MECHANICS OF MATERIALSFourth Edition
Ferdinand P. Beer, E. Russell Johnston, John T. DeWolf
CHAPTER
Stress and Strain – Axial Loading
Lecturer: Nazarena Mazzaro, Ph.DAalborg University
Denmark
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Review of last class
• Concept of stress [Pa] σ =P/A (average)• Static method -> free body diagram
(reaction and internal forces)• Centric and eccentric axial loading
• Shearing stress [Pa] τ=P/A (average)(opposite transverse forces)
• Stresses in an oblique section:
σmax at θ= 0°; τmax at θ=45°
P=1200 N
P=1200 N
A= area
P
P
θθτθσ cossin;cos2
AP
AP
==
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Contents
Stress & Strain: Axial LoadingNormal StrainStress-Strain TestStress-Strain Diagram: Ductile MaterialsStress-Strain Diagram: Brittle Materials Hooke’s Law: Modulus of ElasticityElastic vs. Plastic BehaviorFatigueDeformations Under Axial LoadingExample 2.01
Static IndeterminacyExample 2.04Thermal StressesPoisson’s RatioGeneralized Hooke’s LawShearing StrainExample 2.10Saint-Venant’s PrincipleStress Concentration: Hole, FilletExample 2.12
First Part: 45 min Second Part: 45 min
Third Part: Exercises - 2 hr
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Normal strain
• P-δ diagrams contain information about the particular road of length L and area A under analysis.
• Results from this P-δ diagram cannot be extrapolated to other samples of the same material
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Normal strain
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Normal Strain
L
AP
AP
δε
σ
=
==22
LL
AP
δδε
σ
==
=
22
Strain [unit-less]normal
Stress [N/m2 = Pa]
==
==
L
AP
δε
σ
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Stress-Strain Test
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Stress-Strain Diagram: Ductile Materials
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Determination of σy by offset method
• From ε= 0.2 % a line parallel to the linear portion of the stress-strain curve is drawn. In the point where this line crosses the stress-strain curve, we define the σy.
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Stress-Strain Diagram: Brittle Materials
• Brittle materials: iron, glass, stone, etc.• Brake without previous change in elongation rate: σu = σB, no necking• Rupture occurs along the surface perpendicular to the load due to σ
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True stress and true strain
• Engineering stresses were calculated considering the initial area. In reality the area decreases as P increases.
• True stress: σt=P/A, where A is the real area measured every time P is increased (and consequently A decreases).
• Main difference: σt continues increasing in the necking.
• Engineers job is to determine if σ and ε in a certain configuration are acceptable. They use available data, i.e. engineer σ-εdiagrams.
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Hooke’s Law: Modulus of Elasticity
• Below the yield stress
• Strength is affected by alloying, heat treating, and manufacturing process but the stiffness or ability to resist deformation (Modulus of Elasticity) is not.
• E [Pa] Young’s Modulus or Modulus of Elasticity
σ = E ε
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Hooke’s Law: Modulus of Elasticity
• When E is independent of the direction of loading the material is called isotropic.
• If E depends on the direction of loading the material is called anisotropic. Ex: fiber-reinforced composite materials.
Matrix: soft, weaker material (ex: polymers).
Fibers: strong, stiff material (ex: graphite, glass)
Ex ≠ Ey ≠ Ez; Ex >> Ey and Ez
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Elastic vs. Plastic Behavior
• If the strain disappears when the stress is removed, the material is said to behave elastically.
• When the strain does not return to zero after the stress is removed, the material is said to behave plastically.
• Slip -> stress value reached
• Creep -> time of load
• The largest stress for which this occurs is called the elastic limit.
Elastic limit = σy
Remaining deformation
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Fatigue
• Fatigue properties are shown on S-N diagrams.
• Endurance limit (1) is a level of σfor which fatigue failures do not occur for any number of cycles.
• Fatigue limit (2) is a level of σ that will cause failure after a certain nr of loads, ex. 500 millions.
• Fail due to fatigue may occur at stress levels lower than the σu if subjected to many loading cycles.
(1) Endurance limit
(2) Fatigue limit
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Deformations Under Axial Loading
AEP
EE ===
σεεσ
• From Hooke’s Law:
• From the definition of strain:
Lδε =
• Equating and solving for the deformation,
AEPL
=δ
• With variations in loading, cross-section or material properties,
∑=i ii
iiEALPδ
Pi: internal force, Li: length, Ai: cross sectional area, Ei: Young’s Modulus of each part i
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Example 2.01
Determine the deformation of the steel rod shown under the given loads.
.1942,5811200
22 mmAmmAGPaE
==
=
SOLUTION:• Divide the rod into components at
the load application points.
• Apply a free-body analysis on each component to determine the internal force, considering each part in equilibrium.
• Evaluate the total of the component deflections.
300 kN 180 kN120 kN
300 mm 300 mm400 mm
A1 A2
∑=i ii
iiEALPδ
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SOLUTION:
• Divide the rod into three components:
221
21
580
300
mmAA
mmLL
==
==2
3
3
194
400
mmA
mmL
=
=
• Apply free-body analysis to each component to determine internal forces, considering each part in equilibrium
kNPkNP
kNP
12060120180
240180120300
3
2
1
=−=+−==−+=
• Evaluate total deflection,
( ) ( ) ( )
mm
ALP
ALP
ALP
EEALP
i ii
ii
73.110190
4.01012010580
3.0106010580
3.01024010200
1
1
6
3
6
3
6
3
9
3
33
2
22
1
11
=
⎥⎦
⎤⎢⎣
⎡×
×+
××−
+×
××
=
⎟⎟⎠
⎞⎜⎜⎝
⎛++==∑δ
73.1 mm=δ
300 kN 180 kN120 kN
120 kN
120 kN
120 kN
180 kN
180 kN300 kN