Magnitude fo the compressive stress

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Homework Help Overview

The problem involves two solid cylindrical rods, AB and BC, which are welded together and subjected to a load. The objective is to determine the force P such that the tensile stress in rod AB is twice the compressive stress in rod BC.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of stress, noting that it is typically defined as load divided by area. There is uncertainty about how to apply this to the entire rod versus specific points.
  • Some participants explore the relationship between tensile and compressive stresses, questioning the correct proportions and calculations involved.
  • One participant expresses confusion over their calculations and the expected outcome, indicating a potential misunderstanding of the forces involved.

Discussion Status

The discussion includes various attempts to establish relationships between the forces and stresses in the rods. Some participants have provided hints and guidance, while others are still grappling with the calculations and assumptions. Multiple interpretations of the load conditions are being explored.

Contextual Notes

There are indications of confusion regarding the total load acting on the rods, with participants questioning whether the load should be considered as 30 kips or 60 kips. Additionally, the original poster has noted a specific requirement for the relationship between the tensile and compressive stresses.

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Homework Statement


two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude fo the compressive stress in rod BC.

attachment.php?attachmentid=17276&stc=1&d=1232744490.jpg

edit: p is supposed to be pointing outward (left)

Homework Equations


The Attempt at a Solution



im not sure how to do this because i thought stress was calculated at a certain point/area. I am not sure how to calculate stress for the whole rod.
stress = load/area

how would i do this?
 

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does anybody know how to do this?
 
jaredmt said:
two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude fo the compressive stress in rod BC.

im not sure how to do this because i thought stress was calculated at a certain point/area. I am not sure how to calculate stress for the whole rod.
stress = load/area

Hi jaredmt! :smile:

(best to use links, not attachments, because attachments take too long to get approved, and by then your thread is too far down the list :wink:)

tensile stress = tension force per area,

compressive stress = compression force per area,

so if the tensile stress in rod AB is twice the magnitude fo the compressive stress in rod BC, what is the proportion of tension force over compression force? :smile:
 


Tbc = 2Tab
is that what u mean?
im pretty sure I am calculating at least one of the tensions incorrectly. here's what i have:
Tbc = load/area = (60,000 - P)/(pi*1.5^2)
Tab = load/area = (60,000 + P)/(pi*1^2)

do they look right at all? then i set Tbc = 2Tab to find P but i get a negative answer and its wrong. the answer should be 28.2kips
 


I have a hunch that at the step, that is just 30 kips total, not 60 kips acting there.

sigma_ab=P/(pi*2^2)/4
sigma_bc=(B-P)/(pi*3^2)/4
where B is the load acting at the step, either 30 or 60 kips.

Then
abs(sigma_ab)/abs(sigma_bc) = 2
so that
abs(P/(pi*2^2)/4) = 2*abs((B-P)/(pi*3^2)/4)
P/pi = 2*abs(B-P)/(9*pi/4)
and so forth ...
 
Hi jaredmt! :smile:

(have a pi: π :wink:)
jaredmt said:
Tab = load/area = (60,000 + P)/(pi*1^2)

Nooo …

Hint: if P was 0, how much would the tension force be in ab? :wink:
 


ah ok got it, thanks :)
 

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