)$$

By symmetry these values also apply to $$F_{5,7}$$ and $$F_{6,7}$$. These values can now be added back into the truss model Figure 5: First step of truss solution, forces in elements around node 1 and 7 solved.

We now have enough information to fully analyse Node 2:

$$\text{By vertical equilibrium}$$ $$F_{1,2} \sin60^{\circ} = F_{2,3} \sin60^{\circ}$$ $$F_{2,3} = 57.74 (tensile)$$ $$\text{By horizontal equilibrium}$$ $$F_{2,4} = F_{1,2} \cos60^{\circ} + F_{2,3} \cos60^{\circ}$$ $$F_{2,4} = 57.74 (compressive)$$

By symmetry these values also apply to $$F_{4,6}$$ and $$F_{5,6}$$. These values can now be added back into the truss model

We now have enough information to fully analyse Node 3:

$$\text{By vertical equilibrium}$$ $$F_{2,3} \sin60^{\circ} = F_{3,4} \sin60^{\circ}$$ $$F_{3,4} = 57.74 (compressive)$$ $$\text{By horizontal equilibrium}$$ $$F_{3,5} = 28.87 + F_{2,3} \cos60^{\circ} + F_{3,4} \cos60^{\circ}$$ $$F_{3,5} = 86.61 (tensile)$$

By symmetry these values also apply to $$F_{4,5}$$. These values can now be added back into the truss model Figure 7: Third step of truss solution, forces in elements around node 3 and 5 solved. All element forces now found.

This solves the truss model, it is easy to see how vertical and horizontal equilibrium is satisfied at Node 4.

#### Indeterminate Trusses and Internal Redundancy.

Sometimes a truss has an extra element (an extra unknown). The extra element adds a level of internal redundancy, the element is not strictly needed for the truss to stand up. The extra element (and extra unknown) means that means a truss cannot be solved through resolving of forces, the truss therefore becomes an indeterminate truss and must be solved using stiffness methods.

Sometimes it is possible to ignore a redundant element and remove it from a truss model, this allows the truss to be analysed as a determinate truss. This is sometimes called a zero force element. If the structure is able to stand with the element removed then it would usually also work with the redundant element taking no load.