ElasticityPolars Complete
ElasticityPolars Complete
ElasticityPolars Complete
Problems in Polar
Coordinates
Many problems are most conveniently cast in terms of polar coordinates. To this end,
first the governing differential equations discussed in Chapter 1 are expressed in terms of
polar coordinates. Then a number of important problems involving polar coordinates are
solved.
53
54
Section 4.1
create cylinder by
revolving a surface
about the axis of
symmetry
axis of
symmetry
Some other axisymmetric geometries are illustrated Fig. 4.1.2; a frustum, a disk on a shaft
and a sphere.
Some features are not only axisymmetric – they can be represented by a plane, which is
similar to other planes right through the axis of symmetry. The hollow cylinder shown in
Fig. 4.1.3 is an example of this plane axisymmetry.
axisymmetric plane
representative of
feature
Axially non-symmetric geometries are ones which have a natural axis associated with
them, but which are not completely symmetric. Some examples of this type of feature,
the curved beam and the half-space, are shown in Fig. 4.1.4; the half-space extends to
“infinity” in the axial direction and in the radial direction “below” the surface – it can be
thought of as a solid half-cylinder of infinite radius. One can also have plane axially non-
symmetric features; in fact, both of these are examples of such features; a slice through
the objects perpendicular to the axis of symmetry will be representative of the whole
object.
θ
z
z = 0 plane
The displacement of a material point can be described by the three components in the
radial, tangential and axial directions. These are often denoted by
u ≡ u r , v ≡ uθ and w ≡ u z
respectively; they are shown in Fig. 4.1.6. Note that the displacement v is positive in the
positive θ direction, i.e. the direction of increasing θ .
The stresses acting on a small element of material in the cylindrical coordinate system are
as shown in Fig. 4.1.7 (the normal stresses on the left, the shear stresses on the right).
σ zz
σ zr
σθ z
σ θθ
σ rr
σ rr
σ θθ
σ rθ
σ zz
z strain at point o
ε rr = unit elongation of oA
C ε θθ = unit elongation of oB
θ
ε zz = unit elongation of oC
B ε rθ = ½ change in angle ∠AoB
o ε θ z = ½ change in angle ∠BoC
The stresses in any particular plane of an axisymmetric body can be described using the
two-dimensional polar coordinates (r ,θ ) shown in Fig. 4.1.9.
θ
r
There are three stress components acting in the plane z = 0 : the radial stress σ rr , the
