Stresses due to Rotation

Bodies like rings, circular discs, cylinders, etc. are often present in machines involving
rotational movement. Large magnitudes of centrifugal forces arise in these bodies due to the
high rotation speeds under which the machines operate. The centrifugal forces give rise to
stresses that are distributed symmetrically about the axes of rotation. The object of this
chapter is the study of these stresses, assuming uniform density of the material from which
the circular members (rings, discs, cylinders…) are made.

I - Rotating ring
Fig. 1 shows a thin ring rotating about its centre of gravity at 0.

Fig. 1

Let,

r = mean radius of the ring (m)

t = thickness of the ring (m)
ρ = density of the material of the ring (kg/m3)
ω = angular speed of the ring (rad/s)
FC = centrifugal force
σC = circumferential of hoop stress

As a result of rotation, each and every element of the ring, like LMPQ, will experience
centrifugal (or inertia) force dFC which will tend to expand the ring radially outwards. This
will in turn induce the circumferential (hoop) stress σ C in the ring. The hoop stress is tensile
in nature.
In order to evaluate this stress, the following assumptions are made:

1. The circumferential stress on the area of the cross-section of the ring is uniform.
2. The dimensions of the cross-section of the ring are small as compared to its mean
radius.
3. The constraining effect of spokes is negligible.

Volume of small element LMPQ per unit length = r.dθ .t

Hence, the centrifugal force acting on the element,

dFC = ρ × ( r.dθ .t ) × ω 2 .r (1.1)

The vertical component of the elemental force dFC is given by:

dFC ( vert.) = dFC × sin θ

(1.2)
= ρ × ( r.dθ .t ) × (ω 2 .r ) × sin θ

The horizontal component of dFC will be cancelled when we consider another small element
L’M’P’Q’ in quadrant number 2 at an angle θ. However, this 2nd element L’M’P’Q’ will
contribute an additional component of centrifugal force in the vertical direction.

Therefore, the total vertical component of centrifugal force (or the bursting force across the
horizontal diameter XX is given by FC , where

π π
FC = ∫ dFC × sin θ = ∫ ρ × ( r.dθ .t ) × (ω 2 .r ) × sin θ
0 0
π
= ρ .ω 2 .r 2 .t × ∫ sin θ .dθ (1.3)
0
π
= ρ .ω 2 .r 2 .t × [ − cos θ ]0
= 2.ρ .ω 2 .r 2 .t

The total resisting force opposing the bursting force FC is denoted by Fresist . , where

Fresist . = 2 × σ C × t × 1 (1.4)

The ring will be in equilibrium, when

Total bursting force = total resisting force

2.ρ .ω 2 .r 2 .t = 2.σ C .t.1

(1.5)
σ C = ρ .ω 2 .r 2
But v = ω.r (where v is the linear velocity of the ring)

therefore,
σ C = ρ .v 2 (1.6)
Example I.1
A wheel 800mm in diameter has a thin rim. If the density of the material
from which the wheel is made is 7700 kg m3 and E = 200GPa , calculate:
i). How many revolutions per minute it may make without the hoop stress
exceeding 130MPa ?
ii). Change in diameter

(neglect
neglect the effect of spokes)

Example I.2
A built-up
up ring (see figure below) rotates at 2000rpm . Find the stresses set
up in the steel and copper rings.
Assume: for steel, ES = 200GPa , ρ S = 7800 kg m3
for copper, ECu = 100GPa , ρCu = 8900 kg m3
II - Thin rotating disc
A simplified model for a component such as a gas turbine rotor is a uniformly thin disc which,
when rotating at constant velocity, is subjected to stresses induced by centripetal
acceleration. This is a problem which produces deformations symmetrical about the rotating
axis. If the disc is thin in section, then it is assumed that plane stress exists. The radial and
hoop stresses are constant through the thickness, and there is no stress in the direction of
the axis of rotation.

Fig. 2 (a) shows a circular disc of inner radius r1 and outer radius r2 rotating about its axis.
Let us assume that the disc is of uniform thickness.

Fig. 2

Consider an element ABCD of the disc, at a radius r , subtending an angle dθ at the

centre, and of radial width dr .

Let,

σr = stress on the face CD

σ r + dσ r = stress on the face AB
σC = stress on the face BC
σC = stress on the face AD
On the flat faces of the disc, there is no normal stress, and hence there is free strain in the
direction of the axis.

Volume of element ABCD = r.dθ .dr.t (where, t = thickness of the disc)

Radial force on ABCD due to rotation = ρ × r.dθ .dr.t × ω 2 r
= ρ × dθ .dr.t × ω 2 r 2
Force on face AB (outward) = ( r + dr ) .dθ .t × (σ r + dσ r )
Force on face CD (inward) = r.dθ .t × σ r
Force on faces BC and AD = dr.t × σ C

Forces acting on the element are shown in Fig.2 (b).

