What Is the Correct Calculation of Stress in Welded Cylindrical Rods AB and BC?

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Homework Help Overview

The discussion revolves around calculating the average normal stress in two welded cylindrical rods, AB and BC, under specific loading conditions. The problem involves determining stress values based on given forces and dimensions of the rods.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the stress in both rods using the formula stress = P/A but questions the accuracy of their calculations, particularly for rod BC.
  • Some participants seek clarification on the loading conditions and the configuration of the rods, asking for details about the forces applied at points A, B, and C.
  • Others suggest that the original poster's description may be unclear, leading to confusion in understanding the problem setup.
  • One participant offers a visual representation to aid in understanding the forces acting on the rods.

Discussion Status

Contextual Notes

Participants note that the original poster's calculations may be affected by assumptions about the loading conditions and the connection of the rods to a mechanical ground. There is also mention of difficulties in conveying the problem visually, which may impact understanding.

freespirit
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Two solid cylinderical rods AB & BC are welded together at B & loaded as shown. Knowing that d(1)=30mm & d(2)=50mm, find the average normal stress in the midsection of (a)rod AB, (b) rod BC.
B____________C
125kN--->| |
A______________| |
60kN<---| | |
| | |
|______________| |
|<- 0.9m------>| |
125kN--->|____________|
|<---1.2m--->|


stress = P/A

(a) A= pie r ^2=pie (30mm/2)^2=.000707 m^2
stress=P/A= (60x10^3 N)/(.000707)=84.9MPa

(b) A= pie r^2)= pie(50mm/2)^2=.001963
stress= P/A= (250x10^3)/.001963=106.1 MPa

the stress for (b) is wrong; it's suppose to be 96.8MPa Can someone show me what I have done wrong and explain why it should be done differently? Thanks
 
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figure for post

......B____________C
.....125kN--->| ...|
...A______________|... |
60kN<---|.... | ...|
...| ....| ...|
...|______________|... |
...|<- 0.9m------>|... |
.....125kN--->|___________|
......|<---1.2m-->|
 
figure for post

I can't seem to get the figure to turn out right sorry about this, I will attempt to explain it in words.

the first rod AB is 0.9m long and has a diameter of d1=(30mm); 60kN<---- from the A end of this rod the second rod BC attaches to the B end of the first rod it is 1.2m long and has a diameter is d2=(50mm); 125 kN is applied both above and below where it is attached to the first rod on the B end 125kN---->
 
Maybe I just have a short attention span, but I don't see the situation. I've got rod AB welded to BC (connected at B, naturally). A is on the lefthand side and C is on the righthand side of my scrap paper. Tell me the following and I'll try to help you:

- Force applied at point A and the direction the force is pointing.

- Force applied at point B (and the direction).

- Force applied at point C (and the direction).

- Whether or not the rods are attached to anything (e.g. mechanical ground) and how.
 
forces

force at A is pointed to the left and is in the center on the end
force at B is pointed to the right and is at the top and bottom of where C attaches. Can I send you a pic of the thing? It keeps telling me here that the attachment is bigger then 400X 400 and i made sure it wasn't
 
OK, I think I get it. The way you're describing it is a little difficult to follow and may be the reason you're having trouble solving it.

In cheesy ASCII art, here's my sketch (sans dimensions):

Code: 
        125kN    __________
            --->|          |/
          ______|          |/
         |      |          |/
60kN  <--|      |          |/
         |______|          |/
                |          |/
            --->|__________|/
          125kN  

         A      B          C

So you've got A B and C going from left to right. At surface C, the bar is up against a wall (for all intents and purposes, connected to mechanical ground). At surface A, there is a force of 60 kN to the left. At surface B, there are two points of loading, each with 125 kN, as sketched.

Member AB is in tension. Since the loading is 60 kN and the area is as you have computed, the normal stress is ~84.9 MPa, as you calculated.

Member BC is in compression. The compressive force is (125+125-60)kN = 190 kN. (There is a reaction force normal to the wall that balances this force). The normal stress is again given by &sigma; = P/A. Use the proper value for P with the area you computed and you will find the answer.
 

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