circumferential (tangential) stress σ θθ and the shear stress σ rθ , as shown in Fig. 4.1.10.
Note the direction of the (positive) shear stress – it is conventional to take the z axis out of
the page and so the θ direction is counterclockwise. The three stress components which
do not act in this plane, but which act on this plane ( σ zz , σ θ z and σ zr ), may or may not
be zero, depending on the particular problem (see later).
σ θθ
σ rθ
σ rr
σ rr
σ rθ
σ θθ
∂σ rθ
σ rθ + Δθ ∂σ rθ
∂θ σ rθ + Δr
∂σ θθ ∂r
σ θθ + Δθ ∂σ
∂θ σ rr + rr Δr
∂r
r σ rr
σ rθ
σ θθ
Δθ
⎛ ∂σ rr ⎞
∑F = ⎜ σ rr + Δr ⎟(r + Δr )Δθ − σ rr rΔθ
∂r
r
⎝ ⎠
Δθ ⎛ ∂σ ⎞ Δθ
− sin ⎜ σ θθ + θθ Δθ ⎟Δr − sin (σ θθ )Δr (4.2.1)
2 ⎝ ∂θ ⎠ 2
Δθ ⎛ ∂σ rθ ⎞ Δθ
+ cos ⎜ σ rθ + Δθ ⎟Δr − cos (σ rθ )Δr ≡ 0
2 ⎝ ∂θ ⎠ 2
For a small element, sin θ ≈ θ , cos θ ≈ 1 and so, dividing through by ΔrΔθ ,
∂σ rr
(r + Δr ) + σ rr − σ θθ − Δθ ⎛⎜ ∂σ θθ ⎞⎟ + ∂σ rθ ≡ 0 (4.2.2)
∂r 2 ⎝ ∂θ ⎠ ∂θ
A similar calculation can be carried out for forces in the tangential direction {▲Problem
1}. In the limit as Δr , Δθ → 0 , one then has the two-dimensional equilibrium equations
in polar coordinates:
∂σ rr 1 ∂σ rθ 1
+ + (σ rr − σ θθ ) = 0
∂r r ∂θ r
Equilibrium Equations (4.2.3)
∂σ rθ 1 ∂σ θθ 2σ rθ
+ + =0
∂r r ∂θ r
∂u r
ε rr =
∂r
1 ∂uθ u r
ε θθ = + 2-D Strain-Displacement Expressions (4.2.4)
r ∂θ r
1 ⎛ 1 ∂u r ∂uθ uθ ⎞
ε rθ = ⎜ + − ⎟
2 ⎝ r ∂θ ∂r r ⎠
ε rr =
1
[σ rr − νσ θθ ], ε θθ = 1 [σ θθ − νσ rr ], ε rθ = 1 + ν σ rθ
E E E
ν
ε zz = − (σ rr + σ θθ )
E
Hooke’s Law (Plane Stress) (4.2.5a)
1 +ν
ε rr = [(1 − ν )σ rr − νσ θθ ], ε θθ = 1 + ν [− νσ rr + (1 − ν )σ θθ ], ε rθ = 1 + ν σ rθ
E E E
Hooke’s Law (Plane Strain) (4.2.5b)
1 ∂φ 1 ∂ 2φ ∂ 2φ ∂ ⎛ 1 ∂φ ⎞ 1 ∂φ 1 ∂ 2φ
σ rr = + 2 , σ θθ = , σ rθ = − ⎜ ⎟= − (4.2.6)
r ∂r r ∂θ 2 ∂r 2 ∂r ⎝ r ∂θ ⎠ r 2 ∂θ r ∂r∂θ
It can be verified that these equations automatically satisfy the equilibrium equations
4.2.3 {▲Problem 2}.
2
⎛ ∂2 1 ∂ 1 ∂2 ⎞
⎜⎜ 2 + + 2 ⎟⎟ φ = 0 (4.2.7)
⎝ ∂r r ∂r r ∂θ 2 ⎠
1 ∂ 2 ε rr ∂ 2 ε θθ 2 ∂ 2 ε rθ 1 ∂ε rr 2 ∂ε θθ 2 ∂ε
+ − − + − 2 rθ = 0 (4.2.8)
r ∂θ
2 2
∂r 2
r ∂r∂θ r ∂r r ∂r r ∂θ
4.2.5 Problems
1. Derive the equilibrium equation 4.2.3b
2. Verify that the stress function relations 4.2.6 satisfy the equilibrium equations 4.2.3.
3. Verify that the strains as given by 4.2.4 satisfy the compatibility relations 4.2.8.
To transform equations from Cartesian to polar coordinates, first note the relations
x = r cosθ , y = r sin θ
(4.2.9)
r = x 2 + y 2 , θ = arctan( y / x)
∂ ∂r ∂ ∂θ ∂ ∂ sin θ ∂
= + = cosθ −
∂x ∂x ∂r ∂x ∂θ ∂r r ∂θ
