Calculating Ultimate Tensile Strength from Stress-Strain Curve

In summary, the question asks for the ultimate tensile strength of a material with a stress-strain curve of sigma=120000(epsilon)^1/2 psi, where necking begins at a true strain of 0.25. After some calculations and assumptions, the answer is determined to be 116,800 psi, taking into account the yield strength of 70,000 psi which was not explicitly given in the problem. The yield strength was likely calculated using the information given and the assumption that the material was loaded and unloaded before necking began.
  • #1
pecosbill
13
0

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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
yes, after necking begins. however, this is the only information i was provided with, no young's mod.
 
  • #4
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.
 
  • #5
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.
 
  • #6
the problem mentions nothing about loading and unloading, so i assume it was one constant stress/strain graph though?
 

1. What is ultimate tensile strength?

Ultimate tensile strength (UTS) is the maximum amount of stress that a material can withstand before breaking or fracturing.

2. How is ultimate tensile strength measured?

UTS is typically measured by subjecting a material to a tensile test, where a sample is pulled until it breaks and the maximum force applied is recorded.

3. What factors affect ultimate tensile strength?

The ultimate tensile strength of a material is influenced by various factors, including the type and quality of the material, its composition, temperature, and manufacturing process.

4. Why is ultimate tensile strength important?

UTS is an important property to consider when selecting materials for engineering applications, as it indicates the maximum stress a material can withstand without failing. It helps ensure the safety and reliability of structures and products.

5. How does ultimate tensile strength differ from yield strength?

While ultimate tensile strength measures the maximum stress a material can withstand before breaking, yield strength measures the stress at which a material begins to permanently deform. In most cases, yield strength is lower than ultimate tensile strength.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
9K
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
963
  • Engineering and Comp Sci Homework Help
Replies
7
Views
23K
Replies
5
Views
738
  • Engineering and Comp Sci Homework Help
Replies
2
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
7K
Back
Top