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## 51 thoughts on “Homework 23.A”

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Would there be reaction forces in the Y direction for both I and Q or could we use just one?

It depends. Use the left hand side instead will be much easier.

Yes, I believe that there is reaction forces in both X and Y direction at I and Q.

Pretty sure they exist but you don't need them.

For each of the joints with a 0 force member, can we just identify the joint type and state which member carries no load, or do we have to do an FBD for each?

It is recommended to have an FBD for each joint.

Yes by having an FBD for each joint, you can better organize equilibrium equations and better visualize which joints should be focused on primarily, to most effectively find the specified values.

Make sure to remember to shift the coordinate axis when looking for the zero force members.

Yes, this tricked me for a little bit and is essential to achieving the correct results and relationships.

Thanks for the heads up. I ended up trying to solve the problem out and was confused at first, but this tip would have definitely helped had I known.

Good tip. I was definitely confused in the beginning about why specific members had no force because I did not shift the coordinate axes.

Would IQ be a zero-force member since both ends are constrained at I and Q?

I don't think IQ is a zero-force member, since at joint I there are unknown reactions at I to be considered other than IQ. Same thing at joint Q too, where there is the unknown force member QH.

Figo is right. IQ need not be a zero force member.

No, since there are outside forces acting on the member.

Do we need to explain each zero force or just identify them?

Just a line or two explaining the reason would be sufficient. No thorough explanation needed.

Are there supposed to be reactionary forces in the x or y direction for both I and Q or just one of them?

Hi Basil. From the diagram, you can tell that I and Q have the same support, which is a force on x and y respectively. In general, If there are more unknowns than available equations, maybe is not a good idea to find the reactions; instead, focus on the structure's opposite section so you don't deal with the reactions.

Would it be possible to solve it the other way including the reaction forces?

Hi Dominic. In general, if you find the reactions first you can choose that side of the truss. However, in this problem, we are dealing with an external indeterminate structure, so we would need more than the rigid body equilibrium equations to find the reactions.

Any tips on what to start with after doing a cut through DE, EN, NH, and NO?

Since that cut exposes 4 members, and you are only looking for 1,2 and 3, you can deduce that one of the members is a 0 force member by drawing a FBD. After that, I did sum of the moments about point N, which leads you to solve for one of the members. Hope that helps!

You cant cut through 4 segments usually. Try a cut through EH, HN, and NO first, and then go from there

Do we need to include the zero force members in our FBD's of each joint after we have identified the zero force member?

If you solved that it was a zero force through calculation, then I always include them in the free-body diagram. But if you can initially tell that they will be 0 based on the configuration of the members, then I don't draw them.

Even though the zero force members have been identified, I would still include them in the FBDs to show that they are still part of the system.

How many zero force members should there be?

Hi Victoria. I was able to find 11 zero force members. Although, if someone else could reply to this to confirm that would be helpful.

I also found there to be 11 zero force members for 23.A

I only found 9. Should members AB and JB be zero force members? If so, where does the reasoning behind this come from?

Joint A acts like an elbow joint because there is no external force on A and it has only 2 truss members. AB and AJ are zero force members as they cannot negate each other on both the axis. JB is not a zero force member.

Can there be multiple shifted x and y axes?

For each joint's FBD, yes, you can shift the x and y axes.

Hey Nick,

Yes you can do multiple shifted axes, you can do it on a case by case basis as you like!

At points on the triangle parts of the Russ on the left can we rotate the members so that they are situated in the x and y coordinate axes better in our fbds? Also, are we supposed to assume that members coming off of points such as S or U like SK and RU are perpendicular?

You should rotate the axes rather than the members themselves for specific FBD's. You cannot assume that certain members are perpendicular. By setting up equations for specific joints, you will see that the presence of a force on certain members in question will result in an impossible system of equations, thus showing that it must be a zero force member.

*Truss

How are we supposed to show work for the zero force members ?

What do you mean by work? When we identify a zero-force member and do our method of sections, we could include those forces in our equations, but since we know that their magnitude is 0, we know that they do not contribute to the sum of our forces.

Hi Isaac. One possibility is to draw the FBDs of the joints and apply the equilibrium equations for each node. Another way is to explain it by using words; for example, "by analyzing joint X I can tell that there is no force counteracting the element AB so it must be a zero force element, and so on for joints Y and Z, so elements BC and CD are zero force elements as well". In general, a joint analysis will show the work for zero force elements.

By sectioning off the left side of the truss to solve for F1, F2, and F3, will it include the entire left side, including the large inverted triangle truss? Or can we just focus on the immediate left area around the members that were cut through?

You should include the entire left side so that the external forces are still considered. When I solved for my F1, I used a moment about N which needed those external loads in its equation. Additionally they can be used for a FBD of the left side to to solve for F3 once you have F2 and F1.

When using the method of sections, how do you decide which side of the system to choose? I know you can solve the problem either way, but I just want to know how to save time on the exams.

I personally like to choose the side of the system which has the least amount of reaction forces. This tends to be easier since you'd have less variables to solve for at a given time. (Less calculations)

Agreed- I found it easier to solve this one by sectioning off the left side including the loads. I tried the other way originally, but couldn't find y components for the reactions at I and Q. Is it possible to solve the problem that way?

You can always solve it both ways, but solving the side with the reactions I and Q can be more stressful and you want to save time.

I was also trying to figure out how to section of this problem and this really helped, thanks.

We don't need to calculate Iy and Qy separately since they're on the same direction along the same line, right?

Use the left hand side and you do not need to calculate the constraint at I and Q.

I'm still a little confused about something. In the truss problems before this one and the other problem for HW 23, the angles were all the same, but I'm a little confused about how to find the angles between the segments in the left part of the structure. Maybe I just need to brush up on my geometry knowledge.