CHAPTER
Axisymmetric Problems
8.1 8. 1
8
INTR IN TROD ODUC UCTIO TION N
Many problems of practical importance are concerned with solids of revolution which are deformed symmetrically with respect to the axis of revolution. Examples of such solids are circular cylinders subjected to uniform internal and external pressures, rotating circular disks, spherical shells subjected to uniform internal and external pressures, etc. In this chapter, a few of these problems will be inv invest estig igate ated. d. Let the axi axiss of rev revolu olutio tion n be the z-axis. The deformation being symmetrical with respect to the z-axis, it is convenient to use cylindrical coordinates. Since the deformation is symmetrical about the axis, the stress components do not depend on q . Further, t q q and t q do not exist. Consequently, the differential equations of equilibrium [Eqs (1.67)–(1.69)] can be reduced to our special case. However, it is instructive to derive the relevant equations applicable to axisymmetric problems from first principles. Consider an axisymmetric body shown in Fig. 8.1. Let an elementary radial element be isolated. The stress vectors acting on its faces are as shown. r
z
z s z
+
Ds z
G H s r
o
Dz
s r
E
D F
B
A r
Fig. 8.1 An axisymmetric body
C
s z Dr
s q
+
Ds r
270 Advanced Mechanics of Solids
On faces ABCD and EFGH , the normal stresses are s q and there are no shear stresses. On face ABFE , the stresses are s z and t zr . On face CDHG, the normal and shear stresses are
s z
+ Ds z =
s z
+
t rz
+ Dt rz =
t rz
+
∂σ z
D z
z ∂ ∂τ rz
D z
∂ z
On face AEHD, the normal and shear stresses are s r and t rz. On face BCGF , the stresses are s r +
∂σ r ∂ r
Dr and and
t rz
∂τ rz
+
∂ r
Dr
For equilibrium in z direction ∂σ z ∆r ⎞ ⎛ ⎞⎛ ⎛ + ∆ + ∆ ∆ + σ θ τ rz + z r r ⎜⎝ ⎟⎠ ⎜⎝ z ⎟ ⎜ ⎠ ⎝ 2 ∂ z
∂τ rz ∂ r
⎞ ∆r ⎟ (r + Dr ) Dq D z ⎠
∆r ⎞ ⎛ t rzr Dq D z – s z ⎜ r + – t ⎟ D q ⎝ 2⎠
D r +
∆r ⎞ ⎛ g z ⎜ r + ⎟ D q Dr D z = 0 ⎝ 2⎠
where g z is the body force per unit volume in z direction. Hence, ∂σ z ∂ z
∆r ⎞ ⎛ r + ⎜⎝ ⎟ ∆r ∆θ ∆z + 2 ⎠ +
∂τ rz ∂ r
(r + Dr ) Dr Dq D z
∆r ⎞ ⎛ t rz Dr Dq D z + g z ⎜ r + ⎟ Dr Dq D z = 0 ⎝ 2⎠
Cancelling Dr Dq D z and going to limits ∂σ z ∂ z
+
∂τ rz ∂ r
+
τ rz
+
r
g z = 0
(8.1)
Similarly, for equilibrium in r direction direction we get ∂σ r ∂r
+
∂τ rz ∂ z
+
σ r
− σ θ r
+
g r = 0
(8.2)
where g r is the body force per unit volume in r direction. direction. Since the stress components are independent of q , the equilibrium equation for q direction direction is identically satisfied. For the problems that we are going to discuss in this chapter, we need expressions for the circumferential strain e q and the radial strain e r . Referring to Fig. 8.2(a), consider an arc AE at at distance r , subtending an angle Dq at the centre. The arc length is r Dq . The radial displacement is ur . Consequently, the length of the arc becomes ( r + ur ) Dq . Hence, the circumferential strain is
e q =
( r + ur ) ∆θ − r ∆θ r ∆θ
=
ur r
(8.3)
The radial strain is, from Fig. 8.2(b),
e r =
∂ ur ∂ r
(8.4)
271
Axisymmetric Problems
Dq r
u r E A B Dr
A
ur
A¢ Dr
+
∂u r ∂r
B
∆r
B ¢
(b)) (b (a)) (a
Fig. 8.2 Displacements along along a radius radius The axial strain is ∂ u z
e z =
(8.5a)
z ∂
where u z is the axial displacement. In subsequent sections we shall consider the following problems: Circular cylinder subjected to internal or external pressure Sphere subjected to internal or external pressure Sphere subjected to mutual gravitational attraction Rotating disk of uniform thickness Rotating disk of variable thickness Rotating shaft and cylinder
8.2 8. 2
THICKTHIC K-WA WALL LLED ED CYL CYLIN INDE DER R SUBJ SUBJEC ECTE TED D TO INTERNAL AND EXTERNAL PRESSURES—LAME'S PROBLEM
Consider a cylinder of inner radius a and outer radius b (Fig. 8.3). Let the cylinder be subj subjecte ected d to an inte internal rnal pres pressure sure pa and an external pressure pb. It is possible to treat this problem either as a plane stress case ( s z = 0) or as a plane strain case (e z = 0). Appropriate solutions will be obtained for each case. p b p b b p a
a
p a
(a)
Fig. 8.3
(b)
Thick-walled cylinder under internal and external pressures
272
Advanced Mechanics of Solids
Case (a) Plane Stress Let the ends of the cylinder be free to expand. We shall assume that s z 0 and our results will justify this assumption. Owing to uniform radial deformation, t rz 0. Neglecti Negl ecting ng body force forces, s, Eq. (8.2 (8.2)) reduc r educes es to =
=
∂σ r ∂ r
− σ θ
σ r
+
=0
r
(8.5b)
Since r is is the only independent variable, the above equation can be written as d dr
( r σ r ) − σ
=0
θ
(8.5c)
Equation (8.1) is identically satisfied. From Hooke’s law εr
= 1 (σ r − νσ θ ) ,
εθ
E
= 1
E
(σ
θ
− νσ r )
or the stresses in terms of strains are
=
σr
Substituting for
E
1− ν
+ νεθ )
(ε r
2
e r and e q from
σ r
=
σ θ
=
E
1 − ν
2
E
1 − ν
2
σθ
=
E 2
1 − ν
(ε
θ
+ νε r )
Eqs (8.3) and (8.4)
ur ⎞ ⎛ dur + ν ⎜⎝ dr r ⎟ ⎠
(8.6a)
dur ⎞ ⎛ ur ν + ⎜⎝ r dr ⎟ ⎠
(8.6b)
Substituting these in the equation of equilibrium given by Eq. (8.5c)
⎛ ⎜⎝ r
d dr
or
dur dr
+ r
d ur 2
du r
+ν
dr
dr
d ur dr
dr
2
2
i.e.
du ⎞ ⎞ ⎛u + ν ur ⎟ − ⎜ r + ν r ⎟ = 0 dr ⎠ ⎠ ⎝ r
dur
2
+1
dur
r dr
−
u r 2
r
−
ur r
− ν
du r dr
= 0
=0
This can be reduced to d dr
or
d dr
⎛ dur + ⎜⎝ dr dr ⎡1 ⎢⎣ r
ur ⎞ r ⎟ ⎠
=0
⎤ ur r ) = 0 ( ⎥ dr d
⎦
(8.7)
If the function ur is found from this equation, the stresses are then determined from Eqs. (8.6a) and (8.6b). The solution to Eq. (8.7) is ur
= C1 r +
C 2 r
(8.8)
Axisymmetric Problems
273
where C 1 and C 2 are constants of integration. Substituting this function in Eqs. (8.6a) and (8.6b) σ r
⎡ 1⎤ = E 2 ⎢C1 (1 + ν ) − C 2 (1 − ν ) 2 ⎥ 1 − ν ⎣ r ⎦
(8.9a)
σ θ
⎡ 1⎤ = E 2 ⎢C1 (1 + ν ) + C 2 (1 − ν ) 2 ⎥ 1 − ν ⎣ r ⎦
(8.9b)
The constants C 1 and C 2 are determined from the boundary conditions. When when Hence,
r = a, r = b,
s r = s r =
E
1 − ν 2 E
1 − ν 2 whence,
– pa – pb
⎡ 1 ⎤ (1 ) (1 ) + − − = − pa C C ν ν 2 ⎢⎣ 1 2⎥ a ⎦ ⎡ 1 ⎤ (1 ) (1 ) C C ν ν + − − = − pb 2 ⎢ 1 2⎥ b ⎦ ⎣
2 2 1 − ν pa a − pb b C 1 = 2 2 E b −a
1 + ν a 2 b 2 C2 = ( pa − pb ) E b 2 − a 2 Substituting these in Eqs (8.8) and (8.9) we get
1 − ν pa a 2 − pb b 2 1 + ν a 2b2 pa − pb ur = r + 2 2 E E r b 2 − a 2 b −a σ r
σ θ
=
=
pa a 2 − pb b 2 2
b −a
2
pa a 2 − pb b 2 2
b −a
2
−
a 2 b 2 pa − pb r
+
2
2
b −a
2
a 2 b 2 pa − pb r
2
2
b −a
2
(8.10)
(8.11)
(8.12)
It is interesting to observe that the sum s r + s q is constant through the thickness of the wall of the cylinder, i.e. independent of r . Hence, according to Hooke’s Law, the stresses s r and s q produce a uniform extension or contraction in z direction, and cross-sections perpendicular to the axis of the cylinder remain plane. If we consider two adjacent cross-sections, the deformation undergone by the element does not interfere with the deformation of the neighbouring element. Hence, the elements can be considered to be in a state of plane stress, i.e. s z = 0, as we assumed at the beginning of the discussion. It is important to note that in Eqs (8.10)–(8.12), pa and pb are the numerical values of the compressive pressures applied.
274
Advanced Mechanics of Solids
Cylinder Subjected to Internal Pressure In this case pb Eqs (8.11) and (8.12) become
=
0 and pa
=
p. Then
⎛ b2 ⎞ 1− 2 ⎟ σ r = 2 2 ⎜ ⎝ b −a r ⎠
(8.13)
⎛ b2 ⎞ 1+ 2 ⎟ = 2 2 ⎜ b −a ⎝ r ⎠
(8.14)
pa
σ θ
pa
2
2
These equations show that s r is always a compressive stress and s q a tensile stress. Figure 8.4 shows the variation of radial and circumferential stresses across the thickness of the cylinder under internal pressure. The circumferential stress is greatest at the inner surface of the cylinder, where
(σ )max =
(
p a 2 + b 2
θ
b −a 2
)
(8.15)
2
p
s q
s r
+
p
–
b 2 + a2 b 2 − a2
p
p
(a)
Fig. 8.4
(b)
Cylinder subjected to internal pressure
Hence, (s q )max is always greater than the internal pressure and approaches this value as b increases so that it can never be reduced below pa irrespective of the amount of material added on the outside.
