# ST14: Identifying Zero-Force Members in Trusses

In lecture ST13, we introduced The Method of Joints

for analyzing determinate trusses. We learned that truss members carry tensile and compressive forces only. Interestingly, in a typical truss structure,

depending on its geometry and the location of the applied load,

some of the members may carry no axial force. We call them zero-force members. For example, in this truss, if the only load on the structure is located at joint C,

CD becomes a zero-force member. Some trusses, such as bridges,

are designed to carry moving loads. This means the bridge needs to be

analyzed several times, one time for each load position. Therefore, while member AB carries an axial force

when the load is acting at or near joint A, the member becomes a zero-force member, when the load moves further down the bridge, away from A. There is another reason for having

zero-force members in trusses. A long, slender, truss member subjected to a large compressive force, could fail prematurely, due to what is known as buckling. To prevent such a failure, additional truss members could be added to the structure, in order to brace the compression member, effectively shortening its buckling length. These members often carry no significant axial force, therefore they are treated as zero-force members. Although we can identify the zero-force

members in the truss by analyzing the entire structure, there is a shortcut for doing the same

without performing any calculations. Let’s demonstrate the technique using this truss. For the structure to be in the state of equilibrium, the sum of the forces acting at each joint must be zero. For example, take this joint

which connects two truss members. Let’s place the origin of the XY coordinate system on the joint, and show the member forces. I’m going to label them P and Q. Since the equilibrium condition is independent

of the orientation of the coordinate system, we can rotate the axes so that either X or Y axis becomes aligned with one of the two forces, like this, and still maintain that the sum of the forces in each direction must be zero. Note that now only one of the forces

has a component along the Y axis. P is the only force with a Y component, and since the sum of the forces

in the Y direction must be zero, P must be zero. And if P is zero, given that the sum of the forces in the X direction must also be zero, we can conclude that Q must be zero too. That is, neither member carries any force. Let’s simplify the truss,

by removing the two zero-force members. Here’s another truss joint with only two forces present. Let’s place the coordinate system at the joint, and show the forces that are acting on it. Here, since the forces are acting along the X and Y

axes already, there is no need to rotate the coordinate system. The sum of the forces in the X direction must be zero, hence Q equals zero, since it is the only force in that direction. Similarly, since P is the only force in the Y direction, hence its magnitude must be zero. Let’s simplify the truss

by removing these zero-force members. Now let’s examine this joint. Here, there are three forces.

Two of them act in the X direction, and one force acts in the Y direction. Since the sum of the forces

in the Y direction must be zero, Q must be zero. In the X direction, we know that

the sum of P and R must be zero, but that does not mean the forces are zero. They just need to add up to zero. So the only conclusion that we can draw here is that force Q equals zero. Removing the zero-force member, we now end up with this simplified truss. Next, let’s consider the top joint. Like the previous joint, here three forces are present. But regardless of the orientation

of the coordinate system, the three forces result in at least two

force components in either direction. Therefore, no conclusion about any member forces being zero can be drawn at this joint. The same is true if we examine this joint. Three forces are acting on the joint, resulting in at least two X and two Y components, so no zero forces can be identified at this joint either. Four forces are acting at this joint,

resulting in multiple components along X and Y axes, therefore none of the forces can be said to be zero. As you just saw, we can identify the zero-force members in the truss

using qualitative analysis. The analysis involves using a simple principle. By placing and rotating the coordinate system at a joint. If we can show that there is only one

force acting in either X or Y direction, that force must be zero, and the member associated with it,

a zero-force member. By repeatedly applying this principle to

the joints of statically determinate trusses, we can identify their zero-force members qualitatively. Let’s put this principle to use, and see how it can help us identify

the zero-force members in a couple of trusses. Consider this statically determinate truss. Suppose we wish to identify the zero-force members when the structure is subjected to a load at joint D, keeping in mind that

there is one reaction force at joint K, and two reaction forces at joint A. We start by examining a truss joint

that is subjected to no more than three forces. Joint C connects three members, since there is no applied load

or support reaction at the joint, it is going to be subjected to three forces only. We place the coordinate system at the joint, show the member forces, then rotate the XY axes, so as to align the X axis with the two collinear forces. Now we can see that only one force, in member CE,

has a component along the Y axis. Since the sum of the forces in the Y direction

must be zero, we can conclude that CE is a zero-force member. Let’s remove the member from the structure. The simplified truss reveals that

there are only three forces acting at joint E, So we place the coordinate system at E, then rotate the axes like this: Now we can see that the force in member BE

is the only one with the component in the Y direction. Therefore BE must be a zero-force member. Next, place the coordinate system at joint B, and rotate the axis,

in order to show that BF is a zero-force member. Similarly, by examining joint F in the simplified truss,

we can see that DF is a zero-force member. The members on the right side of the truss

can be examined in the similar manner. From joint I,

we can conclude that HI is a zero-force member. At joint H,

we can show that HJ is a zero-force member. Also, JG is a zero-force member. And, we can easily show that GD

is a zero-force member. No further simplifications can be made past this point. The simplified truss can be analyzed in order to determine the support reactions, and the remaining member forces. Let’s consider another example. This truss is subjected to two concentrated loads, and given that the structure rests on two pin supports, there are four support reactions. We scan the truss joints with two or three forces only. Joint B is a good candidate. if we place the coordinate system at the joint, and draw the member forces, we can see that the force and member BD

is the only one acting in the X direction. Therefore BD is a zero-force member. Let’s remove it from the structure. Next, consider joint E. Since the sum of the forces

in the Y direction must be zero, DE must be a zero-force member. Furthermore, upon examining joint D, we can conclude that CD is a zero-force member. Why? Because after aligning the X axis along line AF, we can see that CD is the only force

with a Y component, and since the sum of the forces

in the Y direction must be zero, CD must be a zero-force member. Finally, if we place the coordinate system at joint C, since there is only one force

acting in either X or Y direction, we can conclude that both AC and CF

are zero-force members. Now let’s examine the remaining joints. At joint A, there are two reaction forces,

and one member force, regardless of the orientation of the coordinate system, these result in two forces along the X axis,

and two forces along the Y axis, so neither force can be said to be zero. At joint F, the three existing forces yield

two X components, and two Y components, therefore neither force can be said to be zero. Joint H is subjected to four forces, two forces in the X direction,

and two forces in the Y direction, therefore neither of the

member forces can be said to be zero. At K, although there are only three forces present,

they yield two X components, and two Y components, therefore neither member

can be said to be carrying a zero force. Joint J is similar to joint H. There are four forces present, hence no member

can be said to be carrying a zero force. Joint G is subject to five forces. Obviously, none of the members

can be said to be carrying a zero force. Finally, at joint I, the existing four forces create

multiple force components along both axes, therefore, neither of the two member forces, or the support reactions can be said to be zero. Now it’s your turn. Identify the zero-force members

in the following trusses:

I want to learn design of truss for industry, workshop, store so how can I identify the size of sections like angle, rectangular hollow tube or circular hollow tube which can bear the coming load can you teach please

Perfect and simple lecture. Thanks

Thank you for your helpful video… please make a video for displacement analysis….

How can I find the answers of last questions?

what happened to Dr.structure team?? we are awaiting for your videos..

thanks for you nice efforts.

Your lesson is precise and concise. I appreciate all your efforts. Please keep up. Thanks.

This is very nice thank you. Although I enjoy the animations, I don't think that they are that necessary. I think you would save a lot on time and effort. The content is what truly matters here.

Thanks again for the video

Really thanks. Your technique is much simpler and understandable than other videos. Video helped me alot Thanks