Homework 22.A

Correction: "All VERTICAL members of length 3d"

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51 thoughts on “Homework 22.A”

  1. 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?

    1. 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.

  2. 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?

    1. 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.

  3. 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?

    1. 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.

    1. 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.

  4. 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?

  5. 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?

  6. 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.

    1. 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.

  7. 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?

  8. 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?

    1. 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.

  9. 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?

    1. 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

  10. 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?

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