notes6

4 CONTENTS

6.3 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.3.1 Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.3.2 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.3.3 Examples of Minimum Potential Energy . . . . . . . . . . . . . . . . 1226.3.4 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.4 Strain Energy for Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1256.5 Complementary Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . 126

6.5.1 Principle of Virtual Forces . . . . . . . . . . . . . . . . . . . . . . . . 1266.5.2 Principle of Virtual Forces for 3-D body . . . . . . . . . . . . . . . . 1286.5.3 Principle of Minimum Complementary Potential Energy . . . . . . . 1296.5.4 Castigliano’s Second Theorem . . . . . . . . . . . . . . . . . . . . . . 1326.5.5 Unit Dummy Load Method . . . . . . . . . . . . . . . . . . . . . . . 136

6.6 Strain Energy for Thermal Loads . . . . . . . . . . . . . . . . . . . . . . . . 1386.6.1 Thermal Strain Energy for Bars . . . . . . . . . . . . . . . . . . . . . 139

6.7 Reciprocal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1406.7.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

6.8 Rayleigh–Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7 Torsion 145

7.1 Circular Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.2 Arbitrary Solid Sections with No Holes . . . . . . . . . . . . . . . . . . . . . 146

7.2.1 Example: Elliptical cross section . . . . . . . . . . . . . . . . . . . . 1497.2.2 Example:Rectangular x-section . . . . . . . . . . . . . . . . . . . . . 151

7.3 Prandtl Theory for Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1527.3.1 Example: Elliptical cross section . . . . . . . . . . . . . . . . . . . . 1557.3.2 Example:Rectangular cross section . . . . . . . . . . . . . . . . . . . 1567.3.3 Membrane Analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1587.3.4 Hydrodynamic Analogy . . . . . . . . . . . . . . . . . . . . . . . . . 159

7.4 Multiple Connected Domains . . . . . . . . . . . . . . . . . . . . . . . . . . 1607.4.1 Example:Elliptical cross section . . . . . . . . . . . . . . . . . . . . . 1627.4.2 Example:General cross section with hole . . . . . . . . . . . . . . . . 163

7.5 Closed Thin Wall Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1637.5.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

7.6 Multi Cell Tubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1687.6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

7.7 Open Thin Wall Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1707.7.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1717.7.2 Shear Center and Center of Twist . . . . . . . . . . . . . . . . . . . . 1727.7.3 Sectorial Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1727.7.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7.8 Warping of Thin Wall Cross Sections in Pure Torsion . . . . . . . . . . . . . 177

Chapter 7

Torsion

7.1 Circular Bars

x1

x2

x1

x2T T

L

x3

β

θ

θ

Consider a prismatic bar with a circular cross section that is being acted upon by a torsionalforce T . Due to the action of the torque T the cross sections will be rotated through anangle of twist θ(x3). The rate of twist β is given by

β =dθ(x3)

dx3

and the applied torsion T satisfies the following relationship

T = µβJ

where J is the polar moment of inertia of the cross section and µ is the shear modulus of thematerial. The only nonzero stresses are the shear stresses that develop on the cross section.These shear stresses increase linearly in magnitude with radial distance from the center ofthe cross section and are perpendicular to the radial lines.

τ =Tr

Jτ

ττ

τ

145

146 CHAPTER 7. TORSION

It is important to note that the cross sections remain plane and do not warp under theinfluence of the torsional load. In general for noncircular cross sections warping of the crosssection will occur and thus the cross sections will not remain plane.

7.2 Arbitrary Solid Sections with No Holes

Consider a prismatic beam with a non-circular cross section acted upon by a torque T . Apoint P located at (x1,x2) will move to a new location P ′ due to the action of the torque byrotating through and angle θ(x3) with respect to the centroid of the cross section.

x

x

PP

θαr

1

2

If the radial distance of point P from the centroid is denoted by r and the radial linethat connects point P to the centroid makes an angle α with the x1 axis then the in-planedisplacements of the point P are given by

u1 = −(rθ) sinα = −θr sinα = −θx2

u2 = (rθ) cosα = θr cosα = θx1

For non circular cross sections warping occurs and point P will also displace in the x3

direction. Using St. Venant’s Theory for warping, a warping function ψ(x1, x2) is assumedto exist such that

u3 = βψ(x1, x2)

The corresponding engineering strains are thus given by

ε11 =∂u1

∂x1

= 0 ε22 =∂u2

∂x2

= 0

ε33 =∂u3

∂x3

=∂β

∂x3

ψ = 0 if β = constant

γ12 = u1,2 + u2,1 = −θ + θ = 0 γ13 = u1,3 + u3,1 = −x2θ,3 + βψ,1 = β(ψ,1 − x2)

γ23 = u2,3 + u3,2 = θ,3x1 + βψ,2 = β(ψ,2 + x1)

7.2. ARBITRARY SOLID SECTIONS WITH NO HOLES 147

The corresponding stresses are given by

σ11 = σ22 = σ33 = 0 σ12 = 0

σ13 = µβ(∂ψ

∂x1

− x2) σ23 = µβ(∂ψ

∂x2

+ x1)

