elasticity problems in polar coordinates … · we are now left with the problem of how to...

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Stephen Martel 10-1 University of Hawaii

ELASTICITY PROBLEMS IN POLAR COORDINATES (10)

I Main topics A Motivation B Cartesian Approach C Transformation of coordinates D Equilibrium equations in polar coordinates E Biharmonic equation in polar coordinates F Stresses in polar coordinates

II Motivation A Many key problems in geomechanics (e.g., stress around a borehole,

stress around a tunnel, stress around a magma chamber) involve cylindrical geometries. Our reference frame should fit the features we examine.

B Introduces the concept of stress concentration due to factors inside a body (as opposed to concentrated boundary loads).

III Approach A Transform elastic equations from xy form to polar form B Alternative: vector and tensor approaches (see C&P, Ch. 11)

IV Transformation of coordinates (See Fig. 34.1) A Rectangular to polar:

θ = tan-1(y/x) r = (x2 + y2)1/2 (10.1) B Polar to rectangular:

x = r cosθ y = r sinθ (10.2) C Stress convention: Still use on-in convention (θrr ,θ rθ ,θθr ,θθθ )

rθ

y

x

rθ

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V Equil ibrium equations in polar coordinates

θ

σ − Δσ

(r, )θτ + Δτ

σ + Δσθ θ

θ θ

τ + Δτ

τ − Δτ

τ − Δτ

σ + Δσr r

σ − Δσr r

δθ/2

δθ/2

dr

dr

δθ/2

Fr∑ = 0 (10.3) We will assume body forces are negligible. General form of the terms is σij = {σij ref + (σij gradient)(distance)}.

σ rr +∂σ rr∂r

dr2

r +

dr2

dθ − σ rr +

∂σ rr∂r

dr2

r +

dr2

dθ +

cos dθ2

σrθ +∂σ rθ∂θ

dθ2

dr − cos dθ

2σ rθ −

∂σ rθ∂θ

dθ2

dr +

− sin dθ2

σθθ +∂σθθ∂θ

dθ2

dr − sin dθ

2σθθ −

∂σθθ∂θ

dθ2

dr = 0. (10.4)

This equation reduces to

2 ∂σrr∂r

dr2rdθ

+ 2σ rr

dr2dθ

+ 2 ∂σ rr

∂θdr2dr2dθ

+ 2 ∂σ rr

∂θdθ2dr cos dθ

2

− 2σθθ dr sin

dθ2

= 0

. (10.5) Dividing through by dr, and noting that for small angles sin (dθ / 2) ≈ dθ / 2 and cos(dθ / 2) ≈ 1, equation (10.5) reduces to ∂σrr∂r

rdθ

+ σrrdθ( ) +

∂σrr∂r

dr2

2dθ

+

∂σ rr∂θ

dθ

− 2σθθ

dθ2

= 0 . (10.5)

The third term drops out because dr2 is tiny relative to the other terms. Dividing this (10.5) through by r dθ yields

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∂σrr∂r

+σ rrr

+1r∂σrr∂θ

−σθθr

= 0. (10.6)

This expression is commonly rearranged as (see eq. 5.41 in C&P): ∂σrr∂r

+1r∂σ rr∂θ

+σrr −σθθ

r= 0. (10.7)

By summing the forces in the θ-direction one can obtain 1r∂σ rθ∂θ

+∂σrθ∂r

+2σ rθr

= 0. (10.8)

Equations (10.7) and (10.8) are the equilibrium equations in polar coordinates.

VI Biharmonic equation in polar coordinates Our starting point is the biharmonic equation

∇4φ = ∇2∇2φ =∂ 2

∂x 2+∂ 2

∂y 2

∂ 2

∂x2+∂ 2

∂y2

φ = 0 , (10.9)

and the stresses in terms of the stress function φ:

(a) σxx =∂2φ

∂y2, (b)

€

σ yy =∂2φ∂x2

, (c)

€

σ xy =−∂2φ∂x∂y

. (10.10)

In order to transform these equations to polar form, we need to know how to express derivatives with respect to x and y in terms of r and θ. We get these relationships from the chain rule:

(a) ∂φ∂x

=∂φ∂r

∂r∂x

+∂φ∂θ

∂θ∂x

(b) ∂φ∂y

=∂φ∂r

∂r∂y

+∂φ∂θ

∂θ∂y

(10.11)

We therefore need to know how r and θ relate to x and y. (a) θ = tan-1(y/x) (b) r = (x2 + y2)1/2 (10.12) (a) x = r cosθ (b) y = r sinθ (10.13)

Now to the derivatives. Starting with eq. (10.12b) ∂r∂x

=12x2 + y2( )−1/2 2 x( ) = x

x2 + y2( )1/2=xr= cosθ (10.14)

∂r∂y

=12x2 + y2( )−1/ 2 2y( ) = y

x2 + y2( )1/2=yr= sin θ (10.15)

Now take derivatives of eq. (10.12a). Recalling that d(tan-1u) = du/(1+u2)

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∂θ∂x

=∂ tan−1(y / x)( )

∂x=

1

1+yx

2

2∂ (y / x)∂x

=1

1+yx

2

2−yx2

=− yr2

=−sin θr

(10.16) Similarly (16)

∂θ∂y

=∂ tan−1(y / x)( )

∂y=

1

1+yx

2

2∂ (y / x)∂y

=1

1+yx

2

21xxx

=xr2

=cosθr

(10.17) Now we can find the derivatives of φ in terms of r and θ by substituting eqs. (10.16) and (10.17) into chain rule equations (10.11):

(a) ∂φ∂x

=∂φ∂rcosθ − ∂φ

∂θsin θr

, (b) ∂φ∂y

=∂φ∂rsin θ + ∂φ

∂θcosθr

. (10.18)

Second derivatives are found by operating on first derivatives.

