Section 4.2
4.2 Differential Equations in Polar Coordinates
Here, the two-dimensional Cartesian relations of Chapter 1 are re-cast in polar
coordinates.
4.2.1
Equilibrium equations in Polar Coordinates
One way of expressing the equations of equilibrium in polar coordinates is to apply a
change of coordinates directly to the 2D Cartesian version, Eqns. 1.1.8, as outlined in the
Appendix to this section, §4.2.6. Alternatively, the equations can be derived from first
principles by considering an element of material subjected to stresses σ rr , σ θθ and σ rθ ,
as shown in Fig. 4.2.1. The dimensions of the element are Δr in the radial direction, and
rΔθ (inner surface) and (r + Δr )Δθ (outer surface) in the tangential direction.
σ rθ +
σ θθ +
∂σ rθ
Δθ
∂θ
∂σ rθ
Δr
∂r
∂σ
σ rr + rr Δr
∂r
σ rθ +
∂σ θθ
Δθ
∂θ
r
σ rr
Δθ
σ rθ
σ θθ
Figure 4.2.1: an element of material
Summing the forces in the radial direction leads to
∑F
r
∂σ rr
⎛
⎞
= ⎜ σ rr +
Δr ⎟(r + Δr )Δθ − σ rr rΔθ
∂r
⎝
⎠
∂σ
Δθ ⎛
Δθ
⎞
(σ θθ )Δr
− sin
⎜ σ θθ + θθ Δθ ⎟Δr − sin
2 ⎝
∂θ
2
⎠
∂σ rθ
Δθ ⎛
Δθ
⎞
(σ rθ )Δr ≡ 0
+ cos
Δθ ⎟Δr − cos
⎜ σ rθ +
2 ⎝
∂θ
2
⎠
(4.2.1)
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
∂r
2 ⎝ ∂θ ⎠ ∂θ
(4.2.2)
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:
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Section 4.2
∂σ rr 1 ∂σ rθ 1
+ (σ rr − σ θθ ) = 0
+
r ∂θ
r
∂r
∂σ rθ 1 ∂σ θθ 2σ rθ
+
=0
+
r ∂θ
r
∂r
4.2.2
Equilibrium Equations (4.2.3)
Strain Displacement Relations and Hooke’s Law
The two-dimensional strain-displacement relations can be derived from first principles by
considering line elements initially lying in the r and θ directions. Alternatively, as
detailed in the Appendix to this section, §4.2.6, they can be derived directly from the
Cartesian version, Eqns. 1.2.5,
∂u r
∂r
1 ∂uθ u r
=
+
2-D Strain-Displacement Expressions
r ∂θ
r
1 ⎛ 1 ∂u r ∂uθ uθ ⎞
= ⎜
+
− ⎟
∂r
r ⎠
2 ⎝ r ∂θ
ε rr =
ε θθ
ε rθ
(4.2.4)
The stress-strain relations in polar coordinates are completely analogous to those in
Cartesian coordinates – the axes through a small material element are simply labelled
with different letters. Thus Hooke’s law is now
ε rr =
ε rr =
4.2.3
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
[(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)
Stress Function Relations
In order to solve problems in polar coordinates using the stress function method, Eqns.
3.2.1 relating the stress components to the Airy stress function can be transformed using
the relations in the Appendix to this section, §4.2.6:
σ rr =
∂ 2φ
∂ ⎛ 1 ∂φ ⎞ 1 ∂φ 1 ∂ 2φ
1 ∂φ 1 ∂ 2φ
σ
σ
−
+ 2
=
=
−
,
,
⎜
⎟=
θθ
rθ
r ∂r r ∂θ 2
∂r ⎝ r ∂θ ⎠ r 2 ∂θ r ∂r∂θ
∂r 2
(4.2.6)
It can be verified that these equations automatically satisfy the equilibrium equations
4.2.3 {▲Problem 2}.