(4.2.10)
∂ ∂r ∂ ∂θ ∂ ∂ cosθ ∂
= + = sin θ +
∂y ∂y ∂r ∂y ∂θ ∂r r ∂θ
∂2 ⎛ ∂ sin θ ∂ ⎞⎛ ∂ sin θ ∂ ⎞
= ⎜ cos θ − ⎟⎜ cos θ − ⎟
∂x 2
⎝ ∂r r ∂θ ⎠⎝ ∂r r ∂θ ⎠
∂ ⎛ ∂ ⎞ ∂ ⎛ sin θ ∂ ⎞ sin θ ∂ ⎛ ∂ ⎞ sin θ ∂ ⎛ sin θ ∂ ⎞
= cos θ ⎜ cos θ ⎟ − cos θ ⎜ ⎟− ⎜ cos θ ⎟ + ⎜ ⎟
∂r ⎝ ∂r ⎠ ∂r ⎝ r ∂θ ⎠ r ∂θ ⎝ ∂r ⎠ r ∂θ ⎝ r ∂θ ⎠
∂2 ⎛1 ∂ 1 ∂2 ⎞ ⎛ 1 ∂ 1 ∂2 ⎞
= cos 2 θ 2 + sin 2 θ ⎜⎜ + 2 ⎟
2 ⎟
+ sin 2θ ⎜⎜ 2 − ⎟⎟
∂r ⎝ r ∂r r ∂θ ⎠ ⎝ r ∂θ r ∂r∂θ ⎠
(4.2.11)
Similarly,
∂2 ∂2 2 ⎛1 ∂ 1 ∂2 ⎞ ⎛ 1 ∂ 1 ∂2 ⎞
= sin 2
θ + cos θ ⎜ +
⎜ r ∂r r 2 ∂θ 2 ⎟⎟ − sin 2θ ⎜⎜ 2 − ⎟⎟
∂y 2 ∂r 2 ⎝ ⎠ ⎝ r ∂θ r ∂r∂θ ⎠
(4.2.12)
∂2 ⎛ ∂2 1 ∂ 1 ∂2 ⎞ ⎛ 1 ∂ 1 ∂2 ⎞
= − sin θ cos θ ⎜ − 2 +
⎜ + ⎟⎟ − cos 2θ ⎜⎜ 2 − ⎟⎟
∂x∂y ⎝ ∂r r ∂r r 2 ∂θ 2 ⎠ ⎝ r ∂θ r ∂r ∂θ ⎠
Equilibrium Equations
The Cartesian stress components can be expressed in terms of polar components using the
stress transformation formulae, Part I, Eqns. 3.4.7. Using a negative rotation (see Fig.
4.2.2), one has
Applying these and 4.2.10 to the 2D Cartesian equilibrium equations 3.1.3a-b lead to
⎡ ∂σ 1 ∂σ rθ 1 ⎤ ⎡ ∂σ 1 ∂σ θθ 2σ rθ ⎤
cos θ ⎢ rr + + (σ rr − σ θθ )⎥ − sin θ ⎢ rθ + + =0
⎣ ∂r r ∂θ r ⎦ ⎣ ∂r r ∂θ r ⎥⎦
(4.2.14)
⎡ ∂σ 1 ∂σ rθ 1 ⎤ ⎡ ∂σ 1 ∂σ θθ 2σ rθ ⎤
sin θ ⎢ rr + + (σ rr − σ θθ )⎥ + cos θ ⎢ rθ + + =0
⎣ ∂r r ∂θ r ⎦ ⎣ ∂r r ∂θ r ⎥⎦
y
θ
r
θ x
Noting that
u x = u r cosθ − uθ sin θ
, (4.2.15)
u y = u r sin θ + uθ cosθ
the strains in polar coordinates can be obtained directly from Eqns. 1.2.5:
∂u x
ε xx =
∂x
⎛ ∂ sin θ ∂ ⎞
= ⎜ cos θ − ⎟(u r cos θ − uθ sin θ ) (4.2.16)
⎝ ∂r r ∂θ ⎠
∂u ⎛ 1 ∂uθ u r ⎞ 1 ⎛ 1 ∂u r ∂uθ uθ ⎞
= cos 2 θ r + sin 2 θ ⎜ + ⎟ − sin 2θ ⎜ + − ⎟
∂r ⎝ r ∂θ r ⎠ 2 ⎝ r ∂θ ∂r r ⎠
One obtains similar expressions for the strains ε yy and ε xy . Substituting the results into
the strain transformation equations Part I, Eqns. 3.8.1,
The stresses in polar coordinates are related to the stresses in Cartesian coordinates
through the stress transformation equations (this time a positive rotation; compare with
Eqns. 4.2.13 and Fig. 4.2.2)
Using the Cartesian stress – stress function relations 3.2.1, one has
∂ 2φ ∂ 2φ ∂ 2φ
σ rr = cos 2
θ + sin 2
θ − sin 2θ (4.2.19)
∂y 2 ∂x 2 ∂x∂y
Beginning with the Cartesian relation 1.3.1, each term can be transformed using 4.2.11-12
and the strain transformation relations, for example
∂ 2 ε xx ⎛ ∂2 2 ⎛1 ∂ 1 ∂2 ⎞ ⎛ 1 ∂ 1 ∂2 ⎞⎞
= ⎜ cos 2
θ + sin θ ⎜ + ⎟ + sin 2θ ⎜⎜ 2 − ⎟⎟ ⎟ ×
⎜ ⎜ r ∂r r 2 ∂θ 2 ⎟ ⎟
∂x 2 ⎝ ∂r 2 ⎝ ⎠ ⎝ r ∂θ r ∂r∂θ ⎠ ⎠ (4.2.20)
(ε rr cos 2 θ + ε θθ sin 2 θ − ε rθ sin 2θ )
After some lengthy calculations, one arrives at 4.2.8.
