Magnitude fo the compressive stress

[jaredmt]

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.

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?

 

[jaredmt]

does anybody know how to do this?

 

[tiny-tim]

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!

(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 )

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?

 

[jaredmt]

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

 

[Dr.D]

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 ...

 

[tiny-tim]

Hi jaredmt!

(have a pi: π )

Tab = load/area = (60,000 + P)/(pi*1^2)

Nooo …

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

 

[jaredmt]

ah ok got it, thanks :)