Cylinder Subjected to External Pressure In this case, pa 0 and pb (8.11) and (8.12) reduce to =
p
b
pb 2
s r
p. Equations
⎛ a2 ⎞ 1− 2 ⎟ σ r = − 2 2 ⎜ b −a ⎝ r ⎠
(8.16)
⎛ a2 ⎞ 1+ 2 ⎟ =− 2 2 ⎜ ⎝ b −a r ⎠
(8.17)
a σ θ
–
–
=
pb 2
s q
The variations of these stresses across the thickness are shown in Fig. 8.5. If there is no inner hole, i.e. if a 0, the stresses are uniformly distributed in the cylinder with – p . s r s q =
Fig. 8.5 Cylinder subjected to external pressure
=
=
Axisymmetric Problems
275
Example 8.1 Select the outer radius b for a cylinder subjected to an internal pressure p = 500 atm with a factor of safety 2. The yield point for the material (in tension as well as in compression) is s yp = 5000 kgf/cm 2 (490000 kPa). The inner radius is 5 cm. Assume that the ends of the cylinder are closed. Solution
The critical point lies on the inner surface of the cylinder, where s r =
− p,
σ θ
=p
b2 + a 2 b −a 2
2
,
σ z
=p
a2 b −a 2
2
(assumed)
In the above expressions, it is assumed that away from the ends, s z caused by p is uniformly distributed across the thickness. The maximum and minimum principal stresses are s 1 = s q and s 3 = s r . Hence, t max
=
2 b 1 (σ − σ 3 ) = p 2 2 2 1 b −a
Substituting the numerical values (1 atm = 98.07 kPa), b
=
5 a = 6.45 cm 3
Example 8.2 A thick-walled steel cylinder with radii a = 5 cm and b = 10 cm is subjected to an internal pressure p. The yield stress in tension for the material is 350 MPa. Using a factor of safety of 1.5, determine the maximum working pressure p according to the major theories of failure. E = 207 ¥ 106 kPa, v = 0.25. Solution
(i) Maximum normal stress theory Maximum normal stress = s q at r = a
= p
\
p
(b 2 + a 2 ) (b 2 − a 2 )
or
p
=
(b 2 + a 2 ) (b 2 − a 2 )
σ y
N
350 × 106 100 − 25 × = 140 × 103 kPa 1.5 100 + 25
=
(ii) Maximum shear stress theory 1 Maximum shear stress = (σ θ − σ r ) at r = a 2 1 ⎛ 2b 2 ⎞ = p ⎜ 2 2 ⎝ b − a 2 ⎟⎠ \
p
b
2
2 2 b −a
=1
σ y
2 N
276
Advanced Mechanics of Solids
350 × 106 100 − 25 or p = × = 87.5 × 103 kPa 3 100 (iii) Maximum strain theory Maximum strain = e q = 1 (σ θ − ν σ r ) at r = a E p a2 = E (b 2 − a 2 )
⎡⎛ ⎛ b2 ⎞ b2 ⎞ ⎤ ⎢⎜⎝1 + 2 ⎟⎠ − ν ⎜⎝1 − 2 ⎟⎠ ⎥ a a ⎦ ⎣ σ
p
\
⎡(a 2 + b2 ) − ν (a 2 − b 2 ) ⎤ = y ⎦ NE E (b2 − a 2 ) ⎣
or
350 × 106 p ⎡125 + ( 0.25 × 75)⎤⎦ = (100 − 25) ⎣ 1.5
\
p
=
350 × 106 × 75 = 121.7 × 103 kPa 1.5 × 143.75
(iv) Octahedral shear stress theory t oct
=
1/ 2 1⎡ 2 2 2⎤ σ + σ r + (σ r − σ θ ) ⎦ 3⎣ θ
at r = a
1/ 2 1⎡ 2 ⎤ = ⎣ 2 (σ r − σ θ ) + 2σ r σ θ ⎦ 3
1/ 2
2 ⎧⎪⎡ 2 2 ⎤ 2 2 ⎫ p (b + a ) 2 2 (b + a ) ⎪ = ⎨ − p − ⎥ −p 2 2 2 2 ⎬ 3 ⎢ ( ) ( )⎪ b a b a − − ⎦ ⎪⎩⎣ ⎭
=
2 σ y 3 N 1/ 2
\
or \
(b 2 + a 2 ) ⎤ 2 ⎡ 4b 4 p ⎢ 2 − 2 2 2 2 ⎥ 3 ( ) ( )⎦ b a b a − − ⎣
(
40000 125 p − 5625 75
1/ 2
)
2 σ y = 3 N
= 350 1.5
p = 100 ¥ 103 kPa
(v) Energy of distortion theory This will give a value identical to that obtained based on octahedral shear stress theory, i.e. p = 100 ¥ 10 3 kPa.
Example 8.3 A pipe made of steel has a tensile elastic limit s y = 275 MPa and E = 207 ¥ 106 kPa. If the pipe has an internal radius a = 5 cm and is subjected to an internal pressure p = 70 ¥ 103 kPa, determine the proper thickness for the pipe wall 4 according to the major theories of failure. Use a factor of safety N = . 3
Axisymmetric Problems Solution
(i) Maximum principal stress theory Maximum principal stress s q at r = a =
= p
(b 2 + a 2 ) 2
2
(b − a )
=
70 × 106 ⎡⎣ (25 × 10− 4 ) + b 2 ⎤⎦
\
⎡b2 − (25 × 10−4 ) ⎤ ⎣ ⎦
σ y
N
1750 × 10−4 + 70b 2 = 825b 2 −
or
6
275 × 10 × 3 = 4 20625 × 10−4 4
136.25b2 = 6906.25 ¥ 10 –4 b = 7.12 ¥ 10 –2 m = 7.12 cm \ Wall thickness t = 2.12 cm \ (ii) Maximum shear stress theory or
t max
=
=
1 (σ − σ r ) at r = a 2 θ pb
2
2
2
(b − a )
=
σ y
N
70 × 106 b 2
\
[b 2 − (25 × 10
−4
)]
= 3 × 275 × 106 8
70b2 = 103.13 b2 – 2578.1 ¥ 10 –4 b = 8.82 ¥ 10 –2 m = 8.82 cm \ Wall thickness t = 3.82 cm \ (iii) Maximum strain theory (with n = 0.25) or
e max
=
1 (σ − νσ r ) at r = a E θ σ
p
⎡(a 2 + b 2 ) − ν (a 2 − b 2 ) ⎤ = y = 2 2 ⎦ NE E (b − a ) ⎣ 70 × 106
⎡ (0.75 × 25 × 10 −4 ) [b − (25 × 10 )] ⎣
\
−4
2
+ (1.25 × b 2 ) ⎦⎤ = 3 × 275 × 106 4
1312.5 ¥ 10 –4 + 87.5b2 = 206.25b2 – 5156.25 ¥ 10 –4
or \ \
b = 7.38 ¥ 10 –2 m = 7.38 cm
Wall thickness t = 2.38 cm
(iv) Maximum distortion energy theory From Eq. (4.12) with s 1
=
s q ,
s 2
=
0,
s 3
=
s r = – p
277
278 Advanced Mechanics of Solids
U *
=
1 [σ θ2 + σ r2 + (σ r − σ θ ) 2 ] 12G
=
(1 + ν ) (2σθ2 + 2 σ r2 − 2σ r σθ ) 6 E
=
(1 + ν ) 2 1 + ν σ y 2 (σ θ + σ r − σ θ σ r ) = 3 E E N 2
2
2
2
σθ
\
+ σ r2 − σ θ σ r =
σ y
2
N
2 2 2 2 2 2 ⎤ σ y ⎡ ( ) ( ) + − b a b a 2 i.e. p ⎢ 2 +1+ 2 = 2 2 2 2 ⎥ (b + a ) ⎦ N ⎣ (b − a ) ⎛ σ y ⎞ Putting ⎜ = f y and simplifying one gets ⎟ p N ⎝ ⎠
(3 − f y2 ) b 4 + 2a 2 f y2 b 2 + (1 − f y2 ) a 4 = 0 b2
\
−2a 2 f y2 ±
⎡ 4a 4 f y4 − 4a 4 (1 − f y2 ) (3 − f y2 ) ⎤ ⎣ ⎦ 2 2 (3 − f y )
2 2 a ⎡ − f y ±
(4 f y4 − 3) ⎤
=
=
With
⎣⎢
⎦⎥
2 (3 − f y2 )
a = 5 ¥ 10 –2 fy =
275 × 106 × 3 6
70 × 10 × 4
= 2.946
b 2 = (63 or 13.4) 10 –4 Wall thickness t = 2.9 cm \
or b = 7.9 ¥ 10 –2 m = 7.9 cm
Case (b) Plane Strain When the cylinder is fairly long, sections that are far from the ends can be considered to be in a state of plane strain and we can assume that s z does not vary along the z-axis. As in the case of plane stress, the equation is d ( r σ r ) − σ θ = 0 dr
From Hooke's law e r = e q = e z
=
1 [σ − ν (σθ + σ z ) ] E r 1 [σ − ν (σ r + σ z ) ] E θ 1 [σ − ν (σ r + σθ ) ] E z
Axisymmetric Problems
Since
e z
=
279
0 in this case, one has from the last equation s z = n (s r + s q ) e r =
e q =
Solving for
1 + ν E
1 + ν E
[(1 − ν ) σ r − νσ θ ]
[ (1 − ν ) σ
θ
(8.18)
− νσ r ]
s q and s r s q =
s r =
On substituting for
E
(1 − 2ν) (1 + ν ) E
(1 − 2ν ) (1 + ν )
e r and e q from
[νε r + (1 − ν ) ε ]
(8.19a)
[(1 − ν ) ε r + νε ]
(8.19b)
θ
θ
Eqs (8.3) and (8.4), the above equations become
s q =
⎡ ν (1 − 2ν ) (1 + ν ) ⎢⎣
s r =
dur ur ⎤ ⎡ 1 ν) ν − + ( (1 − 2ν ) (1 + ν ) ⎢⎣ dr r ⎥ ⎦
E
dur dr
ur ⎤
+ (1 − ν )
r
(8.20)
⎥ ⎦
E
(8.21)
Substituting these in the equation of equilibrium, Eq. (8.5c)
⎡ ⎢(1 − ν ) r ⎣
d dr dur
or
dr d dr
i.e.