∂ 2φ

∂x2=∂ ∂φ

∂x

∂x=∂ ∂φ

∂x

∂rcosθ −

∂ ∂φ∂x

∂φsin θr

. (10.19)

Substituting eq. (10.18a) into eq. (10.19) yields

∂ 2φ

∂x2=∂ ∂φ

∂x

∂x=∂ ∂φ

∂rcosθ − ∂φ

∂θsin θr

∂rcosθ −

∂ ∂φ∂rcosθ − ∂φ

∂θsin θr

∂θsin θr .

(10.20) This can be expanded as

∂ 2φ

∂x2=∂ 2φ

∂r2cos2 θ − ∂ 2φ

∂θ∂rsin θ cosθ

r+∂φ∂θsin θ cosθ

r2

−∂ 2φ∂θ∂r

sin θ cosθr

+∂φ∂rsin 2 θr

+∂ 2φ

∂θ 2sin 2 θr2

+∂φ∂θsin θ cosθ

r2

(10.21)

The expression for ∂2φ

∂y2 can be found by the same procedure:

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∂ 2φ

∂y2=∂ 2φ

∂r2sin 2 θ + ∂ 2φ


r−∂φ∂θsin θ cosθ

r2

+∂ 2φ∂θ∂r

sin θ cosθr

+∂φ∂rcos 2 θr

+∂ 2φ

∂θ 2cos2 θr2

−∂φ∂θsin θ cosθ

r2

(10.22)

Equation (10.21) is the expression for σyy and eq. (10.22) is the expression for σxx. Notice the term-by-term "symmetry" between these two equations.

These equations can be added to give the expression for ∂2φ

∂x2+∂ 2φ

∂y2.

Recalling that sin2θ + cos2θ =1, we get ∂ 2φ

∂x2+∂ 2φ

∂y2= ∂

2φ

∂r2+1r∂φ∂r

+1r2

∂ 2φ

∂θ2 = ∇2φ. (10.23)

The biharmonic equation is obtained by allowing the harmonic equation (10.23) to operate on itself.

∇4φ = ∇2∇2φ =∂ 2

∂r 2+1r∂∂r

+1r2

∂ 2

∂θ 2

∂ 2

∂r2+1r∂∂r

+1r2

∂ 2

∂θ2

φ = 0 . (10.24)

VII Stresses in polar coordinates We are now left with the problem of how to determine the stresses in polar

coordinates from the stress function φ. We know that the mean normal stress (and hence twice the mean stress) is an invariant term - it does not depend on the choice of the system of coordinates. As a result

2σmean = σxx + σyy = σ rr +σθθ =∂ 2φ

∂x 2+∂ 2φ

∂y2

= ∇

2φ . (10.25)

By comparing equations (10.23 and 10.25) we know that if one of the terms on the right side of eq. (10.24) equals σθθ , then the other terms must sum to equal σ rr. If we can show this for any particular position (r,θ), then the result will hold for all positions. We therefore choose a simple case. Let the r-direction be along the x-axis and the θ-direction be parallel to the y-axis, so σ rr = σxx and σθθ = σyy . The x-axis corresponds to θ = 0° and the y-direction to θ = 90°. So

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∂ 2φ

∂x2= σyy = σθθ =

∂ 2φ

∂r2cos2 θ − ∂ 2φ


r+∂φ∂θsin θ cosθ

r2

−∂ 2φ∂θ∂r

sin θ cosθr

+∂φ∂rsin 2 θr

+∂ 2φ

∂θ 2sin 2 θr2

+∂φ∂θsin θ cosθ

r2

.

(10.26) For our case of θ = 0°, all terms with sin θ are zero and this simplifies to

σθθ =∂ 2φ

∂r 2 (10.27)

Similarly

∂ 2φ

∂y2= σxx = σ rr =

∂ 2φ

∂r2sin 2 θ + ∂ 2φ


r−∂φ∂θsin θ cosθ

r2

+∂ 2φ∂θ∂r

sin θ cosθr

+∂φ∂rcos 2 θr

+∂ 2φ

∂θ 2cos2 θr2

−∂φ∂θsin θ cosθ

r2

.

(10.28) Again θ = 0°, so all terms with sin θ are zero, and this simplifies to

σ rr =1r∂φ∂r

+1r2

∂ 2φ

∂θ 2. (10.29)

Note that σθθ + σrr does indeed equal

€

∇ 2φ =∂2φ

∂r2+1r∂φ∂r

+1r2

∂ 2φ

∂θ2.

The shear stress θrθ can be determined from σxy and is given by

(a) σ rθ =−∂∂r

1r∂φ∂θ

or σ rθ =

−1r∂ 2φ∂r∂θ

+1r2

∂φ∂θ

. (10.30)

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elasticity problems in polar coordinates … · we are now left with the problem of how to...

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