CHAPTER 4 - Mechanics of metarial UMP.pdf

MOM

1. MECHANICS OF
MATERIALS
CHAPTER 4: AXIAL LOAD

2. LEARNING OUTCOME
•Definetheelasticdeformationofanaxiallyloadedprismatic
bar
•Definethemultipleprismaticbars
•Definetheprincipleofsuperposition
•DefinetheSaint–Venant’sprinciple
•Calculatethedeformationsofmemberunderaxialload
•Calculatethedeformationsofmemberforsteppedcomposite
bar
•Analysethedeformationsinsystemsofaxiallyloadedbars
•Analysethedeformationsofmemberforstatically
indeterminateassemblies.
•Calculatethedeformationofmemberduetotemperature
effect.

3. INTRODUCTION
•InChapter1,theconceptofstresswasdevelopedas
ameanofmeasuringtheforcedistributionwithina
body
•InChapter2,theconceptofstrainwasintroducedto
describethedeformationproducedinabody
•InChapter3,discussedthebehavioroftypical
engineeringmaterialsandhowthisbehaviorcanbe
idealizedbyequationthatrelatestressandstrain
•Inthischapter,discussedtwogeneralapproaches
usedtoinvestigatedawidevarietyofstructural
membersubjectedtoaxialloadinganddeformation

4. SAINT-VENANT’S PRINCIPLE
•Saint-Venant’sprinciplestatesthatbothlocalized
deformationandstresstendto“evenout”ata
distancesufficientlyremovedfromtheseregions.

5. ELASTIC DEFORMATION OF AN AXIALLY
LOADED MEMBER
•UsingHooke’slawandthedefinitionsofstressand
strain,weareabletodeveloptheelastic
deformationofamembersubjectedtoaxialloads.
•Supposeanelementsubjectedtoloads,()
() dx
dδ
ε
xA
xP
== and  ()
()

=
L
ExA
dxxP
0

= small displacement
L = original length
P(x) = internal axial force
A(x) = cross-sectional area
E= modulus of elasticity

6. Constant Load and Cross-Sectional Area
•When a constant external force is applied at each
end of the member,
Sign Convention
•Force and displacement is positive when tension and
elongation and negative will be compression and
contraction.AE
PL
=
Displacement

7. Example 1

8. Try this
Take diameter as 25 mm and modulus of elasticity is 200 GPa.

9. Example 2
TheassemblyconsistsofanaluminumtubeABhavingacross-
sectionalareaof400mm
2
.Asteelrodhavingadiameterof10mmis
attachedtoarigidcollarandpassesthroughthetube.Ifatensile
loadof80kNisappliedtotherod,determinethedisplacementof
theendCoftherod.TakeE
st=200GPa,E
al=70GPa

10. Try this
TheassemblyconsistsofsteelroadCBandanaluminiumrodBA,each
havingadiameterof12mm.IftherodissubjectedtotheaxialloadingsatA
andatthecouplingB,determinethedisplacementofthecouplingBandthe
endA.theunstretchedlengthofeachsegmentisasshown.Neglectthesize
oftheconnectionsatBandC,andassumethattheyarerigid.Est=200GPa,
Eal=70GPa

11. Try This
2 2 2
Thecompositeshaft,consistingofaluminium,copper,andsteelsections,is
subjectedtotheloadingshown.DeterminethedisplacementofendAwith
respecttoendDandthenormalstressineachsection.Thecross-sectional
areaandmodulusofelasticityforeachsectionareshowninthefigure.
NeglectthesizeofthecollarsatBandC.

12. •Principleofsuperpositionistosimplifystressand
displacementproblemsbysubdividingtheloading
intocomponentsandaddingtheresults
•Twoconditions:
PRINCIPLE OF SUPERPOSITION
The loading must be linearlyrelated to the stress
or displacement that is to be determine1
The loading must not significantly change the
original geometry or configuration of the member2

13. STATICALLY INDETERMINATE AXIALLY
LOADED MEMBER
•Amemberisstaticallyindeterminatewhen
equationsofequilibriumarenotsufficientto
determinethereactionsonamember.
•Example:Thebarfixedatbothends
•Inordertoestablish,specifiestheconditionto
compatibilityorkinematiccondition
P
F
A F
B

14. Example 3
The steel rod has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is
loaded, there is a gap between the wall at and B’ and the rod of 1 mm. Find the reactionsat
A and B’ if the rod is subjected to an axial force of P = 20 kN.Neglect the size of the collar at
C. (E
st= 200 GPa)

15. Example 4
The bolt is made of 2014-T6 aluminum alloy and is tightened so it compresses a
cylindrical tube made of Am 1004-T61 magnesium alloy. The tube has an outer
radius of 10 mm, and both the inner radius of the tube and the radius of the bolt are
5 mm. The washers at the top and bottom of the tube are considered to berigid and
have a negligible thickness. Initially the nut is hand-tightened slightly; then, using a
wrench, the nut is further tightened one-half turn. If the bolt has 20 threads per inch,
determine the stress in the bolt.

16. Solution:(1) 0 ;0 =−=+ tby FFF ()
bt
 −=+ 5.0
Equilibrium requires
When the nut is tightened on the bolt, the tube will shorten.

17. Taking the 2 modulus of elasticity,
Solution:()
 () 
()
() 
() (2) 911251255
10755
60
5.0
1045510
60
32322
bt
bt
FF
FF
−=
−=
−

 kN 56.3131556===
tb
FF
The stresses in the bolt and tube are therefore
Solving Eqs. 1 and 2 simultaneously, we get()
( )
(Ans) MPa 9.133N/mm 9.133
510
31556
(Ans) MPa 8.401N/mm 8.401
5
31556
2
22
2
==
−
==
====




t
t
s
b
b
b
A
F
A
F

18. Example 5

19. •Change in temperature cause a material to change its
dimensions
•Since the material is homogeneous and isotropic
THERMAL STRESS
= linear coefficient of thermal expansion, property of the material
T = algebraic change in temperature of the member
T = original length of the member
= algebraic change in length of the memberTL
T
−=

20. Example 4.12
TherigidbarisfixedtothetopofthethreepostsmadeofA-36steeland2014-T6
aluminum.Thepostseachhavealengthof250mmwhennoloadisappliedtothe
bar,andthetemperatureisT
1=20°C.Determinetheforcesupportedbyeach
postifthebarissubjectedtoauniformdistributedloadof150kN/mandthe
temperatureisraisedtoT
2=80°C.

21. Solution:() (2)
alst
=+ () (1) 010902 ;0
3
=−+=+ alsty FFF
From free-body diagram we have
The top of each post is displaced by an equal amount and hence,

22. The final position of the top of each post is equal to its displacement caused by the
temperature increase and internal axial compressive force.
Solution:()()()()
FalTstFstTst
 +−=+− () ()()
() ()()
FalTalal
FstTstst


+−=+
+−=+


Applying Eq. 2 gives
With reference from the material properties, we have() ( )()
()
()() 
() ( )()
()
() () 
()(3) 109.165216.1
101.7303.0
25.0
25.020801023
1020002.0
25.0
25.020801012
3
92
6
92
6
−=
+−−=+−−
−−
alst
alst
FF
FF

Solving Eqs. 1 and 3 simultaneously yields(Ans) kN 123 and kN 4.16 =−=
alst
FF

23. THANK YOU