Equilibrium requires that the following equations hold over the cross section

σ11,1 + σ12,2 + σ13,3 = 0 −→ 0 = 0

σ21,1 + σ22,2 + σ23,3 = 0 −→ 0 = 0 if β = constant

σ31,1 + σ32,2 + σ33,3 = 0

µβψ,11 + µβψ,22 = 0

µβ(ψ,11 + ψ,22) = 0

O2ψ = 0

Turning our attention to the lateral surface of the beam we require that t = σn = 0 on theouter edges of the cross section. Therefore the following must hold

t1 = σ11n1 + σ12n2 = 0

t2 = σ21n1 + σ22n2 = 0

t3 = σ31n1 + σ32n2

= µβ(ψ,1 − x2)n1 + µβ(ψ,2 + x1)n2 = 0

Therefore in order to have traction free lateral surfaces on the beam the following equationmust be satisfied

∂ψ

∂x1

n1 +∂ψ

∂x2

n2 = x2n1 − x1n2 on Γ (7.1)

Now consider the shear forces acting in the x1 and x2 directions denoted by V1 and V2,respectively. We require that both V1 and V2 be equal to zero since no net shear forces are


applied to the cross section. Therefore

V1 =

∫

A

σ31 dx1dx2 = µβ

∫

A

(ψ,1 − x2) dA

= µβ

∫

A

[

(x1(ψ,1 − x2)),1 − x1ψ,11]

dx1dx2

= µβ

∫

A

[

(x1(ψ,1 − x2)),1 + x1ψ,22]

dx1dx2 since O2ψ = 0

= µβ

∫

A

[

(x1(ψ,1 − x2)),1 + (x1(ψ,2 + x1)),2]

dx1dx2

= µβ

∮

Γ

n1[x1(ψ,1 − x2)] + n2[x1(ψ,2 + x1)]

ds

= µβ

∮

Γ

x1

n1(ψ,1 − x2) + n2(ψ,2 + x1)

ds

= 0

The last result is obtained by assuming that Equation (7.1) is satisfied. Similarly V2 = 0can be shown.

The applied torque T must satisfy the following conditions

T =

∫

A

(σ32x1 − σ31x2) dx1dx2

=

∫

A

(µβ(ψ,2 + x1)x1 − µβ(ψ,1 − x2)x2) dx1dx2

= µβ

∫

A

(x21 + x2

2 + x1ψ,2 − x2ψ,1) dx1dx2

T = µβJ (7.2)

where J is the effective torsional constant for the cross section and is given by

J = J +

∫

A

(x1ψ,2 − x2ψ,1) dx1dx2

Here J is the polar moment of inertia for the cross section. The quantity µJ is called thetorsional rigidity of the beam. It is important to note that for circular cross sections nowarping exists due to torsion thus ψ = 0 and J = J . Obtaining the warping function ψ iskey to being able to obtain the stresses and displacements caused by torsion.


7.2.1 Example: Elliptical cross section

x

x

a

b

2

1

n

Consider an elliptical cross section as shown above. The warping function for such a crosssection is given by

ψ(x1, x2) =b2 − a2

a2 + b2x1x2

One can easily verify that the above function satisfies equilibrium by showing that O2ψ = 0.

To show that the traction free lateral surface condition, i.e. Equation (7.1), is satisfied weneed to obtain an expression for the outward pointing unit normal n to the boundary of thecross section. The equation for the boundary of the cross section is given by

b2x21 + a2x2

2 − a2b2 = 0

On the boundary we therefore have the following relationship

0 = 2b2x1dx1 + 2a2x2dx2

leading to the following

dx1 ∝ a2x2 ∝ n2

dx2 ∝ −b2x1 ∝ −n1

n1 = γb2x1 n2 = γa2x2

dx2dx1

dx2dx1

n

where γ is a scaling factor to ensure that n has length equal to 1. Checking Equation (7.1)we obtain

ψ,1n1 + ψ,2n2 = x2n1 − x1n2(

b2 − a2

a2 + b2

)

x2n1 +

(

b2 − a2

a2 + b2

)

x1n2 = x2n1 − x1n2

(b2 − a2)x2n1 + (b2 − a2)x1n2 = x2n1(a2 + b2) − x1n2(a

2 + b2)

2b2x1n2 = 2a2x2n1


Substituting in the values for n1 and n2 we obtain

2b2x1γa2x2 = 2a2x2γb

2x1

Therefore Equation (7.1) is satisfied.The torsional constant is given by

J =

∫

A

(x21 + x2

2 + x1ψ,2 − x2ψ,1) dx1dx2

=

∫

A

(x21 + x2

2) dx1dx2 +

∫

A

(x1

[b2 − a2

a2 + b2x1

]

− x2

[a2 − b2

a2 + b2x2

]

) dx1dx2

=πa3b3

a2 + b2

Note that the polar moment of inertia for the cross section is given by J = π4ab(a2 + b2).