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Section 4.2
The biharmonic equation 3.2.3 becomes
⎛ ∂2 1 ∂
1 ∂2
⎜⎜ 2 +
+ 2
r ∂r r ∂θ 2
⎝ ∂r
4.2.4
2
⎞
⎟⎟ φ = 0
⎠
(4.2.7)
The Compatibility Relation
The compatibility relation expressed in polar coordinates is (see the Appendix to this
section, §4.2.6)
1 ∂ 2 ε rr ∂ 2 ε θθ 2 ∂ 2 ε rθ 1 ∂ε rr 2 ∂ε θθ
2 ∂ε
+
−
−
+
− 2 rθ = 0
2
2
2
r ∂r∂θ r ∂r
r ∂r
∂r
r ∂θ
r ∂θ
4.2.5
(4.2.8)
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.
4.2.6
Appendix to §4.2
From Cartesian Coordinates to Polar Coordinates
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)
Then the Cartesian partial derivatives become
∂ ∂r ∂ ∂θ
=
+
∂x ∂x ∂r ∂x
∂ ∂r ∂ ∂θ
=
+
∂y ∂y ∂r ∂y
∂
∂ sin θ
= cosθ
−
r
∂θ
∂r
∂
∂ cosθ
= sin θ
+
r
∂θ
∂r
∂
∂θ
∂
∂θ
(4.2.10)
The second partial derivatives are then
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Section 4.2
∂ sin θ ∂ ⎞
∂ sin θ ∂ ⎞⎛
∂2 ⎛
−
= ⎜ cos θ
−
⎟
⎟⎜ cos θ
2
r ∂θ ⎠⎝
r ∂θ ⎠
∂r
∂r
∂x
⎝
∂ ⎞ sin θ ∂ ⎛ sin θ ∂ ⎞
∂ ⎛ sin θ ∂ ⎞ sin θ ∂ ⎛
∂ ⎛
∂ ⎞
= cos θ ⎜ cos θ ⎟ − cos θ ⎜
⎟
⎜
⎜ cos θ ⎟ +
⎟−
r ∂θ ⎝
r ∂θ ⎝ r ∂θ ⎠
∂r ⎠
∂r ⎝ r ∂θ ⎠
∂r ⎝
∂r ⎠
⎛1 ∂
⎛ 1 ∂ 1 ∂2 ⎞
1 ∂2 ⎞
∂2
⎟
⎜⎜ 2
⎟⎟
+ 2
+
−
θ
sin
2
= cos 2 θ 2 + sin 2 θ ⎜⎜
2 ⎟
∂r
⎝ r ∂r r ∂θ ⎠
⎝ r ∂θ r ∂r∂θ ⎠
(4.2.11)
Similarly,
1 ∂2
∂2
∂2
2
2 ⎛1 ∂
⎜
sin
cos
θ
θ
=
+
+
⎜ r ∂r r 2 ∂θ 2
∂y 2
∂r 2
⎝
⎞
⎛ 1 ∂ 1 ∂2 ⎞
⎟⎟ − sin 2θ ⎜⎜ 2
⎟⎟
−
⎠
⎝ r ∂θ r ∂r∂θ ⎠
⎛ ∂2 1 ∂
1 ∂2
∂2
⎜
= − sin θ cos θ ⎜ − 2 +
+
∂x∂y
r ∂r r 2 ∂θ 2
⎝ ∂r
⎞
⎛ 1 ∂ 1 ∂2 ⎞
⎟⎟ − cos 2θ ⎜⎜ 2
⎟⎟
−
∂
∂
∂
θ
r
r
θ
r
⎠
⎝
⎠
(4.2.12)
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
σ xx = σ rr cos 2 θ + σ θθ sin 2 θ − σ rθ sin 2θ
σ yy = σ rr sin 2 θ + σ θθ cos 2 θ + σ rθ sin 2θ
(4.2.13)
σ xy = sin θ cos θ (σ rr − σ θθ ) + σ rθ cos 2θ
Applying these and 4.2.10 to the 2D Cartesian equilibrium equations 3.1.3a-b lead to
1 ∂σ rθ 1
1 ∂σ θθ 2σ rθ ⎤
⎡ ∂σ
⎤
⎡ ∂σ
cos θ ⎢ rr +
+ (σ rr − σ θθ )⎥ − sin θ ⎢ rθ +
+
=0
r ∂θ
r
r ∂θ
r ⎥⎦
⎣ ∂r
⎦
⎣ ∂r
1 ∂σ rθ 1
1 ∂σ θθ 2σ rθ ⎤
⎡ ∂σ
⎤
⎡ ∂σ
sin θ ⎢ rr +
+ (σ rr − σ θθ )⎥ + cos θ ⎢ rθ +
+
=0
r ∂θ
r
r ∂θ
r ⎥⎦
⎣ ∂r
⎦
⎣ ∂r
(4.2.14)
which then give Eqns. 4.2.3.
