PROBLEM 9.55
The aluminum rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 6000 psi to the 12–in. portion BC of the rod.
Knowing that
6
10.1 10
E= ×
psi and
0.36,v=
determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
SOLUTION
6
6
6
6
6000 psi 0
1()
1[ 6000 (0.36)(0) (0.36)( 6000)]
10.1 10
380.198 10
1()
1[ (0.36)( 6000) 0 (0.36)( 6000)]
10.1 10
427.72 10
xz y
x xyz
y xy z
P
vv
E
vv
E
ss s
ε sss
ε ss s
−
−
= =−=− =
= −−
= −− − −
×
=−×
=− +−
= − − +− −
×
= ×
Length subjected to strain
: 12 in.
x
L
ε
=
(a)
6
(12)(427.72 10 )
yy
L
δε
−
= = ×
3
5.13 10 in.
l
δ
−
= ×
(b)
6
(1.5)( 380.198 10 )
xx
d
δε
−
==−×
3
0.570 10 in.
d
δ
−
=−×
consent of McGraw–Hill Education.
PROBLEM 9.56
For the rod of Prob. 9.55, determine the forces that should be applied to
the ends A and D of the rod (a) if the axial strain in portion BC of the rod
is to remain zero as the hydrostatic pressure is applied, (b) if the total
length AD of the rod is to remain unchanged.
PROBLEM 9.55 The aluminum rod AD is fitted with a jacket that is
used to apply a hydrostatic pressure of 6000 psi to the 12–in. portion BC
of the rod. Knowing that
6
10.1 10
E= ×
psi and
0.36,v=
determine
(a) the change in the total length AD, (b) the change in diameter at the
middle of the rod.
SOLUTION
Over the pressurized portion BC,
1
() ( )
1(2 )
xz yy
y BC x y z
y
p
vv
E
vp
E
σσ σσ
ε σσ σ
σ
==−=
=− +−
= +
(a)
22 2
() 0 2 0
2 (2)(0.36)(6000)
4320 psi
(1.5) 1.76715 in
44
(1.76715)( 4320) 7630 lb
εσ
σ
pp
σ
= +=
=−=−
= −
= = =
== −=−
y BC y
y
y
vp
vp
Ad
FA
i.e., 7630 lb compression
(b) Over unpressurized portions AB and CD,
0
xz
σσ
= =
() () y
y AB y CD E
σ
εε
= =
For no change in length,
() () () 0
( )() () 0
12
(20 12) (2 ) 0
24 (24)(0.36)(6000) 2592 psi
20 20
(1.76715)( 2592) 4580 lb
δε ε ε
εε
σσ
σ
σ
=++=
+ +=
− + +=
=−=− =−
== −=−
AB y AB BC y BC CD y CD
AB CD y AB BC y BC
yy
y
y
LLL
LL L
vp
EE
vp
PA
4580 lb compressionP=
consent of McGraw–Hill Education.
PROBLEM 9.57
A 30–mm square was scribed on the side of a large steel pressure
vessel. After pressurization, the biaxial stress condition at the square
is as shown. For
200 GPaE=
and
0.30,v=
determine the change
in length of (a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
Given:
80 MPa 40 MPa
xy
σσ
= =
Using Eq’s (2.28):
6
3
1 80 0.3(40)
( ) 340 10
200 10
x xy
v
E
ε σσ
−
−
= −= =×
×
6
3
1 40 0.3(80)
( ) 80 10
200 10
y yx
v
E
ε σσ
−
−
= −= =×
×
(a) Change in length of AB.
63
( ) (30 mm)(340 10 ) 10.20 10 mm
AB x
AB
δε
−−
= = ×=×
10.20 m
AB
δm
=
(b) Change in length of BC.
63
( ) (30 mm)(80 10 ) 2.40 10 mm
BC y
BC
δε
−−
= = ×=×
2.40 m
BC
δm
=
(c) Change in length of diagonal AC.
