Truss Analysis • This is the Analysis of forces on a single beam in a system.
• There are 2 ways to do this: the Method of Joints and the Method of Sections

Method of Joints

• This isolates each joint and analyses the forces on the beams connected to it.
• With standard force analysis, work out the unknown forces acting on each joint.
• Best methods for analysis are the force polygon, or by splitting it up into horizontal and vertical components.
• HINT: It's best to start on a joint with only one or two unknown forces.
• Determining the state of the beam can be done by looking at the force's direction.
• If the forces on the beam point away from each other, it is resisting compression (by being in tension) and vice versa.
• Remember that all forces are reaction forces
• HINTS:
• Remember that the beams carry axial loads. This means that if the force on one joint is in one direction, the force on the 2nd joint it's connected to is opposing that direction.
• If a force is pointing away from the joint, the beam is most likely in tension
• If a force is pointing towards the joint, the beam is most likely in compression

Method of Sections • This is a way to quickly find the forces on a single member.
• A "Section" is taken through the beam that is being analysed. (In this case, it's Y)
• Then, choose the section that is to be used.
• It is optimal to find a section which contains a joint that allows the desired beam to be the ONLY unknown force/moment.
• If that is impossible, choose the one with the least unknown forces.
• Once chosen, take moments around a chosen joint to find any other unknown forces that could aid in finding the moment.
• Remember that a force going through a joint cannot have a moment around that joint.
• In the case below, we are trying to find X around the 20kN joint. Let each truss have a length of 2m.
As $\Sigma M = 0:$

(1)
\begin{align} X \times \sqrt{3} + 35 kN \times 1 = 49.17kN \times 2 \end{align}
(2)
\begin{align} X \times \sqrt{3} = 98.34kN - 35kN \end{align}
(3)
\begin{align} X = \frac{63.34kN}{\sqrt{3}} \end{align}
(4)
\begin{align} \therefore X = 36.57kN \end{align}
• Finally, take moments about a (second) chosen moment to find Y.
• For Y's direction, first assume its direction. If it ends up as a negative answer, it's facing the opposite direction. (This is demonstrated below where we take Y to be opposite its direction in the picture)
• Below, we are finding Y by taking the moment about the support. As $\Sigma M = 0:$

(5)
\begin{align} Y \sin{60} \times 2 + 35kN \times 1 + 20kN \times 2 = 36.57kN \times \sqrt{3} \end{align}
(6)
\begin{align} Y \times \sqrt{3} = 63.34kN - 40kN - 35kN \end{align}
(7)
\begin{align} \therefore Y = -6.73kN \end{align}
• Alternatively, we can take the horizontal/vertical component of Y and use the fact that $\Sigma F_{h/v} = 0$ to find out the force on Y.
page revision: 2, last edited: 06 Jul 2011 07:58