The stresses are given by

σ13 = µβ(ψ,1 − x2) = µβ[

(

b2 − a2

a2 + b2

)

x2 − x2

]

= µβ

(

−2a2

a2 + b2

)

x2 = − 2x2

πab3

[

πa3b3

a2 + b2

]

µβ

= − 2x2

πab3Jµβ = − 2x2

πab3T

σ23 = µβ(ψ,2 + x1) = µβ[b2 − a2

a2 + b2x1 + x1

]

= µβ

(

2b2x1

a2 + b2

)

=2x1

πa3bT

πab22T

πab22Tπba2

2T

πba22T

x

x 2

1

Shear stresses −1.5 −1 −0.5 0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

x1/a

x2/b

Fringe plot of (a2+b2)ψ(ab)(b2−a2)

and shear force

vectors for a/b=1.5


−1

−0.5

0

0.5

1

−1

−0.5

0

0.5

1−0.5

0

0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

x1/ax2/b

(a2+b2

)ψ(ab)

(b2−a2)

warping function (a/b=1.5)

7.2.2 Example:Rectangular x-section

Consider a rectangular cross section for which the width is equal to 6 and the height is equalto 8. A closed form solution for the warping function ψ is not possible therefore numericaltechniques must be employed to solve for the warping function. One way to obtain thewarping function is to solve the steady state heat conduction problem with a conductivityκ = 1 and applying zero temperature at the centroid of the cross section. Furthermore, theheat flux q on the boundary of the cross section must satisfy the following condition

q(x1, x2) = x2n1 − x1n2 x ∈ Γ

The resulting temperatures will be equal to the warping function value.

The mesh shown below consists of 829 nodes and 778 quadrilateral elements. The result-ing plots for the warping function and shear stress vectors are also shown below.

−3 −2 −1 0 1 2 3−4

−3

−2

−1

0

1

2

3

4

x1

x2

Finite element mesh for rectangular crosssection

−4

−2

0

2

4 −4−2

02

4

−4

−2

0

2

x1x2

ψ

warping function for rectangular crosssection


−3 −2 −1 0 1 2 3−4

−3

−2

−1

0

1

2

3

4

−3

−2

−1

0

1

2

3

x1

x2

Fring plot of warping function and shearstress vectors for rectangular cross section

The torsional rigidity for the cross section is given by

J = J +

∫

A

(x1ψ,2 − x2ψ,1) dx1dx2 = 312.2

In fact if one takes b to be equal to the larger dimension of the rectangular cross section andt to be equal to the smaller dimension of the rectangular cross section then the torsionalrigidity is given by

J = c1bt3

and the largest shear stress due to torsion on the cross section is given by

τmax = c2

(

T

bt2

)

where c1 and c2 are given in the following table:

b/t ∞ 10 5 3 2.5 2.0 1.5 1.2 1.0

c1 0.333 0.312 0.291 0.263 0.249 0.229 0.196 0.166 0.141

c2 3.00 3.20 3.44 3.74 3.88 4.06 4.33 4.57 4.80

See Timoshenko, S.P. and Goodier, J.N., “Theory of Elasticity”, 2nd edition.

7.3 Prandtl Theory for Torsion

Another approach to the torsion of solid sections is to introduce a Prandtl Stress function

φ(x1, x2) such that the following relationships are satisfied

σ13 = µβφ,2 σ23 = −µβφ,1

7.3. PRANDTL THEORY FOR TORSION 153

Substituting the above relationships into the equilibrium equation for the x3 direction oneobtains

σ31,1 + σ32,2 = 0

µβφ,21 − µβφ,12 = 0

Therefore equilibrium is automatically satisfied provided that φ,21 = φ,12. Note that we haveassumed the following stress state in the beam

σ11 = σ22 = σ33 = σ12 = 0

Comparing the Prandtl approach to the St. Venant approach we have the following relation-ships between the warping function and the Prandtl Stress function

φ,2 = ψ,1 − x2 (7.3)

φ,1 = −ψ,2 − x1 (7.4)

We can eliminate ψ by taking the partial derivative of Equation (7.3) with respect to x2 andadding it to the partial derivative of Equation (7.4) with respect to x1, i.e.

φ,22 + φ,11 = −1 − 1 = −2

O2φ = −2 (7.5)

On the outer boundary of the cross section we require traction free lateral surfaces therefore

t1 = σ31n1 + σ32n2 = 0

φ,2n1 − φ,1n2 = 0

where n1 and n2 are the components of the outward pointing unit normal vector to theboundary of the cross section. Letting s denote the distance measured along Γ, i.e. theouter edge of the cross section, we can compute the components of n as

n1 =dx2

dsn2 = −dx1

ds -dx1

dx2

n

β

ds

Therefore on the outer edge of the cross section we have

∂φ

∂x2

dx2

ds+∂φ

∂x1

dx1

ds= 0

dφ

ds= 0 on Γ


Thus in order to have traction free lateral surfaces the Prandtl stress function must be equalto a constant on the outer edge of the cross section. Typically for convenience one sets φ = 0on Γ.

Calculating the resultant shear forces on the cross section, V1 and V2 we get

V1 =

∫

A

σ31 dx1dx2 =

∫

A

µβφ,2 dx1dx2 = µβ

∫

A

φ,2 dx1dx2

using Green’s theorem

V1 = µβ

∮

Γ

φn2 dΓ

but φ = 0 on Γ thus

V1 = 0

Similarly one can shown that V2 = 0. The applied torque T is given by

T =

∫

A

[

− σ31x2 + σ32x1

]

dx1dx2 = µβ

∫

A

[

− x2φ,2 − x1φ,1]

dx1dx2

= µβ

∫

A

[

− (x2φ),2 − (x1φ),1]

dx1dx2 +

∫

A

(φ+ φ) dx1dx2

= µβ

∫

Γ

[

− x2φn2 − x1φn1

]

ds+ 2

∫

A

φ dx1dx2

T = 2µβ

∫

A

φ dx1dx2 (7.6)