From geometry,
222
()()()AC AB BC= +
Differentiate:
2( ) ( ) 2( ) ( ) 2( ) ( )AC AC AB AB BC BC∆= ∆+ ∆
But
() () ()
AC AB BC
AC AB BC
δδδ
∆= ∆= ∆=
Thus,
2( ) 2( ) 2( )
AC AB BC
AC AB BC
δδδ
= +
11
(10.20 m)+ (2.40 m)
22
AC AB BC
AB BC
AC AC
δδδ m m
=+=
8.91 m
AC
δm
=
consent of McGraw–Hill Education.
consent of McGraw-Hill Education.
PROBLEM 9.58
A fabric used in air-inflated structures is subjected to a biaxial loading
that results in normal stresses
120
x
MPa and
160
z
MPa.
Knowing that the properties of the fabric can be approximated as
E 87 GPa and v 0.34, determine the change in length of (a) side AB,
(b) side BC, (c) diagonal AC.
SOLUTION
6
125 125
AC
PROBLEM 9.59
In many situations, it is known that the normal stress in a given direction is zero.
For example,
0
z
σ
=
in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains
ε
x and
ε
y have been determined
experimentally, we can express
,,
σσ
xy
and
z
ε
as follows:
22
()
1
11
xy yx
x y z xy
vv
v
EE v
vv
εε εε
σ σ ε εε
++
= = =−+
−
−−
SOLUTION
0
1()
z
x xy
v
E
σ
ε σσ
=
= −
(1)
1()
y xy
v
E
ε σσ
=−+
(2)
Multiplying (2) by v and adding to (1),
2
2
1or ( )
1
xy x x xy
vE
vv
Ev
εε σ σ εε
−
+= = +
−
Multiplying (1) by v and adding to (2),
2
2
1or ( )
1
yx y y yx
vE
vv
Ev
εε σ σ εε
−
+= = +
−
1()
z xy
v
vv
EE
ε σσ
=−− =− E
⋅
2
2
()
1
(1 ) () ()
1
1
x yy x
xy xy
vv
v
vv v
v
v
ε εε ε
εε εε
+ ++
−
+
=− +=− +
−
−
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 9.60
In many situations, physical constraints
prevent strain from occurring in a given
direction. For example,
0
z
ε
=
in the case
shown, where longitudinal movement of
the long prism is prevented at every point.
Plane sections perpendicular to the
longitudinal axis remain plane and the
same distance apart. Show that for this
situation, which is known as plane strain,
we can express
,
σ
z
,
x
ε
and
ε
y
as follows:
2
2
()
1[(1 ) (1 ) ]
1[(1 ) (1 ) ]
z xy
x xy
y yx
v
v vv
E
v vv
E
σ σσ
ε σσ
ε σσ
= +
= − −+
= − −+
SOLUTION
1
0 ( ) or ( )
z x yz z xy
vv v
E
ε σ σσ σ σσ
==−− + = +
2
2
1()
1[ ( )]
1[(1 ) (1 ) ]
x xyz
x y xy
xy
vv
E
vv
E
v vv
E
ε σσσ
σ σ σσ
σσ
= −−
= −− +
= − −+
2
2
1()
1[ ( )]
1[(1 ) (1 ) ]
y xy z
xy xy
yx
vv
E
vv
E
v vv
E
ε σσ σ
σσ σσ
σσ
=− +−
=− +− +
= − −+
consent of McGraw–Hill Education.
PROBLEM 9.61
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55–kip load P is applied. Knowing that for the plastic used
150G=
ksi,
determine the deflection of the plate.
SOLUTION
2
3
3
3
3
3
(3.2)(4.8) 15.36 in
55 10 lb
55 10 3580.7 psi
15.36
150 10 psi
3580.7 23.871 10
150 10
2 in.
A
P
P
A
G
G
h
τ
τ
γ
−
= =
= ×
×
= = =
= ×
= = = ×
×
=
3
3
(2)(23.871 10 )
47.7 10 in.
h
δγ
−
−
= = ×
= ×
0.0477 in.
δ
= ↓
PROBLEM 9.62
A vibration isolation unit consists of two blocks of hard rubber bonded to
a plate AB and to rigid supports as shown. Knowing that a force of
magnitude
25 kNP=
causes a deflection
1.5 mm
δ
=
of plate AB,
PROBLEM 9.63
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity
19 MPaG=
bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by
δ
the corresponding deflection, determine the
effective spring constant,
/,kP
δ
=
of the system.
