Presentation is loading. Please wait.

Presentation is loading. Please wait.

Copyright © 2012 Pearson Education Inc. Stress, Strain, and elastic moduli Physics 7C lecture 16 Tuesday November 26, 8:00 AM – 9:20 AM Engineering Hall.

Similar presentations


Presentation on theme: "Copyright © 2012 Pearson Education Inc. Stress, Strain, and elastic moduli Physics 7C lecture 16 Tuesday November 26, 8:00 AM – 9:20 AM Engineering Hall."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. Stress, Strain, and elastic moduli Physics 7C lecture 16 Tuesday November 26, 8:00 AM – 9:20 AM Engineering Hall 1200

2 Copyright © 2012 Pearson Education Inc. 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 Copyright © 2012 Pearson Education Inc. Strain, stress, and elastic moduli Stress Elastic modulus Strain = force per unit dimension fractional deformation molecular basis for Hook’s law:

4 Copyright © 2012 Pearson Education Inc. Tensile and compressive stress and strain Tensile stress = F  /A and tensile strain =  l/l 0. 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)(l 0 /  l).

5 Copyright © 2012 Pearson Education Inc. Tensile and compressive stress and strain Young’s modulus = tensile stress / tensile strain, Y = (F  /A)(l 0 /  l). This is pressure! tiny! setup for measuring Young’s modulus

6 Copyright © 2012 Pearson Education Inc. Some values of elastic moduli

7 Copyright © 2012 Pearson Education Inc. 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! Y steel = 20 X P the elongation is dL = L * F / (A Y) = only 0.5 mm!

8 Copyright © 2012 Pearson Education Inc. Tensile stress and strain In many cases, a body can experience both tensile and compressive stress at the same time.

9 Copyright © 2012 Pearson Education Inc. 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. Q11.5 Two rods are made of the same kind of steel and have the same diameter. FF length 2L FF length L 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

10 Copyright © 2012 Pearson Education Inc. 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. A11.5 Two rods are made of the same kind of steel and have the same diameter. FF length 2L FF length L 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

11 Copyright © 2012 Pearson Education Inc. 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. Q11.6 Two rods are made of the same kind of steel. The longer rod has a greater diameter. FF length 2L FF length L 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

12 Copyright © 2012 Pearson Education Inc. 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. A11.6 Two rods are made of the same kind of steel. The longer rod has a greater diameter. FF length 2L FF length L 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

13 Copyright © 2012 Pearson Education Inc. Bulk stress and strain

14 Copyright © 2012 Pearson Education Inc. 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/V 0. Bulk modulus is bulk stress divided by bulk strain and is given by B = –  p/(  V/V 0 ).

15 Copyright © 2012 Pearson Education Inc. Anglerfish 1000 m deep sea

16 © 2012 Pearson Education, Inc. Table / 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 = % ! water is very hard to compress!

17 Copyright © 2012 Pearson Education Inc. Sheer stress and strain

18 Copyright © 2012 Pearson Education Inc. 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 Copyright © 2012 Pearson Education Inc. 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 = shear stress = shear strain * S = * 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 Copyright © 2012 Pearson Education Inc. Elasticity and plasticity Hooke’s law applies up to point a in Figure below. Table 11.3 shows some approximate breaking stresses.

21 Copyright © 2012 Pearson Education Inc. Midterm problem A solid, uniform cylinder (radius r; mass m; moment of inertia around central axis I = ½ m r 2 ) rolls without slipping up a hill, as shown in the figure. The initial speed is v 0 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. (a)How far from the foot of the cliff does the cylinder land? (20 points) (b) Calculate the linear speed v 2 and angular speed ω 2 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 v 0 ? (20 points)

22 Copyright © 2012 Pearson Education Inc. Midterm problem

23 Copyright © 2012 Pearson Education Inc. Midterm problem


Download ppt "Copyright © 2012 Pearson Education Inc. Stress, Strain, and elastic moduli Physics 7C lecture 16 Tuesday November 26, 8:00 AM – 9:20 AM Engineering Hall."

Similar presentations


Ads by Google