Forces in a staically determinate cantilever Truss, why some experiences compression while others experience tension?

Question

The Redundant Truss Apparatus is constructed with units of basic triangular pin-jointed.

At joint 1, it is fixed firmly due to a support.

At joint 2, it is fixed to a movable support.

Suppose at joint 4, $F = 250N$

Q1) Using pin-jointed Theory, show the full working to calculate theoretical member forces for the framework.

For qn1, I am able to obtain the magnitude of the forces for each members using free body diagrams. A small part of my calculation based on free-body diagram is at joint 4 $$\frac{1}{\sqrt2}F_7 - F = 0$$ $$F_7 = 250\sqrt2 N$$

Q2) From your results and the theoretical member forces, identify which members are in compression and which are in tension. Explain your choices.

Even if I answer qns 1 correctly, I have no idea how to explain why some of them experience compression while some experience tension.

Q3) Explain the reading of member 5.

While I understand that member 5 does not apply any force, I have no idea how to explain it.

Answer 1

Tension on the vertical member 5 is zero for the following two reasons:

• Member 4 can only transfer horizontal forces.
• Joint 2 is free to move vertically and thus carries no vertical reaction forces.

So if you think joint 1 could impart a vertical force on member 5, then there would be no way to react upon it, and thus the structure would not be statically stable.

Answer 2

To get a better understanding, maybe it helps to look at this from an intuitive perspective: The truss holds up some weight which wants to fall down.

What ways are there for the weight to lower itself? Well, you could elongate member 7 (in the following, blue members are of the same length, just possibly rotated, the red members are elogented, green ones are shortened):

Or you could shorten member 3:

Thus, member 7 is in tension and member 3 is in compression, and you can continue this for the others.

Now for member 5, you'll see that both shortening and elongating would lower the weight. Thus, you can (at least heuristically) conclude that member 5 is in an unstable equilibrium, hence neither in tension nor compression, consistent with the other answer.

Answer 3

Whether an object is under tension or compression is a matter of what direction the forces are pointing. To solve the problem, you probably drew free-body-diagrams for each joint. It may be helpful to draw free-body-diagrams for the "members" too.

Take "member 3". There are two forces acting on it:

• force of joint 3 on member 3
• force of joint 4 on member 3

From Newton's second law the net force on it is zero (it's not accelerating). That means one must point to the left and and the other to the right, in order to cancel out. If the forces point inward toward member 3, then member 3 is under compression. If they point outward away from member 3, then member 3 is under tension.

Answer 4

Pin connected trusses---all members are 1-force members. No dimensions given but from your response I assume M7 is at a 45 degree angle. (I notice there is no M6.)

First solve for the exterior reactions. Draw a FBD of the whole truss with the reactions shown as forces. The pin at J1 has x and y reactions, R1x & R1y. The roller at J2 has only a component in the x direction, Rx2. Ry2 = 0.

Sum forces in the y direction, Ry1 = 250N. Sum moments around J1. For equal panels the distance between J2 & J4 is twice the distance between J1 & J2 (bit easier w/ real numbers) so R2x = 500N directed to the right. Sum forces in x direction, R1x = 500N directed to the left.

Isolate Joint 4 with F7 resolved to x and y components. Sum of the forces in the y direction = 0, F7y = the applied force = 250N. From geometry the x and y components of F7 are equal so F7x = 250N. F7 is found by the square root of the sum of the squares = 353.6N T.

When you isolate a joint and draw the force diagram (FBD), a force pointing away from the joint shows the member is in tension (pulling on the joint). F7y is pointing up to balance the applied load so F7 is in tension.

From there you can work through the rest of the truss. F3 = 250N C (from the sum of the forces in the x direction). Isolate Joint 5 with the known component forces F7y & F7x. Sum forces in each direction. F2 is 250N C and F1 is 250N T. Isolating joint 3 shows that F8y = 250N T. F8 = 353.6N T. Summing forces in the x direction at joint 3 gives 500N for F4 (C).

From the sum of the forces you can now see that F5 = 0 but you can also see that by inspection of the truss. Whenever you see no diagonal at a joint with no applied load and no reaction component in the direction of a member, that member is a zero force member. There is no diagonal at J2 to impart a y component force to M5, there is no applied load at J2 and the roller eliminates a vertical member force in M5.

You could solve the truss by the method of joints without solving the reactions but knowing the reactions gives you a check on your member forces. The horizontal component of F8 combines with F1 to equal the 500N horizontal reaction at J1.