Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 –...

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Hooke's Law and Modulus of Elasticity 1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing

Transcript of Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 –...

Page 1: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 1

Hooke’s Law and Modulus of Elasticity (2.1-2.7)

MAE 314 – Solid Mechanics

Yun Jing

Page 2: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 2

Introduction to Normal Strain

L

A

P

stress

L

A

P

A

P

2

2

LL

A

P

2

2

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Hooke's Law and Modulus of Elasticity 3

Introduction to Normal Strain Normal strain (ε) is defined as the deformation per unit length of a member under axial loading. Normal strain is dimensionless but can be expressed in several ways. Let us assume L = 100 mm and δ = 0.01 mm.

ε = 0.01 mm / 100 mm = 1 x 10-4 or 100 x 10-6 ε = 100 μ (read as 100 microstrain) ε = 1 x 10-4 in/in (if using English units) ε = 1 x 10-4 * 100 = 0.01%

L

Page 4: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 4

Mechanical Properties of Materials It is preferable to have a method of analysis that is characteristic of the properties of materials (σ and ε) rather than the dimensions or load (δ and P) of a particular specimen. Why?

σ & ε are truly material properties P & δ are specimen properties

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Hooke's Law and Modulus of Elasticity 5

Mechanical Properties of Materials Stress and strain can be measured, so we want to develop a relationship between the two for a given material. How do we calculate the elongation of a bar due to loading?

Apply force P Calculate σ = P/A Use material relation ε = f(σ) to calculate ε Calculate δ = εL

Page 6: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 6

Stress-Strain Diagram Material behavior is generally represented by a stress-strain diagram, which is obtained by conducting a tensile test on a specimen of material.

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Hooke's Law and Modulus of Elasticity 7

Stress-Strain Diagram Stress-strain diagrams of various materials vary widely. Different tensile tests conducted on the same material may yield different results depending on test conditions (temperature, loading method, etc.). Divide materials into two broad categories:

Ductile material - Material that undergoes large permanent strains before failure (e.g. steel, aluminum) Brittle material - Material that fails with little elongation after yield stress (e.g. glass, ceramics, concrete)

Let’s examine the stress-strain diagram for a typical ductile material (low-carbon steel) region by region.

Page 8: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 8

Stress-Strain Diagram (Low-carbon steel)

Linear region Stress-strain response is linear Slope = Modulus of Elasticity (Young’smodulus) = E E has units of forceper unit area (same asstress) We get a relationbetween stress andstrain known as Hooke’s Law.

E

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Hooke's Law and Modulus of Elasticity 9

Stress-Strain Diagram (Low-carbon steel)

Yielding region Begins at yield stress σY Slope rapidly decreasesuntil it is horizontal ornear horizontal Large strain increase,small stress increase Strain is permanent

Page 10: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 10

Stress-Strain Diagram (Low-carbon steel)

Strain Hardening After undergoing largedeformations, the metalhas changed itscrystalline structure. The material hasincreased resistanceto applied stress(it appears to be“harder”).

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Hooke's Law and Modulus of Elasticity 11

Stress-Strain Diagram (Low-carbon steel)

Necking The maximum supportedstress value is called theultimate stress, σu. Loading beyond σuresults in decreasedload supported andeventually rupture. Breaking strength, σB

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Hooke's Law and Modulus of Elasticity 12

Stress-Strain Diagram (Low-carbon steel)

Why does the stress appear to drop during necking? If we measure the true area, the graph looks like: The difference is in thearea: true stress takesinto account the decreased cross-section area. Thus, at the same stresslevel, the load drops.

true stress x

Page 13: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

For some materials (e.g. aluminum) there is not a clear yield stress. We can use the offset method to determine σY. Choose the offset (0.002 is shown here).

Draw a line with slope E, throughthe point (0.002, 0). σY is given by the intersection ofthis line with the stress-strain curve.

Hooke's Law and Modulus of Elasticity 13

Offset Method

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Hooke's Law and Modulus of Elasticity 14

Elastic vs. Plastic A material is said to behave elastically if the strain caused by the application of load disappear when the load is removed – it returns to its original state. The largest value of stress for which the material behaves elastically is called the elastic limit (basically the same as σY in materials with a well-defined yield point). Once the yield stress has been obtained, when the load is removed, the stress and strain decrease linearly but do not return to their original state. This indicates plastic deformation. When a material does not have a well-defined yield point, the elastic limit can be closely approximated using the offset method.

Page 15: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Hooke's Law and Modulus of Elasticity 15

Elastic vs. Plastic

Y

Y

Reload

Plastic deformation (Permanent strain)

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Deformations 16

Deformations (2.8-2.10)

MAE 314 – Solid Mechanics

Yun Jing

Page 17: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Deformations 17

Deformations Under Axial Loading Conditions

Solid bars, cables, coil springs, etc. Axial tension or compression Prismatic

Recall Hooke’s Law ( ) andequations for stress ( ) and strain ( ).

E

A

P

L

L

EA

P AE

PL

Page 18: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Deformations 18

Deformations Under Axial Loading What about non-prismatic bars?

Discrete Changes: total change in lengthis simply the summation of the change inlength of each portion.

Important: Each time the internalforce, area, or material changesyou need a new free-body diagram!i ii

ii

AE

LP

EA

LP

EA

LP

EA

LP

3

33

2

22

1

11

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Deformations 19

Deformations Under Axial Loading What about non-prismatic bars?

