2. 1 2
1
0
0
_
0
X
A B
F
P R R
Super position
P
Stress
A
P
Strain
EA
Pa
EA
3. Thermal Expension coefficient
1
2
1 1 2
1 2
( )
0
T
T
T
L T L L
1
1
2
2
1 2
2
1 2
B
B
B B
R
EA
R
EA
R L R L
EA EA
1 2
1 2
1 2
1 2
1 2
1 2
( ) 0
(
B
B
R L L
T L L
E A A
R L L
T L L
E A A
4. 2
1 2 0 0
( 1)
( )
0
B B
B
B
a
B a B
B
B
A B
R R
A EA
R L
EA
R LPa
EA EA
P
P P R L
L
R P
a
R
L
Pa a
R R P P P
L L
P a L
a L
L
5. A250mm bar of 15x30mm rectangular cross-section of two aluminum layers .5mm thick
,brazed to a center brass layer of the same thickness . If it is subjected to centric forces of
magnetude P=30KN. And Knowing that E of a =70GPa &E of B=105GPa . Determine the
normal stress (a ) in the aluminum layers (b) in the brass layer
3
9 6
8
3
9 6
: Is the force in the aluminium layers
F : Is the force in the steel layers
30 (1)
(250)(10 )
70(10 )(10)(30)(10 )
1.19 10
(250)(10 )
105(10 )(10)(5)(10 )
a
S
a S
a a a
a
a a
a a
S S S
S
S S
S
F
F F KN
F L F
E A
x F mm
F L F
E A
8
1.587 10
_
1.334
12.85
17.15
85.67
57.17
S
S a
a S
S
a
S
a
x F mm
Combatibility conditon
F F
F KN
F KN
MPa
MPa
6. Compressive centric forces of 160KN are applied at both ends of the assembly
shown by means of rigid plates Knowing that E of S=200GPa and E of a=70GPa
determine (a) the normal stresses in the steel core and the aluminum shell,
(b) the deformation of the assembly
3
9 2 2 6
9
3
9 2
: Is the force in the aluminium layers
F : Is the force in the steel layers
160 (1)
(250)(10 )
70(10 )( )(62 25 )(10 )
4
1.4126 10
(250)(10 )
200(10 )( )(25
4
a
S
a S
a a a
a
a a
a a
S S S
S
S S
F
F F KN
F L F
E A
x F mm
F L F
E A
6
9
4
)(10 )
2.5465 10
_
1.8027
57.1
102.9
116.32
40.7
1.454(10 )
S S
S a
a S
S
a
S
a
S a
x F mm
Combatibility conditon
F F
F KN
F KN
MPa
MPa
m
7. Three steel rods (E=200GPa) support a 36-KN load P Each of the rods AB and CD
has a 200mm2 cross-sectional area .
Neglecting the deformation of rod BED . Determine (a) the change in length of
rod EF
(b) the stress in each rod
3
9 6
6
3
36
(2 ) ( ) 36( ) 0
0.5(36 )
_
0.5(36 )(1000)(500)(10 )
200(10 )(200)(10 )
6.25(36 )(10 )
(1000)(400)(10 )
20
Y CD BA EF
B CD EF
CD EF
AB CD
D D EF
B D
D D
D EF
E E EF
E
E E
F F F F KN
M F a F a a
F F
from symmetry F F
F L F
E A
F m
F L F
E A
9 6
6
0(10 )(625)(10 )
3.2 10
sin _ _ _ _
_
23.81
6.095
38.096
30.47
0.0762
E EF
E D
EF
CD AB
EF
AB
E
F x m
ce the BED is rigid
Combatibility conditon
F KN
F F KN
MPa
MPa
mm
8. Two cylindrical rods , one of steel and the other of brass ,are joined at c and restrained by
rigid support at Aand E . For the loading shown and Knowing that E of s=200GPa and E of
b=105GPa , determine
(a) the reactions atAand E
(b) the deflection of point C