dur dr
2
+ r
d ur 2
−
dr
⎛ dur ⎜⎝ dr +
ur ⎞ r ⎟ ⎠
⎤ + ν u r ⎥ − ν ⎦
u r r
dur
− (1 − ν )
dr
u r r
=
0
=0
=0
This is the same as Eq. (8.7) for the plane stress case. The solution is the same as in Eq. (8.8). u r = C1 r +
C 2 r
where C 1 and C 2 are constants of integration. From Eqs (8.20) and (8.21) s q =
⎡ C + (1 − 2ν ) (1 − 2ν ) (1 + ν ) ⎢⎣ 1
s r =
C 2 ⎤ ⎡ C (1 2 ) ν − − 2 ⎥ (1 − 2ν ) (1 + ν ) ⎢⎣ 1 r ⎦
E
E
Once again, we observe that s r + axial stress from Eq. (8.18) is s z
=
−
C 2 ⎤ 2
r
⎥⎦
(8.22a)
(8.22b)
s q is
a constant independent of r . Further, the
2ν E C (1 − 2ν ) (1 + ν ) 1
(8.22c)
280 Advanced Mechanics of Solids
Applying the boundary conditions s r = – pa when r = a,
Solving,
and
C 1
C 2
s r = – pb
when r = b
E (1 − 2ν) (1 + ν )
C 2 ⎤ ⎡ C ν ) (1 2 − − = − pa ⎢ 1 2 ⎥ a ⎣ ⎦
E (1 − 2ν ) (1 + ν )
C 2 ⎤ ⎡ C ν ) − − = − pb (1 2 ⎢ 1 2 ⎥ b ⎦ ⎣
=
2 2 (1 − 2ν ) (1 + ν ) pbb − pa a 2 2 E a −b
=
2 2 1 + ν ( pb − pa ) a b 2 2 E a −b
Substituting these, the stress components become s r =
s q =
s z =
pa a 2 − pb b 2 2
b −a
2
pa a 2 − pb b 2 b −a 2
2ν
2
−
+
pa − pb a 2 b 2 2
b −a
2
2
r
pa − pb a 2 b 2 b −a 2
2
2
r
pb a 2 − pa b 2 b −a 2
2
(8.23)
(8.24)
(8.25)
It is observed that the values of s r and s q are identical to those of the plane stress case. But in the plane stress case, s z = 0, whereas in the plane strain case,s z has a constant value given by Eq. (8.25). 8.3
STRESSES IN COMPOSITE TUBES—SHRINK FITS
The problem which will be considered now, involves two cylinders made of two different materials and fitted one inside the other. Before assembling, the inner cylinder has an internal radius a and an external radius c. The internal radius of the outer cylinder is less than c by D, i.e. its internal radius is c – D. Its external radius is b. If the inner cylinder is cooled and the outer cylinder is heated, then the two cylinders can be assembled, one fitting inside the other. When the cylinders come to room temperature, a shrink fit is obtained. The problem lies in determining the contact pressure pc between the two cylinders. The above construction is often used to obtain thick-walled vessels to withstand high pressures. For example, if we need a vessel to withstand a pressure of say 15000 atm, the yield point of the material must be at least 30000 kgf/cm2 (2940000 kPa). Since no such high-strength material exists, shrink-fitted composite tubes are designed. The contact pressure pc acting on the outer surface of the inner cylinder reduces its outer radius by u1. On the other hand, the same contact pressure increases the inner radius of the outer cylinder by u 2 . The sum of these two quantities,
Axisymmetric Problems
281
i.e. (– u 1 + u2) must be equal to D, the difference in the radii of the cylinders. To determine u1 and u2, we make use of Eq. (8.10), assuming a plane stress case. For the inner tube
or
u1
=
1 − ν1 E1
u1
=
−
2 ⎛ ⎞ 1 + ν 1 a 2 c 2 c c+ ⎜ − pc 2 2⎟ E1 c c −a ⎠ ⎝
pc ⎞ ⎛ − ⎜⎝ 2 2⎟ c −a ⎠
cpc
⎡(1 − ν1 ) c 2 + (1 + ν 1 ) a 2 ⎤ 2 2 ⎦ E1 (c − a ) ⎣
For the outer tube
or
u2
=
1 − ν2 − E2
u2
=
−
2 1 + ν 2 c 2 b 2 ⎛ pc ⎞ ⎛ ⎞ c c+ ⎜⎝ pc 2 2⎟ E2 c ⎜⎝ b 2 − c 2 ⎟⎠ b −c ⎠
cpc
⎡(1 − ν 2 ) c 2 + (1 + ν 2 )b 2 ⎤ ⎦ E2 (b 2 − c 2 ) ⎣
In calculating u2, we have neglected D since it is very small as compared to c. Noting that u1 is negative and u2 is positive, we should have – u1 + u2 = D cpc
⎡ (1 − ν1 )c 2 + (1 + ν 1 ) a 2 ⎤ 2 2 ⎦ E1 ( c − a ) ⎣
i.e.
+
cpc
⎡(1 − ν2 )c2 + (1 + ν 2 )b2 ⎤ = ∆ 2 2 ⎦ E2 (b − c ) ⎣
(8.26a)
Regrouping, the contact pressure pc is given by p c
=
∆ /c 2 2 ⎤ 1 1 ⎡c + a − ν1 ⎥ + E1 ⎢ c 2 − a 2 ⎣ ⎦ E 2
2
If the two cylinders are made of the same material, then E 1 Equation (8.26) will then reduce to p c
=
2 2 2 2 E ∆ (c − a ) (b − c )
2c
3
2
2
(b − a )
(8.26b)
⎡b + c ⎤ + ν 2 ⎥ ⎢ 2 2 b c − ⎣ ⎦ 2
=
E 2 and n 1
= n 2.
(8.27)
It is important to note that in Eqs (8.26) and (8.27), D is the difference in radii between the inner cylinder and the outer jacket. Because of shrink fitting, therefore, the inner cylinder is under external pressure pc. The stress distribution in the assembled cylinders is shown in Fig. 8.6. If the composite cylinder made up of the same material is now subjected to an internal pressure p, then the two parts will act as a single unit and the additional stresses induced in the composite can be determined from Eqs (8.13) and (8.14). At the inner surface of the inner cylinder, the internal pressure p causes a tensile tangential stress s q , Eq. (8.14), but, the contact pressure pc causes at the same points a compressive tangential stress, Eq (8.17). Hence, a composite cylinder can
282 Advanced Mechanics of Solids
s q
a
b
c
–
s r
D
c
p c
Streses in composite tubes
Fig. 8.6
support greater internal pressure than an ordinary one. However, at the inner points of the jacket or the outer cylinder, the internal pressure p and the contact pressure pc both will induce tensile tangential (i.e. circumferential) stresses s q . For design purposes, one can choose the shrink-fit allowance D such that the strengths of the two cylinders are equal. To determine this value of D, one can proceed as follows. Let a and c be the radii of the inner cylinder, and c and b the radii of the jacket (see Fig. 8.7) c is the common radius of the two cylinders at the contact surface when the composite cylinder is s q a + experiencing an internal pressure p and the shrinkA B s r – fit pressure pc. If the strengths of the two cylinc b ders are the same, then according to the maximum shear stress theory, ( s 1 – s 3) at point A of the inner cylinder should be equal to ( s 1 – s 3) at point B of the outer cylinder. s 1 and s 3 are the Fig. 8.7 Equal strength maximum and minimum normal stresses, which are composite tube respectively equal to s q and s r . At point A, due to internal pressure p, from Eqs (8.13) and (8.14), (s q –
s r ) A =
=
p
b2 + a2 b
2 p
2
−
a
b b
2
−
2
(− p )
2
−
a
2
Because of shrink-fitting pressure pc, at the same point, from Eqs (8.16) and (8.17), (s q –
s r ) A =
−
2 pc
c c
2
2
−
a
2
Hence, the resultant value of (s q – (s q –
s r ) A =
2 p
b b
2
s r )
at A is
2
−
a
2
−
2 pc
c c
2
2
−
a
2
(8.28)
At point B of the outer cylinder, since the composite involves the same material, due to the pressure p, from Eqs (8.13) and (8.14), and observing that r = c in these equations,
Axisymmetric Problems
(s q –
s r ) B =
283
⎡ a 2 (c 2 + b 2 ) a 2 ( c 2 − b 2 ) ⎤ p ⎢ 2 2 − 2 2 2 2 ⎥ c b a c b a − − ( ) ( )⎦ ⎣ 2 2
=
2 p
a b
c (b − a ) 2
2
2
At the same point B, due to the contact pressure pc, from Eqs (8.13) and (8.14), with internal radius equal to c and external radius b, (s q –
s r ) B =
=
⎡ c2 + b2 c 2 − b2 ⎤ pc ⎢ 2 − 2 2 2⎥ b c b c − − ⎣ ⎦ b
2 pc
b −c 2
The resultant value of (s q – (s q –
s r ) B =
2 p
2 2
s r )
at B is therefore
a 2b 2
+ 2 pc
c (b − a ) 2
2
2
b2
(8.29)
( b2 − c2 )
For equal strength, equating Eqs (8.28) and (8.29) 2 p
or
b
2
2 2 (b − a )
− 2 pc
c
2
2 2 (c − a )
= 2p
2 2
a b
c (b − a ) 2
2
2
+ 2 pc
b
2
2 2 (b − c )
⎡ b2 ⎤ ⎡ b2 ⎤ c2 a 2b 2 + = − pc ⎢ 2 p ⎢ 2 2 2 2 ⎥ 2 2 2 2 ⎥ − − − − b c c a b a c b a ( ) ( ) ( )⎦ ⎣ ⎦ ⎣
(8.30)
The shrink-fitting pressure pc is related to the negative allowance D through Eq. (8.27) and it is this value of D that is required now for equal strength. Hence, substituting for pc from Eq. (8.27), Eq. (8.30) becomes 2 2 2 2 ∆ E (c − a ) (b − c ) 3 2 2 2c (b − a )
or
2 2 2 ⎡ b2 ⎤ ⎡ b2 ⎤ c a b + = − p ⎢ 2 ⎢ 2 2 2 2 ⎥ 2 2 2 2 ⎥ b c c a b a c b a − − − − ( ) ( ) ( ) ( )⎦ ⎣ ⎦ ⎣
2 2 2 2 4 2 2 2 ∆ E (2b c − b a − c ) = pb (c − a ) 2 2 2 c (b − a ) 2c 3 (b 2 − a 2 )
or
D
=
b 2 c (c 2 − a 2 ) 2 p E b 2 (c 2 − a 2 ) − c 2 (b 2 − c 2 )
(8.31a)
Also, from Eq. (8.30), p c
=
The value of (s q – s q
–
p
s r )
s r =
b 2 (c 2 − a 2 ) 2 (b 2 − c 2 ) c (b − a ) ⎡ b (c − a ) + c (b − c ) ⎤ 2
2
2
⎣
2
2
2
2
2
2
⎦
either at A or at B, from Eqs (8.28) and (8.31), is
2b 2 p 2 (b − a 2 )
2 2 2 2 ⎡ ⎤ ( c − a ) (b − c ) − 1 ⎢ 2 2 2 2 2 2 ⎥ b c a c b c − + − ( ) ( )⎦ ⎣
(8.31b)
284 Advanced Mechanics of Solids
or
s q
–
s r =
⎡ ⎢ 2b 2 1 ⎢1 − p 2 2 2 b c (b − a 2 ) ⎢ + 2 ⎢ 2 2 b −c c − a2 ⎣
⎤ ⎥ ⎥ ⎥ ⎥ ⎦
(8.32)
Therefore, for composites made of the same material, in order to have equal strength according to the shear stress theory, the shrink-fit allowance D that is necessary is given by Eq. (8.31a), and this depends on the internal pressure p. Further, D depends upon the difference between the external radius of the inner cylinder and the internal radius of the jacket. In other words, this depends on c(+) and c(–). With a, b and p fixed, one can determine the optimum value of c for minimum (s q – s r ) at A and B. From Eq. (8.32), the minimum value of ( s q – s r ) is obtained when the denominator of the second expression within the square brackets is a maximum, i.e. when D is a maximum, where b
D =
2
+
c
2
b −c c −a Differentiating with respect to c and equating the differential to zero,
dD dc
=
2
2
2cb2 2
2 2
(b − c )
2
+
2
2c (c 2 − a 2 ) − 2c3 2
2 2
(c − a )
=0
Simplifying, one gets c = ab The corresponding value of ( s q –
(s q –
s r )min
=
=
or
(s q –
s r )min
=
s r ),
from Eq. (8.32), is
⎡ ⎤ ⎢ ⎥ 2b2 1 ⎢1 − ⎥ p 2 2 2 ⎥ (b − a ) ⎢ b ab + ⎢ b(b − a ) a (b − a ) ⎥⎦ ⎣ p p
⎡ (b − a ) ⎤ 1− ⎥ 2 2 ⎢ 2b ⎦ (b − a ) ⎣ 2b2 b
(b − a)
(8.33)
Also, the optimum value of D is from Eq. (8.31a), D op t
=
p 1 pc = ab E E
(8.34)
Example 8.4 Determine the diameters 2c and 2b and the negative allowance D for a two-layer barrel of inner diameter 2a = 100 mm. The maximum pressure the barrel is to withstand is p max = 2000 kgf/cm2 (196000 kPa). The material is steel with E = 2(10) 6 kgf/cm2 (196 ¥ 105 kPa); s yp in tension or compression is 6000 kgf/cm 2 (588 ¥ 10 3 kPa). The factor of safety is 2.
