Principles of Engineering Page 1 of 6 Virtual Tensometer Homework Exercise
Introduction
The relationship between the stress and strain that a particular material displays is known as that particular material's stress–strain curve. It is unique for each material and is found by recording the amount of deformation (strain) at distinct intervals of tensile or compressive loading (stress). These curves reveal many of the properties of a material, including the Modulus of Elasticity, E.
For a thorough discussion, see Wikipedia Stress-Strain Curves.
In this exercise, you will analyze 4 common engineering materials using a Virtual Tensometer.
Figure 1: A typical stress-strain curve for a material such as steel
Procedure 1. Go to the Virtual Tensometer. Principles of Engineering Page 2 of 6 Virtual Tensometer Homework Exercise http://lrrpublic.cli.det.nsw.edu.au/lrrSecure/Sites/Web/tensile_testing/index.htm
2. Click ‘Enter’ to begin the simulation.
3. Choose ‘Want to try’? followed by “Use the virtual tensometer”.
4. This should bring you to the Virtual Tensometer for testing.
5. Test each of the 4 materials by dragging it to the clamps and pressing ‘start’. Principles of Engineering Page 3 of 6 Virtual Tensometer Homework Exercise 6. When each test is complete, click on the print icon to print the graph.
Part 1: Questions 1-3 You may do the following work on separate paper, or create space below and type your answers directly in this Word document. When you are done, submit your work as an email attachment.
NOTE: The original diameter of each specimen is 5mm and the original length is 75mm.
1. Yield Strength The stress point at which a material begins to deform plastically. Prior to the yield point the material will deform elastically – i.e. it will return to its original shape when the applied stress is removed. ‘Yield’ does not necessarily mean ‘Failure’, as illustrated by Figure 1, above.
Question: What is the yield strength of each material, in Pascals? 1 Pascal = 1 N/m2 (Newton per square meter)
a. Steel b. Aluminum c. Copper d. Cast Iron
Example: On its load-extension diagram, an unknown metal sample yields under a load of 16 kN. The sample had an original diameter of 5mm and an original length of 75mm.
Step 1: Find the cross sectional area in meters2, from A = (3.14) x (.0025m)2 = 0.00002 m2
Step 2: Solve for the yield strength in Pascals (N/m2). Yield Strength = Force/Area = 16,000 N / 0.00002 m2 = 800 x 106 Pascals = 800 MPa
2. Ultimate Strength Principles of Engineering Page 4 of 6 Virtual Tensometer Homework Exercise Ultimate strength, also called ultimate tensile strength (UTS), is the maximum stress that a material can withstand while being stretched or pulled before failing or breaking. Beyond this point, elongation of the sample may continue for a time, but the force exerted decreases.
What is the ultimate strength of each material, in Pascals? 1 Pascal = 1 N/m2 (Newton per square meter)
a. Steel b. Aluminum c. Copper d. Cast Iron
Example: On its load-extension diagram, an unknown metal sample reaches ultimate strength at a load of 20 kN. The sample had an original diameter of 5mm and an original length of 75mm.
Step 1: Find the cross sectional area in meters2, from A = (3.14) x (.0025m)2 = 0.00002 m2
Step 2: Solve for the ultimate strength in Pascals (N/m2). Ultimate Strength = Force/Area = 20,000 N / 0.00002 m2 = 1,000 x 106 Pascals = 1,000 MPa
3. Breaking/Failure Point The maximum amount of stress that can be applied before rupture occurs. The specimen fractures in the necking region where the material reduces in diameter as it elongates.
What is the failure point of each material, in Pascals? 1 Pascal = 1 N/m2 (Newton per square meter)
a. Steel b. Aluminum c. Copper d. Cast Iron
Example: On its load-extension diagram, an unknown metal sample reaches the failure point at a load of 14 kN. The sample had an original diameter of 5mm and an original length of 75mm. Principles of Engineering Page 5 of 6 Virtual Tensometer Homework Exercise
Step 1: Find the cross sectional area in meters2, from A = (3.14) x (.0025m)2 = 0.00002 m2
Step 2: Solve for the failure stress in Pascals (N/m2). Failure Stress = Force/Area = 14,000 N / 0.00002 m2 = 700 x 106 Pascals = 700 MPa
Part 2: Questions 4-11
4. Which of the 4 materials is the most brittle?
5. Which of the 4 materials is the most ductile?
6. Rank the 4 materials in terms of Ultimate Strength.
7. Rank the 4 materials in terms of Stiffness (highest modulus of elasticity).
8. From an engineering standpoint, why do you think cast iron has been largely replaced with other materials (such as ductile iron and/or steel)?
9. Copper is 5 times more expensive than steel. Yet mild steel and copper have similar yield strength and ultimate strength. Considering this fact, why would copper be used as an engineering material? List 3 good reasons, with an example of each.
10.Plain aluminum is 2 times more expensive than steel. Furthermore, its yield strength and ultimate strength are significantly less than steel. Given these facts, list 3 good reasons – with an example of each - why aluminum is used for engineering applications.
11.If steel is so strong, why isn’t your entire skeleton – all 206 bones - designed out of steel? Wouldn’t that be an improvement? List 5 good engineering reasons why it is not. This will take some research on your part. Principles of Engineering Page 6 of 6 Virtual Tensometer Homework Exercise