Shear stress without fracture?

In summary: I have an equation for polymers which describes the relationship between the shear yield point and uniaxial yield point based on the Von Mises criterion:τy = ( ((1+μ)/sqrt(3)) /sqrt(3) ) * σyFrom a diagram plotted by a stress test, I have the value for uniaxial yield point σy; the point where the material becomes subject to plastic deformation.μ is a material parameter which determines the change in yield point with respect to a change in pressure. But I'm not sure how I can acquire this parameter, nor can I find any tables for this parameter. Neither do I know the name of
  • #1
risecolt
38
0
Material: PVC
E-modulus: 1300 N/mm^2
Poisson's ratio: 0.35
Applied shear force: 100 N
Resisting area: 28 mm^2

Shear modulus: E = 2G(1+v) --> G = E/(2(1+v)) = 1300/(2(1+0.35) = 481 N/mm^2

Does this mean that a force which can be applied to the area without fracturing is 481 N/mm^2 * 28 mm^2?

I just need to know. How can I use this information to calculate how much shear force the area can be exposed to?
 
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  • #2
G is the ratio of shear stress to shear strain, analogous to E being the ratio of tensile stress to tensile strain.

In order to answer your question about maximum shear stress of PVC, you will have to find out more information about the material (e.g., max. tensile stress) The max. shear stress is usually expressed as a factor * ult. tensile stress for ductile materials like metals. For plastics like PVC, there may be a different relationship.
 
  • #3
SteamKing said:
G is the ratio of shear stress to shear strain, analogous to E being the ratio of tensile stress to tensile strain.

In order to answer your question about maximum shear stress of PVC, you will have to find out more information about the material (e.g., max. tensile stress) The max. shear stress is usually expressed as a factor * ult. tensile stress for ductile materials like metals. For plastics like PVC, there may be a different relationship.

I have an equation for polymers which describes the relationship between the shear yield point and uniaxial yield point based on the Von Mises criterion:

τy = ( ((1+μ)/sqrt(3)) /sqrt(3) ) * σy

From a diagram plotted by a stress test, I have the value for uniaxial yield point σy; the point where the material becomes subject to plastic deformation.

μ is a material parameter which determines the change in yield point with respect to a change in pressure. But I'm not sure how I can acquire this parameter, nor can I find any tables for this parameter. Neither do I know the name of this parameter. Is it effective viscosity?
 

1. What is shear stress without fracture?

Shear stress without fracture is the maximum amount of stress a material can withstand before it begins to deform and change shape without breaking or fracturing.

2. How is shear stress without fracture different from tensile stress?

Tensile stress measures the maximum amount of stress a material can withstand before it breaks or fractures, while shear stress without fracture measures the maximum amount of stress a material can withstand before it begins to deform without breaking.

3. What factors affect shear stress without fracture?

Some factors that can affect shear stress without fracture include the type and composition of the material, the temperature and environment it is in, and any external forces or loads acting on the material.

4. Why is understanding shear stress without fracture important?

Understanding shear stress without fracture is important for engineers and scientists in designing and selecting materials for various applications. It helps ensure the structural integrity and safety of buildings, bridges, and other structures, as well as the reliability of mechanical components.

5. How is shear stress without fracture measured?

Shear stress without fracture is typically measured using a shear stress test, where a sample of the material is subjected to a gradually increasing force until it reaches its maximum shear stress. This can also be calculated using the material's shear modulus and the amount of shear strain it experiences.

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