Chapter08 Deformation of Solids Ss
Chapter08 Deformation of Solids Ss
Chapter08 Deformation of Solids Ss
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Behaviour (
k : force(Hook e) constant
Force-Extension Graphs Extension-Force Graphs
CDE C : yield point
D : point of maximum force (stress)
F
- This region is known
E : fracture (breaking) point
as plastic deformation. Steel
- - When the force Force, F
(stress) increases, the D
E
elongation (strain) Glass
increases BC
rapidly. A Rubber
e
OT extension, e
O
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Y @ E) Y @ E)
1
strain energy ( stress)( strain) Al 0 Volume
2
strain energy 1
( stress)(strain)
volume 2
Volume
Solution: Example 2
Solution: Solution:
m2=0.10 kg, m1=0.25 kg, lt=1.00 m, dt=0.22 x 10 -3
m, lc=1.50 m, dc=0.50x10-3 m.
Yt After the tungsten and copper wire stop to Yt
modulus,
stretch, the system is in equilibrium, hence
Tt lt Tc lc d2
Yt and Yc but A
m2 0 .10 kg F 0 m2 0 .10 kg 4
Free body diagram :
Ae t Ae c
Tc Tt 4Tt lt 4Tc lc
hence Yt and Yc
Yc
Tc m1 g m2 Tt m 2 g Tc
Yc d t2 et d c2 ec
m1
Tt 3 .43 N
Tc 2 . 45 N Tc
m1 0 .25 kg m1 g m2 g m1 0 .25 kg
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Solution: Summary
Deformation of Solids
Therefore
et 2 .51x10 4 m and ec 1.56 x10 4 m
Elasticity Behaviour of a wire
The total strain (elastic potential) e
Steel
energy is given by F = ke
1 1 Glass
Total strain energy Tc ec Tt et Rubber
2 2
O F
Total strain energy 6 . 22 x10 4 J =Stress/Strain
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= Fl/Ae Elastic P.E.