3
9 2 6
5
3
9
1-Eliminating the support reaction_at_E
0, 40
40 , 100
(100)(1000)(180)(10 )
200(10 )( )(40 )(10 )
4
7.162(10 )
(40)(1000)(120)(10 )
200(10 )( )(
4
ED CB
CD AB
BA BA
BA
BA BA
BA
CB CB
CB
CB CB
P P KN
P KN P KN
F L
E A
m
F L
E A
2 6
5
3
9 2 6
5
4
1
40 )(10 )
1.91 10
(40)(1000)(100)(10 )
105(10 )( )(30 )(10 )
4
5.3894 10
01.446(10 )
CB
DC DC
DC
DC DC
DC
E BA CB DC
x m
F L
E A
F x m
m
9. 2
6
2
2 2
9
2
2 1
2-Deflection_at_E due_to the support reaction
300 200
(10 )
200( )(40 ) 105( )(30 )
4 4
3.888 10
0
37.19
100 37.19 62.8
AC ED
E E
AC AC ED ED
E E
E E
E E
E
A
E
C BA BC
L L
R
E A E A
R
x R
R KN
R KN
R
3 3
5 5
9 2 6
5
(37.19)(10 )(300)(10 )
7.162(10 ) 1.91(10 )
200(10 )( )(40 )(10 )
4
4.633(10 )
AC
AC AC
C
C
L
E A
m
10. A rod consiststing of two cylindrical portions AB and BC is restrained at both
ends potion AB is made of steel and portion BC is made of
brass Knowing that the rod is initially unstressed
determine the comperssive force induced in ABC when there is a temperature
rise of 50C
6 3
4
6 3
4
1
1-Eliminate_the_ reaction_force_at_A
11.7(10 )(50)(250)(10 )
1.4625(10 )
20.9(10 )(50)(300)(10 )
3.135(10 )
4
_ _
_ _
.597
A AB BC
AB S AB
AB
BC B BC
BC
A
USING THE PRINICIPLE
OF SUPPER
T
PO
L
m
TL
SI N
m
TIO
4
5(10 )m
6
200 , 11.7 10s sE GPa x C
6
105 , 20.9 10b bE GPa x C
11. 2
3 3
2
9 2 6 9 2 6
9
2
1 2
2-Deflection_at_A due_to the support reaction
250(10 ) 300(10 )
200(10 )( )(30 )(10 ) 105(10 )( )(50 )(10 )
4 4
3.224 (10 )
142.6
BCAB
A A
AB AB BC BC
A A
A A
A A A
A
LL
R
E A E A
R
R m
R KN
12. Knowing that a 0.5mm gap exists when the temperature is24C, determine (a)
the temperature at which the normal stress in the aluminum bar will be equal to
75MPa (b) the corresponding exact length of the aluminum bar
2
6
1500
105
21.6 10
Bronze
A mm
E GPa
x C
2
6
min
1800
73
23.2 10
Alu um
A mm
E GPa
x C
13. 6 3
5
t:Is the bar total displacement due to temperature rise
T:Is the change in the temperature
1-The elong
23
atio
.2(1
n of the
0 )(4
bar due to
50)(10 )
temperature r
1
is
.044 10
21.
e
a a a
a
b b b
TL
Tx m
TL
6 3
6
5
6(10 )(350)(10 )( )
7.56 (10 )
1.8 (10 ) (1
b
t a b
T
T m
Tx m
14. 6
9
2
2-Deflection of the due to the reaction
450 350
(10 )
73(1800) 105(1500)
5.647 (10 )
555.56 /
Forin ducing -75 MPa of in the alumium bar
,the reaction
a b
R
a a b b
R
R
a
a
L L
R
E A E A
R
R m
R
RN m
A
force must be
135R KN
15. 3
3
_
_ Re
_
Due to the 0.5mm GPa ,the compatibility equation will be :
0.5 10
1.2623 10
70
24 94
3-The delection in the aluminum rod
0.7308
0.4623
t R
t
Surrounding
a thermal
a action
a a ther
x m
x m
T C
T T C
mm
mm
_ Re 0.2685
450.2685
mal a action
a
mm
L mm