Axisymmetric Problems Solution
285
From Eq. (8.33),
Therefore,
6000 = 2000 b b−a 2 b 3 a =
Since c = ab, c = 3 a. The numerical values are therefore, 2 a = 100 mm, 2b = 300 mm, 2c = 173 mm. With c = ab, the value of D is, from Eq. (8.34), p 2000 ab = (50 × 150) = 0.866 mm D = 6 E 2 × 10
Example 8.5 A steel shaft of 10 cm diameter is shrunk inside a bronze cylinder of 25 cm outer diameter. The shrink allowance is 1 part per 1000 (i.e. 0.005 cm difference between the radii). Find the circumferential stresses in the bronze cylinder at the inside and outer radii and the stress in the shaft. E steel = 2.18 ¥ 106 kgf/cm2 (214 ¥ 106 kPa) E bron ze = 1.09 ¥ 106 kgf/cm2 (107 ¥ 106 kPa) n = 0.3
and Solution
for both metals.
In Eq. (8.26), a = 0,
c = 5,
b = 12.5,
D = 0.005,
n 1 = n 2 = 0.3
Substituting in Eq. (8.26a), 5 pc 2.18 × 106 × 25
(0.7 × 25) +
5 pc 1.09 × 106 × (156.25 − 25)
× ( 0.7 × 25 + 1.3 × 156.25) = 0.005 pc = 610 kgf/cm2 (59780 kPa) or, For the bronze tube, the circumferential stress is, from Eq. (8.14), s q =
610 × 25
⎛ 156.25 ⎞ 1+ ⎟ (156.25 − 25) ⎜⎝ r 2 ⎠
When r = 5 cm and r = 12.5 cm s q = 842.4
kgf/cm2 (82555 kPa)
kgf/cm2 (22775 kPa) The shaft experiences equal s r and s q at every point, from Eqs (8.16) and (8.17). Hence, 2 s r = s q = –610 kgf/cm (59780 kPa) s q = 232.4
Example 8.6 A compound cylinder made of copper inner tube of radii a = 10 cm and c = 20 cm is snug fitted (D = 0) inside a steel jacket of external radius b = 40 cm. If the compound cylinder is subjected to an internal pressure p = 1500 kgf/cm2 (147009 kPa), determine the contact pressure pc and the values of s r and s q at the inner and external points of the inner cylinder and of the jacket. Use the following data: E st = 2 ¥ 106 kgf/cm2 (196 ¥ 106 kPa), E cu = 1 ¥ 106 kgf/cm2 (98 ¥ 106 kPa),
n st = 0.3,
n cu = 0.34
286
Advanced Mechanics of Solids
Solution
Since the initial shrink-fit allowance D is zero, the initial contact pressure is zero. When the compound cylinder is subjected to an internal pressure p, the increase in the external radius of the copper cylinder under p and contact pressure pc should be equal to the increase in the internal radius of the jacket under the contact pressure pc, i.e.
⎡ (ur ) p + (ur ) p at r = c ⎤ = ⎡(ur ) p at r = c ⎤ ⎣ ⎦ cu ⎣ ⎦ st c
c
For copper cylinder, from Eq. (8.10), (ur ) p
1 − ν cu pa 2 c 1 + ν cu a 2 c 2 p + Ecu (c 2 − a 2 ) Ecu c (c 2 − a 2 )
=
pc 1 − ν cu pc c 3 1 + ν cu a 2 c 2 − − Ecu (c 2 − a 2 ) Ecu c (c 2 − a 2 )
(u ) pc =
(u r ) total
2 pa 2 c
=
2
2
Ecu (c − a )
pc c
− Ecu
⎡ c 2 (1 − ν cu ) + a 2 (1 + ν cu ) ⎤ ⎦ (c 2 − a 2 ) ⎣
For steel jacket, from Eq. (8.10),
(u ) pc
pc 1 − ν st pc c 3 1 + ν st c 2b 2 + Est (b 2 − c 2 ) Est c (b 2 − c 2 )
=
=
pc c
⎡ c 2 (1 − ν st ) + b 2 (1 + ν st ) ⎤ ⎦ Est (b 2 − c 2 ) ⎣
Equating the (ur )s 2 pa 2 c Ecu (c − a ) 2
2
=
pc c
− Ecu
⎡c 2 (1 − ν cu ) + a 2 (1 + ν cu ) ⎤ ⎦ (c 2 − a 2 ) ⎣
pc c
⎡c 2 (1 − ν st ) + b2 (1 + ν st ) ⎤ 2 2 ⎦ Est (b − c ) ⎣
⎡ (c 2 + a 2 ) − νcu ( c 2 − a 2 ) (b 2 + c 2 ) + ν st ( b 2 − c 2 ) ⎤ or pc ⎢ + ⎥ 2 2 2 2 − − ( ) ( ) E c a E b c cu st ⎣ ⎦ = p
2
Ecu
2a (c 2 − a 2 )
With p = 1500 kgf/cm2, a = 10, c = 20, b = 40, n st = 0.3, n cu = 0.34,
⎡ 500 − 300 × 0.34 2000 + 1200 × 0.3 ⎤ 3000 × 100 pc ⎢ + = 6 6 ⎥ 6 × × × 300 10 2 1200 10 ⎣ ⎦ 300 × 10 \
p c = 433 kgf/cm2 (42453 kPa)
Axisymmetric Problems
287
Now, pc will act as an external pressure on the copper tube and as an internal pressure on the steel jacket. For copper tube, from Eqs (8.11) and (8.12), (i) Inner surface:, 2 s r at r a is –1500 kgf/cm and, s q at r a is =
=
(1500 × 100) − (433 × 400) 100 × 400 (1500 − 433) + × (400 − 100) 100 (400 − 100)
=
1357 kgf/cm2 (compressive) (ii) Outer surface: 2 s r at r c is – 433 kgf/cm and, s q at r c is =
=
=
(1500 × 100) − (433 × 400) 100 × 400 (1500 − 433) + (400 − 100) 400 (400 − 100)
=
279 kgf/cm2 For steel jacket, from Eqs (8.11) and (8.12), (i) Inner surface: 2 s r at r c is – 433 kgf/cm and, s q at r c is =
=
=
(433 × 400) 400 × 1600 433 = × (1600 − 400) 400 (1600 − 400)
=
722 kgf/cm2 (ii) Outer surface: s r at r b is zero and, s q at r b is =
=
=
(433 × 400) 400 × 1600 433 + × (1600 − 400) 1600 (1600 − 400)
=
289 kgf/cm2
=
8.4
SPHERE WITH PURELY RADIAL DISPLACEMENTS
Consider a uniform sphere or spherical shell subjected to radial forces only, such as internal or external pressures. The sphere or the spherical shell will then undergo radial displacements only. Consider a particle situated at radius r before deformation. After deformation, the spherical surface of radius r becomes a surface of radius ( r + u r ) and the particle undergoes a displacement ur . Similarly, another particle at distance ( r + Dr ) along the same radial line will undergo a
∂u ⎛ ⎞ displacement ⎜ ur + r ∆r ⎟ . ∂r ⎠ ⎝ Hence, the radial strain is e r =
∂ ur ∂ r
288 Advanced Mechanics of Solids
Before deformation, the circumference of any great circle on the surface of radius r is 2 p r . After deformation, the radius becomes ( r + u r ) and the circumference of the great circle is 2 p (r + ur ) Hence, the circumferential strain is
e f =
2π
( r + u ) − 2π r r
ur
=
2π r r This is the strain in every direction perpendicular to the radius r . Because of complete symmetry, we can choose a frame of reference, as shown in Fig. 8.8. s f s +
q
r
s
f (e f )
Ds r
r
r
r
(c)
(e ) r
s f
s q
2 q
q (e q )
s f
s + r
s
Ds r
r
(a)
2 q s q
s f
(b)
Fig. 8.8
Sphere with purely radial displacement
Thus, the three extensional strains along the three axes are
∂ ur u u (8.35) , εθ = r , ε φ = r ∂ r r r Because of symmetry, there are no shear stresses and shear strains. Let g r be the body force per unit volume in the radial direction. The stress equations of equilibrium can also be derived easily. Consider a spherical element of thickness D r at distance r , subtending a small angle 2 q at the centre. Because of spherical symmetry, s q = s f . For equilibrium in the radial direction, e r =
– sr (2q r )(2q r ) + (s r + Ds r ) (r + Dr )2q (r + Dr )2q
(
−2 r +
∆r
2
)
2θ∆rσ φ sin θ − 2
Putting s q = s f and r
2
∂σ r ∂ r
+
Ds r =
2rσ r
−
σ r r
2rσ φ
(
r +
∆r ,
+r
2
)
∆r
2
2θ
∆rσ θ
sin θ
2 2
+ γ r 4θ r ∆r =
0
the above equation reduces in the limit to
γ r = 0
Since r is the only independent variable, the above equation can be rewritten as d dr
( r 2σ r ) − 2rσ φ
2
+ r γ r =
0
(8.36)
Axisymmetric Problems
289
If body force is ignored,
1 d 2 2 ( σ ) σ φ = r r 2 dr r r From Hooke’s law e r =
or
dur dr
(8.37)
1 ⎡ σ r − ν (σ θ + σ φ ) ⎤⎦ ⎣ E
=
1 (σ − 2νσ φ ) E r
=
1 ⎡ σ − ν (σ φ + σ r ) ⎤⎦ E ⎣ φ
(8.38)
and e f
or ur 1 ⎡ (1 − ν ) σ φ − νσ r ⎤⎦ ⎣ E r Equations (8.37)–(8.39) can be solved. From Eq. (8.39) =
(8.39)
1 ⎡ (1 − ν ) r σ φ − ν r σ r ⎤⎦ E ⎣ Differentiating with respect to r u r
dur
=
=
dr
1 E
d (r σ φ ) ⎡ d ( r σ r ) ⎤ (1 ν ) ν − − ⎢ dr dr ⎥⎦ ⎣
Subtracting the above equation from Eq. (8.38) 0
=
− (1 − ν )
d (r σ φ ) dr
+ν
d (r σ r ) + σ r − 2νσ φ dr
Substituting for s f from Eq. (8.36)
If r 2 s r
=
2 d 2 ( r 2σ r ) d (rσ r ) 1 ν d ( r σ r ) (1 − ν ) −ν − σ r + =0 2 dr r dr 2 dr y,
(8.40)
1 d d ⎛ y ⎞ 1 dy ( rσ r ) = = − y dr dr ⎜⎝ r ⎟⎠ r dr r 2 Therefore, Eq. (8.40) becomes 2 d y ν dy ν y y 1 ν dy 1 − ν ) − + − + =0 ( 2 2 2 r dr r dr 2 dr r r
or
2
d y 2
−2
y
y
Ar 2 +
2
(8.41)
=0
dr r This is a homogeneous linear equation with the solution =
B r
290
Advanced Mechanics of Solids
where A and B are constants. Hence, s r = A +
B
(8.42)
3
r
And from Eq. (8.37) s f =
(
)
B B 1 d Ar 2 + = A− 3 r 2r dr 2r
(8.43)
The constants A and B are determined from the boundary conditions.
