Physics 71 Lecture Notes

Chapter 11 Equilibrium and Elasticity

Topic 3.03: Elasticity

Date: April 21, 2014

Stress Strain and Elastic Moduli Stress:

The strength of the forces causing the deformation.

Strain: The resulting deformation.

Hooke's Law: When the stress and strain are small enough, they

are directly proportional to each other.

StressStrain

=Elastic Modulus

Tensile Stress and StrainObject in tension Forces of the same magnitude but opposite in directions are applied at the object's ends (pulling the object from its ends).

Tensile stress=F A

S.I. unit of tensile stress is 1 [N/m2] = 1[Pa]

Tensile strain=ll0l 0

= ll0

Y=Tensile stressTensile strain

=F l 0A l

Young's modulus (Y)

Tensile Stress and StrainObject in compression Forces on the ends of a bar pushes rather than pulls.

Compressive strain=ll 0l0

= ll 0

Y=F l 0A l

Tensile Stress and Strain

A material with higher value of Y is relatively unstretchable.

For most materials, the Young's modulus for both the tensile and compressive stresses are the same.

Tensile Stress and Strain Bodies can experience tensile and compressive

stresses at the same time. For most materials, the Young's modulus for both the

tensile and compressive stresses are the same.

Tensile Stress and StrainA steel rod 2.0 [m] long has a cross-sectional area of 0.30 cm2 . It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the stress on the rod and the resulting strain and elongation.

Stress=F A

=550 [kg ]9.8 [m / s2]

0.30104 [m2]=1.8108[Pa ]

E1. Calculate the corresponding strain on the rod.

Y= StressStrain

Strain= StressY

= 1.8108[Pa ]

201010 [Pa ]=9.0104

Tensile Stress and StrainA steel rod 2.0 [m] long has a cross-sectional area of 0.30 cm2 . It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the stress on the rod and the resulting strain and elongation.

Stress=F A

=1.8108[Pa ]

Strain= ll0

l=Strainl 0

To solve for the elongation:

=(9.0104)(2.0 [m])

=18101(103 [m])=1.8[mm]

Tensile Stress and StrainA circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?

Y steel=201010 [Pa ]

l0=2.00 [m ] l=0.25[cm]=0.25102 [m ]

F=400 [N ]A=?d=?

Y steel=F A

l0 l

A=F Y steel

l0 l

E2. Calculate the diameter of the wire.

d=1.4 [mm]

d2

4=

F Y steel

l0 l

d= 4 F Y steel l0 l

Bulk Stress and StrainThe stress is now a uniform pressure on all sides, and the resulting deformation is a volume change.

p=F A

S.I. unit of pressure: 1 [N/m2]=1[Pa]

1atmosphere=1[atm ]=1.013105 [Pa ]

The force per unit area that the fluid exerts on the surface of an object immersed object is called the pressure p:

The approximate pressure of the earth's atmosphere at sea level:

Bulk Strain=VV 0

Where V0 is the object's initial volume and V is the change in volume.

Bulk Stress and Strain

B= pV /V 0

;The bulk modulus is defined as

p= p p0

Bulk Stress and Strain

Compressibility (k): k=1B= 1

V 0V p

Material with higher compressibility (lower bulk modulus) is easier to compress than those with lower k (higher B).

Bulk Stress and StrainA hydraulic press contains 0.25 m3 (250 L) of oil. Find the decrease in the volume of the oil when it is subject to a pressure increase of 1.6 x 107 [Pa]. The bulk modulus of the oil is 5.0 x 109 [Pa].

B= pV /V 0

V= pV 0B

=(1.6107[Pa ])(0.25[m3])

5.0109 [Pa ]=8.0104 [m3]=0.8[L ]

Note: Pressure is not the same as force. It is a scalar quantity with no particular direction.

Shear Stress and StrainForces of equal magnitude but opposite direction act tangent to the surfaces of opposite ends of the object

Shear Stress=FA

Shear Strain= xh

Shear Modulus=S= Shear StressShear Strain =

FAhx

In lab tests on a 9.25-cm cube of a certain material, a force of 1375 N directed at 8.50 to the cube causes the cube to deform through an angle of 1.24. What is the shear modulus of the material?

E3. What is the shear stress on the cube?

Shear Stress and Strain

s=9.25[cm]=9.25102[m ]A=s2=8.556103[m2]

Shear stress=159[k Pa ]

Shear strain= xh=tan (1.24) =0.02165

S= Shear StressShear Strain

=159103 [Pa ]

0.02165=7.34106 [Pa ]

Shear stress=F A= 1375 cos8.50(9.25102[m ])2

Elasticity and Plasticity

Hooke's law is applicable only in the small stress-strain region.

Hysteresis materials follow different curves for increasing and decreasing stress.

Elasticity and PlasticityA brass wire is to withstand a tensile force of 350 [N] without breaking. What minimum diameter must the wire have? The breaking stress of the brass is 4.7 x 108 [Pa].

Breaking Stress=F A

; A= d2

4

A=F

Breaking Stress d 2

4=

F Breaking Stress

d= 4 F Breaking Stress =0.97 [mm]

A 5.0 x 102-N weight is hung from the end of a wire of cross-sectional area 0.010 cm2. The wire stretches from its original length of 200.00 cm to 200.50 cm.1. What is the stress on the wire? 5 x 108 [Pa]2. What is the elongation strain on the wire? 2.5 x 10-3

3. Determine the Young's modulus of the wire. 2 x 1011 [Pa]

4. A cable stretches by an amount d when it supports a crate of mass M. The cable is cut in half. What is the mass of the load that can be supported by either half of the cable if the cable stretches by an amount d? (a) M/4 (b) M/2 (c) M (d) 2M (e) 4M 5. A cable stretches by an amount d when it supports a crate of mass M. The cable is cut in half. If the same crate is supported by either half of the cable, by how much will the cable stretch? (a) d (b) d/2 (c) d/4 (d) 2d (e) 4d

Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18