Correction: **"All VERTICAL members of length 3d"**

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Correction: **"All VERTICAL members of length 3d"**

We are encouraging you to communicate with us and with your colleagues in the class through the threaded discussions on the course blog. If you have questions on this homework, please ask here.

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Since A has no reactionary forces in the x direction, should the reactionary force in the x direction at K be equal and opposite to P + 2P?

Very good observation! Yes, you are right.

What does the statement saying all horizontal members of lengths 3d and 4d mean? Does it mean we should calculate the moments for both lengths?

Just give you the information of length of each truss in case you need moment equation.

Oh that was a typo. Should be "All VERTICAL members of length 3d"

Thank you for clearing this up as I too had the same concern when first examining the problem.

I believe it was an error. It should say all vertical members are length 3d and all horizontal members are length 4d.

We only need to find the load in the members 1, 2, and 3, correct? We do not have to find the loads in each member of the truss to label them with either tension or compression?

Yes Olivia, we only find the loads in members 1, 2 and 3 and then label these as compression, tension, or zero force members.

Radhika, do we only have to label members 1, 2, and 3 as in compression, tension, or zero members? Or is it all of the members of the truss?

I believe we only label members 1, 2, and 3!

Does A have a vertical reactionary force?

Yes, because A is a roller there will be a y component at A

There will be force acting in the y direction because of the fact that we are dealing with a roller.

So can we assume AB and AC are zero force members or just that they cancel eachother out?

Yes. There should be a reaction normal to the roller at A, which in this case is in the vertical direction.

When finding the reaction forces, do we have to consider the whole truss system or can we use the sections we broke into?

Followup: It is the whole truss system

Do I need to solve the external reaction first or can I just solve the internal reaction?

Hi Alexa, I found it helpful to solve for the external reactions first! This way, you can solve for Ay, Kx, and Ky and then use these values in equations for your internal reactions.

Splitting up the truss and solving it from either side should give you the same answer, correct? Is there a specific side that is better to pick than the other?

Yes, splitting up the truss from either side will work but for me it was easier to use the upper right section so you can minimize the values you are working with and target the specific members the problem asks for.

I agree, using the right side is advantageous because you only need to consider two external forces instead of 4 if you were to use the left side.

I understand that this problems requires the use of the Method of Sections. Do we have to solve for the forces from both directions after we slice it or can we just select one section and find all the forces from that?

Selecting one section would allow you to find all loads on the members that needed to be found. I suggest that starting by finding the vertical reaction force at point A by using the sum of Fx, Fy, and Mk. With Ay it should be pretty easy to find the three loads. FBD on the left side section would help.

I want to clarify that A is roller and K is pinned, right??

Yes, there will be x and y reaction forces at K and a reaction force normal to A in the y direction.

Should the reaction force in the x direction at K be equal and opposite to P or P+2P? Since A has no reaction forces that is.

To solve for the reaction forces at K, you need to look at the entire truss. So to find Kx you would sum all the forces in the x direction and set it equal to 0.

Yeah, since A doesn't have the ability to react in the X direction all the external forces in the x direction have to be reacted to at K.

Is the reaction Ay equal to P in the opposite direction?

I would think so, since the reaction Kx is equal to P+2P.

I also agree with Mehal. This should be case after a summation of the forces in the y direction.

If we can mentally deduce which side we want to use for our method of sections do we have to draw a free body diagram for the other side that we are not using?

Draw FBD for the side that you are using.

Does the point B create a moment that can be shown in the sum of the moments in K? I am not sure because B is not directly connect to K through its x length?

If B is in your free body, include it.

Is the reaction force at K in the y- direction supposed to be negative (pointing downwards)? I got a negative number from my calculations but that doesn't seem right.

why do you think that is not right?

Actually now that I've looked at it more, it's making sense to me.

I'm confused on how to find the angle to solve for Fbe. Could anyone help?

you can set up a right angle triangle where the vertical is 3 and the horizontal is 4. Then you can use arctan of 3/4 to find the angle between section BE and the horizontal.

I was able to reach the correct load carried by member 3 using equilibrium of forces. However, I should also be able to find this value using a moment equation about D. When I solved the moment equation, I somehow got a different answer of (-20/3)P using 0 = Fce(3d) + Fbe(3/5)d - P(4d) + Ay(8d), where Fce was (-8/3)P. Where did I make a mistake?

Would you guys suggest to use multiple moment equations to find the desired member forces, or just stick to one moment and the sum of Fx and Fy?

Hi, for this question I didn't have to use multiple moment equations, just sum of Fx, sum of Fy and a moment equation. I think it's because there weren't many forces in the desired section and using those 3 equations are sufficient.

For the lengths given, is one of them supposed to say vertical instead of horizontal?

@Jake Patrick yes, it says on the front of this homework discussion! On the assignment, horizontal (4d) and vertical (3d)

I can't tell if I'm setting up my equations properly or not, but I am getting 4 unknowns (the three members plus the reaction at A) and 3 equations (sum of x-components, sum of y-components, and sum of moments). Is there a 4th equation I should use? Or are there things getting cancelled out that I am not thinking of?

I think I may have found a workaround by finding the sum of moments about a different joint to go along with the first sum of moments I was using

For the FBD's on this problem i was thinking of doing one for each side of the system after i split it, is that what you guys did or is there a better way?

That's what I did! I split it and I thought it was easier that way.