"Calculating Stress-Strain Graph Homework

In summary, the student is trying to plot a stress-strain graph for a given question. They have calculated the strain by dividing the extensions by the initial length and now need to calculate the stress using the force and area. They are unsure whether to use the original diameter or the diameter at fracture, and the conversation includes a discussion about the maximum load and yield point. Ultimately, they decide to use the original diameter for all points except the last two, where they will use the stated diameter.
  • #1
TyErd
299
0

Homework Statement


okay i have attached the question and what i simply have to do is plot the stress-strain graph.

Homework Equations


strain = extension/initial length
stress=force/area

The Attempt at a Solution


Okay, i calculated the strain by dividing each of the extensions by 30mm which is correct right??
But now i need to calculate the stress in order to plot the graph which is simply the force divided by the area however it says the diameter reduces down from 12mm to 11.74mm. at fracture. So then which one do i use to calculate the stress the 12 or 11.74. how do i know exactly at what force the diameter changes??or do i just ignore that and find the stress by dividing the force by 12mm---0.0012metres?
 

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  • #2
TyErd said:

Homework Statement


okay i have attached the question and what i simply have to do is plot the stress-strain graph.

Homework Equations


strain = extension/initial length
stress=force/area

The Attempt at a Solution


Okay, i calculated the strain by dividing each of the extensions by 30mm which is correct right??
But now i need to calculate the stress in order to plot the graph which is simply the force divided by the area however it says the diameter reduces down from 12mm to 11.74mm. at fracture. So then which one do i use to calculate the stress the 12 or 11.74. how do i know exactly at what force the diameter changes??or do i just ignore that and find the stress by dividing the force by 12mm---0.0012metres?

To calculate the stress I would ignore the diameter changes - but I would not divide by the diameter - I think you should be dividing by the cross-sectional area.

Your proposal from stress was correct.
 
  • #3
okay thankyou.
 
  • #4
okay i have another similar question to this except its given original diameter, diameter at maximum load and diameter at fracture. Would i still proceed to find stress by using original diameter??
 
  • #5
TyErd said:
okay i have another similar question to this except its given original diameter, diameter at maximum load and diameter at fracture. Would i still proceed to find stress by using original diameter??

How much difference in the diameters?

When you say diameter at maximum load - was the maximum load near fracture or near the yield point?

Certainly in the last bit before a sample fails we see the applied force drop as the sample distorts - the stress of course remaining the same.
 
  • #6
well the diameters are:
original diameter-12.8mm
diameter at max load - 11.5mm
diameter at fracture - 7.21mm

the max load is near the fracture but I am not sure because i can't find what the yield point is without knowing which diameters to use to find stress. I've attached the table given.
 

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  • #7
TyErd said:
well the diameters are:
original diameter-12.8mm
diameter at max load - 11.5mm
diameter at fracture - 7.21mm

the max load is near the fracture but I am not sure because i can't find what the yield point is without knowing which diameters to use to find stress. I've attached the table given.

I would certainly use original diameter for all points up to the last two, then the stated diameter fro those two.
 
  • #8
okay thanks.
 

What is a stress-strain graph?

A stress-strain graph is a graphical representation of the relationship between the amount of stress applied to a material and the resulting strain or deformation of the material. It shows how a material responds to external forces and can provide valuable information about its strength, stiffness, and ductility.

How do you calculate stress and strain?

To calculate stress, you divide the force applied to a material by its cross-sectional area. The resulting unit is usually expressed in units of force per unit area, such as pounds per square inch (psi) or newtons per square meter (N/m^2). To calculate strain, you divide the change in length of the material by its original length. This is typically expressed as a decimal or percentage.

What affects the shape of a stress-strain graph?

The shape of a stress-strain graph is affected by several factors, including the type of material, its composition, and its physical properties. Different types of materials, such as metals, polymers, and ceramics, have unique stress-strain curves due to their varying structures and bonding. The composition of a material can also affect its behavior under stress, as different elements can alter its strength and ductility. Finally, physical properties like temperature and strain rate can also impact the shape of a stress-strain graph.

What can a stress-strain graph tell you about a material?

A stress-strain graph can provide valuable information about a material's mechanical properties. The slope of the initial linear portion of the graph, known as the elastic region, represents the material's stiffness or Young's modulus. The point at which the graph deviates from linearity, known as the yield point, indicates the material's yield strength. The maximum stress the material can withstand before failure is shown by the ultimate or tensile strength. The area under the curve represents the total energy required to fracture the material, known as toughness. Overall, a stress-strain graph can help determine a material's strength, stiffness, and ductility.

How can a stress-strain graph be used in engineering and material selection?

In engineering, a stress-strain graph is often used to evaluate the mechanical properties of different materials and determine their suitability for specific applications. By comparing the stress-strain curves of various materials, engineers can select the most appropriate material for a particular project based on its required strength, stiffness, and ductility. It can also help engineers predict the behavior of a material under different conditions, such as varying stress levels or temperatures.

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