Young Modulus of Elasticity
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1 Phy 2053 Announcements Good luck on Exam 2! Exam 2 grades – grades should be available by Monday, April 6 on UF e- Learning web site. Exam 2 solutions will be posted early next week HW Assignment 9 next Wednesday, April 8 Stress - force per unit area causing the deformation Strain - measure of the amount of deformation The elastic modulus is the constant of proportionality between stress and strain Three types of moduli: Young’s modulus, shear modulus, bulk modulus Elastic Properties of Solids o F L Y A L Δ = Young’s Modulus: Y Strain Stress = Stress = Y x Strain or Stress (F/A) has units of N/m 2 1 N/m 2 = 1 Pascal (Pa) Young’s Modulus: Elasticity in Length, Hooke’s Law redux F A L 0 ΔL Stress = F/A Strain = ΔL/L 0 L L Y A F L L Y A F Δ = ⇒ Δ = 0 0 k Determine the elongation of the rod shown below if it is under a tension of 5.8 × 10 3 N. Example #9-11 Y Al = 7 x 10 10 Pa Y Al = 11 x 10 10 Pa Shear Modulus S is the shear modulus A material having a large shear modulus is difficult to bend F shear stress A x shear strain h F x S A h = Δ = Δ = -F F Δx h
Transcript of Young Modulus of Elasticity
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1Phy 2053 Announcements
Good luck on Exam 2!
Exam 2 grades grades should be available by Monday, April 6 on UF e-Learning web site.
Exam 2 solutions will be posted early next week
HW Assignment 9 next Wednesday, April 8
Stress - force per unit area causing the deformation
Strain - measure of the amount of deformation
The elastic modulus is the constant of proportionality between stress and strain
Three types of moduli: Youngs modulus, shear modulus,
bulk modulus
Elastic Properties of Solids
o
F LY
A L=
Youngs Modulus:
YStrainStress =Stress = Y x Strain or
Stress (F/A) has units of N/m2 1 N/m2 = 1 Pascal (Pa)
Youngs Modulus: Elasticity in Length,Hookes Law redux
F
A
L0
L
Stress = F/A
Strain = L/L0
LLYAF
LLY
AF ==
00
k
Determine the elongation of the rod shown below if it is under a tension of 5.8 103 N.
Example #9-11
YAl = 7 x 1010 Pa YAl = 11 x 1010 Pa
Shear Modulus
S is the shear modulus
A material having a large shear modulus is difficult to bend
Fshear stress
Ax
shear strainh
F xS
A h
==
=
-F
Fx
h
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2Bulk Modulus
A material with a large bulk modulus is difficult to compress
The negative sign is included since an increase in pressure will produce a decrease in volume B is always positive
The compressibility is the reciprocal of the bulk modulus
VP B
V =
Notes on Moduli
Solids have Youngs, Bulk, and Shear moduli
Liquids and gases have only bulk moduli, they will not undergo a shearing or tensile stress The liquid or gas would flow instead
Pressure The force exerted
by a fluid on a submerged object at any point if perpendicular to the surface of the object
2mNPain
AFP =
Density The density of a substance of
uniform composition is defined as its mass per unit volume:
Units are kg/m3 (SI) or g/cm3(cgs)
1 g/cm3 = 1000 kg/m3
mV
Pressure
2mNPain
AFP =
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3Pressure and Depth
Examine the darker region, assumed to be a fluid It has a cross-
sectional area A Extends to a depth h
below the surface
Three external forces act on the region
A
Pressure and Depth equation
Po is normal
atmospheric pressure1.013 x 105 Pa
The pressure does not depend upon the shape of the container
Liquid in a U- tubePatm
Patm
h1h2
P = Patm + gh1 P = Patm + gh2
Liquid in a U- tubePatm Patm
h1h2
P = Patm + gh1 P = Patm + gh2
h1 = h2