Problem of Thick Hollow Sphere Consider a spherical body formed by the boundaries of two spherical surfaces of radii a and b respectively. Let the hollow sphere be subjected to an internal pressure pa and an external pressure pb . The boundary conditions are therefore s r = – pa when r = a, and s r = – pb when r = b From Eq. (8.42) – p a
=
A +
B a
and − pb = A +
3
B b
3
Solving, a 3b3
B = 3 , ( pb − pa ) 3 3 3 b −a b −a Thus, the general expressions for s r and s f are A
=
s r = s f =
−
b3 pb − a 3 pa
3 3 ⎡ 3 ⎤ a b 3 b p a p p p − + + − ( ) b a b a ⎥ 3 3 ⎢ 3 b −a ⎣ r ⎦
1
(8.44)
3 3 ⎡ 3 ⎤ a b 3 b p a p p p = 3 − + − − ( ) b a b a ⎥ 3 ⎢ 2r 3 b −a ⎣ ⎦
1
σ θ
(8.45)
If the sphere is subjected to internal pressure only, pb = 0, and s r =
3 ⎛ b ⎞ 1− 3⎟ pa 3 3 ⎜ ⎝ b −a r ⎠
(8.46)
s f =
3 ⎛ b ⎞ 1+ 3⎟ σ θ = pa 3 3 ⎜ ⎝ 2r ⎠ b −a
(8.47)
a
3
a
3
The above two equations can also be written as s r = pa s f =
σ θ
⎛ 1 1⎞ − 3 3 ⎜ 3 3⎟ r ⎠ 1 − (a / b ) ⎝ b a
= pa
3
⎛ 1 1 ⎞ + 3 3 ⎜ 3 3⎟ 1 − (a / b ) ⎝ b 2r ⎠ a
3
In the case of a cavity inside an infinite or a large medium, b Æ equations reduce to s r = − pa
a
•
and the above
3
3
r
(8.48)
Axisymmetric Problems
s f = σ θ = + pa
291
a3
(8.49) 2r 3 The above equations can also be used to calculate stresses in a body of any shape with a spherical hole under an internal pressure pa, provided the outer surface of the body is free from pressure and provided that every point of this outer surface is at a distance greater than four or five times the diameter of the hole from its centre.
Example 8.7 Calculate the thickness of the shell of a bomb calorimeter of spherical form of 10 cm inside diameter if the working stress is s kgf/cm2 (98 s kPa) and the internal pressure is s /2 kgf /cm 2 (49 s kPa). Solution
From equations (8.46) and (8.47), the maximum tensile stress is due to s f , which occurs at r = a. Hence,
⎛ b3 ⎞ σ 53 s f = 1+ ⎟ 2 b3 − 53 ⎜⎝ 2 × 53 ⎠ Equating this to the working stress s ⎛ b3 ⎞ ⎜1 + ⎟=2 b3 − 53 ⎝ 2 × 53 ⎠ 53
b ª 6.3 cm Hence, the thickness of the shell is 1.3 cm. \
Example 8.8 Express the stress equation of equilibrium, i.e. 1 d 2 2 ( σ ) σ φ = 0 − r r 2 dr r r given by Eq. (8.37), in terms of the displacement component u r , using Hooke's law and strain–displacement relations.
Solution
We have e r = 1 ⎡(σ r − 2νσ φ ) ⎤ ⎦ E ⎣
e f = 1 ⎡(1 − ν ) σ φ − νσ r ⎤ ⎦ E ⎣ Solving for s r and s f , s r =
E ⎡ε r (1 − ν ) + 2νεφ ⎤ ⎦ (1 + ν )(1 − 2ν ) ⎣
(8.50)
s f =
E (νε r + ε φ ) (1 + ν )(1 − 2ν )
(8.51)
Using the strain–displacement relations e r =
dur dr
and ε φ =
ur r
292
Advanced Mechanics of Solids
and substituting in the equilibrium equation, we get E (1 − ν ) (1 + ν )(1 − 2ν )
ur ⎞ E (1 − ν ) d ⎛ dur 2 + =0 (1 + ν )(1 − 2ν ) dr ⎜⎝ dr r ⎟⎠
or
8.5
⎡ d 2 ur 2 dur 2 ⎤ ⎢ 2 + r dr − 2 ur ⎥ = 0 r ⎢⎣ dr ⎥⎦ (8.52)
STRESSES DUE TO GRAVITATION
When body forces are operative, the stress equation of equilibrium is, from Eq. (8.36),
1 d 2 2 σ σ + γ r = 0 r − (8.53) r 2 r φ r dr where g r is the body force per unit volume. The problem of a sphere strained by the mutual gravitation of its parts will now be considered . It is known from the theory of attractions
(
)
r a where a is the radius of the sphere, r is the mass density, r is the radius of any point from the centre and g is the acceleration due to gravity. Expressing the equations of equilibrium in terms of displacement ur [Eq. (8.52)], we have g r
=
− ρ g
ur ⎞ E (1 − ν ) d ⎛ dur r 2 + − ρ g = 0 ⎜ ⎟ r ⎠ a (1 + ν ) (1 − 2ν ) dr ⎝ dr
(8.54)
The complementary solution is u r
=
u r
=
Cr +
C 1
r 2 and the particular solution is 1 (1 + ν ) (1 − 2ν ) ρ gr 3 10 E (1 − ν ) a
Hence, the complete solution is u r
=
Cr +
1 (1 + ν ) (1 − 2ν ) 3 ρ + gr r 2 10 E (1 − ν )a
C 1
For a solid sphere, C 1 should be equal to zero as otherwise the displacement will become infinite at r 0. The remaining constant is determined from the boundary condition s r 0 at r a. From the general solution =
=
dur dr
=
C+
=
3(1 + ν ) (1 − 2ν ) ρ gr 2 , 10 E (1 − ν )a
and from Eq. (8.50) s r
=
E ⎡ε (1 − ν ) + 2νεφ ⎤ ⎦ (1 + ν ) (1 − 2ν ) ⎣ r
ur r
=
C+
(1 + ν ) (1 − 2ν ) ρ gr 2 10 E (1 − ν ) a
Axisymmetric Problems
=
ur ⎤ ⎡ dur E (1 ν ) 2 ν − + r ⎥⎦ (1 + ν ) (1 − 2ν ) ⎢⎣ dr
=
E (1 + ν ) (1 − 2ν )
3(1 + ν ) (1 − 2ν ) ⎡ 2 ν ρ (1 ) − + + 2ν C C gr ⎢⎣ 10 Ea
+ =
E (1 + ν ) (1 − 2ν )
=
Hence,
=
s r
=
−
ν (1 + ν ) (1 − 2ν ) ⎤ ρ gr 2 ⎥ 5 E (1 − ν )a ⎦
(3 − ν ) (1 + ν ) (1 − 2ν ) ⎡ 2⎤ (1 + ν ) + ρ C gr ⎢⎣ ⎥⎦ 10 Ea(1 − ν )
From the boundary condition s r 0 C
293
at
r
=
a,
(3 − ν ) (1 − 2ν ) ρ ga 10 E (1 − ν )
(3 − ν ) 2 ρ g ( a − r 2 ) − 1 10 (1 − ν ) a
(8.55)
2 2 1 (3 − ν )a − (1 + 3ν ) r ρ g σ θ = − a 10 (1 − ν )
(8.56)
and from Eq. (8.53)
s f
=
It will be observed that both stress components s r and s q are compressive at every point. At the centre ( r 0), they are equal and have a magnitude =
s r
=
σ φ = σ θ =
Further, dur dr
=
C+
1 3 − ν ρ ga (compressive) 10 1 − ν
1 3(1 + ν ) (1 − 2ν ) ρ gr 2 10 E (1 − ν )a
(3 − ν ) (1 − 2ν ) 1 3(1 + ν ) (1 − 2ν ) ρ ga + ρ gr 2 =− 1 E (1 − ν ) E (1 − ν )a 10 10
(1 + ν ) (1 − 2ν ) ρ g ⎡ 2 (3 − ν ) 2 ⎤ 3r − = 1 a ⎥ 10 (1 + ν ) E (1 − ν ) a ⎢⎣ ⎦ The above value is zero when r 2
=
(3 − ν ) 2 a 3(1 + ν )
Hence, if n is positive (which is true for all known materials), there is a definite surface outside which the radial strain is an extension. In other words, for 1/ 2
⎡ (1 + ν ) ⎤ r > a⎢ ⎣ 3(1 + ν ) ⎥⎦
the radial strain e r is positive though the radial stress s r is compressive everywhere. This result is due, of course, to the ‘Poisson effect’ of the large circumferential stress, i.e. hoop stress, which is compressive.