Thus the applied torque is equal to 2µβ times the volume under the φ stress function. SinceT = µβJ , the effective torsional constant J is given by

J = 2

∫

A

φ dx1dx2 (7.7)

ddsφ

φ = constant

n

s


Consider the contour lines of constant φ. Along these lines we have

dφ

ds=

∂φ

∂x1

dx1

ds+∂φ

∂x2

dx2

ds= 0

0 = − ∂φ

∂x1

n2 +∂φ

∂x2

n1

Denoting the shear stress normal to the contour line as τn we have

τn = σ31n1 + σ32n2

τn = µβ[

φ,2n1 − φ,1n2

]

τn = µβdφ

ds= 0

Therefore there is no shear stress normal to the contour lines. Instead the shear stress istangential to the contour lines. The shear stress tangential to the contour line denoted byτs is given by

τ = τs = −σ31n2 + σ32n1

= −µβ[

φ,1n1 + φ,2n2

]

= −µβ(

Oφ · n)

Thus the largest shear stress is observed where the gradient of the Prandtl stress function islargest.

7.3.1 Example: Elliptical cross section

Consider once again the elliptic cross section. The Prandtl stress function is given by

φ = A[

(

x1

a

)2

+

(

x2

b

)2

− 1]

where A is a constant to be determined. One can easily show that on the outer boundary ofthe cross section φ = 0. To determine the value of A we require the Prandtl stress functionto satisfy the Poisson’s equation given by

O2φ = −2

A( 2

a2+

2

b2)

= −2

A = − a2b2

a2 + b2

The stresses are given by

σ31 = µβφ,2 = −µβ(

a2b2

a2 + b2

)(

2x2

b2

)

= −µβ 2a2x2

a2 + b2

σ32 = −µβφ,1 = µβ

(

2b2x1

a2 + b2

)


which are equal to the stresses obtained from the St. Venant theory. The effective torsionalconstant is given

J = 2

∫

A

φ dx1dx2 = −(

a2b2

a2 + b2

)∫

A

[

(

x1

a

)2

+

(

x2

b

)2

− 1)]

dx1dx2 =πa3b3

a2 + b2

which again is equal to the value obtained before.

Contour lines for Prandtl function(a/b=3)

Prandtl function (a/b=3)

7.3.2 Example:Rectangular cross section

a

1

x 2

b

x

Consider again a rectangular cross section of width b and height a. Solving for the Prandtlfunction can be done using a double Fourier series of the form

φ(x1, x2) =∞

∑

m=1

∞∑

n=1

φmn cos

(

mπx1

a

)

cos

(

nπx2

b

)

where φmn are the Fourier coefficients of the Prandtl function. It should be noted that theabove expression for φ is equal to zero along all four outer edges of the cross section. In


general, any function f(x1, x2) over the domain; −b/2 ≤ x1 ≤ b/2, −a/2 ≤ x2 ≤ a/2, canbe written as a double cosine series as

f(x1, x2) =∞

∑

m

∞∑

n

fmn cos

(

mπx1

b

)

cos

(

nπx2

a

)

where the Fourier coefficients are given by

fmn =4

ab

∫ a/2

−a/2

∫ b/2

−b/2f(x1, x2) cos

(

mπx1

b

)

cos

(

nπx2

a

)

For the case when f(x1, x2) = −2 we obtain for the Fourier coefficients

fmn =

(

−32ab

mnπ2

)

sin

(

mπ

2

)

sin

(

nπ

2

)

which will give values of 0 for fmn whenever m or n are even integers.Substituting the Fourier series for φ and -2 into Equation (7.5) one can solve for the

Fourier coefficents of φ as:

φ,11 + φ,22 = −2

−(

mπ

b

)2 ∞∑

m=1

∞∑

n=1

φmn cos

(

mπx1

a

)

cos

(

nπx2

b

)

−(

nπ

a

)2 ∞∑

m=1

∞∑

n=1

φmn cos

(

mπx1

a

)

cos

(

nπx2

b

)

=∞

∑

m

∞∑

n

fmn cos

(

mπx1

b

)

cos

(

nπx2

a

)

φmn = − fmn(

mπb

)2+

(

nπa

)2

Since the Fourier coefficients for φ will be 0 for even integers of m and n one can rewrite theFourier series to only include nonzero terms as

φ(x1, x2) =∞

∑

m=1

∞∑

n=1

φmn cos

(

(2m− 1)πx1

b

)

cos

(

(2n− 1)πx2

b

)

where

φmn =32ab sin

(

(2m−1)π2

)

sin(

(2n−1)π2

)

(2m− 1)(2n− 1)π4[

(

2m−1b

)2+

(

2n−1a

)2]

=32ab(−1)(m+1)(−1)(n+1)

(2m− 1)(2n− 1)π4[

(

2m−1b

)2+

(

2n−1a

)2]


As the values of m and n grow the magnitude of the Fourier coefficients φmn decreases thusone can obtain a very good estimate for the Prandtl function by including only a finitenumber of terms in the Fourier series, i.e.