It should be noted that many problems involve axisymmetric geometries but non-
axisymmetric loadings, and vice versa. These problems are not axisymmetric. An
example is shown in Fig. 4.3.3 (the problem involves a plane axisymmetric geometry).
1
the rotation induces a stress in the disk
2
the rest of the cylinder is coming out of, and into, the page
axisymmetric plane
representative of
feature
The important characteristic of these axisymmetric problems is that all quantities, be they
stress, displacement, strain, or anything else associated with the problem, must be
independent of the circumferential variable . As a consequence, any term in the
differential equations of §4.2 involving the derivatives / , 2 / 2 , etc. can be
immediately set to zero.
u r
rr
r
ur
(4.3.1)
r
1 u u
r
2 r r
Here, it will be assumed that the displacement u 0 . Cases where u 0 but where the
stresses and strains are still independent of are termed quasi-axisymmetric problems;
these will be examined in a later section. Then 4.3.1 reduces to
u r u
rr , r , r 0 (4.3.2)
r r
It follows from Hooke’s law that r 0 . The non-zero stresses are illustrated in Fig.
4.3.4.
rr
rr
1
rr rr rr
E
rr
E 1 2
or (4.3.3)
1
rr rr
E
E 1 2
with zz rr , zr z 0 and zz 0 .
E
In the plane strain case, the strains zz , z and zr are zero. This will occur, for
example, in a hollow cylinder under internal pressure, with the ends fixed between
immovable platens, Fig. 4.3.6.
rr
1
(1 ) rr rr
E
1 rr
E 1 1 2
or (4.3.4)
1
(1 ) rr
E
1
E 1 1 2 rr
with zz rr .
Shown in Fig. 4.3.7 are the stresses acting in the axisymmetric plane body (with zz zero
in the plane stress case).
zz
rr
rr
zz
rr 1
rr 0 , (4.3.5)
r r
Taking the plane stress case, substituting 4.3.2 into the second of 4.3.3 and then
substituting the result into 4.3.5 leads to (with a similar result for plane strain)
d 2 u 1 du 1
u0 (4.3.6)
dr 2 r dr r 2
1
u C1 r C 2 (4.3.7)
r
With the displacement known, the stresses and strains can be evaluated, and the full
solution is
1
u C1 r C 2
r
1 1
rr C1 C 2 2
, C1 C 2 2 (4.3.8)
r r
E E 1 E E 1
rr C1 C 2 2 , C1 C2 2
1 1 r 1 1 r
For problems involving stress boundary conditions, it is best to have simpler expressions
for the stress so, introducing new constants A EC 2 / 1 and C EC1 / 21 , the
solution can be re-written as
1 1
rr A
2
2C , A 2 2C
r r
rr
1 A 1 21 C
,
1 A 1 21 C
, zz
4C
2 2
(4.3.9)
E r E E r E E
u
1 A 1 21 C r
E r E
Plane stress axisymmetric solution
Similarly, the plane strain solution turns out to be again 4.3.8a-b only the stresses are now
{▲Problem 1}
1 1
1 2 C 2 2 C1 , 1 2 C 2 2 C1
E E
rr
1 1 2 r 1 1 2 r
(4.3.10)
Then, with A EC 2 / 1 and C EC1 / 21 1 2 , the solution can be written
as
1 1
rr A 2
2C , A 2 2C , zz 4C
r r
1 1 1 1
rr A 2 21 2 C , A 2 21 2 C (4.3.11)
E r E r
1 1
u A 21 2 Cr
E r
Plane strain axisymmetric solution
The solutions 4.3.9, 4.3.11 involve two constants. When there is a solid body with one
boundary, A must be zero in order to ensure finite-valued stresses and strains; C can be
determined from the boundary condition. When there are two boundaries, both A and C
are determined from the boundary conditions.