294
8.6
Advanced Mechanics of Solids
ROTATING DISKS OF UNIFORM THICKNESS
We shall now consider the stress distribution in rotating circular disks which are thin. We assume that over the thickness, the radial and circumferential stresses do not vary and that the stress s z in the axial direction is zero. The equation of equilibrium given by Eq. (8.5b) can be used, provided we add the inertia force term rw 2r , i.e. in the general equation of equilibrium [Eq. (8.2)] we put the body force term equal to the inertia term rw 2r , where w is the angular velocity of the rotating disk and r is the density of the disk material. The z -axis is the axis of rotation. Then;
∂σ r ∂ r
σ r − σ θ r
+
+
ρω 2 r = 0
(8.57a)
d rσ r ) − σ θ + ρω 2 r 2 =0 ( dr The strain components are, as before,
or
dur and dr From Hooke’s law, with s z 0,
e r
ε θ
=
=
(8.57b)
ur r
(8.58)
=
e r
=
1 (σ − νσ θ ) E r
e q
=
1 (σ E θ
e r
=
νσ r )
−
From Eq. (8.58) d (r ε θ ) dr
From Hooke’s law 1 (σ − νσ θ ) E r Let
e r
=
d (r ε θ ) dr
=
=
1 d (rσ θ − ν r σ r ) E dr
r s r y
(8.59) (8.60a)
=
Then, from Eq. (8.57b) σ θ
=
dy 2 2 + ρω r dr
(8.60b)
Substituting these in Eq. (8.59) and rearranging r
2
2
d y 2
r
+
dy 2 3 − y + (3 + ν ) ρω r dr
=
dr The solution of the above differential equation is y
=
Cr
s r
=
C
+
C1
1 (3 + ν ) ρω 2 r3 − r 8
0
(8.61)
(8.62)
From Eq. (8.60) +
C1
1 2 r
−
(3 + ν ) ρω 2 r 3 8
(8.63)
Axisymmetric Problems s q
C
=
(1 + 3ν ) ρω 2 r 3 − C1 12 −
(8.64)
8
r
295
The integration constants are determined from boundary conditions. Solid Disk
For a solid disk, we must take C 1 0, since otherwise the stresses s r and s q become infinite at the centre. The constant C is determined from the condition at the periphery (r b) of the disk. If there are no forces applied there, then, =
=
(s r )r
=
b
=
C
−
3 + ν ρω 2b 2 = 0 8
Hence, 3 + ν ρω 2b 2 8 and the stress components become C
s r
=
=
3 + ν ρω 2 (b 2 − 8
2
r
(8.65a)
)
3+ν 1 + 3ν (8.65b) ρω 2b 2 − ρω 2r 2 8 8 These stresses attain their maximum values at the centre of the disk, where s q
=
s r
=
σ θ =
3 + ν ρω 2b 2 8
(8.66)
Circular Disk with a Hole of Radius
a
If there are no forces applied at the boundaries a and b, then (s r )r a 0, (s r )r b 0 from which we find that =
=
=
=
3+ν ρω 2 b 2 + a 2 , 8 Substituting these in Eqs (8.63) and (8.64) C
s r
s q
The radial stress
(
=
=
=
)
⎛ 3 + ν ρω 2 ⎜ b2 + a 2 − 8 ⎝ ⎛ 3+ν ρω 2 ⎜ b 2 + a 2 + 8 ⎝
s r reaches
C1
2 2
a b 2
r
2 2
a b 2
r
its maximum at r
=
=−
3 + ν ρω 2a 2b 2 8
⎞ − r 2 ⎟ ⎠ −
1 + 3ν 2 ⎞ r 3 + ν ⎟⎠
ab
(8.68)
where
3 + ν ρω 2 (b − a ) 2 8 The maximum circumferential stress is at the inner boundary, where 3+ν 1 − ν 2 ⎞ ⎛ (s q )max ρω 2 ⎜ b 2 + a 4 3 + ν ⎟⎠ ⎝ It can be seen that (s q )max is greater than (s r )max. (s r )max
(8.67)
=
(8.69)
=
(8.70)
296
Advanced Mechanics of Solids
When the radius a of the hole approaches zero, the maximum circumferential stress approaches a value twice as great as that for a solid disk [Eq. (8.66)]. In other words, by making a small circular hole at the centre of a solid rotating disk, we double the maximum stress. The displacement ur for all the cases considered above can be calculated from Eq. (8.58), i.e. u r = r εθ = r (σ θ − νσ r ) E
(8.71)
Example 8.9 A flat steel disk of 75 cm outside diameter with a 15 cm diameter hole is shrunk around a solid steel shaft. The shrink-fit allowance is 1 part in 1000 (i.e. an allowance of 0.0075 cm in radius). E = 2.18 ¥ 106 kgf/cm 2 (214 ¥ 106 kPa). (i) What are the stresses due to shrink-fit? (ii) At what rpm will the shrink-fit loosen up as a result of rotation? (iii ) What is the circumferential stress in the disk when spinning at the above speed? Assume that the same equations as for the disk are applicable to the solid rotating shaft also. Solution
(i) To calculate the shrink-fit pressure, we have from Eq. (8.27) pc
=
2.18 × 106 × 0.0075 2 × 7.53
×
(7.5 2 − 0) (37.5 2 − 7.5 2 ) (37.52 − 0)
pc = 1044 kgf/cm2 (102312 kPa) or The tangential stress at the hole will be the largest stress in the system and from Eq. (8.24) s q
=
1044 × 7.52 ⎛
37.52 ⎞ 1+ ⎟ 2 2 ⎜ ⎝ (37.5 − 7.5 ) 7.52 ⎠
= 1131
kgf/cm2 (110838 kPa)
(ii) When the shrink-fit loosens up as a result of rotation, there will be no radial pressure on any boundary. When the shaft and the disk are rotating, the radial displacement of the disk at the hole will be greater than the radial displacement of the shaft at its boundary. The difference between these two radial displacements should equal D = 0.0075 cm at 7.5 cm radius. From Eqs (8.71), (8.67) and (8.68) u disk = r (σ θ − νσ r ) E
3+ν = r ρω 2 E 8
2 2 ⎡ 2 1 + 3ν 2 a b 2 b a r + + − ⎢ 2 + 3 ν r ⎣
2 2 ⎛ 2 ⎞⎤ a b 2 −ν ⎜ b + a − 2 − r 2 ⎟ ⎥ ⎝ ⎠⎦ r
Axisymmetric Problems
297
1 + ν b 2 a 2 1 + ν 2 ⎞ 2 r ( 3 + ν ) (1 − ν ) 2 ⎛ 2 r − = ρω ⎜ b + a + 1 − ν r 2 3 + ν ⎟⎠ ⎝ 8 E
3.3 × 0.7 7.5 2 × ρω 8 2.18 × 106
=
¥
2 2 ⎛ 1.3 37.5 × 7.5 1.3 2 2 2⎞ + + × − × 37.5 7.5 7.5 ⎜⎝ ⎟⎠ 0.7 3.3 7.52
4052 ¥ 10 –6 rw 2 From equations (8.71), (8.65a) and (8.65b) =
ushaft
=
r (σ − νσ r ) E θ
=
1 − ν ρω 2 r ⎡⎣ (3 + ν ) b 2 − (1 + ν ) r 2 ⎤⎦ 8 E
=
0.7 2 2 2 × × − × ρω 7.5 (3.3 7.5 1.3 7.5 ) 6 8 × 2.18 × 10
=
34 ¥ 106 rw 2
Therefore, (4052 – 34) ¥ 10 –6 rw 2 = 0.0075 or
w 2
=
=
0.0075 × 106 ×
1 × 981 4018 0.0081
226066 (rad/s)2
Therefore, w 475 rad/s or 4536 rpm (iii) The stresses in the disk can be calculated from Eq. (8.68) =
s r
=
(
3.3 1.9 ρω 2 37.52 + 7.52 + 37.52 − × 7.52 8 3.3
)
1170 rw 2
=
= 1170 × 0.0081 × 226066
981 2 = 2184 kgf/cm (214024 kPa)
Example 8.10 A flat steel turbine disk of 75 cm outside diameter and 15 cm inside diameter rotates at 3000 rpm, at which speed the blades and shrouding cause a tensile rim loading of 44kgf/cm2 (4312 kPa). The maximum stress at this speed is to be 1164 kgf/cm2 (114072 kPa). Find the maximum shrinkage allowance on the diameter when the disk and the shift are rotating. Solution
Let c be the radius of the shaft and b that of the disk. From Eq. (8.70), the maximum circumferential stress due to rotation alone is (s q )1
=
3+ν ρω 2 4
⎛ 2 1 − ν 2 ⎞ ⎜⎝ b + 3 + ν c ⎟⎠
298
Advanced Mechanics of Solids
(
= 3.3 ρω 2 37.52 + 0.7 × 7.52 4
3.3
)
1170 rw 2
=
Owing to shrinkage pressure pc, and the tensile rim loading pb, from Eq. (8.12) (s q )2
=
2 2 pbb ⎛ b ⎞ 1+ 2 ⎟ + 2 2 2 2 ⎜ 2 b −c ⎝ c ⎠ b −c
pc c
2
7.52 = pc 37.52 − 7.52
⎛ 44 × 37.52 37.52 ⎞ +2 ⎜⎝1 + 2 ⎟ ⎠ 7.5 37.52 − 7.52
1.08 pc + 91.7
=
Hence, the combined stress at 7.5 cm radius is
s q = 1170rw 2 + 1.08 pc + 91.7 This should be equal to 1164 kgf/cm 2. Hence, 1.08 pc 1164 – 1170rw 2 – 91.7 =
2 = 1164 − 1170 × (100π ) × 0.0081 − 91.7
981
1164 – 953.5 – 91.7 118.8
=
=
Hence, pc = 110 kgf/cm2
The corresponding shrink-fit allowance is obtained from Eq. (8.27), i.e. 110
=
E ∆ × 3 2 × 7.5
(
7.52 37.52 − 7.52
)
37.52
0.064 E D
=
D
or
8.7
=
110 × 10 −6 = 0.0008 cm 0.064 × 2.18
DISKS OF VARIABLE THICKNESS
Assuming that the stresses do not vary over the thickness of the disk, the method of analysis developed in the previous section for thin disks of constant thickness can be extended also to disks of variable thickness. Let h be the thickness of the disk, varying with radius r . The equation of equilibrium can be obtained by referring to Fig. 8.9. For equilibrium in the radial direction
⎛ ⎞ ∂ ( hσ r ) ∆r ∆θ h + ∆h ω 2 r + ∆ r h r r r r σ θ ρ + ∆ + ∆ ∆ + + ( ) ⎜⎝ r ⎟⎠ ∂ r 2 2 2
(
) (
− hσ r r ∆θ − 2σ θ h + ∆h ∆r sin ∆θ = 0
(
2
)
2
) (
)
299
Axisymmetric Problems
Dr
r
Dq
Fig 8.9 Rotating disk of variable thickness Simplifying and going to the limit r
d (hσ r ) + hσ r + ρω 2 r 2 h − σ θ h = 0 dr
d ( r hσ r ) − σ θ h + ρω 2 r 2 h = 0 dr y = rhs r Putting
or
hs q =
dy + hρω 2 r 2 dr
(8.72) (8.73a) (8.73b)
The strain components remain as in Eq. (8.58), i.e. dur u and ε θ = r dr r Hence, e r = d (r ε θ ) dr From Hooke’s law and Eq. (8.59)
e r =
1 1 d (σ r − νσ θ ) = ( rσ θ − ν r σ r ) E E dr Substituting for s r and s q from Eqs (8.73a) and (8.73b) r
2
d2y 2
dr
+r
dy ⎛ dy ⎞ − y + (3 + ν ) ρω 2 hr 3 − r dh ⎜ r − ν y ⎟ = 0 dr h dr ⎝ dr ⎠
(8.74)
In the particular case where the thickness varies according to the equation y
=
Cr n
(8.75)
in which C is a constant and n any number, Eq. (8.74) can easily be integrated. The general solution has the form y
in which
=
m=
mr n + 2 + Ar a + Br b
(3 + ν ) ρω 2 c − (ν n + 3n + 8)
and a and b are the roots of the quadratic equation x2 – nx
+
rn – 1 = 0
A and B are constants which are determined from the boundary conditions.
