Transcript CM 197 Mechanics of Materials Chap 10: Strength of Materials Strains

CM 197
Mechanics of Materials
Chap 10: Strength of Materials
Strains
Professor Joe Greene
CSU, CHICO
Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill,
Westerville, OH (1997)
CM 197
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Chap 10: Strains
• Objectives
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Introduction
Linear Strain
Hooke’s Law
Axial Deformation
Statically Indeterminate Problems
Thermal Stresses
Poisson’s ratio
Shear Strain
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Mechanical Test Considerations
• Normal and Shear Stresses
P
– Force per unit area
• Normal force per unit area
P
– Forces are normal (in same direction) to the surface
• Shear force per unit area
– Forces are perpendicular (right angle) to the surface
A
P P
P
 
A
P
P
• Direct Normal Forces and Primary types of loading
– Prismatic Bar: bar of uniform cross section subject to equal and opposite pulling
forces P acting along the axis of the rod.
– Axial loads: Forces pulling on the bar
– Tension= pulling the bar; Compression= pushing; torsion=twisting; flexure=
bending; shear= sliding forces
Normal Forces
tension
compression
Shear Forces
shear
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Strain
• Strain: Physical change in the dimensions of a specimen that results from
applying a load to the test specimen.
• Strain calculated by the ratio of the change in length, , and the original
length, L. (Deformation)

 ( Lf  L )

L

L
L
• Where,
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 = linear strain ( is Greek for epsilon)
 = total axial deformation (elongation of contraction) = Lfinal –Linitial = Lf - L
L = Original length
Strain units (Dimensionless)
• Units
– When units are given they usually are in/in or mm/mm. (Change in dimension
divided by original length)
• % Elongation = strain x 100%
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Strain
• Example
– Tensile Bar is 10in x 1in x 0.1in is mounted vertically in
test machine. The bar supports 100 lbs. What is the
strain that is developed if the bar grows to 10.2in? What
is % Elongation?
•  =Strain = (Lf - L0)/L0 = (10.2 -10)/(10) = 0.02 in/in
0.1 in
1 in
10in
100 lbs
• Percent Elongation = 0.02 * 100 = 2%
• What is the strain if the bar grows to 10.5 inches?
• What is the percent elongation?
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Stress-Strain Diagrams
Forces
• Equipment
– Tensile Testing machine
• UTM- Universal testing machine
• Measures
Test Sample
– Load, pounds force or N
– Deflection, inches or mm
• Data is recorded at several readings
– Results are averaged
– e.g., 10 samples per second during the test.
• Calculates
– Stress, Normal stress or shear stress
– Strain, Linear strain
– Modulus, ratio of stress/strain
Fixed
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Stress-Strain Diagrams
• Stress-strain diagrams is a plot of stress with the
corresponding strain produced.
• Stress is the y-axis
• Strain is the x-axis
Stress

Linear
(Hookean)
Non-Linear
(non-Hookean)
Strain

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Hooke’s Law
• Hooke’s Law relates stress to strain by way of modulus
– Hooke’s law says that strain can be calculated as long as the stress is lower than
the maximum allowable stress or lower than the proportional limit.
• If the stress is higher than the proportional limit or max allowable stress than the
part will fail and you can’t use Hooke’s law to calculate strain.
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Stress = modulus of elasticity, E, times strain
Stress=  = load per area, P/A
Strain=  = deformation per length,  /L
Rearrange Hooke’s law
Solving for deformation is Equation 10-5
  E
P

( Lf  L )
E E
A
L
L
• With these equations you can find
PL
– How much a rod can stretch without breaking.

AE
Eqn 10-3
Eqn 10-4
Eqn 10-5
– What the area is needed to handle load without breaking
– What diameter is needed to handle load without breaking
• Example 10-1
• Example 10-3
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Statically Indeterminate Problems
• Sometimes you might have forces in structural members
that can’t be found with stress-strain equations
– Statically indeterminate
• Those involving axially loaded members can be analyzed by
– Introducing conditions of axial deformations
– Example 10-5
– Example 10-6
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Thermal Stresses
• Most materials expand when heated as the temperature increases.
– As the temperature goes up, the material expands and results in forces that cause
stress in the part. As temperature increases the stresses increase in part.
• Examples,
– Cast iron engine block heat up to 500F and expands the cast iron block which
causes stresses at the bolts. The bolts must be large enough to withstand the
stress.
– Aluminum heats up and expands and then cools off and contracts.
» Sometimes the stresses causes cracks in the aluminum block.
– Space shuttle blasts off and heats up, goes into space and cools down (-200F),
and reenters Earths atmosphere and heats up (3000F)
» Aluminum melts at 1300F so need ceramic heat shields
» Aluminum structure expands and cools.
– The amount the material expands is as follows:
• Change in length that is causes by temperature change (hot or cold)
  LT
– Where,
Eqn 10-6
»  = change in length
»  = the CLTE (coefficient of linear thermal expansion
» T = change in temperature (Thot – Tcold)
» L = length of member
• Example 10-7 and 10-8
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Strain and Poisson’s Ratio
• Axial strain is the strain that occurs in the same direction
as the applied stress.
Transverse
• Transverse strain is the strain that occurs perpendicular
Strain
to the direction of the applied stress.
Axial
• Poisson’s ratio is ratio of lateral strain to axial strain.
Strain
Poisson’s ratio = lateral strain   transverse _ strain   t
axial _ strain
a
P, Load
axial strain
– Example
• Calculate the Poisson’s ratio of a material with lateral
strain of 0.002 and an axial strain of 0.006
• Poisson’s ratio = 0.002/0.006 = 0.333
– Example 10-10
Note: For most materials, Poisson’s ratio is between 0.25 and 0.5
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Shear Strain
• Shear strain - occurs when a shear stress is applied
– Shear stress is sliding force or tangential
– Shear strain is deformation in tangential or side direction
Shear strain = lateral deformation
length of side
• Hooks Law for shear strain
– Remember: Normal stress = normal modulus * normal strain
– Shear stress = shear modulus * shear strain
Eqn 10-10
=G
– Shear Modulus is found from the Normal Modulus, E, by
dividing it by Poisson’s ratio,  .
• G = __E_
Eqn 10-11
2(1+ )
• For example, if the Normal modulus (Modulus of Elasticity) is 30
million psi (steel) and the shear modulus is 11.6 million psi, then the
Poisson’s ratio is found from Equation 10-11 after you rearrange it.
• Poisson’s ratio,   E / 2G  1  30Mpsi /( 2 *11.6Mpsi )  0.29
• Example 10-11
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