2 Diapositivas Cap Deformaci On

8/8/2019 2 Diapositivas Cap Deformaci On

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CHAPTER

Ferdinand P. Beer

E. Russell Johnston, Jr. Stress and Strain.

Lecture Notes: Axial Loading

J. Walt Oler

Texas Tech University


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MECHANIC OF MATERIAL Beer Johnston DeWolfContents

Stress & Strain: Axial Loading

Stress-Strain Test

Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials

Hookes Law: Modulus of Elasticity

Fatigue

Deformations Under Axial Loading

xamp e .

Sample Problem 2.1

Static Indeterminac

Example 2.04

Thermal Stresses

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MECHANIC OF MATERIAL Beer Johnston DeWolfStress & Strain: Axial Loading

Suitability of a structure or machine may depend on the deformations in

the structure as well as the stresses induced under loading. Statics

analyses alone are not sufficient.

Considering structures as deformable allows determination of memberforces and reactions which are statically indeterminate.

Determination of the stress distribution within a member also requires

cons era on o e orma ons n e mem er.

Cha ter 2 is concerned with deformation of a structural member underaxial loading. Later chapters will deal with torsional and pure bendingloads.

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MECHANIC OF MATERIAL Beer Johnston DeWolfNormal Strain

stress==A

P

AA ==

2

2

A =

2 - 4

strainnormal==

L

L =

LL ==

2


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MECHANIC OF MATERIAL Beer Johnston DeWolfStress-Strain Test

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MECHANIC OF MATERIAL Beer Johnston DeWolfStress-Strain Diagram: Ductile Materials

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MECHANIC OF MATERIAL Beer Johnston DeWolfStress-Strain Diagram: Brittle Materials

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MECHANIC OF MATERIAL Beer Johnston DeWolfHookes Law: Modulus of Elasticity

Below the yield stress

= E

ElasticityofModulus

oro u usoungs=

Strength is affected by alloying,

heat treating, and manufacturing

process but stiffness (Modulus of

Elasticity) is not.

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MECHANIC OF MATERIAL Beer Johnston DeWolfElastic vs. Plastic Behavior

stress is removed, the material is

said to behave elastically.

The largest stress for which this

When the strain does not return

.

to zero after the stress is

removed, the material is said to

.

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MECHANIC OF MATERIAL Beer Johnston DeWolfFatigue

Fati ue ro erties are shown on

S-N diagrams.

A member may fail due tofatigue

at stress levels significantly below

the ultimate stren th if sub ected

to many loading cycles.

When the stress is reduced below

the endurance limit, fatigue

failures do not occur for anynumber of cycles.

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MECHANIC OF MATERIAL Beer Johnston DeWolfDeformations Under Axial Loading

From Hookes Law:

EEE ===

L

=

Equating and solving for the deformation,

PL=

E

With variations in loading, cross-section ormater a propert es,

= iiE

LP

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MECHANIC OF MATERIAL Beer Johnston DeWolfExample 2.01

Divide the rod into components at

the load application points.

psi1029 6= E Apply a free-body analysis on each

component to determine the

in.618.0in.07.1 == dD nterna orce

Evaluate the total of the component

the steel rod shown under the

given loads.

e ec ons.

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MECHANIC OF MATERIAL Beer Johnston DeWolfSOLUTION: Apply free-body analysis to each

Divide the rod into three

components:

component to determine internal forces,

lb1060 31 =P

lb1030

lb1015

33

32

=

=

P

P

Evaluate total deflection,

13

33

2

22

1

11

++==

ALP

ALP

ALP

EEALP

i ii

ii

3.0

161030

9.0

121015

9.0

121060

1029

16

+

+

=

21 in.12== 3 in.16=L

in.109.75=

in.109.75 3=

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21 in9.0== AA2

3 in3.0=A


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MECHANIC OF MATERIAL Beer Johnston DeWolfSample Problem 2.1

SOLUTION:

Apply a free-body analysis to the bar

BDEto find the forces exerted by

linksAB andDC.

The rigid barBDEis supported by two

Evaluate the deformation of linksAB

andDCor the displacements ofB

n s an .

LinkAB is made of aluminum (E= 70

-

.

Work out the geometry to find the

deflection at E iven the deflections

mm2. LinkCD is made of steel (E= 200

GPa) and has a cross-sectional area of (600

at B and D.

.

For the 30-kN force shown, determine the

deflection a of b of D and c of E.

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Displacement ofB:

=PL

Free bod : Bar BDE

SOLUTION:

( )( )m3.0N1060

926-

3=

E

m10514 6=

.B

Displacement ofD:

=PL

( ) FM

CD

B

m2.0m6.0kN3000

+=

=

( )( )m4.0N1090926-

3

=

AEtensionFCD

0M

kN90

D =

+=

m10300

am

6=ncompressioFB

B

kN60

..

=

=

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= mm300.0D


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Displacement of D:

mm200mm514.0

=

x

HD

BH

DD

BB

mm7.73

mm0.300

=

=

x

x

=

HD

HE

DD

EE

( )mm7.73

mm7.73400

mm300.0

=

+=E

= mm928.1

.

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MECHANIC OF MATERIAL Beer Johnston DeWolfStatic Indeterminacy

Structures for which internal forces and reactions

cannot be determined from statics alone are said

to bestatically indeterminate.

A structure will be statically indeterminatewhenever it is held by more supports than are

required to maintain its equilibrium.

Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

Deformations due to actual loads and redundant

reactions are determined separately and then added

0=+= RL

orsuperposed.

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Determine the reactions atA andB for the steelbar and loading shown, assuming a close fit at

.

SOLUTION: Consider the reaction atB as redundant, release

the bar from that support, and solve for the

dis lacement atB due to the a lied loads.

Solve for the displacement atB due to the

redundant reaction at B.

Require that the displacements due to the loads

,

i.e., require that their sum be zero.

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and the reaction found atB.


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SOLUTION: Solve for the displacement atB due to the applied

,

PPPP

2626

34

3321

m10250m10400

N10900N106000

====

====

P

LLLL

9

4321

10125.1

m150.0

====

EEAi iiL ==

Solve for the displacement atB due to the redundant

constraint,

==

BRPP 21

==

==

LL 21

21

m300.0

mm

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==i

B

ii

iiR

EEA

.

C C O


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Require that the displacements due to the loads and due to

the redundant reaction be compatible,

( )1095.110125.1

0

39

=+= RL

R

kN577N10577 3 ==BR

EE

Find the reaction atA due to the loads and the reaction atB

kN323

kN577kN600kN3000

=

+== Ay

R

RF

kN577

kN323

=

=A

R

R

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MECHANIC OF MATERIAL


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MECHANIC OF MATERIAL Beer Johnston DeWolfThermal Stresses

A temperature change results in a change in length or

thermal strain. There is no stress associated with the

thermal strain unless the elongation is restrained by

the supports.

PL

Treat t e a itional support as re un ant an apply

the principle of superposition.

coef.expansionthermal=

==

AE

PT

0=+=

e erma e orma on an e e orma on rom

the redundant support must be compatible.

=+= 0

( ) 0=+E

PLLT

( )

( )TEP

TAEP

==

=

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MECHANIC OF MATERIAL


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MECHANIC OF MATERIAL Beer Johnston DeWolfPoissons Ratio

For a slender bar subjected to axial loading:

0=== zyxE

T e elongation in t e x- irection is

accompanied by a contraction in the other

directions. Assuming that the material is

isotropic (no directional dependence),0= zy

Poissons ratio is defined as

strainlateral

xx ===

strainaxial

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