300 Advanced Mechanics of Solids
Example 8.11 Determine the shape for a disk with uniform stress, i.e. s r = s q . Solution
From Hooke's law e r
if
s r
s q then e r
=
=
e r
Since
e r
=
d ε r dr
i.e.
dur dr
=
and ε θ =
ur r
d d r εθ ) = ( r ε ) ( dr dr r e q , the above equation gives e r
we get
1 1 σ r − νσ θ ) , εθ = (σ θ − νσ r ) ( E E e q . From strain–displacement relations
=
e r
=
0
=
=
constant
Hence, from Hooke's law, s r and s q are not only equal but also constant throughout the disk. Let s r s q s . Equilibrium Eq. (8.73) gives =
hs
=
=
or
1 dh h dr
=
=
d (rhσ ) + hρω 2 r 2 dr dh hσ + rσ 2 + ρω 2 hr dr − 1 ρω 2 r σ
which upon integration gives log h
or
8.8
h
=
=
ρω 2 2 − r + C 1 2σ 2 ⎡ ρω 2 2 ⎤ ⎛ 2 r ⎞ r + C1 ⎥ = C exp ⎜ − ρω exp ⎢ − 2σ 2σ ⎠⎟ ⎝ ⎣ ⎦
ROTATING SHAFTS AND CYLINDERS
In Sec. 8.5 and 8.6, we assumed that the disk was thin and that it was in a state of plane stress with s z 0. It is also possible to treat the problem as a plane strain problem as in the case of a uniformly rotating long circular shaft or a cylinder. Let the z-axis be the axis of rotation. The equation of equilibrium is the same as in Eq. (8.57): =
d (rσ r ) − σ θ + ρω 2 r 2 =0 dr The strain components are, as before, e r
=
dur , dr
εθ =
ur , r
(8.76)
ε z =
∂ u z =0 ∂ z
(8.77)
Axisymmetric Problems
301
From Hooke's law
Since
e z
=
e r
=
e q
=
1 [σ − ν (σθ + σ z )] E r 1 [σ − ν (σ r + σ z )] E θ
1 [σ − ν (σ r + σθ ) ] E z 0 (plane strain), e z
=
s z
=
ν (σ r + σ θ )
and hence, substituting in equations for e r
=
e r and e q
1 + ν [ (1 − ν ) σ r − νσ θ ] E
1 + ν [ (1 − ν ) σ θ − νσ r ] E From strain–displacement relations given in Eq. (8.77) e q
=
d (r ε ) dr θ and using the above expressions for e r and e q , we get e r
=
(1 − ν ) σ r − νσ θ = With r s r y, Eq. (8.76) gives for =
s q
Substituting for (1 − ν )
or
r
2
(8.78)
s q
dy + ρω 2 r 2 dr
s r and s q in
Eq. (8.78)
y dy ⎡ ⎤ ⎛ dy ⎞ −ν − νρω 2 r 2 = d ⎢(1 − ν ) ⎜ r + ρω 2 r 3 ⎟ − ν y ⎥ r dr dr ⎣ ⎝ dr ⎠ ⎦
d2y 2
=
d [ (1 − ν )r σθ − ν r σ r ] dr
+r
3 − 2ν dy ρω 2 r 3 = 0 − y+ 1 − ν dr
dr The solution for this differential equation is
Hence,
and
1 (3 − 2ν ) ρω 2 r 3 − r 8(1 − ν )
y
=
Cr + C1
s r
=
C + C1
(3 − 2ν ) 1 − ρω 2 r 2 2 8(1 − ν ) r
(8.79a)
s q
=
C − C1
(1 + 2ν ) 1 − ρω 2 r 2 2 8(1 − ν ) r
(8.79b)
s z
=
ν ⎢ 2C −
⎡ ⎣
1
2(1 − ν )
⎤ ⎦
ρω 2 r 2 ⎥
(8.79c)
302 Advanced Mechanics of Solids
(i) For a hollow shaft or a long cylinder, inner and outer radii. From these
where
C
=
K
=
K (a
2
s r
=
0 at r
=
a and r
=
b which are the
1 = − Ka 2b 2 + b2 ) and C
(3 − 2ν ) 2 ρω 8 (1 − ν )
Hence, from Eqs (8.79a–c) s r
s q assumes
=
s q
=
s z
=
(3 − 2ν ) 8 (1 − ν )
⎡ 2 2 a b + ( )− ⎢ ⎣
(3 − 2ν ) 8 (1 − ν )
⎡ 2 2 ⎢ (a + b ) + ⎣
2 2
a b 2
r
2 2
a b 2
r
⎤ − r 2 ⎥ ρω 2 ⎦ −
1 + 2ν 2 ⎤ 2 r ⎥ ρω 3 − 2ν ⎦
⎡ ( a 2 + b2 ) (3 − 2ν ) − 2r 2 ⎤ ρω 2 ⎦ 4 (1 − ν ) ⎣ ν
(8.80)
(8.81) (8.82)
a maximum value at r a and its value is =
(s q )max
=
(3 − 2ν ) 8 (1 − ν )
1 + 2ν 2 ⎞ ⎛ 2 2 2 2 + − b a a ⎟ ρω ⎜⎝ 3 − 2ν ⎠
If a2/b2 is very small, we find that (s q )max ≈
(3 − 2ν ) 2 2 b ρω 4 (1 − ν )
(8.83)
(ii) For a long solid shaft, the constant C 1 must be equal to zero, since otherwise the stresses would become infinite at r 0. Using the other boundary condition that s r 0 when r b, the radius of the shaft, we find that =
=
=
C
=
(3 − 2ν ) 2 2 ρω b 8 (1 − ν )
Hence, the stresses are
The value of
s r
=
s q
=
s z
=
s q
(3 − 2ν ) 2 (b − r 2 ) ρω 2 8 (1 − ν )
(8.84)
(3 − 2ν ) 8 (1 − ν )
(8.85)
⎡b2 (3 − 2ν ) − 2r 2 ⎤ ρω 2 ⎦ 4 (1 − ν ) ⎣ ν
at r
(s q )max
=
⎛ 2 1 + 2ν 2 ⎞ 2 ⎜⎝ b − 3 − 2ν r ⎟⎠ ρω
=
(8.86)
0 is
(3 − 2ν ) 2 2 b ρω 8 (1 − ν )
(8.87)
Comparing Eq. (8.87) with Eq. (8.83), we find that by drilling a small hole along the axis in a solid shaft, the maximum circumferential stress is doubled in its magnitude.
Axisymmetric Problems
303
Example 8.12 A solid steel propeller shaft, 60 cm in diameter, is rotating at a speed of 300 rpm. If the shaft is constrained at its ends so that it cannot expand or contract longitudinally, calculate the total longitudinal thrust over a cross-section due to rotational stresses. Poisson’s ratio may be taken as 0.3. The weight of steel may be taken as 0.0081 kgf/cm3 (0.07938 N/cm3). Solution
The total axial force is b
F z
=
∫ σ z 2π r dr 0
and from Eq. (8.86), substituting for F z
=
s z,
ν ⎡b2 (3 − 2ν) π b 2 − π b4 ⎤ ρω 2 ⎦ 4(1 − ν ) ⎣
= ν π b4 ρω 2
2 Substituting the numerical values 2
0.3 × π × 304 × 0.0081 × 3002 × π 42 F z 2 981 60 3120 kgf (31576 N) Tensile force =
=
8.9
SUMMARY OF RESULTS FOR USE IN PROBLEMS
(i) For a tube of internal radius a and external radius b subjected to an internal pressure pa and an external pressure pb, the radial and circumferential stresses are given by (according to plane stress theory) 2
s r
=
pa a − pbb b −a 2
=
−
2
2
s q
2
pa a − pbb
a 2b2 pa − pb r
2
b2 − a 2
2
2 2
+ a b2 r
b −a 2
2
pa − pb b2 − a 2
0 The stress s r < 0 for all values of pa and pb, whereas s q can be greater or less than zero depending on the values of pa and pb. s q is greater than zero if s z
=
pb ⎛ b 2 ⎞ + 1 p a > ⎟⎠ 2 ⎜⎝ a 2
The maximum and minimum stresses are (s r )max
=
(s r )min
=
(s q )max
=
σ r (at r = b) = − pb σ r (at r = a ) = − pa
σ θ (at r = a) =
pa (a 2 + b 2 ) − 2 pbb 2 2
b −a
2
304 Advanced Mechanics of Solids
(s q )min = σ θ (at r = b) =
If then,
pa
=
2 pa a 2 − pb ( a 2 + b 2 ) 2
b −a
2
⎛ b2 ⎞ 1 pb ⎜ 2 + 1⎟ 2 ⎝a ⎠
(s q )min = 0 (ur )r = a
=
2 ⎤ ⎛ b2 + a 2 ⎞ a ⎡ b + ν ⎟ − 2 pb 2 ⎢ p 2⎥ E ⎢ a ⎜⎝ b2 − a 2 b − a ⎥⎦ ⎠ ⎣
=
⎛b + a ⎞⎤ b ⎡ a2 2 − − ν p p ⎢ b⎜ 2 ⎟⎠ ⎥ 2 E ⎢ a b2 − a 2 − b a ⎝ ⎥⎦ ⎣ 2
(ur )r = b
2
(ii) Built-up cylinders: When the cylinders are of equal length, the contact pressure pc due to difference D between the outer radius of the inner tube and the inner radius of the outer tube is given by pc
=
∆ /c 2 2 ⎡ 1 ⎛ c2 + a2 ⎞ ⎞⎤ 1 ⎛b + c − ν + − ν ⎢ E ⎜ 2 1⎟ 2⎟ ⎥ 2 ⎠ E 2 ⎜⎝ b2 − c2 ⎠ ⎥⎦ ⎢⎣ 1 ⎝ c − a
where E 1, n 1, a and c refer to the inner tube’s modulus, Poisson’s ratio, inner radius and outer radius respectively. E 2,n 2, c and b are the corresponding values for the outer tube. If E 1 = E 2 and n 1 = n 2, then pc
=
2 2 2 2 ∆ E (b − c )(c − a ) 2 2 2c 3 b −a
(iii) For a sphere subjected to an internal pressure pa and an external pressure p b, the radial and circumferential stresses are given by s r =
3 3 ⎡ 3 ⎤ a b 3 b p a p p p − + + − − ( ) b a b a ⎥ 3 3 ⎢ 3 b −a ⎣ r ⎦
s q =
3 3 ⎡ 3 ⎤ a b 3 σ φ = 3 ( ) b p a p p p − + − − − b a b a ⎥ 3 ⎢ b −a ⎣ 2r 3 ⎦
1
1
(iv) For a thin solid disk of radius b rotating with an angular velocity stresses are given by s r =
3 + ν ρω 2 (b2 − r 2 ) 8
3+ν 1 + 3ν ρω 2b 2 − ρω 2 r 2 8 8 These stresses attain their maximum values at the centre r = 0, where s q =
s q =
σ r =
3 + ν ρω 2b 2 8
w ,
the
Axisymmetric Problems
305
The radial outward displacement at r = b is 1 − ν ρω 2b3 4 E (v) For a thin disk with a hole of radius a, rotating with an angular velocity stresses are (ur )r = b
and
=
s r =
3 + ν 2 ρω 8
2 2 ⎛ 2 a b 2 2⎞ b a r + − − ⎜⎝ ⎟⎠ 2 r
s q =
3+ν ρω 2 8
2 2 ⎛ 2 1 + 3ν 2 ⎞ a b 2 b a r + + − ⎜⎝ 2 3 + ν ⎟⎠ r
(
(s r )max
=
σ r at r =
(s q )max
=
σ θ (at r = a) =
)
ab =
w ,
the
3 + ν ρω 2 (b − a ) 2 8
3+ν ρω 2 4
⎛ 2 1 + ν 2 ⎞ ⎜⎝ b + 3 + ν a ⎟⎠
(s q )max > (s r )max
The radial displacements are (ur )r = a
=
3+ν 1 − ν 2 ⎞ ⎛ ρω 2 a ⎜ b 2 + a 4 E 3 + ν ⎟⎠ ⎝
(ur )r = b
=
3+ν 1 − ν 2 ⎞ ⎛ ρω 2b ⎜ a 2 + b 4 E 3 + ν ⎟⎠ ⎝
8.1 A thick-walled tube has an internal radius of 4 cm and an external radius of 8 cm. It is subjected to an external pressure of 1000 kPa (10.24 kgf/cm 2). If E = 1.2 ¥ 10 8 kPa (1.23 ¥ 10 6 kgf/cm 2) and n = 0.24, determine the internal pressure according to Mohr’s theory of failure, which says that
(σ )max − n (σ )min ≤ σ tenslie strength where n is the ratio of s -tensile strength to s -compressive strength. For the present problem, assume s -tensile strength = 30000 kPa (307.2 kgf/cm2) and 2 s -compressive strength = 120000 kPa (1228.8 kgf/cm ). [ Ans. p = 17000 kPa (174 kgf/cm2)] 8.2 In the above problem, determine the changes in the radii.