Consider the problem of Fig. 4.3.8. The two unknown constants A and C are obtained
from the boundary conditions
rr (a) p
(4.3.12)
rr (b) 0
which lead to
A A
rr (a) 2
2C p, rr (b) 2 2C 0 (4.3.13)
a b
so that
b2 / r 2 1 b2 / r 2 1
rr p 2 2 , p 2 2 , zz rr (4.3.14)
b / a 1 b / a 1
Cylinder under Internal Pressure
a r
b
The stresses through the thickness of the cylinder walls are shown in Fig. 4.3.9a. The
maximum principal stress is the stress and this attains a maximum at the inner face.
For this reason, internally pressurized vessels often fail there first, with microcracks
perpendicular to the inner edge been driven by the tangential stress, as illustrated in Fig.
4.3.9b.
Note that by setting b a t and taking the wall thickness to be very small, t , t 2 a ,
and letting a r , the solution 4.3.14 reduces to:
r r
rr p, p , zz p (4.3.15)
t t
b2 / a 2 1
p 2 2
b / a 1
2 zz
p
b / a 1
2 2
r
p rr
ra r b
(a ) ( b)
Figure 4.3.9: (a) stresses in the thick-walled cylinder, (b) microcracks driven by
tangential stress
The solution for a pressurized cylinder in plane strain was given above, i.e. where zz was
enforced to be zero. There are two other useful situations:
(1) The cylinder is free to expand in the axial direction. In this case, zz is not forced to
zero, but allowed to be a constant along the length of the cylinder, say zz . The zz
stress is zero, as in plane stress. This situation is called generalized plane strain.
(2) The cylinder is closed at its ends. Here, the axial stresses zz inside the walls of the
tube are counteracted by the internal pressure p acting on the closed ends. The force
acting on the closed ends due to the pressure is p a 2 and the balancing axial force is
zz b 2 a 2 , assuming zz to be constant through the thickness. For equilibrium
p
zz (4.3.16)
b / a2 1
2
Returning to the full three-dimensional stress-strain equations (Part I, Eqns. 6.1.9), set
zz zz , a constant, and xz yz 0 . Re-labelling x, y, z with r , , z , and again with
r 0 , one has
E
rr (1 ) rr zz
(1 )(1 2 )
E
(1 ) rr zz (4.3.17)
(1 )(1 2 )
E
zz (1 ) zz rr
(1 )(1 2 )
Substituting the strain-displacement relations 4.3.2 into 4.3.16a-b, and, as before, using
the axisymmetric equilibrium equation 4.3.5, again leads to the differential equation 4.3.6
and the solution u C1r C2 / r , with rr C1 C2 / r 2 , C1 C2 / r 2 , but now the
stresses are
E 1
rr C1 1 2 C2 2 zz
1 1 2 r
E 1
C1 1 2 C2 2 zz (4.3.18)
1 1 2 r
E
zz 2 C 1 zz
1 1 2 1
As before, to make the solution more amenable to stress boundary conditions, we let
A EC 2 / 1 and C EC1 / 21 1 2 , so that the solution is
1 E 1 E
rr A 2C zz , A 2 2C
r 2
1 1 2 r 1 1 2 zz
zz 4 C
1 E
1 1 2 zz
(4.3.19)
1 1 1 1
rr A r 2 2 1 2 C , E A r 2 2 1 2 C
E
1 1
u A 2 1 2 Cr
E r
Generalised axisymmetric solution
For internal pressure p, the solution to 4.3.19 gives the same solution for radial and
tangential stresses as before, Eqn. 4.3.14. The axial displacement is u z z zz (to within a
constant).
In the case of the cylinder with open ends (generalized plane strain), zz 0 , and one
finds from Eqn. 4.3.19 that zz 2 p / E b 2 / a 2 1 0 . In the case of the cylinder
with closed ends, one finds that zz 1 2 p / E b 2 / a 2 1 0 .
Here, take zz zz , a constant. Then, using the strain-displacement relations and the
equilibrium equation, one again arrives at the differential equation 4.3.6 so the solution
for displacement and strain is again 4.3.8a-b. With A C 2 / C12 C11 and
C C1 / 2C11 C12 , the stresses can be expressed as
1
rr A 2C C13 zz
r2
1
A 2 2C C13 zz (4.3.21)
r
C13
zz 4C C 33 zz
C11 C12
The plane strain solution then follows from zz 0 and the generalized plane strain
solution from zz 0 . These solutions reduce to 4.3.11, 4.3.19 in the isotropic case.
