Consider P is load applied on a tensile specimen. Why dP=0 at maximum load and what does this dP=0 physically mean?

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# Consider P is load applied on a tensile specimen. Why dP=0 at maximum load and what does this dP=0 physically mean?

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I think you are asking in Engineering stress strain curve. So dP=0 is the situation where the change in load is 0. Generally for Engineering stress strain curve it's assumed that area is constant. So the maxiumum load you subject to the specimen, is the maximum stress the material can withstand,whicRead more

I think you are asking in Engineering stress strain curve. So dP=0 is the situation where the change in load is 0. Generally for Engineering stress strain curve it’s assumed that area is constant. So the maxiumum load you subject to the specimen, is the maximum stress the material can withstand,which is UTS. Afterwards necking starts in the sample.

Mathematically speaking , the question you asked is why slope is 0 at maximum point ofÂ the curve.

I think physical meaning of dP=0 is formation of necking in the sample

Correct me, if I am wrong.

See lessYes, I am asking for Engineering Stress Strain Curve. Explain mathematically why dP=0.

Yes, I am asking for Engineering Stress Strain Curve. Explain mathematically why dP=0.

See lessP=Stress*Area.=S*A dP=SdA+AdS dP/(A*S)=-(-dA/A)+(dS/S) (dP/P)=-(-dA/A)+(dS/S) Before UTS dP/P is >0 as change in stress is higher than change in area After UTS dP/P is <0 as change in area is higher. So the load required for the deformation is also low. In engineering stress strain curve we coRead more

P=Stress*Area.=S*A

dP=SdA+AdS

dP/(A*S)=-(-dA/A)+(dS/S)

(dP/P)=-(-dA/A)+(dS/S)

Before UTS

dP/P is >0 as change in stress is higher than change in area

After UTS dP/P is <0 as change in area is higher. So the load required for the deformation is also low.

In engineering stress strain curve we consider current load/Intial area. So the the stress decreases which means P decreases.

Mathematically when the curveÂ has slope changing from positive to negative, that means there should be a maximum where change is 0. Here the mxiumum is UTS where dP is 0.

Adittional:

You can try in this way. Draw a typical engineering stress strain curve. Consider any two points on the curve immediately before UTS, and two points after UTS.

projectÂ all the 4points on X axis and Y axis. It is clear that before UTS, more change in X-axis leads to less change in Y-axis but after UTS, slight change in X-axis leads to large change in Y-axis. This can happen when curve is concave in nature which has maximum.

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