φ(x1, x2) ≈N1∑

m=1

N2∑

n=1

φmn cos

(

(2m− 1)πx1

b

)

cos

(

(2n− 1)πx2

b

)

Here N1 and N2 are integers and N1N2 will be equal to the number of nonzero terms includedin the Fourier series. The maximum shear stress on the cross section will occur on theboundary of the rectangle and will be given by

τmax =

µβ∑N1

m=1

∑N2

n=1

(

(2n−1)πa

)

φmn(−1)(n+1) a ≤ b; at midpoint of side with length b

µβ∑N1

m=1

∑N2

n=1

(

(2m−1)πb

)

φmn(−1)(m+1) b < a; at midpoint of side with length a

For the case when a = 6 and b = 8 one obtains the following plots for the Prandtl functionfor the case when N1 = N2 = 20:

−4−2

02

4

−4

−2

0

2

40

2

4

6

8

x1x2

φ

Prandtl function for rectangular x-section(a = 6, b = 8)

−3 −2 −1 0 1 2 3−4

−3

−2

−1

0

1

2

3

4

1

2

3

4

5

6

x1

x2

Contour lines for Prandtl function for rec-tangular x-section (a = 6, b = 8)

The torsional constant is equal to 311.8 (N1 = N2 = 20) and the maximum shear stresswhich occurs at x1 = 0;x2 = ±3 is equal to 4.79µβ (N1 = N2 = 50).

7.3.3 Membrane Analogy

Consider a thin membrane stretched over a wire such that the edges of the membrane arefirmly attached to the wire. If a uniform pressure p is applied to the membrane, the verticaldeflection w(x1, x2) satisfies the following relationship

w,11 + w,22 = − p

T


where T is the tension that develops in the membrane which assumed to be constant through-out the membrane. This equation is analogous to the equation that the Prandtl stressfunction must satisfy, therefore one can observe that the deflection of the membrane isproportional to the Prandtl stress function and J is proportional to the volume under themembrane.

p

7.3.4 Hydrodynamic Analogy

For irrotational incompressible flow we use stream functions ψ such that

u1 = ψ,2 u2 = −ψ,1

where u1 and u2 are fluid velocity components in the x1 and x2 directions, respectively.Conservation of mass requires that the following equation be satisfied

u1,1 + u2,2 = 0

which is automatically satisfied provided the stream function satisfies the following

ψ,21 = ψ,12

Since the flow is assumed to be irrotational the vorticity in the fluid w is assumed to beconstant and

u1,2 − u2,1 = −2w

therefore the stream functions must satisfy

ψ,11 + ψ,22 = −2w

Thus the streamlines are analogous to lines of constant Prandtl stress function and representthe shear stress directions.


Fluid Flow Examples

Flow around a circle

B

A

uA = uB = 2U∞

Flow around an ellipse

B

A

2a

2b

uA = uB = U∞(1 +a

b)

One can use the hydrodynamic flow analogy to get a solution for the shear stresses thatdevelop in a shaft with a small elliptical hole. Consider pt A at the tip of the elliptical hole.The shear stress at point A will be given by

τA =TrAJ

(1 +a

b)

where T is the applied torsion, rA is the radial distance of point A from the centroid, and aand b are the major and minor axis of the elliptical notch. The above solution will only bevalid if the notch is small in comparison to the dimensions of the cross section.

A

7.4 Multiple Connected Domains

The membrane analogy can be employed to determine the shear stresses resulting from anapplied torsion on a cross section that has multiple connected domains. Consider the crosssection shown below in which domain Ω1 has a hole denoted by domain Ω2. Let Γ1 be theboundary of Ω1 and Γ2 be the boundary of Ω2. On both Γ1 and Γ2, the Prandtl stressfunction is constant. For convenience we set φ=0 on Γ1 and φ = c0 on Γ2.

7.4. MULTIPLE CONNECTED DOMAINS 161

φ

ΩΩ

Γ

12

1

Γ2n

φ=0

c0φ=

To obtain the value of c0 we make use of the fact that the integration of the warping functionaround Γ2 should sum up to zero, i.e.

∮

Γ1

dψ = 0

∮

Γ1

(ψ,1dx1 + ψ,2dx2) = 0

∮

Γ1

[

(φ,2 + x2)dx1 − (φ,1 + x1)dx2

]

= 0

∮

Γ1

(φ,2dx1 − φ,1dx2) =

∮

Γ1

(x1dx2 − x2dx1)

∮

(σ31

µβdx1 +

σ32

µβdx2) =

∮

Γ1

(x1n1 + x2n2)ds

1

µβ

∮

Γ1

(σ31dx1

ds+ σ32

dx2

ds) ds =

∮

Ω1

2dx1dx2

1

µβ

∮

Γ1

(σ31n2 + σ32n1) ds = 2A2

∮

Γ2

τ(s) ds = 2A2µβ

where τ(s) is the shear stress along the boundary curve Γ2 and A2 is the area containedwithin the domain Ω2.