1 2
rr , 2 , r 0 (4.3.22)
r r r
One can check that these equations satisfy the axisymmetric equilibrium equation 4.3.4.
The biharmonic equation in polar coordinates is given by Eqn. 4.2.7. Specialising this to
the axisymmetric case, that is, setting / 0 , leads to
2
2 1 2 1 2 1
2 2 0 (4.3.23)
r r r r r r r 2 r r
or
d 4 2 d 3 1 d 2 1 d
0 (4.3.24)
dr 4 r dr 3 r 2 dr 2 r 3 dr
Alternatively, one could have started with the compatibility relation 4.2.8, specialised that
to the axisymmetric case:
2 1 rr 2
0 (4.3.25)
r 2 r r r r
and then combine with Hooke’s law 4.3.3 or 4.3.4, and 4.3.22, to again get 4.3.24.
Eqn. 4.3.24 is an Euler-type ODE and has solution (see Appendix to this section, §4.3.8)
A ln r Br 2 ln r Cr 2 D (4.3.26)
B1 2 ln r 2C
A
rr
r2
(4.3.27)
2 B3 2 ln r 2C
A
r
The strains are obtained from the stress-strain relations. For plane strain, one has, from
4.3.4,
1 A
rr 2 B1 4 21 2 ln r 2C 1 2
E r
(4.3.28)
1 A
2 B3 4 21 2 ln r 2C 1 2
E r
Comparing these with the strain-displacement relations 4.3.2, and integrating rr , one has
1 A
u r rr dr Br 1 21 2 ln r 2C 1 2 r F
E r
(4.3.27)
1 A
u r r Br 1 21 2 ln r 2C 1 2 r
E r
To ensure that one has a unique displacement u r , one must have B 0 and the constant
of integration F 0 , and so one again has the solution 4.3.113.
4.3.7 Problems
1. Derive the solution equations 4.3.11 for axisymmetric plane strain.
2. A cylindrical rock specimen is subjected to a pressure pover its cylindrical face and is
constrained in the axial direction. What are the stresses, including the axial stress, in
the specimen? What are the displacements?
3
the biharmonic equation was derived using the expression for compatibility of strains (4.3.23 being the
axisymmetric version). In simply connected domains, i.e. bodies without holes, compatibility is assured
(and indeed A and B must be zero in 4.3.26 to ensure finite strains). In multiply connected domains,
however, for example the hollow cylinder, the compatibility condition is necessary but not sufficient to
ensure compatible strains (see, for example, Shames and Cozzarelli (1997)), and this is why compatibility
of strains must be explicitly enforced as in 4.3.25
3. A long hollow tube is subjected to internal pressure pi and external pressures p o and
constrained in the axial direction. What is the stress state in the walls of the tube?
What if pi p o p ?
4. A long mine tunnel of radius a is cut in deep rock. Before the mine is constructed the
rock is under a uniform pressure p. Considering the rock to be an infinite,
homogeneous elastic medium with elastic constants E and , determine the radial
displacement at the surface of the tunnel due to the excavation. What radial stress
rr (a) P should be applied to the wall of the tunnel to prevent any such
displacement?
5. A long hollow elastic tube is fitted to an inner rigid (immovable) shaft. The tube is
perfectly bonded to the shaft. An external pressure p is applied to the tube. What are
the stresses and strains in the tube?
6. Repeat Problem 3 for the case when the tube is free to expand in the axial direction.
How much does the tube expand in the axial direction (take u z 0 at z 0 )?
4.3.8 Appendix
1 dt
r e t , log r t , (4.3.28)
r dr
du du dt 1 du
dr dt dr r dt
(4.3.29)
d 2 u d du dt d 2 u dt dt du d 2 t 1 d 2 u 1 du
dr 2 dr dt dr dt 2 dr dr dt dr 2 r 2 dt 2 r 2 dt
d 2u
u 0 (4.3.30)
dt 2
u C1e t C 2 e t
1 (4.3.31)
C1 r C 2
r
The solution procedure for 4.3.24 is similar to that given above for 4.3.6. Using the
substitution r e t leads to the differential equation with constant coefficients
d 4 d 3 d 2
4 4 0 (4.3.32)
dt 4 dt 3 dt 2
At Bte 2t Ce 2t D (4.3.33)
r 2
The accelerations lead to an inertial force (per unit volume) Fa r 2 which in turn
leads to stresses in the disc. The inertial force is an axisymmetric “loading” and so this is
an axisymmetric problem. The axisymmetric equation of equilibrium is given by 4.3.5.
