Stress, Strain, and elastic moduli

1. Physics 7C lecture 16 • Stress, Strain, and elastic moduli Tuesday November 26, 8:00 AM – 9:20 AM Engineering Hall 1200

2. Strain, stress, and elastic moduli • Stretching, squeezing, and twisting a real body causes it to deform, as shown in figure below. We shall study the relationship between forces and the deformations they cause. • Stress is the force per unit area and strain is the fractional deformation due to the stress. Elastic modulus is stress divided by strain. • The proportionality of stress and strain is called Hooke’s law.

3. Strain, stress, and elastic moduli force per unit dimension Stress = Elastic modulus Strain fractional deformation molecular basis for Hook’s law:

4. Tensile and compressive stress and strain • Tensile stress = F /A and tensile strain = l/l0. Compressive stress and compressive strain are defined in a similar way. • Young’s modulus is tensile stress divided by tensile strain, and is given by Y = (F/A)(l0/l).

5. Tensile and compressive stress and strain Young’s modulus = tensile stress / tensile strain, Y = (F/A)(l0/l). setup for measuring Young’s modulus tiny! This is pressure!

6. Some values of elastic moduli

7. Tensile and compressive stress and strain setup for measuring Young’s modulus 1 meter long steel string with 1 mm diameter, pulled by 10 N weight! Ysteel = 20 X 1010 P the elongation is dL = L * F / (A Y) = only 0.5 mm!

8. Tensile stress and strain • In many cases, a body can experience both tensile and compressive stress at the same time.

9. Q11.5 length L F F Two rods are made of the same kind of steel and have the same diameter. length 2L F F A force of magnitude F is applied to the end of each rod. Compared to the rod of length L, the rod of length 2L has A. more stress and more strain. B. the same stress and more strain. C. the same stress and less strain. D. less stress and less strain. E. the same stress and the same strain.

10. A11.5 length L F F Two rods are made of the same kind of steel and have the same diameter. length 2L F F A force of magnitude F is applied to the end of each rod. Compared to the rod of length L, the rod of length 2L has A. more stress and more strain. B. the same stress and more strain. C. the same stress and less strain. D. less stress and less strain. E. the same stress and the same strain.

11. Q11.6 length L F F Two rods are made of the same kind of steel. The longer rod has a greater diameter. length 2L F F A force of magnitude F is applied to the end of each rod. Compared to the rod of length L, the rod of length 2L has A. more stress and more strain. B. the same stress and more strain. C. the same stress and less strain. D. less stress and less strain. E. the same stress and the same strain.

12. A11.6 length L F F Two rods are made of the same kind of steel. The longer rod has a greater diameter. length 2L F F A force of magnitude F is applied to the end of each rod. Compared to the rod of length L, the rod of length 2L has A. more stress and more strain. B. the same stress and more strain. C. the same stress and less strain. D. less stress and less strain. E. the same stress and the same strain.

13. Bulk stress and strain

14. Bulk stress and strain • Pressure in a fluid is force per unit area: p = F/A. • Bulk stress is pressure change p and bulk strain is fractional volume changeV/V0. • Bulk modulus is bulk stress divided by bulk strain and is given by B = –p/(V/V0).

15. Anglerfish 1000 m deep sea

16. 1/ bulk modulus = compressibility k = 1/B In 1000 meter deep sea, how much is water compressed? dV/V = k * p = 46.4E-6 atm-1* 100 atm = 0.464 % ! water is very hard to compress! Table 11.2

17. Sheer stress and strain

18. Sheer stress and strain • Sheer stress is F||/A and sheer strain is x/h, as shown in figure. • Sheer modulus is sheer stress divided by sheer strain, and is given by S = (F||/A)(h/x).

19. Sheer stress and strain of brass The left piece experience shear force in an earthquake. It is 0.8 m square and 0.5 cm thick. What is the force exerted on each of its edges if the displacement is x = 0.16 mm? sol: shear strain= x/h = 0.16 mm / 0.8 m = 0.0002 shear stress = shear strain * S = 0.0002 * 3.5E10 Pa = 7E6 Pa force F = shear stress * A = 7E6 Pa * 0.8 m* 0.5 cm = 2.8E4 N This is a huge force, even the displacement (0.16) is tiny!

20. Elasticity and plasticity • Hooke’s law applies up to point a in Figure 11.18 below. • Table 11.3 shows some approximate breaking stresses.

21. Midterm problem A solid, uniform cylinder (radius r; mass m; moment of inertia around central axis I = ½ m r2) rolls without slipping up a hill, as shown in the figure. The initial speed is v0 and the height of the hill is h. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. How far from the foot of the cliff does the cylinder land? (20 points) (b) Calculate the linear speed v2 and angular speed ω2 when the cylinder lands. Is the total kinetic energy at that moment equal or smaller than the initial kinetic energy when the cylinder was at the initial speed v0? (20 points)

22. Midterm problem

23. Midterm problem