y
θ
r
θ
x
Figure 4.2.2: rotation of axes
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Section 4.2
The Strain-Displacement Relations
Noting that
u x = u r cosθ − uθ sin θ
u y = u r sin θ + uθ cosθ
,
(4.2.15)
the strains in polar coordinates can be obtained directly from Eqns. 1.2.5:
∂u x
∂x
∂ sin θ ∂ ⎞
⎛
= ⎜ cos θ
−
⎟(u r cos θ − uθ sin θ )
∂r
r ∂θ ⎠
⎝
∂u
1 ⎛ 1 ∂u r ∂uθ uθ ⎞
⎛ 1 ∂uθ u r ⎞
= cos 2 θ r + sin 2 θ ⎜
+ ⎟ − sin 2θ ⎜
+
− ⎟
2 ⎝ r ∂θ
∂r
r ⎠
∂r
r ⎠
⎝ r ∂θ
ε xx =
(4.2.16)
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,
ε rr = ε xx cos 2 θ + ε yy sin 2 θ + ε xy sin 2θ
ε θθ = ε xx sin 2 θ + ε yy cos 2 θ − ε xy sin 2θ
ε rθ = sin θ cos θ (ε yy − ε xx ) + ε xy cos 2θ
(4.2.17)
then leads to the equations given above, Eqns. 4.2.4.
The Stress – Stress Function Relations
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)
σ rr = σ xx cos 2 θ + σ yy sin 2 θ + σ xy sin 2θ
σ θθ = σ xx sin 2 θ + σ yy cos 2 θ − σ xy sin 2θ
σ rθ = sin θ cos θ (σ yy − σ xx ) + σ xy cos 2θ
(4.2.18)
Using the Cartesian stress – stress function relations 3.2.1, one has
σ rr =
∂ 2φ
∂ 2φ
∂ 2φ
2
2
cos
θ
sin
θ
sin 2θ
+
−
∂x∂y
∂y 2
∂x 2
(4.2.19)
and similarly for σ θθ , σ rθ . Using 4.2.11-12 then leads to 4.2.6.
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Section 4.2
The Compatibility Relation
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
⎛ 1 ∂ 1 ∂2 ⎞⎞
∂ 2 ε xx ⎛
∂2
1 ∂2 ⎞
2
2 ⎛1 ∂
⎜
⎜
⎟
⎜⎜ 2
⎟⎟ ⎟ ×
=
+
+
+
−
cos
sin
sin
2
θ
θ
θ
⎜ r ∂r r 2 ∂θ 2 ⎟
⎜
⎟
∂x 2
∂r 2
⎝
⎠
⎝ r ∂θ r ∂r∂θ ⎠ ⎠
⎝
(ε rr cos 2 θ + ε θθ sin 2 θ − ε rθ sin 2θ )
(4.2.20)
After some lengthy calculations, one arrives at 4.2.8.
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