7.4.1 Example:Elliptical cross section

b kb

a

ka

φ =

φ

−φ

1

2

Consider the elliptical cross section with minor axis length of b and major axis length ofa with an elliptical hole cut in the middle with minor axis length kb and major axis lengthak where k < 1. The Prandtl stress function φ is set equal to zero at the outer edge and setequal to a constant on the inner edge. Using the membrane analogy one can determine that

φ = φ1 − φ2

where φ1 is the Prandtl stress function for a solid ellipse with minor axis b and major axisa, and φ2 is the Prandtl stress function for a solid ellipse with minor axis kb and major axiska. The values of φ1 and φ2 are given by

φ1 =a2b2

a2 + b2µβ(1 −

(

x1

a

)2

−(

x2

b

)2

)

φ2 =a2b2k2

a2 + b2µβ(1 −

(

x1

ka

)2

−(

x2

kb

)2

)

The applied torque T is given by

T = 2

∫

A

φ dx1dx2 = 2

∫

A1

φ1 dx1dx2 − 2

∫

A2

φ2 dx1dx2

=πa3b3

a2 + b2µβ − πk4a3b3

a2 + b2µβ

T = µβπ

[

a3b3

a2 + b2

]

(1 − k4)

J = π

(

a3b3

a2 + b2

)

(1 − k4)

If a > b then the maximum shear stress occurs at (0, b) and is given by

τmax = Oφ · e2 =∂φ

∂x2

=a2b2

a2 + b2µβ

(

−2x2

b2

)

∣

∣

∣

x2=b= − 2a2b

a2 + b2µβ

Rewriting in terms of the applied torque T we obtain

τmax =2T

πab2(1 − k4)

7.5. CLOSED THIN WALL SECTIONS 163

7.4.2 Example:General cross section with hole

Consider the cross section shown above with a circular hole. One can solve for the Prandtlstress function by treating the problem as a steady state heat conduction problem with theconductivity κ = 1 and employing a finite element analysis. Furthermore, a temperature of0 is prescribed on the outer boundary of the cross section while on the edge of the hole oneconstrains all the nodes to have the same temperature. The resulting temperature field isequivalent to the Prandtl stress function φ. Plots of φ and the corresponding shear stressvectors due to torsion are shown below for the cross section.

5

10

15

20

25

30

35

40

45

50

7.5 Closed Thin Wall Sections

Consider the case of a closed thin wall section as shown below. The Prandtl stress functionwould be equal to zero on the outer edge and equal to a constant value on the inner edge.The midline of the wall is taken as a reference line and a coordinate system with one directiontangent to the reference line and the other coordinate perpendicular to the tangent line isemployed, i.e. s and η respectively. If the thickness of the wall is small then the Prandtlstress function changes little in the direction of s and most of the change occurs in the η


direction. It is important to note here that the thickness of the wall is not assumed to beconstant but is assumed to be small in comparison to the other dimensions of the crosssection.

φ=constant

t

referenceline

.

η

s

φ=0

η

φ

t/2-t/2

For convenience one defines the shear flow q as the integration of the shear stressesthrough the thickness of the wall, i.e.

q ≡∫ t/2

−t/2τ(s, η) dη = τ(s)t (7.8)

where the average shear stress τ is assumed to be given by τ(s, η = 0).

Consider the equilibrium of a slice of the body taken between two arbitrary points 1 and2 as shown below. Equilibrium in the x3 direction requires

τ1t1∆x3 = τ2t2∆x3

τ1t1 = τ2t2

q1 = q2

Since the two points are arbitrary one can conclude that the shear flow q must be constantthroughout the cross section.

x3

x3

x3τ ∆t1 1

x3τ ∆t2 2

2

1 1

2

x2

x1

∆


From the discussion of multiple connected domains we obtain the following equation∮

C0

τ ds = 2A0µβ

Here C0 is the curve defined by the midline of the thin wall section and A0 is the areaenclosed by the midline. Since at the midline we have τ = q

tthe above equation becomes

∮

C0

q

tds = 2A0µβ

q =2A0µβ∮

C0

dst

(7.9)

To calculate the torsion T one arbitrarily selects a point O enclosed by the the midline totake the moment of the shear stresses about. Denoting by p the moment arm of a smallshear force given by τ t ds = qds we obtain for torsion T

T =

∮

C0

p(s)q ds = q

∮

C0

pds

O

ds

qds

p

O

dsp

dA

Denoting by dA the infinitesimal sectorial area enclosed by ds one can show the followingholds

dA =1

2p ds

ThusT = 2A0q (7.10)

The largest stresses occur where the thickness is smallest as

q =T

2A0

=⇒ τ =T

2A0t(7.11)

Employing Equations (7.9) and (7.11) one obtains

T

2A0

=2A0µβ∮

dst

T = µβ

(

4A20

∮

dst

)

= µβJ


J =4A2

0∮

dst

(7.12)

7.5.1 Examples

ta

ta

tb tb

2a

2bA0 = (2a− tb)(2b− ta)

∮

ds

t=

4b− 2tatb

+4a− 2tb

ta

Consider a hollow circular cross section with constant thickness t. From section 7.4.1 wehave the effective torsional constant equal to

r

t

r

J =π

2(r4 − (r − t)4) (7.13)

where we have substituted k = 1− tr

and a = b = r. Here r is the outside radius of the hollowcircular cross-section. The midline of the cross section is given by the following equation

x21 + x2

2 = r2

where the radius of the midline r is equal to r− t2. According to the thin wall approximation

the effective torsional constant is given by

A0 = πr2 = π(r − t

2)2

∮

ds

t=

1

t

∫ 2π

0

r dθ =1

t2πr =

2π(r − t2)

t

J =4A2

0∮

dst

=π

4r4(2 − t

r)3 t

r(7.14)

