1. Problem

2. Draw

Sketch:

Free-body Diagram:

3. Knowns and Unknowns

Knowns:

• P = 500 N
• Width = 4 m
• Total height = 3 m

Unknowns:

• R_Ax
• R_Ay
• R_By
• F_CG

4. Approach

Part a: determine reaction forces using equilibrium equations

Part b: calculate F_CG using method of sections. Make a cut, then solve internal forces using equilibrium equations

5. Analysis

Part a:

Solving for R_Ax:

$$\sum F_x=0=P-R_{Ax}\\R_{Ax}=P\\R_{Ax}=500N$$

Solving for R_By:

$$\sum M_A=0=(r_{BA}\cdot R_{By})-(r_{GA}\cdot P)\\r_{BA}\cdot R_{By}=r_{GA}\cdot P\\R_{By}=\frac{r_{GA}\cdot P}{r_{BA}}\\R_{By}=\frac{2m\cdot 500N}{4m}\\R_{By}=250N$$

Solving for R_Ay:

$$\sum F_y=0=R_{By}+R_{Ay}\\R_{Ay}=-R_{By}\\R_{Ay}=-250N$$

The answer we got is a negative number. All this means is that the direction of this vector is drawn wrong on our original diagram (in reference to our coordinate frame). This makes sense as RAy and RBy are the only external forces in the y direction, so they have to cancel each other for the equilibrium equations to be true. Therefore, one of them should have a negative direction. We will leave this answer as is for now, but the next time we draw the system, we will change the direction of the arrow.

$$ \underline{R_{Ax}=500N,\; R_{Ay}=-250N, \; R_{By}=250N}$$

Part b:

Firstly, we re-draw the diagram, changing the direction of R_Ay. Then, since we are using method of sections, we make a cut so that the member F_CG (the one we want to find) is cut.

Now we re-draw, choosing one of the pieces from the cut. Here the top half is chosen, but you could also chose the bottom half and get the right answer. The only reason the top half was chosen here is because there are less external forces to consider for the top.

Solve for the member we are looking for:

 $$\sum F_x=0=P+\frac{4}{\sqrt{17}}F_{CG}\\\frac{4}{\sqrt{17}}F_{CG}=-P\\F_{CG}=-P(\frac{\sqrt{17}}{4})\\F_{CG}=-500N(\frac{\sqrt{17}}{4})\\F_{CG}=-515.388N$$

Again, the number we get is negative. The way we drew F_CG originally was as if the member was in tension. The negative number just means that it is actually compression, not tension.

$$ \underline{F_{CG}=515 \text{N (Compression)}}$$

Part c:

For part b I used the method of sections, as it would be the fastest method. The method of joints would require the lower joints to be solved first which would be a much slower process, whereas with this method a simple cut can be made and the member;s load can be quickly solved using equilibrium equations. Had the question asked for all member loads to be solved however, the method of joints would have been the faster approach.

6. Review

Part a:

R_Ax is equal and opposite to P so we know it is correct, and the value of R_By should also be correct as its moment about A (250 * 4m = 1000 Nm), is equal and opposite to the moment of P about A, (500 N * 2 m = 1000 Nm). as R_Ay is equal and opposite to R_By it is also correct.

Part b:

The x component of the calculated value of F_CG is equal in magnitude to P (see equation below), and it is the only cut member acting in the x direction. Therefore, it must be correct.

$$\frac{4}{\sqrt{17}}(515 N)=500 N$$

Part c:

The method of sections allows you to solve a very specific area of the systems internal forces (the members that are cut), whereas the method of joints usually requires you to solve most if not all of the internal forces of the system. Therefore, the method of sections is the most efficient for finding the internal forces of specific parts of the system, whereas the method of joints is more efficient for solving the whole system.

1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• All of the known values do not mean anything – we only need to know where the forces exist

Unknown:

• Which members are zero-force

4. Approach

Look at each joint and determine how many forces are in each direction. If there is only one force in a direction, that member is zero-force.

5. Analysis

Part a:

Let’s start with joint C. If we think of the forces acting in the x and y directions as shown below by the coordinate frame, we see that there are two forces acting in the y direction, and only one in the x direction. Therefore, assuming the joint is in static equilibrium, member CE is zero-force.

If we do the same type of analysis for the other joints and remove the zero-force members, the structure now looks like this:

After one more analysis of the joints, we find one more zero-force member, as shown below.

Answer: CE, DG, and FG are zero-force members.

Part b:

Zero-force members exist to provide stability to the truss, to keep the shape rigid.

6. Review

Although the new truss (without zero-force members) looks strange, there are no joints where there’s only one force in one direction, therefore there are no more zero-force members.