Question about stress and strain of a fiber

In summary, the homework statement says that if a tension is applied to a wire then the diameter will decrease. stressShear= E×I. The given equation states that stressShear= (10×109×7.85×10-21) which is 78.5 pPa.
  • #1
billy722
19
0

Homework Statement


2vbmvm1.jpg

i can not find the N.
what i can only found is:
D0=0.02mm;V=0.25;E=10G Pa;A=π(0.02mm/2)2
ΔD=εshear×D0
εshearaxial×V
εaxialaxial / E
σaxial=N/A

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
  • #2
I don't see anything in the question about a tension being applied, so why should it change diameter?
 
  • Like
Likes Chestermiller
  • #3
stressShear= E×I
R=0.01mm,E=10GPa, I0= (π/4)×(0.01×10(-3))4=7.85 zm4
Stressshear=(10×109×7.85×10-21)
=78.5p Pa

Shear modulus=E/[2(1+V)] = 10G/[2(1+0.25)]=4GPa
Shear modulus= Stressshearshear
εshear=78.5p/4G=19.625z

ΔD=εshear×D0
=19.625z×0.02m
=0.3925y m
Is it right?
 
  • #4
billy722 said:
stressShear= E×I
R=0.01mm,E=10GPa, I0= (π/4)×(0.01×10(-3))4=7.85 zm4
Stressshear=(10×109×7.85×10-21)
=78.5p Pa

Shear modulus=E/[2(1+V)] = 10G/[2(1+0.25)]=4GPa
Shear modulus= Stressshearshear
εshear=78.5p/4G=19.625z

ΔD=εshear×D0
=19.625z×0.02m
=0.3925y m
Is it right?
You seem to be using the given Young's modulus as though it is the applied stress. They happen to have the same dimension, pressure, but they are not the same thing. Specifically, the modulus is that stress which would, in theory, double the wire's length.
If the problem statement does not specify an applied load then there is no way to answer it.

Even then, I am unable to follow the rest of your calculation. Not sure what all those ys and zs mean. It makes no sense that the change in diameter can be 20 times the initial diameter.
 
  • #5
Stressshear=E×I=F/A
V=strainshear/strainaxial
⇒strainshear=Vχstrainaxial=ΔD/D0
⇒ΔD=D0χVχStrainaxial
=D0χVχ(stressaxial/E)
=D0χVχ(1/E)χ(FχA)
F in shear and F in axial are same?
 
  • #6
However, I calculate out 3.925*10^-26 again
I think that's wrong
 
  • #7
billy722 said:
Stressshear=E×I=F/A
V=strainshear/strainaxial
⇒strainshear=Vχstrainaxial=ΔD/D0
⇒ΔD=D0χVχStrainaxial
=D0χVχ(stressaxial/E)
=D0χVχ(1/E)χ(FχA)
F in shear and F in axial are same?
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
Am I getting through?
 

What is stress and strain?

Stress and strain are two important properties used to describe the mechanical behavior of a fiber. Stress refers to the amount of force applied to a material, while strain refers to the deformation or change in shape of the material as a result of that force.

How are stress and strain related?

Stress and strain are directly proportional to each other. This means that as the amount of stress applied to a material increases, the strain or deformation also increases.

What is the difference between tensile and compressive stress and strain?

Tensile stress and strain occur when a material is stretched or pulled, while compressive stress and strain occur when a material is compressed or pushed together. Both types of stress and strain can affect the behavior of a fiber in different ways.

What factors can affect the stress and strain of a fiber?

The stress and strain of a fiber can be influenced by various factors such as the type of material, the length and diameter of the fiber, the amount of force applied, and the temperature of the environment. The properties of the fiber itself, such as its elasticity and strength, also play a role.

Why is understanding stress and strain important for scientists?

Stress and strain are crucial concepts in materials science and engineering, as they help scientists and engineers predict and understand the behavior of a material under different conditions. This knowledge is essential for developing and designing new materials for various applications.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top