Fall 2002 Axially Loaded Bars 1

Axial Deformation

Chapter 5Lecture 2

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Fall 2002 Axially Loaded Bars 3

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Strain Distribution

Axial deformation is characterized by extensional strain that is not a function of

position in the cross section.

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Strain-Displacement Relationship

Length Initial

Length Initial Length Final -ε

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Strain-Displacement Relationship

x*

x x 0

0

x

x

x xlim

x

u x x u xlim

du

dx

x

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Strain-Displacement Equation

dx

duε

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Strain-Displacement Equation

L

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Material BehaviorAssume linear elastic behavior:

x xE

EE is a material property called the modulus of elasticity or Young’s modulus.

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Total Elongation

u L u

x dxL

0

0

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Summary

Axial Stress Formula

Axial Strain Formula

Axial Force-Deformation

Equation

x

x

L

0

x

0

P xx

A x

P xdu

dx EA x

P xe dx

EA x

Pu u 0 d

EA

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Uniform Axial Deformation

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Uniform Axial Deformation

constant

L L

0 0

F x 0 P x P

P x P PLdx dx

EA x EA EA

Force-Deformation Behavior.

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Uniform Axial Deformation

PL

EA

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Flexibility

f flexibility coefficie t

f

n

P

L

LP

EA

EA

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Stiffness

k stiffness coefficien

k

t

PLP

EA

EA

L

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Definitions1. The flexibility coefficient, f, is the

elongation produced when a unit force is applied to the member. It has dimensions of Length/Force (L/F )

2. The stiffness coefficient, k, is the force required to produce a unit elongation of the member. It has dimensions of Force/Length (F/L)

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Lateral Strain

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Lateral Strain

lat

dL dL

dL

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Lateral Strain

lat

lat lat

L 1 L

L L L

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Lateral Strain

lat

lat

0 0

0 0

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latD 1 D

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Poisson’s Ratio

Greek letter nu is

the Poisson's Ratio

lat

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Poisson’s Ratio

Poisson's Ratio is a material property

just as the modulus of elasticity is.

0 0.5

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Lateral Strain

lat

P

EA

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Example

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A prismatic bar with length L = 200 mm and a circular cross section with a diameter D = 10 mm is subjected to a tensile load P = 16 kN. The length and diameter of the deformed bar are measured and determined to be L’ = 200.60 mm and D’ = 9.99 mm. What are the modulus of elasticity and the Poisson’s ratio for the bar?

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Example

Stress:

Strain:

Lateral Strain

2

lat

P 16000N203.7 MPa

A .01m

4

L L 200.60mm 200mm0.003

L 200mm

D D 9.99mm 10mm0.001

D 10mm

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Example

lat

lat

203.7 MPa

0.003

0.001

203.7 MPaE 67.9 GPa

0.003

0.0010.333

0.003

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Example

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Example

The bar has cross sectional area of A = 0.4 in2

and a modulus of elasticity of E = 12 x 106 psi. If a 10 kip force is applied downward at B how far downward does point B move?

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Example

10 kip

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yF 10kip Psin 0

P 12.5kip

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6 22

P 12.5kip

L 20 in

12500lb 20inPL0.0521in

lbEA12 10 0.4in

in

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20 in

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small deformation approximation

2 2 2

2 2

2 2

12 16 v 20

32v v 40

v v

32v 40

v 0.0651 in

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Example

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Bars AB and AC each have a cross sectional area A = 60 mm2 and a modulus of elasticity E = 200 GPa The dimensions h = 200 mm If a downward force F = 40 kN is applied at A what is the resulting horizontal and vertical displacements of point A?

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40 kN

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o ox AB AC

o oy AB AC

AB

AC

F 0 P cos60 P cos45 0

F 0 P sin60 P sin45 40kN 0

P 29.28kN

P 20.70kN

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AB

AC

oAB AB

AB9 5 2

2

oAC AC

AC9 5 2

2

P 29.28kN

P 20.70kN

0.2m29280N

P L sin60.000563m

NEA200 10 6 10 m

m

0.2m20700N

P L sin45.000487m

NEA200 10 6 10 m

m

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2 2 2o oAB AB AB AB

o 2 o 2 2AB AB AB AB AB

o oAB

L sin60 v L cos60 u L

2vL sin60 v 2L ucos60 u 2 L

vsin60 ucos60

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2 2 2o oAC AC AC AC

o 2 o 2 2AC AC AC AC

o oAC

L sin45 v L cos45 u L

2vL sin45 v 2L ucos45 u 2 L

vsin45 ucos45

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mm

mm

mm

mm

AB

AC

o oAB

o oAC

0.563

0.487

vsin60 ucos60

vsin45 ucos45

u 0.0250

v 0.6652