Resolving the forces in the radial direction and considering the equilibrium of forces, we get

 dθ 
r.dθ .t.σ r + 2.σ C .t.dr × sin   = (σ r + dσ r )( r + dr ) .dθ .t + ρ .dθ .dr.t × ω 2 r 2 (1.7)
 2 

 dθ  dθ
Since dθ is very small, sin  ≅
 2  2

Therefore, equation (1.7) reduces to

σ r .r + σ C .dr = σ r .r + σ r .dr + r.d σ r + d σ r .dr + ρ .ω 2 r 2 .dr (1.8)

Neglecting second order term ( dσ r × dr ) and on further simplification, the expression

becomes

σ C .dr = σ r .dr + r.d σ r + ρ .ω 2 r 2 .dr (1.9)

Dividing both sides by dr , we get

dσ r
σ C = σ r + r. + ρ .ω 2 r 2
dr
or (1.10)
dσ r
σ C − σ r = r. + ρ .ω 2 r 2
dr

Let us now consider circumferential and radial strains.

When the disc is rotating at high speed, let the disc dimensions change such that r
becomes ( r + u ) and ( r + dr ) becomes ( r + dr + du ) .

Therefore, circumferential strain ε C is given by

2π ( r + u ) − 2π r u
εC = = (1.11)
2π r r
and the radial strain ε r is given by

( r + dr + du ) − ( r + dr ) du
εr = = (1.12)
dr dr

Also,
1 u
εC = (σ C − νσ r ) =
E r
(1.13)
r
∴u = (σ C − νσ r )
E

1 du
εr = (σ r − νσ C ) = (1.14)
E dr

(see theory on stress/strain relationships and Poisson’s ratio for equations (1.13) & (1.14) )

From equations (1.10, 1.13 & 1.14) we obtain expressions for the hoop (circumferential)
stress and the radial stress.

C1 C2  3 + ν  2 2
σr = + −  .ρ .ω r (1.15)
2 r2  8 

C1 C2  1 + 3ν  2 2
σC = − −  .ρ .ω r (1.16)
2 r2  8 

Depending on the boundary conditions, the integration constants C1 and C2 can be

evaluated and will yield the corresponding radial and circumferential stress expressions for
i). a solid disc
ii). a disc with a central hole
Example II.1
Determine the intensities of principal stresses ( σ C , σ r ) in a flat steel disc
of uniform thickness having a diameter of 1m and rotating at 2400rpm .
What will be the maximum principal stresses if the disc has a central hole of
0.2m diameter?
Take Poisson’s ratio= 1 and ρ = 7850 kg m3
3
III - Rotating long cylinders
The analysis of a rotating long cylinder is similar to that of a thin disc, the only difference
being that the length of the cylinder along the axis is large as compared to the radius and
axial stress is taken into account along the length of the cylinder. Longitudinal strain is
assumed to be constant.

Let,

σr = radial stress
σC = circumferential stress
σl = longitudinal (or axial) stress

Let these stresses ( σ r , σ C , σ l ) act on any element of a section of the cylinder of radius r
(Fig. 2(b)).

Then, radial strain,

1 du
εr = σ r − ν (σ C + σ l )  = (1.17)
E dr
Circumferential strain,
1 u
εC = σ C − ν (σ r + σ l )  = (1.18)
E r

Longitudinal strain (assumed constant),

1
εl = σ l − ν (σ r + σ C )  (1.19)
E

From equations (1.10, 1.17, 1.18 & 1.19) we obtain expressions for the hoop
(circumferential) stress, the longitudinal stress and the radial stress.

σ l = C1 + ν (σ r + σ C ) (1.20)
where C1 = E.ε l

C2 C3 ρ .ω 2 r 2  3 − 2ν 
σr = + − .  (1.21)
2 r2 8  1 −ν 

C2 C3 ρ .ω 2 r 2  1 + 2ν 
σC = − − .  (1.22)
2 r2 8  1 −ν 

Equations (1.21) and (1.22) are the governing equations for a rotating cylinder. C2 and C3
are integration constants which are evaluated with the help of boundary conditions.
Example III.1

A long cylinder of 300mm radius is rotating at 4500rpm . The density of the

material is 7800 kg m3 and Poisson’s ratio = 0.3 .
i). Calculate the maximum stress in the cylinder;
ii). Draw the variations of radial and circumferential stresses along the
radius.

Example III.2

A hollow cylinder, 200mm external radius and 100mm internal radius is

rotating at 3000rpm . The density of the material is 7800 kg m3 and
Poisson’s ratio = 0.3 .
i). Calculate the maximum stress in the cylinder;
ii). Draw the variations of radial and circumferential stresses in the cylinder.