Plotting Jr4

versus tr

for Equations (7.13) and (7.14) one obtains


J vs. tr

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.2 0.4 0.6 0.8 1

phi forumulationthin wall approx

tr

J r4

For values of tr< 0.373 the difference in the values of the effective torsional constant J pre-

dicted by the two formulations is less than 5%. Plotting the ratio of the constant shear stresspredicted by the thin wall approximation (τt) versus the maximum shear stress predicted bythe more accurate φ formulation (τ0) we obtain

τt =T

2A0t=

T

2π(r − t/2)2t

τ0 =Tr

J=

Trπ2(r4 − (r − t)4)

Ratio of Shear stresses vs tr

0.82

0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.98

1

0 0.2 0.4 0.6 0.8 1

ylab

el

tr

For all values of tr

the difference in the shear stresses predicted by the two formulas is lessthan 20%. However for the difference to be less than or equal to 10% we require that t

r≤ 0.24


or tr≥ 0.875.

7.6 Multi Cell Tubes

q - q

q1

1 2

2q

A

A01

02

A

q3

q2

q1

03

Let us consider the case of a thin wall cross section made of two or more closed cells. Eachcell will aid in resisting the torque applied to the cross section. If we denote the torqueresisted by cell i as Ti, the area enclosed by the midline of cell i as A0i, and the shear flowthat develops in cell i as qi then we have

T = T1 + T2 + · · · + Tn = 2A01q1 + 2A02q2 + · · · + 2A0nqn

T = 2n

∑

i=1

A0iqi

Here n is the total number of cells in the cross section. Compatibility of strains requires thatall cells undergo the same rate of twist β thus

β1 = β2 = · · · = βn = β

2A01µβ =

∮

1

qds1

t

2A02µβ =

∮

2

qds2

t...

2A0nµβ =

∮

n

qdsnt

where the total shear flow q must be determined as the vectorial sum of the individual shearflows qi in walls that are shared by two or more cells.

7.6. MULTI CELL TUBES 169

7.6.1 Example

21a aa

a

a 2a

2a

q q1 2

t = constant

A01 = a2 A02 = 2a2

T = 2n

∑

i=1

A0iqi

T = 2a2q1 + 4a2q2 (7.15)

2A0iµβ =

∮

qds

t

from cell 1

2a2µβ =3aq1t

+(q1 − q2)a

t

2aµβt = 4q1 − q2 (7.16)

from cell 2

4a2µβ =5aq2t

+(q2 − q1)a

t

4µβat = 6q2 − q1 (7.17)

Solving Equations (7.16) and (7.17) for q1 and q2 we obtain

q1 =16

23µβat q2 =

18

23µβat

substituting the values of q1 and q2 into Equation (7.15) we obtain for the effective torsionalconstant

T = 2a2(16

23µβat) + 4a2(

18

23µβat)

=104

23µβa3t = µβJ

J =104

23a3t

τ =q

t


2

1623

aτ= µβ

1623

aτ= µβ

1623

aτ= µβ

2318τ= µβ

2318τ= µβ

2318τ= µβ23

τ= µβ

7.7 Open Thin Wall Section

η

φ

s

=0

φ=0φ=0

φ=0

b

b

bb=b + b

1

21 2

ηs

φ=0

φ=0

Applying the membrane analogy to a thin wall open cross section results in shear flows asshown above. If we denote η as the direction perpendicular to the midline of the cross sectionand η = 0 corresponding to the midline then one can see that sufficiently far away from theends and any corners, the Prandtl stress function becomes independent of s the distancealong the midline. Therefore we have

φ(s, η) = φ(η)

O2φ = −2 = φ,ηη

φ = −η2 + c1η + c2

where c1 and c2 are constants of integration. Since we require that φ = 0 on η = ± t2

thenwe have

φ =t2

4− η2

τ =∂φ

∂ηµβ = −2µβη

η= -t/2

η= t/2

η= t/2

η= -t/2

µβ tφ τ

t−µβ

7.7. OPEN THIN WALL SECTION 171

J = 2

∫

A

φ dA = 2

∫ b

0

∫ t/2

−t/2

(t2

4− η2

)

dηds

= 2

∫ b

0

1

6t3 ds =

1

3

∫ b

0

t3 ds

If the thickness t is constant then the above simplifies to

J =t3b

3

7.7.1 Examples

bt

t

11

2

b 2

J =1

3b1t

31 +

1

3b2t

32

a

t

Jclosed =4A2

0∮

dst

= 2πa3t

at

Jopen =1

3bt3 =

2πat3

3

βopen

βclosed

=Jclosed

Jopen

=3a2

t2


7.7.2 Shear Center and Center of Twist

The shear center of a cross section is defined as the point at which the transverse loadingapplied to a prismatic member must pass in order for no twisting to result. A prismaticmember that is not loaded through its shear center will bend and twist. The center of twist

of a cross section is defined as the point about which the entire cross section rotates whensubjected to a pure twist. If a cross section has two lines of symmetry then the centroid,shear center, and center of twist are all coincident. It is generally agreed that the center oftwist and the shear center are coincident for unsymmetric cross sections.

shear center

=centroid

d

FdF

F

7.7.3 Sectorial Properties

One defines the sectorial area ω(s) of a curve that is a function of distance along the curve,s, as

ω(s) =

∫ s

0

p(s) ds (7.18)

Here p(s) is the perpendicular distance from a point P , termed the pole, to a tangent lineto the curve at point s. The area swept as s goes from 0 to s is given by 1

2

∫ s

0p(s) ds thus

the sectorial area w(s) is twice the area swept as s goes from 0 to s. If the area swept iscounterclockwise then w(s) is considered positive. If the area swept is clockwise then w(s)is considered negative.

ds

p(s)

P

2d (s)ω

P

(s)/2

s=0

s

ω

It should be noted that the selection of point P and the selection of the point I where s=0will both change the value of the sectorial areas. If the pole is fixed and the point I ischanged the sectorial areas will change by a constant value.


