# Homework 7.1

## 23 thoughts on “Homework 7.1”

1. Daniel Toth says:

There is a lot of confusion on where the weight of the sign acts. Is the sign and bar separate rigid bodies (I.e. weight acts at 6L) or are they separate bodies (ie weight acts at 5L)?

2. Aniket Roy Chowdhury says:

Because the sign is rigid and the beam is not, would we assume that there is a reaction moment acting at the attachment point b/w the sign and the beam that we also have to account for?

1. Lukas Karim Hessini says:

Yes, you should first do an FBD of the sign with its weight and reaction force and moment to find those reactions; then an FBD of the beam with reactions at A and the end of the bar to find rxns at A.

2. Lucas Eichenberger says:

You can do this as Lukas described, or you can also do equilibrium of the beam-sign system (with the weight force acting 6L away from the fixed end) to find the reactions at A. Both methods should result in the same answer.

3. Scott Jay Shelton says:

Are we supposed to say that the sign generates a moment at the end of the bar due to there being a horizontal distance "L" between the center of mass of the sign and its connection to the bar?

1. Bruno Dantas De Maio Carrilho says:

I think so. In the end, I believe you might have a force and a moment (both due to the sign) in the end of the beam

4. Lukas Karim Hessini says:

Do we solve this in terms of w? Since I is in terms of w and w is not given.

1. Lukas Karim Hessini says:

Nevermind I see we are finding w

5. Richard Lu says:

For part (b), what value do we assume to take as w?

1. Micah Milton Larsen says:

I don't know definitively, but the answer from part (a) makes the most sense.

6. Chibuzor Vincent Nwaobi says:

Hey, are we supposed to say that the sign generates a moment at the end of the bar due to there being a horizontal distance "L" between the COM of the sign and its connection to the bar?

1. Micah Milton Larsen says:

I believe so. You can either treat the beam as 5L long with a moment at the end, or 6L long and analyze it at 5L.

7. Peter James Agostino says:

I just want to appreciate the LOTR reference.

8. Mason Gumin says:

I am able to find equations for slope and deflection using the second order method, but I am unsure how to relate this to finding w. Does it have to do with the second area moment for the beam?

1. Samuel Stephen Robertson says:

I believe so, EI should be in your equation for v and w would be in your equation for second moment area. So you can get w from those relations

9. ayeverin says:

Is there an example from the book or on the website that is similar to this?

10. Samuel Stephen Robertson says:

Are the dimensions of the sign, used at all for this question?

1. Mason Gumin says:

I only used the fact that the point where the mg force occurs is L meters away from the 5L beam in the x direction.

11. Donglin says:

Should we consider the moment that the sign generates at the attached point or from the center of mass of the mass of the sign which x = 6L?

1. ayeverin says:

I believe focusing on the moment that the weight of the sign generates at the attached point on the beam is the correct approach.

12. Samuel Stephen Robertson says:

Will the moment found from the FBD of the sign have the opposite sign when viewed on the beam?

1. purdue_exclamation_mark says:

The moment would act in a CW direction with respect to the wall, as the weight of the sign is a downward force.

13. James Sun says:

How should I deal with the part that the sign exceed the beam. Should I add a moment at the connection?