# Magnitude fo the compressive stress

• jaredmt
No, you haven't got it yet :wink:The 60,000 lb load is spread over the whole of ab, so the force per unit area in ab is 60,000/π.
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)

## 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.

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.

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?

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!

(have a pi: π )
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?

ah ok got it, thanks :)

## What is magnitude of the compressive stress?

The magnitude of the compressive stress is the amount of force per unit area acting on an object that is being compressed. It is a measure of the internal resistance of an object to being squeezed or compacted.

## How is magnitude of the compressive stress calculated?

The magnitude of the compressive stress is calculated by dividing the force applied on an object by its cross-sectional area. It is typically measured in units of pressure, such as pascals (Pa) or newtons per square meter (N/m^2).

## What factors affect the magnitude of compressive stress?

The magnitude of compressive stress is affected by the force applied, the cross-sectional area of the object, and the material properties of the object. Objects with larger cross-sectional areas and stronger materials will have a higher magnitude of compressive stress.

## What are the implications of high magnitude of compressive stress?

High magnitude of compressive stress can cause deformation or failure of an object. It can also lead to structural instability and collapse. Therefore, it is important to consider the magnitude of compressive stress in the design and construction of structures.

## How is magnitude of compressive stress related to tensile stress?

Magnitude of compressive stress is essentially the opposite of tensile stress. While compressive stress squeezes or compresses an object, tensile stress pulls or stretches it. The magnitude of both stresses is important in determining the strength and stability of an object.

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