Poisson's Ratio Calculation for a Tie Bar Under Tensile Force

In summary: There are 2 'e''s, big E and little e. Big E usually denotes the modulus of elasticity, in units of force/length^2, and little e denotes the strain, in dimensionless units of length/length. E is the slope of the stress strain curve, that is Stress/Strain = E, as you noted. Since axial stress is Force/Area, calculated above, and e is delta L/L, as was calculated above, solve for E.
  • #1
johnboy14
22
0
My question is a bout this question below (Bear with me people)

A tie bar 25mm in diameter and 1m long extends in length by 1.2mm, when subjected to an axial tensile force of 80KN. If the diameter decreases by 0.007mm, determine the values of poisson's ration, and E,G and K.

My question is how do you solve the Poissions ratio if it is equal to lateral strain divided by Axial strain. Can someone help me to determine the lateral strain and do I need to covert 80KN into strain.

How can I figure out lateral strain that what I really need to know?
 
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  • #2
johnboy14 said:
My question is a bout this question below (Bear with me people)

A tie bar 25mm in diameter and 1m long extends in length by 1.2mm, when subjected to an axial tensile force of 80KN. If the diameter decreases by 0.007mm, determine the values of poisson's ration, and E,G and K.

My question is how do you solve the Poissions ratio if it is equal to lateral strain divided by Axial strain. Can someone help me to determine the lateral strain and do I need to covert 80KN into strain.

How can I figure out lateral strain that what I really need to know?
axial strain (e) is 1.2mm/1000mm = .0012; lateral strain is .007mm/25mm = .00028; Then poissons ratio is, as you noted, .00028/.0012 = .23.
The 80kN force creates an axial stress of P/A = 80,000N/((pi)(.025)^2/4) = 163MPa; the since stress is eE, solve E = 136GPa, and then G is .23(E) = 37.5MPa.
 
  • #3
PhanthomJay said:
axial strain (e) is 1.2mm/1000mm = .0012; lateral strain is .007mm/25mm = .00028; Then poissons ratio is, as you noted, .00028/.0012 = .23.
The 80kN force creates an axial stress of P/A = 80,000N/((pi)(.025)^2/4) = 163MPa; the since stress is eE, solve E = 136GPa, and then G is .23(E) = 37.5MPa.


Your going to have to clarify this bit, I am at a loss where you got the figures to solve E. Your answer of 136GPa is right though, I want to clarify to solve E isn't it stress/strain. How did you get this answer.
 
  • #4
johnboy14 said:
Your going to have to clarify this bit, I am at a loss where you got the figures to solve E. Your answer of 136GPa is right though, I want to clarify to solve E isn't it stress/strain. How did you get this answer.
There are 2 'e''s, big E and little e. Big E usually denotes the modulus of elasticity, in units of force/length^2, and little e denotes the strain, in dimensionless units of length/length. E is the slope of the stress strain curve, that is Stress/Strain = E, as you noted. Since axial stress is Force/Area, calculated above, and e is delta L/L, as was calculated above, solve for E.
 

1. What is Poisson ratio?

Poisson ratio is a material property that describes the relationship between the strain (deformation) that occurs in one direction when a material is stretched or compressed in another direction. It is typically denoted by the Greek letter μ, and is defined as the negative ratio of the transverse strain to the axial strain.

2. How is Poisson ratio calculated?

Poisson ratio can be calculated by dividing the negative transverse strain by the axial strain. It is also equal to the ratio of the lateral stress to the axial stress under uniaxial loading conditions. Poisson ratio can also be determined experimentally by measuring the lateral and axial strains and calculating their ratio.

3. What is a typical range for Poisson ratio?

Poisson ratio ranges from -1 to 0.5, with most common materials having a value between 0 and 0.5. Rubber and other highly compressible materials can have a Poisson ratio close to 0.5, while most metals and ceramics have a Poisson ratio around 0.3. Some materials, such as cork, have a negative Poisson ratio, meaning they expand laterally when compressed axially.

4. How does Poisson ratio affect a material's behavior?

Poisson ratio affects a material's behavior by determining how it deforms under stress. Materials with a low Poisson ratio tend to be stiffer and more brittle, while materials with a high Poisson ratio are more flexible and have greater resistance to fracture. Poisson ratio also plays a role in determining a material's ability to withstand different types of loads, such as tensile, compressive, and shear stresses.

5. Can Poisson ratio be altered?

Yes, Poisson ratio can be altered through various methods, such as changing the material's microstructure, adding fillers or reinforcements, or applying heat or pressure during manufacturing. It can also be affected by the type of loading conditions a material is subjected to. However, for most materials, Poisson ratio is considered to be a constant property that cannot be significantly altered without changing the material's composition or structure.

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