2 axial loading - nazarena.mazzaro.dk

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MECHANICS OF MATERIALS Fourth Edition Ferdinand P. Beer, E. Russell Johnston, John T. DeWolf CHAPTER Stress and Strain – Axial Loading Lecturer: Nazarena Mazzaro, Ph.D Aalborg University Denmark

Transcript of 2 axial loading - nazarena.mazzaro.dk

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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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MECHANICS OF MATERIALS

Nazarena Mazzaro, AAU

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

• 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