Topic 2.2: Rigid Body Structures - Trusses
A very common structure used in construction is a truss. An ideal truss is a structure which is composed completely of (weightless) Axial Members that lie in a plane, connected by pinned (hinged) joints, forming triangular substructures (within the main structure), and with the external loads applied only at the joints. See Diagram 1. In real trusses, of course, the members have weight, but it is often much less than the applied load and may be neglected with little error. Or the weight maybe included by dividing the weight in half and allowing half the weight to act at each end of the member. Also in actual trusses the joints may be welded, riveted, or bolted to a gusset plate at the joint. However as long as the centerline of the member coincide at the joint, the assumption of a pinned joint maybe used. In cases where there are distributed loads on a truss, these may be transmitted to a joint by use of a support system composed of stringers and cross beams, which is supported at the joints and transmits the load to the joints.
The procedure for determining the external support reactions acting on a truss is exactly the same as the procedure for determining the support forces in non-truss problems, however the method for determining the internal forces in members of a truss is not the same. The procedures for finding internal forces in truss members are Method of Sections and Method of Joints (either of which may be used), and in fact, one must be very careful not to use these methods with non-truss problems as they will not give correct results. Perhaps the best way to clarify these concepts is to work very slowly and carefully through a truss example.
Example 1: In Diagram 1 we have a truss supported by a pinned joint at Point A and supported by a roller at point D. A vertical load of 500 lb. acts at point F, and a horizontal load of 800 lb. acts at Point C. For this structure we wish to determine the values of the support reactions, and the force (tension/compression) in members BE, BC, and EF.
For the first part, determining the external support reactions, we apply the normal static equilibrium procedure:
I. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces, include any needed dimensions and angles. Note that at the pinned support point A, the best we can do is to put both an unknown x and y support force, however at point D we only need a unknown y support force since a roller can only be in compression and so must support vertically in this problem.
II. Resolve (break) all forces into their x and y-components.
III. Apply the Equilibrium Equations (
) and solve for the unknown forces.
(Sum of x-forces)
(Sum of y-forces)
(Sum of Torque about A.)
Solving we obtain: Ax = 800 lb, Ay = -33 lb, Dy = 533 lb (The negative sign for force Ay means that we initially chose it in the incorrect direction, Ay acts downward, not upward as shown in FBD.)
Part 2
Once we have determined the values of the external support reactions, we may proceed to determining the values of the forces in the members themselves, the internal forces. In this first example, we will use Method of Joints to determine the force in the selected members. In Method of Joints, rather then analyze the entire structure, or even a member of the structure, we rather examine the joint (pin or hinge) where members come together. As the structure is in static equilibrium, so the pin or joint will be in static equilibrium, and we may apply the static equilibrium conditions (and procedure) to solve for the forces on the joint(s) due to the members, which will also equal the forces in the axial members - due to Newton's third law of equal and opposite reactions (forces).
There are several points to keep in mind as we use method of joints. One is that since we are analyzing a point (joint) rather then an extended body, our sum of torque equation will be of no help. That is, since all the forces pass through the same point, they have no perpendicular distance to that point and so produce no torque. This means that to solve completely for the forces acting on a joints we must have a joint which has, at most, two unknown forces acting. In our example (Diagram 2), we notice that there are only two joints which initially have only two unknowns acting - Joint A and Joint D. Thus, we start our process at one of these. We will begin with Joint A.
Step 1. FBD of the Joint A, showing and labeling all forces acting on the joint. Include needed angles. In Diagram 3 we have shown Joint A with the all the forces which act on the joint. The forces on Joint A, due to members AB and AE, act along the directions of the members (since the members are axial). We choose directions for AB and AE (into or out of the joint). When we solve for the forces AB and AE, if these values are negative it means that our chosen directions were incorrect and the forces act in the opposite direction. A force acting into the joint due to a member means that member is in compression. That is, if a member of a truss is in compression, it will push outward on it ends - pushing into the joint. And likewise, if a truss member is in tension, it will pull outward on the joint. In Diagram 3, we have assumed member AB is in compression, showing its direction into the joint, and that member AE is in tension, showing it acting out of the joint.
[A little consideration will show that we have actually chosen AB in an incorrect direction. We can see this if we consider the y-component of AB, which clearly will be in the - y direction. However, the y-support reaction, 33 lb., is also in the -y direction. There are no other y-forces on the joint, so it can not be in equilibrium is both forces act in the same direction. We will leave AB as chosen to see, if indeed, that the solution will tell us that AB is in the wrong direction.]
Step 2: Resolve (Break) all forces into their x and y components.
Step 3: Apply the Equilibrium Conditions: 
(Sum of x-forces)
(Sum of y-forces)
Solving: AE = 756 lb. (Tension), AB = -55 lb. (The negative sign indicates we selected an incorrect initial direction for AB, AB is in Tension, not Compression.)
Remember, we are trying to find the forces in members BE, BC, and EF. Now that we have the forces in members AE (756 lb. tension) and AB (55 lb. tension), we can move unto a second joint (joint B) and find two of the unknowns we are looking for. We could not have solved for the forces acting at joint B initially, since there were three unknowns (initially) at joint B (AB, BC, and BE). However now that we have analyzed joint A, we have the value of the force in member AB, and can proceed to joint B where we will now have only two unknowns to determine (BE & BC).
Joint B: Procedure - Method of Joints
Step 1. FBD of the Joint B, showing and labeling all forces acting on the joint. Include needed angles.
Step 2: Resolve (Break) all forces into their x and y components.
Step 3: Apply the Equilibrium Conditions:
In Diagram 5, we have on the left the FBD of joint B with all external forces acting on the joint shown, and our initial direction for the forces. If the directions we chose for the unknowns are correct, their values will be positive in the solution. If a value is negative it means the force acts in the opposite direction. On the right side of Diagram 5 is the FBD with all forces resolved into x and y-components. We now apply the Equilibrium Conditions (for joints).
(Sum of x-forces)
(Sum of y-forces)
Solving : BC = 44 lb. (Tension) BE = 33 lb. (Compression)
Finally, we can now proceed to analyze joint E and determine the force in member EF.
Joint E: Procedure - Method of Joints
Step 1. FBD of the Joint E, showing and labeling all forces acting on the joint. Include needed angles.
Step 2: Resolve (Break) all forces into their x and y components.
Step 3: Apply the Equilibrium Conditions:
In Diagram 6, we have on the left, the FBD of joint E with all external forces acting on the joint shown, and our initial direction for the forces. If the directions we chose for the unknowns are correct, their values will be positive in the solution. If a value is negative it means the force acts in the opposite direction. The right hand drawing in Diagram 6 is the FBD of joint E with all forces resolved into x and y-components. We now apply the Equilibrium Conditions (for joints).
(Sum of x-forces)
(Sum of y-forces)
Solving : EC = 55 lb. (Tension) EF = 712 lb. (Tension) Since both force values came out positive in our solution, this means that the initial directions selected for the forces were correct.
We have now solved our problem, finding both the external forces and forces in members BE, BC, and EF (and along the way, the forces in several other members - See Diagram 7 )
To see an example of finding internal forces in a truss using Method of Sections, select Example 1
For additional Examples of Truss using Method of Joints, select Example 2
For additional Examples of Truss using Method of Sections, select Example 3
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