Cylinder Theory Derivation
Andrew Ning
We will find the stress in a cylinder by using a free-body diagram approach (Fig. 1). σr is called the
radial stress, σθ is the tangential stress or hoop stress, and σz is the longitudinal or axial stress. There are
no shear stresses in this coordinate system.
Figure 1: Stress on a radial slice within the cylinder, and then on a small wedge within that slice.
A force balance yields
(σr + dσr )(r + dr)dθdz − σr (rdθdz) = 2σθ (drdz) sin(dθ/2)
(1)
Because we are looking at a differential element sin θ ' θ, and after simplification yields
σr dr + dσr (r + dr) = σθ dr
(2)
The product of differentiables drdσr can be considered negligible compared to the other terms. This gives
σr dr + rdσr = σθ dr
1
(3)
which can be written as
dσr
(4)
dr
This same result can be found purely mathematically by starting with the compatibility equations for a
linearly elastic structure (Eq. (5)), which is just a statement of static equilibrium.
(σθ − σr ) = r
∇·σ =0
(5)
We will expand this equation in cylindrical coordinates
1 ∂τrθ
1
∂τrz
∂σr
+
+
+ (σr − σθ ) = 0
∂r
r ∂θ
∂z
r
∂τθz
1 ∂(r2 τrθ ) 1 ∂σθ
+
+
=0
r2
∂r
r ∂θ
∂z
1 ∂(rτrz ) 1 ∂τθz
∂σz
+
+
=0
r ∂r
r ∂θ
∂z
(6)
(7)
(8)
We assume symmetry such that deformations are independent of θ and assume we are far from an end so
that deformations are independent of z (thus, all partial derivatives with respect to θ and z drop out). Then,
the second and third equation result in
C1
r2
C2
=
r
τrθ =
τrz
(9)
(10)
Since the shear stress is zero at the free surface, the constants C1 and C2 must be zero, and thus both of
these shear stresses are zero throughout the cylinder. The first equation then simplifes to give the the same
result as Eq. (4).
Just as we is done in deriving beam bending equations, we will assume that plane sections remain plane.
This means that the longitudinal strain z must be constant across the cross-section. We also assume that
the longitudinal stress is constant (at least away from the walls). Then from the stress-strain relationships
z =
1
[σz − ν(σr + σθ )]
E
(11)
we know that σr + σθ must also be a constant. Let us set that constant as 2A
σr + σθ = 2A
(12)
If we sub σθ from Eq. (12) into Eq. (4) we have
(2A − 2σr ) = r
dσr
dr
(13)
We now multiply through by r to get
2rσr + r2
dσr
− 2Ar = 0
dr
(14)
That was done so that we can pull out a differential
d(r2 σr − Ar2 ) = 0
(15)
This implies that r2 σr − Ar2 is a constant which we set to B. This gives
σr = A +
2
B
r2
(16)
and substituting back into Eq. (12) we have
σθ = A −
B
r2
(17)
These equations are called the Lamé equations and are the basis for our equations on cylindrical stress. As
we will see the longitudinal stress is also related
σz = A
(18)
The boundary conditions of the problem will determine the unknown constants.
We can solve for the radial and tangential stress in a pressurized thick cylinder by applying the boundary
conditions. At σr (ro ) = −po and σr (ri ) = −pi . In both cases, the sign is negative because the pressure
causes compression. If we plug in our boundary conditions into Eq. (16) we get
B
= −pi
ri2
B
A + 2 = −po
ro
A+
(19)
This gives us two equations to solve for the unknown constants. The result is
pi ri2 − po ro2
ro2 − ri2
(po − pi )ri2 ro2
B=
ro2 − ri2
A=
(20)
We now have the radial and tangential stress in a pressurized cylinder
pi ri2 − po ro2 + ri2 ro2 (po − pi )/r2
ro2 − ri2
pi ri2 − po ro2 − ri2 ro2 (po − pi )/r2
σθ (r) =
ro2 − ri2
σr (r) =
(21)
(22)
The longitudinal stress can be found from a simple force balance (see Fig. 2):
pi ri2 − po ro2 = σz (ro2 − ri2 )
⇒ σz =
pi ri2 − po ro2
ro2 − ri2
σz = A
Note that σz is halfway between σr and σθ .
3
(23)
Figure 2: Longitudinal stress balances by pressure acting on cylinder.
4