Transcript Normal Stress (1.1-1.5)

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
Introduction to Normal Strain


P
 st ress
A

L


2P P

2A A

L
Hooke's Law and Modulus of Elasticity
P
A
2 


2L L

2
Introduction to Normal Strain

Normal strain (ε) is defined as the deformation per unit length
of a member under axial loading.



L
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%
Hooke's Law and Modulus of Elasticity
3
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
Hooke's Law and Modulus of Elasticity
4
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
Hooke's Law and Modulus of Elasticity
5
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.
Hooke's Law and Modulus of Elasticity
6
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.
Hooke's Law and Modulus of Elasticity
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Stress-Strain Diagram (Low-carbon steel)

Linear region




Stress-strain response
is linear
Slope = Modulus of
Elasticity (Young’s
modulus) = E
E has units of force
per unit area (same as
stress)
We get a relation
between stress and
strain known as
Hooke’s Law.
  E
Hooke's Law and Modulus of Elasticity
8
Stress-Strain Diagram (Low-carbon steel)

Yielding region




Begins at yield stress σY
Slope rapidly decreases
until it is horizontal or
near horizontal
Large strain increase,
small stress increase
Strain is permanent
Hooke's Law and Modulus of Elasticity
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Stress-Strain Diagram (Low-carbon steel)

Strain Hardening


After undergoing large
deformations, the metal
has changed its
crystalline structure.
The material has
increased resistance
to applied stress
(it appears to be
“harder”).
Hooke's Law and Modulus of Elasticity
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Stress-Strain Diagram (Low-carbon steel)

Necking



The maximum supported
stress value is called the
ultimate stress, σu.
Loading beyond σu
results in decreased
load supported and
eventually rupture.
Breaking strength, σB
Hooke's Law and Modulus of Elasticity
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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:
true stress

The difference is in the
area: true stress takes
into account the
decreased crosssection area.

Thus, at the same stress
level, the load drops.
Hooke's Law and Modulus of Elasticity
x
12
Offset Method

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, through
the point (0.002, 0).

σY is given by the intersection of
this line with the stress-strain curve.
Hooke's Law and Modulus of Elasticity
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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.
Hooke's Law and Modulus of Elasticity
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Elastic vs. Plastic
Y
Y
Reload
Plastic deformation (Permanent strain)
Hooke's Law and Modulus of Elasticity
15
Deformations (2.8-2.10)
MAE 314 – Solid Mechanics
Yun Jing
Deformations
16
Deformations Under Axial Loading


Conditions
 Solid bars, cables, coil springs, etc.
 Axial tension or compression
 Prismatic
Recall Hooke’s Law (   E ) and

P
equations for stress (  ) and strain (  ).
A
L
P

PL
E   
A
L
AE
Deformations
17
Deformations Under Axial Loading

What about non-prismatic bars?

Discrete Changes: total change in length
is simply the summation of the change in
length of each portion.
Pi Li
 
i Ei Ai

Important: Each time the internal
force, area, or material changes
you need a new free-body diagram!
P1 L1 P2 L2 P3 L3



A1 E A2 E A3 E
Deformations
18
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 length
dx can be expressed as:
P( x)dx
d 
EA( x)

Integrating this over the length of the
rod:
L
P( x)
 
dx
EA( x)
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.
Deformations
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Example problem
The rigid bar BDE is supported by two links AB and CD. Link AB has a crosssectional 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.
Hooke's Law and Modulus of Elasticity
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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
Deformations
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Statically Indeterminate Problems
R1
R1
R = 1 : R1
R = 2 : R1, R2
N = 1 : ΣFY = R1 – P = 0
N = 1 : ΣFY = R1 – P – R2 = 0
M=1-1=0
M=2-1=1
Deformations
R2
Statically indeterminate structure
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Temperature Changes

Changes in temperature produce expansion or compression,
which cause strain.
 T  T




α = coefficient of thermal expansion
ΔT = change in temperature
Sign convention: expansion is positive (+), contraction is
negative (-)
For a bar that is completely free to deform (one or both ends
free):
T   T L   (T ) L

In this case, there is thermal strain but no thermal stress!
Deformations
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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.
Deformations
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3 Types of Displacement Problems

Statically Determinate

Statically Indeterminate

Temperature Change
Deformations
25
More Mechanical
Properties (2.11-2.15)
MAE 314 – Solid Mechanics
Yun Jing
More Mechanical Properties
26
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.
lateral strain

Poisson’s ratio (υ) is the ratio
of lateral strain to axial strain.
y
z
 
 
axial strain
x
x
lateral strain
Minus sign needed to obtain a positive value

axial strain
υ is a material specific property and is dimensionless.
More Mechanical Properties
27
Poisson’s Ratio

Let’s generalize Hooke’s Law (σ=Eε).

Assumptions: linear elastic material, small deformations
x 
 x  y  z
E

E

z  

y  
E
 x  y
E

E

 x
E

 y  z
E

E
z
E
So, for the case of a homogenous isotropic bar that is axially
loaded along the x-axis (σy=0 and σz=0), we get
x 
x
E
y  z  
 x
E
Even though the stress in the y and z axes
are zero, the strain is not!
More Mechanical Properties
28
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
P
P
 x   y   z  P

P
Can write an expression for the change
P
in volume of the cube
V  1   x 1   y 1   z   1
P
P
0
0
 1  1   x   y   z   x y   x z 
0
0
 y  z   x y  z
εx, εy, εz are very small, so we
can neglect the terms of order
ε2 or ε3
More Mechanical Properties
29
Poisson’s Ratio



ΔV simplifies to V   x   y   z
Plug σ=P/A=P into our generalized equations for strain.
 x  y  z
P P P P
x 


 

 2  1
E
E
E
E E E E
 x  y  z P P P P
y  



 
 2  1
E
E
E
E E E E
 x  y  z P P P P
z  




  2  1
E
E
E
E E E E
Plug these values into the expression for ΔV.
V 
3P
2  1
E
More Mechanical Properties
30
Poisson’s Ratio

Since the cube is compressed, we know ΔV must be less than zero.
P
3P
2  1  0
E
2  1  0
0  
1
2
More Mechanical Properties
P
P
P
P
P
31
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 angle
measured in degrees or radians
(dimensionless)

Sign convention
More Mechanical Properties
32
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).
 xy  G xy
 xz  G xz
 yz  G yz
More Mechanical Properties
33
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.
E
G
2(1  )

An isotropic material has two independent properties. Once you
know two of them (E, G, or υ), you can find the third.
More Mechanical Properties
34
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.
More Mechanical Properties
35