1. Physics 6A Stress, Strain and Elastic Deformations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. When a force is applied to an object, it will deform. If it snaps back to its original shape when the force is removed, then the deformation was ELASTIC. We already know about springs - remember Hooke’s Law : F spring = -kΔx Hooke’s Law is a special case of a more general rule involving stress and strain. The constant will depend on the material that the object is made from, and it is called an ELASTIC MODULUS. In the case of tension (stretching) or compression we will call it Young’s Modulus*. So our basic formula will be: *Bonus Question – who is this formula named for? Click here for the answerClick here for the answer Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5. To use our formula we need to define what we mean by Stress and Strain. STRESS is the same idea as PRESSURE. In fact it is the same formula: STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain: Now we can put these together to get our formula for the Young’s Modulus: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Young’s modulus for this nylon? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

7. ΔL=1.1m L 0 =45m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Young’s modulus for this nylon?

8. ΔL=1.1m L 0 =45m A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Young’s modulus for this nylon?

9. ΔL=1.1m L 0 =45m 7mm A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Don’t forget to cut the diameter in half. EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Young’s modulus for this nylon?

10. ΔL=1.1m L 0 =45m 7mm A couple of quick calculations and we can just plug in to our formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Don’t forget to cut the diameter in half. EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a 65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is Young’s modulus for this nylon?

11. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

12. L 0 =2m ΔL=0.25cm 400N diam=? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

13. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

14. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

15. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

16. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

17. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

18. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

19. L 0 =2m ΔL=0.25cm 400N diam=? We have most of the information for our formula. We can look up Young’s modulus for steel in a table: The only piece missing is the area – we can rearrange the formula One last step – we need the diameter, and we have the area: double the radius to get the diameter: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum diameter must this wire have?

20. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

21. We can do this one just by staring at the formula for stress: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

22. We can do this one just by staring at the formula for stress: The force is the same in both cases because it says they use the same weight. The area is related to the square of the radius (or diameter), so when the diameter doubles the area goes up by a factor of 4. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

23. We can do this one just by staring at the formula for stress: The force is the same in both cases because it says they use the same weight. The area is related to the square of the radius (or diameter), so when the diameter doubles the area goes up by a factor of 4. Thus the stress should go down by a factor of 4 (area is in the denominator) Answerc) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a tensile stress X in the wire. If the same weight is hung from a wire having twice the diameter as the first one, the tensile stress in this wire will be

24. When pressure is applied to an object from all directions, its volume will change accordingly. Think of squishing a foam ball, or inflating a balloon. In this case we use a 3-dimensional version of Young’s modulus. We call it BULK MODULUS, and it is defined in a similar way: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Bulk Modulus and Volume Changes Bulk Modulus Volume change Pressure change

25. Example: When water freezes into ice it expands in volume by 9.05 percent. Suppose a volume of water is in a household water pipe or a cavity in a rock. If the water freezes, what pressure must be exerted on it to keep its volume from expanding? (If the pipe or rock cannot supply this pressure, the pipe will burst or the rock will split.) The bulk modulus for ice is 8x10 9 N/m 2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Bulk Modulus and Volume Changes Answser: 6.6x10 8 N/m 2