How to Calculate Young's Modulus from a Non-Linear Stress-Strain Graph

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To calculate Young's modulus from a non-linear stress-strain graph, it's essential to focus on the linear portion of the graph, typically below the elastic limit. The formula for Young's modulus is the change in stress divided by the change in strain. When analyzing the graph, different gradients indicate non-linearity, complicating the calculation. The correct approach is to use the gradient of the initial linear segment for an accurate Young's modulus value. Understanding that the first data point is (0,0) helps clarify the calculation process.
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Homework Statement


Question simply asks to calculate the Youngs modulus. I've attached the question.



Homework Equations


Youngs Modulus = (change in stress)/(change in strain)


The Attempt at a Solution


okay, what i did was simply find the gradient between each of the points. However they were all different gradients, which means the graph isn't linear (i think). So then how do i solve it?
 

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TyErd said:

Homework Statement


Question simply asks to calculate the Youngs modulus. I've attached the question.

Homework Equations


Youngs Modulus = (change in stress)/(change in strain)

The Attempt at a Solution


okay, what i did was simply find the gradient between each of the points. However they were all different gradients, which means the graph isn't linear (i think). So then how do i solve it?

I believe Youngs modulus can only be calculated for stress-strain that remains below the elastic limit - ie the first straight part of the graph.

EDIT: btw, I don't see the question attached, but hopefully this will help anyway.
 
sorry about that, I've attached the question now.
 
TyErd said:
sorry about that, I've attached the question now.

What did you make of the statement in the first sentence?
 
Oh i get it, the first data point would be 0 stress 0 strain and the gradient between 0,0 and the 1st data point is the same as the gradient between the first data point and the second data point but different from the 2nd and 3rd. I get it, thank you!
 
TyErd said:
Oh i get it, the first data point would be 0 stress 0 strain and the gradient between 0,0 and the 1st data point is the same as the gradient between the first data point and the second data point but different from the 2nd and 3rd. I get it, thank you!

well done.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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