⎡ Ans. ⎢⎣
⎤ Dr 2 = 0.007 mm ⎥ ⎦ Dr 1 = 0.01 mm
8.3 In Example 8.1, if one uses the energy of distortion theory, what will be the external radius of the cylinder? The rest of the data remain the same. [ Ans. = 6.05 cm] 8.4 A thick-walled tube with an internal radius of 10 cm is subjected to an internal pressure of 2000 kgf/cm2 (196000 kPa). E = 2 ¥ 10 6 kgf/cm 2
306
Advanced Mechanics of Solids
(196 ¥ 106 kPa) and n = 0.3. Determine the value of the external radius if the maximum shear stress developed is limited to 3000 kgf/cm2 (294 ¥ 106 kPa). Calculate the change in the internal radius due to the pressure. ⎡ Ans. r 2 = 17.3 cm ⎤ ⎢⎣ Dr 1 = 0.023 cm ⎥ ⎦ 8.5 A thick-walled tube is subjected to an external pressure p2. Its internal and external radii are 10 cm and 15 cm respectively, n = 0.3 and E = 200000 MPa (2041 ¥ 10 3 kgf/cm2). If the maximum shear stress is limited to 200000 kPa (2041 kgf/cm2), determine the value of p2 and also the change in the external radius.
⎡ Ans. p 2 = 111 MPa (1133 kgf/cm2) ⎤ ⎢⎣ ⎥⎦ Dr 2 = – 0.19 mm 8.6 Determine the pressure p0 between the concrete tube and the perfectly rigid core. Assume E c = 2 ¥ 106 kgf/cm2, r c = 0.16. Take r 1/r 2 = 0.5 (Fig. 8.10). [ Ans. p0 = 17.4 kgf/cm2]
r 1 r 2
p
= 12 kgf/cm 2 (1126 kPa)
Fig. 8.10 Problem 8.6 8.7 Determine the dimensions of a two-piece composite tube of optimum dimensions if the internal pressure is 2000 kgf /cm2 (196000 kPa), external pressure p2 = 0, internal radius r 1 = 8 cm and E = 2 ¥ 10 6 kgf/cm 2 (196 ¥ 10 6 kPa). The maximum shear stress is to be limited to 1500 kgf/cm 2 (147 ¥ 10 6 kPa). Check the strength according to the maximum shear theory. ⎡ Ans. r 2 ª 14 cm; r 3 = 24 cm ⎤ ⎥ ⎢ D = 0.014 cm ⎥ ⎢ pc = 500 kgf/cm2 ⎥ ⎢ (49030 kPa) ⎦ ⎣ 8.8 Determine the radial and circumferential stresses due to the internal pressure p = 2000 kgf/cm2 (196,000 kPa) in a composite tube consisting of an inner copper tube of radii 10 cm and 20 cm and an outer steel tube of external radius 40 cm.n st = 0.3, n cu = 0.34, E st = 2 ¥ 106 kgf/cm2 (196 ¥106 kPa) and E cu = 10 6 kgf/cm2 (98 ¥ 10 6 kPa). Calculate the stresses at the inner and outer radius points of each tube. Determine the contact pressure also.
Axisymmetric Problems
307
⎡ Ans. For inner tube: ⎤ ⎢ ⎥ 2 s r = –2000 kgf/cm (–196000 kPa) ⎢ ⎥ 2 s r = –577 kgf/cm (–56546 kPa) ⎢ ⎥ 2 ⎢ ⎥ s t = 1800 kgf/cm (176400 kPa) 2 ⎢ ⎥ s t = 371 kgf/cm (36358 kPa) ⎢ ⎥ For outer tube: ⎢ ⎥ 2 –577 kgf/cm (–56546 kPa) = s ⎢ ⎥ r ⎢ ⎥ s r = 0 2 ⎢ ⎥ s t = 962 kgf/cm (94276 kPa) ⎢ ⎥ 2 s t = 385 kgf /cm (37730 kPa) ⎢ ⎥ ⎣ ⎦ p c = 577 kgf/cm2 (56546) 8.9 In problem 8.7, if the inner tube is made of steel (radii 10 cm and 20 cm) and the outer tube is of copper (outer radius 40 cm), determine the circumferential and radial stresses at the inner and outer radii points of each tube. ⎡ Ans. For inner tube: ⎤ 2 ⎢ s r = –2000 kgf/cm (–196000 kPa) ⎥ ⎢ ⎥ 2 s r = –248 kgf/cm (–24304 kPa) ⎢ ⎥ 2 s t = 2672 kgf/cm (262032 kPa) ⎢ ⎥ 2 ⎢ ⎥ s t = 920 kgf/cm (90221 kPa) ⎢ ⎥ For outer tube: ⎢ ⎥ 2 s r = –248 kgf/cm (–24304 kPa) ⎢ ⎥ 0 = s ⎢ ⎥ r 2 ⎢ ⎥ s t = 413 kgf/cm (40474 kPa) 2 ⎢ ⎥ s t = 165 kgf/cm (16170 kPa) ⎢ ⎥ p c = 248 kgf/cm2 (24304 kPa) ⎣ ⎦ 8.10 A composite tube is made of an inner copper tube of radii 10 cm and 20 cm and an outer steel tube of external radius 40 cm. If the temperature of the assembly is raised by 100°C, determine the radial and tangential stresses at the inner and outer radius points of each tube. a cu = 16.5 ¥ 10 – 6; a st = 12.5 ¥ 10 – 6; E st , n st , E cu and n cu are as in Problem 8.
⎡ Ans. For inner tube: ⎢ s r = 0 ⎢ 2 s r = –173 kgf/cm (–16954 kPa) ⎢ 2 s t = – 461 kgf/cm (– 45080 kPa) ⎢ 2 ⎢ s = –288 kgf/cm (–28243 kPa) ⎢ For outer tube: ⎢ 2 s r = –173 kgf/cm (–16954 kPa) ⎢ s r = 0 ⎢ 2 ⎢ s t = 288 kgf/cm (28243 kPa) 2 ⎢⎣ s t = 115 kgf/cm (11270 kPa)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
8.11 Determine for the composite three-piece tube (Fig. 8.11): (a) Stresses due to the heavy-force fits with interferences of D1 = 0.06 mm and D2 = 0.12 mm in diameters (b) Stresses due to the internal pressure p = 2400 kgf/cm2 r 1 = 80 mm, r 2 = 100 mm, r 3 = 140 mm, r 4 = 200 mm, E = 2.2 ¥ 106 kgf/cm2.
308 Advanced Mechanics of Solids
r 4 r 3
r 1 r 2 p
Fig 8.11 Problem 8.11 8.12 The radial displacement at the outside of a thick cylinder subjected to an internal pressure pa is
( 2 p r r ) E ( r − r ) 2 a b a
2 b
2 a
By Maxwell's reciprocal theorem, find the inward radial displacement at the inside of a thick cylinder subjected to external pressure.
⎡ 2 pb ab 2 ⎤ ⎢ Ans. ur = 2 2 ⎥ E b a − ( )⎦ ⎣ 8.13 A thin spherical shell of thickness h and radius R is subjected to an internal pressure p. Determine the mean radial stress, the circumferential stress and the radial displacement.
⎡ Ans. u pR 2 (1 − ν )/(2Eh) r ⎢ ⎢ pR s q σ φ = ⎢ 2h ⎢ ⎢ 1 p (s r ) average ⎢⎣ 2 =
=
=
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
8.14 An infinite elastic medium with a spherical cavity of radius R is subjected to hydrostatic compression p at the outside. Determine the radial and circumferential stresses at point r . Show that the circumferential stress at the surface of the cavity exceeds the pressure at infinity. 2 ⎡ ⎛ R ⎞ − p ⎜1 − 3 ⎟ ⎢ Ans. s r ⎝ r ⎠ ⎢ ⎢ ⎡ R3 ⎤ ⎢ s q s f − p ⎢1 + 3 ⎥ ⎢ ⎣ 2r ⎦ ⎢ ⎢ 3 ⎢ (s q ) at cavity − p 2 ⎣ =
=
=
=
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