Continuous Changes: continuously changing area (as shown) or continuously changing force (such as a rod hanging under its own weight) Deformation of an element of lengthdx can be expressed as: Integrating this over the length of therod:

)(

)(

xEA

dxxPd

L

dxxEA

xP

0 )(

)(• This is an approximation since we made the assumption earlier that the stress distribution is constant over the cross-section.• For small variations this is a good approximation.

Page 20: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Example problem

Hooke's Law and Modulus of Elasticity 20

The rigid bar BDE is supported by two links AB and CD. Link AB has a cross-sectional area of 500mm^2, E=70GPa. Link CD has a cross-sectional area of 600mm^2, E=200GPa. For the 30kN force shown, determine the deflection of B, D and E.

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Deformations 21

Statically Indeterminate Problems Statically determinate structure – reactions and internal forces can be determined uniquely from free-body diagram and equations of equilibrium. Statically indeterminate structure – there are more unknown reactions than equations of equilibrium. Where do the other equations needed to solve the unknown reactions come from?

Equations of compatibility which are based on displacements. Here is a easy method to determine how many compatibility equations you need for any given problem:

M = R – N M = number of compatibility equations needed R = number of unknown reactions (or internal stresses) N = number of equilibrium equations

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Deformations 22

Statically Indeterminate Problems

Statically indeterminate structure

R = 1 : R1N = 1 : ΣFY = R1 – P = 0M = 1 - 1 = 0R = 2 : R1, R2N = 1 : ΣFY = R1 – P – R2 = 0M = 2 - 1 = 1

R1

R2

R1

Page 23: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Deformations 23

Temperature Changes Changes in temperature produce expansion or compression, which cause strain.

α = coefficient of thermal expansion ΔT = change in temperature Sign convention: expansion is positive (+), contraction isnegative (-)

For a bar that is completely free to deform (one or both ends free): In this case, there is thermal strain but no thermal stress!

TT

LTLTT )(

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Deformations 24

Temperature Changes Thermal stresses occur when the bar is constrained such that it cannot deform freely.

In this case there is thermal stress but no thermal strain! Statically Determinate Structures

Uniform ΔT in the members produces thermal strains but no thermal stresses. Statically Indeterminate Structures

Uniform Δ T in the members produces thermal strains and/or thermal stresses.

Page 25: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

Deformations 25

3 Types of Displacement Problems Statically Determinate Statically Indeterminate Temperature Change

Page 26: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 26

More Mechanical Properties (2.11-2.15)

MAE 314 – Solid Mechanics

Yun Jing

Page 27: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 27

Poisson’s Ratio When an axial force is applied to a bar, the bar not only elongates but also shortens in the other two orthogonal directions. Poisson’s ratio (υ) is the ratio of lateral strain to axial strain.

υ is a material specific property and is dimensionless.axial strain

lateral strain

x

z

x

y

strainaxial

strainlateral

Minus sign needed to obtain a positive value

Page 28: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 28

Poisson’s Ratio Let’s generalize Hooke’s Law (σ=Eε).

Assumptions: linear elastic material, small deformations

So, for the case of a homogenous isotropic bar that is axially loaded along the x-axis (σy=0 and σz=0), we get

EEE

EEEEEE

zyxz

zyxy

zyxx

EEx

zyx

x

Even though the stress in the y and z axes are zero, the strain is not!

Page 29: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 29

Poisson’s Ratio What are the limits on υ? We know that υ > 0.

Consider a cube with side lengths = 1 Apply hydrostatic pressure to the cube Can write an expression for the changein volume of the cube

P

P P

PP

P

Pzyx

zyxzy

zxyxzyx

zyxV

11

1111

εx, εy, εz are very small, so we can neglect the terms of order ε2 or ε3

00

00

Page 30: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 30

Poisson’s Ratio ΔV simplifies to Plug σ=P/A=P into our generalized equations for strain.

Plug these values into the expression for ΔV.

zyxV

12

12

12

E

P

E

P

E

P

E

P

EEE

E

P

E

P

E

P

E

P

EEE

E

P

E

P

E

P

E

P

EEE

zyxz

zyxy

zyxx

123

E

PV

Page 31: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 31

Poisson’s Ratio Since the cube is compressed, we know ΔV must be less than zero. P

P P

PP

P

012

0123

E

P

2

10

Page 32: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 32

Shear Strain Recall that

Normal stresses produce a change in volume of the element Shear stresses produce a change in shape of the element

Shear strain (γ) is an anglemeasured in degrees or radians(dimensionless) Sign convention

Page 33: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 33

Shear Strain

Hooke’s Law for shear stress is defined as G = shear modulus (or modulus of rigidity) G is a material specific property with the same units as E (psi or Pa).

yzyzxzxzxyxy GGG

Page 34: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 34

Shear Strain Are the three material properties E, υ,and G related or do we have to determine each separately through testing? There is a relationship between them which is derived in section 2.15 in the textbook.

An isotropic material has two independent properties. Once you know two of them (E, G, or υ), you can find the third.

)1(2

EG

Page 35: Hooke's Law and Modulus of Elasticity1 Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing.

More Mechanical Properties 35

Example Problem A vibration isolation unit consists of two blocks of hard rubber bonded to a plate AB and to rigid supports as shown. Knowing that a force of magnitude P = 6 kips causes a deflection of δ=0.0625 in. of plate AB, determine the modulus of rigidity of the rubber used.