Calculating Ultimate Tensile Strength from Stress-Strain Curve

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SUMMARY

The ultimate tensile strength (UTS) of a material can be calculated from its stress-strain curve, specifically when necking begins at a true strain of 0.25. The stress-strain relationship is defined by the equation sigma = 120000(epsilon)^(1/2) psi. The calculated UTS is 116,800 psi, which includes the yield strength of 70,000 psi added to the stress at necking of 46,800 psi. Understanding the derivation of yield strength is crucial, as it is not explicitly provided in the problem statement.

PREREQUISITES
  • Understanding of stress-strain relationships in materials science
  • Familiarity with true strain and necking concepts
  • Knowledge of yield strength and its significance in material properties
  • Basic grasp of logarithmic functions and exponential equations
NEXT STEPS
  • Study the derivation of yield strength in materials using stress-strain curves
  • Learn about the implications of necking in ductile materials
  • Research the differences between elastic and plastic deformation
  • Explore the use of Young's modulus in calculating material properties
USEFUL FOR

Students in materials science, mechanical engineering, and anyone involved in material testing and analysis, particularly those studying tensile strength and stress-strain behavior.

pecosbill
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Homework Statement



What is the ultimate tensile strength if necking begins at a true strain = 0.25 in a material whose stress strain curve obeys the relation:
sigma=120000(epsilon)^1/2 psi?

Homework Equations




The Attempt at a Solution




I'm kind of stumped on this. I really doesn't understand what's going on, but I know the answer is 116,700 psi.
 
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Is this the relationship after the yield point, i.e. does this apply after the material has yielded, and therefore has elastic strain of YS/E?
 
yes, after necking begins. however, this is the only information i was provided with, no young's mod.
 
this is the solution my teacher handed out in class today, but when i asked him about adding the yield strength, he forgot why he added it.

_____
TRANSLATE INTO S:
if volume is constant,
sigma=l/a=(V(0)L)*s/VL(0))

sigma=(L/L(0))s
ln(L/L(0))=E or L/L(0)=e^0.25

sigma(at UTS)=60000=se^0.25
s=60000*e^/0.25=46800 psi

s(uts)=s+s(yield)=46800+70000 psi=116800 psi

________
This solution cleared nothing up for me. I understand up to the very last line...that all makes sense to me. But why does he add yield strength, and where the hell does he get that value of yield strength from? It's not given in the problem, so I assume there is an easy way to calculate it from the information given?
I know this problem is really confusing, but if anyone can offer an explanation of the yield strength and where it came from, I'd appreciate it.
 
If the material was loaded to necking and then unloaded, the measured true strain is the the permanent strain, since the material unloads along a diagonal parallel with the elastic line.

The stress-strain relationship appears to be just the stress-strain relationship for plasic deformation beyond yield.
 
the problem mentions nothing about loading and unloading, so i assume it was one constant stress/strain graph though?
 

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