elasticity wire

Examples for Chapter 10Fundamentals of College Physics, 3rd.Ed

Dr. Peter J. Nolan

"Fundamentals of College Physics" Third EditionDr. Peter J. Nolan, SUNY Farmingdale

Chapter 10 ElasticityComputer Assisted InstructionInteractive Examples

Example 10.1

10.0-kg mass hung from it. (a)How much will the wire stretch? (b) What is the stresson the wire? (c)What is the strain?

Initial Conditions

1 mm = 10 kg

g = 9.8d = 1 mm

Y = 2.10E+11

Solution.

3.14159 ) x ( 0.001 7.854E-07

We assume that the cross-sectional area of the wire does not change during the stretching process. The force stretching the wire is the weight of the 10.0-kg mass, that is,

F = mg = ( 10 kg) x ( 9.8 98 N

elongation of the wire, found from modifying equation 10.6, is

98 N) x ( 1 m)]

/ [ ( 7.85E-07 2.10E+11

5.94E-04 m = 0.59418 mm

F / A = ( 98 N) / ( 7.85E-07 1.25E+08

Stretching a wire. A steel wire 1.00 m long with a diameter d = 1.00 mm has a

L0 =

m/s2

N/m2

a. The cross-sectional area of the wire is given by

A = pd2 / 4 = [( m)2] / 4 = m2

m/s2) =

Young's modulus for steel is found in table 10.1 as Y = 21 x 1010 N/m2. The

DL = F L0 / A YDL = [(

m2) x ( N/m2)DL =

b. The stress acting on the wire is

m2) = N/m2

c. The strain of the wire is


Dr. Peter J. Nolan

5.94E-04 m) / ( 1 5.94E-04

Example 10.2

3.05 m long and 10.2 cm in diameter. By how much is the column compressed?

Initial Conditions

3.05 m d = 0.102 m

F = 445000 N Y = 2.10E+11

Solution. The cross-sectional area of the column is

3.14159 ) x ( 0.102 0.00817

The change in length of the column, found from equation 10.6, is

445000 N) x ( 3.05 m)]

/ [ ( 8.17E-03 2.10E+11

7.91E-04 m = 0.0791 cm = 0.79095 mm

Note that the compression is quite small 0.790952 mm considering the very large load 445000 N. This is indicative of the very strong molecular forcesin the lattice structure of the solid.

Example 10.3

Initial Conditions

25 cm A = 4

F = 5.00E+05 1.70E+08

Solution. The stress acting on the bone is found from

F / A = 5.00E+05 N) / ( 4.00E-04 1.25E+09

DL / L0 = ( m) =

Compressing a steel column. A 445,000-N load is placed on top of a steel column

L0 = N/m2

A = pd2 / 4 = [( m)2] / 4 = m2

DL = F L0 / A YDL = [(

m2) x ( N/m2)DL =

Exceeding the ultimate compressive strength. A human bone is subjected to a

compressive force of 5.00 x 105 N/m2. The bone is 25.0 cm long and has an

approximate area of 4.00 cm2. If the ultimate compressive strength for a bone is

1.70 x 108 N/m2, will the bone be compressed or will it break under this force?

L0 = cm2

N/m2 (F/A)max = N/m2

m2) = N/m2

Since this stress exceeds the ultimate compressive stress of a bone, 1.70 x 108 N/m2,


Dr. Peter J. Nolan

the bone will break.

Example 10.4

a 0.500-kg mass. Find the elongation of the spring.

Initial Conditions50 N/m

g = 9.8m = 0.5

Solution. The elongation of the spring, found from Hooke's law, equation 10.9, is

x = F / k = mg / k

x = ( 0.5 kg) x ( 9.8 50 N/mx = ( 0.098 m

Example 10.5

is acted on by a tangential force of 50,000 N, as shown in figure 10.9. The value of S

Initial Conditionsbase b = 0.75 m thickness t = 0.5 cm = 5.00E-03 m

height h = 1 m 50000 N

Shear Modulus S = 4.20E+10Solution. a. The area that the tangential force is acting over is

A = bt =

0.75 m) x ( 5.00E-03 3.75E-03

The elongation of a spring. A spring with a force constant of 50.0 N/m is loaded with

k = m/s2

kg

m/s2) / (

Elasticity of shear. A sheet of copper 0.750 m long, 1.00 m high, and 0.500 cm thick

for copper is 4.20 x 1010 N/m2. Find (a) the shearing stress, (b) the shearing strain, and (c)the linear displacement Dx.

tangential force Ft =

N/m2

A = ( m) = m2


Dr. Peter J. Nolan

where b is the length of the base and t is the thickness of the copper sheet shown in figure 10.9. The shearing stress is

5.00E+04 N) / ( 3.75E-03 1.33E+07

b. The shearing strain, found from equation 10.15, is

50000 N / [ ( 3.75E-03 4.20E+10

3.17E-04 rad

1 m) x ( 3.17E-04

3.17E-04 m = 0.31746 mm

Example 10.6

Initial Conditions

0.5

p = 3.00E+05

B = 1.40E+11

Solution. The change in volume, found from equation 10.20, is

0.5 3.00E+05

/ [ ( 1.40E+11

-1.07E-06

The minus sign indicates that the volume has decreased.

Ft / A = m2) = N/m2

f = Ft / A S

f = [( m2) x ( N/m2)]f =

c. The linear displacement Dx, found from equation 10.10, isDx = h f

Dx = ( rad)Dx =

Elasticity of volume. A solid copper sphere of 0.500-m3 volume is placed 30.5 m

below the ocean surface where the pressure is 3.00 x 105 N/m2. What is the change

in volume of the sphere? The bulk modulus for copper is 14 x 1010 N/m2.

V0 = m3

N/m2

N/m2

DV = - V0 p / B

DV = -[( m3) x ( N/m2)]

N/m2)

DV = m3


Dr. Peter J. Nolan

"Fundamentals of College Physics" Third Edition

10.0-kg mass hung from it. (a)How much will the wire stretch? (b) What is the stress


Dr. Peter J. Nolan

3.05 m long and 10.2 cm in diameter. By how much is the column compressed?

. If the ultimate compressive strength for a bone is

, will the bone be compressed or will it break under this force?


Dr. Peter J. Nolan

is acted on by a tangential force of 50,000 N, as shown in figure 10.9. The value of S

. Find (a) the shearing stress, (b) the shearing strain,


Dr. Peter J. Nolan

. What is the change

elasticity wire

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