Adding in the acceleration term gives the corresponding equation of motion:
rr 1
rr r 2 , (4.4.1)
r r
rr 1
rr br 0 , (4.4.2)
r r
where br r 2 . Thus the dynamic rotating disc problem has been converted into an
equivalent static problem of a disc subjected to a known body force. Note that, in a
general dynamic problem, and unlike here, one does not know what the accelerations are
– they have to be found as part of the solution procedure.
Using the strain-displacement relations 4.3.2 and the plane stress Hooke’s law 4.3.3 then
leads to the differential equation
d 2 u 1 du 1 1 2
2
2
u r 2 (4.4.3)
dr r dr r E
This is Eqn. 4.3.6 with a non-homogeneous term. The solution is derived in the
Appendix to this section, §4.4.3:
1 1 1 2 3 2
u C1 r C 2 r (4.4.4)
r 8 E
As in §4.3.4, let A EC 2 / 1 and C EC1 / 21 , and the full general solution is,
using 4.3.2 and 4.3.3, {▲Problem 1}
1 1
rr A 2
2C 3 2 r 2
r 8
1 1
A 2 2C 1 3 2 r 2
r 8
1 3
rr 1 2 21 C 1 2 2 r 2
E
A
8
(4.4.5)
r
1
E
A 1
1 2 21 C 1 2 2 r 2
8
r
1
E
A 1
u 1 21 Cr 1 2 2 r 3
8
r
A Solid Disc
For a solid disc, A in 4.4.5 must be zero to ensure finite stresses and strains at r 0 . C is
then obtained from the boundary condition rr (b) 0 , where b is the disc radius:
1
A 0, C 3 2 b 2 (4.4.6)
16
3
rr (r )
8
2 b 2 r 2
3 1 3 2
(r ) 2 b 2 r (4.4.7)
8 3
3 1 2 1 2
u (r ) 2 r b r
8 E 3
Note that the displacement is zero at the disc centre, as it must be, but the strains (and
hence stresses) do not have to be, and are not, zero there.
Dimensionless stress and displacement are plotted in Fig. 4.4.2 for the case of 0.3 .
The maximum stress occurs at r 0 , where
3
rr (0) (0) 2 b 2 (4.4.8)
8
1
u (b) 2 b 3 (4.4.9)
4E
0.9
8 0.8
3 2b 2
0.7
rr
0.6
8E
u
3 1 2b3 0.5
0.4
u
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
r /b
A Hollow Disc
rr (a ) 0, rr (b) 0 (4.4.10)
where a and b are the inner and outer radii respectively. It follows from 4.4.5 that
A
1
8
3 2 a 2 b 2 , C
1
16
3 2 a 2 b 2 (4.4.11)
3 a 2b 2
rr (r ) 2 a 2 b 2 r 2 2
8 r
3 1 3 2 a 2 b 2
(r ) 2 a 2 b 2 r 2 (4.4.12)
8 3 r
3 1 1 2 1 a 2b 2
u (r ) 2 r a 2 b 2 r
8 E 3 1 r 2
Dimensionless stress and displacement are plotted in Fig. 4.4.3 for the case of 0.3
and a / b 0.2 . The maximum stress occurs at the inner surface, where
3 1
(0) 2 b 2 1
a / b2 (4.4.13)
4 3
2.5
2
8
3 2b2
1.5
8E
u
3 1 2b3
1
rr
0.5
0
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
r /b
4.4.2 Problems
1. Derive the full solution equations 4.4.5 for the thin rotating disc, from the
displacement solution 4.4.4.
d 2u 1 2 3t 2
u e (4.4.14)
dt 2 E
The homogeneous solution is given by 4.3.31. Assume a particular solution of the form
u p Ae 3t which, from 4.4.14, gives
1 1 2
up 2 e 3t (4.4.15)
8 E
Adding together the homogeneous and particular solutions and transforming back to r’s
then gives 4.4.4.