SOLUTION
δ
PROBLEM 9.55
The aluminum rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 6000 psi to the 12–in. portion BC of the rod.
Knowing that
6
10.1 10
E= ×
psi and
0.36,v=
determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
SOLUTION
6
6
6
6
6000 psi 0
1()
1[ 6000 (0.36)(0) (0.36)( 6000)]
10.1 10
380.198 10
1()
1[ (0.36)( 6000) 0 (0.36)( 6000)]
10.1 10
427.72 10
xz y
x xyz
y xy z
P
vv
E
vv
E
ss s
ε sss
ε ss s
−
−
= =−=− =
= −−
= −− − −
×
=−×
=− +−
= − − +− −
×
= ×
Length subjected to strain
: 12 in.
x
L
ε
=
(a)
6
(12)(427.72 10 )
yy
L
δε
−
= = ×
3
5.13 10 in.
l
δ
−
= ×
(b)
6
(1.5)( 380.198 10 )
xx
d
δε
−
==−×
3
0.570 10 in.
d
δ
−
=−×
consent of McGraw–Hill Education.
PROBLEM 9.56
For the rod of Prob. 9.55, determine the forces that should be applied to
the ends A and D of the rod (a) if the axial strain in portion BC of the rod
is to remain zero as the hydrostatic pressure is applied, (b) if the total
length AD of the rod is to remain unchanged.
PROBLEM 9.55 The aluminum rod AD is fitted with a jacket that is
used to apply a hydrostatic pressure of 6000 psi to the 12–in. portion BC
of the rod. Knowing that
6
10.1 10
E= ×
psi and
0.36,v=
determine
(a) the change in the total length AD, (b) the change in diameter at the
middle of the rod.
SOLUTION
Over the pressurized portion BC,
1
() ( )
1(2 )
xz yy
y BC x y z
y
p
vv
E
vp
E
σσ σσ
ε σσ σ
σ
==−=
=− +−
= +
(a)
22 2
() 0 2 0
2 (2)(0.36)(6000)
4320 psi
(1.5) 1.76715 in
44
(1.76715)( 4320) 7630 lb
εσ
σ
pp
σ
= +=
=−=−
= −
= = =
== −=−
y BC y
y
y
vp
vp
Ad
FA
i.e., 7630 lb compression
(b) Over unpressurized portions AB and CD,
0
xz
σσ
= =
() () y
y AB y CD E
σ
εε
= =
For no change in length,
() () () 0
( )() () 0
12
(20 12) (2 ) 0
24 (24)(0.36)(6000) 2592 psi
20 20
(1.76715)( 2592) 4580 lb
δε ε ε
εε
σσ
σ
σ
=++=
+ +=
− + +=
=−=− =−
== −=−
AB y AB BC y BC CD y CD
AB CD y AB BC y BC
yy
y
y
LLL
LL L
vp
EE
vp
PA
4580 lb compressionP=
consent of McGraw–Hill Education.
PROBLEM 9.57
A 30–mm square was scribed on the side of a large steel pressure
vessel. After pressurization, the biaxial stress condition at the square
is as shown. For
200 GPaE=
and
0.30,v=
determine the change
in length of (a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
Given:
80 MPa 40 MPa
xy
σσ
= =
Using Eq’s (2.28):
6
3
1 80 0.3(40)
( ) 340 10
200 10
x xy
v
E
ε σσ
−
−
= −= =×
×
6
3
1 40 0.3(80)
( ) 80 10
200 10
y yx
v
E
ε σσ
−
−
= −= =×
×
(a) Change in length of AB.
63
( ) (30 mm)(340 10 ) 10.20 10 mm
AB x
AB
δε
−−
= = ×=×
10.20 m
AB
δm
=
(b) Change in length of BC.
63
( ) (30 mm)(80 10 ) 2.40 10 mm
BC y
BC
δε
−−
= = ×=×
2.40 m
BC
δm
=
(c) Change in length of diagonal AC.
