Transcript Here are the slides of Chapter 4

Chapter 4
Elasticity
by Ibrhim AlMohimeed
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Video
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Deformation
• When a object crash a car, the car may not move but it
will noticeably change shape.
• A change in the shape due to the application of a force is
a deformation.
• Even very small forces are known to cause some
deformation.
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Cont. Deformation
• For deformations, two important characteristics are
observed:
I.
The object may returns to its original shape when the
force is removed.
II. The size of the deformation is proportional to the force.
𝐹 ∝ ∆𝐿
where F is force, ∆L is the change in length
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Hooke’s law
• where ΔL is the amount of deformation (the change in
length) produced by the force F, and k is a
proportionality constant that depends on the shape and
composition of the object and the direction of the force.
𝐹 = 𝑘 ∆𝐿
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Cont. Hooke’s law
• The proportionality constant k depends upon a number
of factors for the material.
• For example, a guitar string made of nylon stretches
when it is tightened, and the elongation ΔL is
proportional to the force applied. Thicker nylon
strings and ones made of different material (steel)
stretch less for the same applied force, implying they
have a larger k.
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Hooke’s law Examples
• Example 4.1: A force of 600 N will compress a spring
0.5 meters. What is the spring constant of the spring?
• Example 4.2: A spring has spring constant 0.1 N/m.
What force is necessary to stretch the spring by 2
meters?
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Elastic Modulus
• ΔL depends on the material of
the subject.
• ΔL is proportional to the force F.
• ΔL is proportional to the original
length L0.
• ΔL is inversely proportional to
the cross-sectional area of the
subject
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Cont. Elastic Modulus
• All these factors can be combined
into one equation for ΔL :
1 𝐹
∆𝐿 =
𝐿0
𝑌 𝐴
ΔL the change in length.
F the applied force.
Y is Young’s modulus, or elastic modulus
(that depends on the subject material).
A is the cross-sectional area,
L0 is the original length
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Stress and Strain
• The ratio of force to area,
𝐹
,
𝐴
is defined as stress. So the
stress is the force per unit area (pressure!!!)
𝜎=
𝐹
𝐴
𝑁/𝑚 2 or Pa
• The ratio of the change in length to length,
∆𝐿
,
𝐿0
is
defined as strain (a unitless quantity)
∆𝐿
𝜀=
𝐿0
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Cont. Stress and Strain
• So, the equation of the elastic modulus can be
rearranged in term of stress and strain:
𝝈 𝒔𝒕𝒓𝒆𝒔𝒔 = 𝒀 𝒀𝒐𝒖𝒏𝒈′ 𝒔 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝜺 (𝒔𝒕𝒓𝒂𝒊𝒏)
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Stress and Strain Example
• Example 4.3: A steel wire 10 m long
and 2 mm in diameter is attached to
L
the ceiling and a 200-N weight is
attached to the end. What is the
DL
applied stress?
• If the wire stretches 3.08 mm. What
A = π × r2
is the longitudinal strain?
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Stress and Strain Curve
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Cont. Stress and Strain Curve
• Flash
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Cont. Stress and Strain Curve
• Elastic behavior: the object will returns to its original
length (or shape) when the stress acting on it is
removed.
• Plastic behavior: the object will NOT returns to its
original length (or shape) when the stress acting on it is
removed.
• Proportionality limit: the stress above is not longer
proportional to strain.
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Cont. Stress and Strain Curve
• Elastic behavior: the object will returns to its original
length (or shape) when the stress acting on it is
removed.
•
Elastic Limit: The maximum stress that can be
applied without resulting permanent deformation.
•
Yield Stress: at which there are large increases in
strain with little or no increase in stress. steel exhibits
this type of response.
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Cont. Stress and Strain Curve
• Strain Hardening: when yielding has ended, a further
load can be applied to the object.
• Ultimate Strength: the maximum stress the object can
withstand.
• Necking: after the ultimate stress, the cross-sectional
area begins to decrease in a localized region of the
object.
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Stress & Strain Curve Example
Example 4.4: The elastic limit for steel of 3.14 x 10-6 m2
is 2.48 x 108 Pa and The ultimate strength is 4.89 x 108 Pa.
a)What is the maximum weight that can be supported
without exceeding the elastic limit?
b) What is the maximum weight that can be supported
without breaking the wire?
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Stress & Strain Curve Example
Example 4.5:
a stress applied to an object was 6.37 x 107 Pa and its
strain was 3.08 x 10-4. Find the modulus of elasticity for
object?
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Stress & Strain Curve Example
Example 4.6:
Young’s modulus for brass is 8.96 x
1011Pa.
A 120-N weight is attached to
an 8-m length of brass wire; find the
increase in length. The diameter is 1.5
8m
DL
120 N
mm.
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Strain Hardening
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Cont. Strain Hardening
• Strain Hardening
- If the material is loaded again from the
beginning stage of strain hardening, the curve will
be the same Elastic Modulus (slope).
- The material will has a higher yield strength.
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Types of Material
Isotropic materials:
have elastic properties that are independent of direction.
Anisotropic materials:
whose properties depend upon direction.
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Material Behavior
Behavior of materials can be broadly classified into two
categories:
 Brittle
(Example: glass, ceramics)
 Ductile
(Example: Metals; Gold, silver, copper, iron )
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Brittle Behavior
•
Brittle
 \fracture at much lower strains.
 yielding
region
is
nearly
nonexistent.
 often have relatively large Young's
moduli and ultimate stresses.
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Ductile Behavior
•
Ductile
 \withstand large strains before
failure.
 yielding region often takes up the
majority of the stress-strain curve
 capable of absorbing much larger
quantities of energy before failure.
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Resilience & Toughness
•
Resilience:
\Amount of energy stored in material up to elastic
limit up to elastic limit per unit volume.
• Toughness:
Amount of energy stored in material up to fracture per
unit volume
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Cont. Resilience & Toughness
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Shear modulus
The shear modulus (S) is
the elastic modulus we use
for the deformation which
takes place when a force is
applied parallel to one face
of the object while the
opposite face is held fixed
by another equal force.
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Cont. Shear modulus
• Shear deformation behaves similarly to tension and
compression and can be described with similar
equations.
𝜏 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑆 𝑠ℎ𝑒𝑎𝑟 𝑚𝑜𝑑𝑢𝑙𝑒 × 𝜙(𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑖𝑛)
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Bulk Modulus
• An object will be compressed
in all directions if inward
forces are applied evenly on all
its surfaces.
• It is relatively easy to
compress gases and extremely
difficult to compress liquids
and solids.
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Bulk Modulus
𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝐹/𝐴
𝐵=
=−
𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑎𝑖𝑛
Δ𝑉/𝑉
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Bulk Modulus Example
Example 4.6:
A hydrostatic press contains 5 liters of oil. Find the
decrease in volume of the oil if it is subjected to a
pressure of 3000 kPa. (Assume that B = 1700 MPa.)
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End of the Chapter
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