From geometry,
222
()()()AC AB BC= +
Differentiate:
2( ) ( ) 2( ) ( ) 2( ) ( )AC AC AB AB BC BC∆= ∆+ ∆
But
() () ()
AC AB BC
AC AB BC
δδδ
∆= ∆= ∆=
Thus,
2( ) 2( ) 2( )
AC AB BC
AC AB BC
δδδ
= +
11
(10.20 m)+ (2.40 m)
22
AC AB BC
AB BC
AC AC
δδδ m m
=+=
8.91 m
AC
δm
=
consent of McGraw–Hill Education.
consent of McGraw-Hill Education.
PROBLEM 9.58
A fabric used in air-inflated structures is subjected to a biaxial loading
that results in normal stresses
120
x
MPa and
160
z
MPa.
Knowing that the properties of the fabric can be approximated as
E 87 GPa and v 0.34, determine the change in length of (a) side AB,
(b) side BC, (c) diagonal AC.
SOLUTION
6
125 125
AC
PROBLEM 9.59
In many situations, it is known that the normal stress in a given direction is zero.
For example,
0
z
σ
=
in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains
ε
x and
ε
y have been determined
experimentally, we can express
,,
σσ
xy
and
z
ε
as follows:
22
()
1
11
xy yx
x y z xy
vv
v
EE v
vv
εε εε
σ σ ε εε
++
= = =−+
−
−−
SOLUTION
0
1()
z
x xy
v
E
σ
ε σσ
=
= −
(1)
1()
y xy
v
E
ε σσ
=−+
(2)
Multiplying (2) by v and adding to (1),
2
2
1or ( )
1
xy x x xy
vE
vv
Ev
εε σ σ εε
−
+= = +
−
Multiplying (1) by v and adding to (2),
2
2
1or ( )
1
yx y y yx
vE
vv
Ev
εε σ σ εε
−
+= = +
−
1()
z xy
v
vv
EE
ε σσ
=−− =− E
⋅
2
2
()
1
(1 ) () ()
1
1
x yy x
xy xy
vv
v
vv v
v
v
ε εε ε
εε εε
+ ++
−
+
=− +=− +
−
−
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 9.60
In many situations, physical constraints
prevent strain from occurring in a given
direction. For example,
0
z
ε
=
in the case
shown, where longitudinal movement of
the long prism is prevented at every point.
Plane sections perpendicular to the
longitudinal axis remain plane and the
same distance apart. Show that for this
situation, which is known as plane strain,
we can express
,
σ
z
,
x
ε
and
ε
y
as follows:
2
2
()
1[(1 ) (1 ) ]
1[(1 ) (1 ) ]
z xy
x xy
y yx
v
v vv
E
v vv
E
σ σσ
ε σσ
ε σσ
= +
= − −+
= − −+
SOLUTION
1
0 ( ) or ( )
z x yz z xy
vv v
E
ε σ σσ σ σσ
==−− + = +
2
2
1()
1[ ( )]
1[(1 ) (1 ) ]
x xyz
x y xy
xy
vv
E
vv
E
v vv
E
ε σσσ
σ σ σσ
σσ
= −−
= −− +
= − −+
2
2
1()
1[ ( )]
1[(1 ) (1 ) ]
y xy z
xy xy
yx
vv
E
vv
E
v vv
E
ε σσ σ
σσ σσ
σσ
=− +−
=− +− +
= − −+
consent of McGraw–Hill Education.
PROBLEM 9.61
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55–kip load P is applied. Knowing that for the plastic used
150G=
ksi,
determine the deflection of the plate.
SOLUTION
2
3
3
3
3
3
(3.2)(4.8) 15.36 in
55 10 lb
55 10 3580.7 psi
15.36
150 10 psi
3580.7 23.871 10
150 10
2 in.
A
P
P
A
G
G
h
τ
τ
γ
−
= =
= ×
×
= = =
= ×
= = = ×
×
=
3
3
(2)(23.871 10 )
47.7 10 in.
h
δγ
−
−
= = ×
= ×
0.0477 in.
δ
= ↓
PROBLEM 9.62
A vibration isolation unit consists of two blocks of hard rubber bonded to
a plate AB and to rigid supports as shown. Knowing that a force of
magnitude
25 kNP=
causes a deflection
1.5 mm
δ
=
of plate AB,
PROBLEM 9.63
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity
19 MPaG=
bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by
δ
the corresponding deflection, determine the
effective spring constant,
/,kP
δ
=
of the system.
SOLUTION
δ