Consider the case when the point I is fixed and two different poles P1 and P2 are consid-ered. If one denotes two sets of parallel axes as x1, x2 and x′1, x

′2 with origins at P1 and P2

respectively, then

ωP2(s) = ωP1

(s) + a∆x2 − b∆x1

where ωP2(s) is the sectorial area with respect to the pole at point P2, ωP1

(s) is the sectorialarea with respect to the pole at point P1, b is the perpendicular distance between x1 and x′1,a is the perpendicular distance between x2 and x′2, and

∆x1 = x1(s) − x1(s = 0) ∆x2 = x2(s) − x2(s = 0)

x 1

x 1

x2

x2s=0

s

P

P

1

2

a

bx 1

x2

∆

∆

Choosing the x1 axis and the x2 axis as centroidal axes and setting the pole at theshear center one can define the following sectorial properties for thin wall sections where thesectorial areas are calculated using the midline curve:

Sectorial static moment: Sω =

∫

A

ω dA =

∫

C0

ω(s)t(s) ds

Sectorial linear moment: Sx1ω =

∫

A

x1ω dA =

∫

C0

x1(s)ω(s)t(s) ds

Sectorial linear moment: Sx2ω =

∫

A

x2ω dA =

∫

C0

x2(s)ω(s)t(s) ds

Sectorial moment of inertia: Jω =

∫

A

ω2 dA =

∫

C0

ω(s)2t(s) ds

The principal sectorial areas are obtained when the point I is chosen such that the sectorialstatic moment Sω=0. One can choose an arbitrary point I where s = 0 and calculate thesectorial areas ω(s). Since the sectorial areas are changed by a constant value when the poleis fixed and the point I is changed then we have the following:

ωP (s) = ω(s) − ω0 and

∫

A

ωP dA =

∫

C0

ωP (s)t(s) ds = 0


thus

ω0 =1

A

∫

A

ω dA =1

A

∫

C0

ω(s)t(s) ds

where ωP corresponds to the principal sectorial areas. If the principal sectorial areas areused and x1 and x2 correspond to the centroidal axes then the following holds

Sω = 0 Sx1ω = 0 Sx2ω = 0

and Jω will be equal to the warping constant for the cross section which plays a role in caseswhere the warping of the cross section is restrained.

7.7.4 Example

1x

x

s

a

a

t

a a

2

s=0s=a s=3.828a

s=2.414a

s=−a

s=−2.414a

s=−3.828a

Consider the cross section shown above with constant thickness t. Since the cross sectionhas two lines of symmetry the center of twist, the shear center, and the centroid all coincideat point S. Also the axis x1 and x2 are centroidal axes since they pass through the centroid.In order to calculate the principal sectorial areas we require that the pole be placed at theshear center which in this example coincides with point S. Arbitrarly we select a point onthe midline where we set s = 0. In general this will result in sectorial areas which are not

principal sectorial areas. However the sectorial areas will differ from the principal sectorialareas by a constant ω0.

For the line segment from s = 0 to s = a we have

p(s) = 0 =⇒ ω(s) =

∫ s

0

0 ds = 0 0 ≤ s ≤ a


For the line segment from s = a to s = a(1 +√

2) = 2.414a the area swept is clockwise thuswe have

p(s) =a√2

ω(s) = ω(s = a) −∫ s

a

a√2ds

ω(s) = 0 − as√2

∣

∣

∣

s

a= − as√

2+

a2

√2

a ≤ s ≤ a(1 +√

2)

ω(a) = 0 ω(2.414a) = −a2

For the line segment from to s = a(1 +√

2) = 2.414a to s = a(1 + 2√

2) = 3.828a the areaswept is clockwise thus we have

p(s) =a√2

ω(s) = ω(s = 2.414a) −∫ s

2.414a

a√2ds

ω(s) = −a2 − as√2

∣

∣

∣

s

2.414a= −a2 +

a√2(1 +

√2)a− as√

2

=a2

√2− as√

2a(1 +

√2) ≤ s ≤ a(1 + 2

√2)

ω(2.414a) = −a2 ω(3.828a) = −2a2

For the line segment from s = −a to s = 0 we have

p(s) = 0 ω(s = 0) − ω(s) =

∫ 0

s

0 ds

ω(s) = 0 − a ≤ s ≤ 0

For the line segment from to s = −a(1 +√

2) = −2.414a to s = −a the area swept iscounterclockwise thus we have

p(s) =a√2

ω(−a) − ω(s) = +

∫ −a

s

a√2ds

ω(s) = ω(−a) − as√2

∣

∣

∣

−a

s= 0 +

as√2

+a2

√2

ω(s) =a√2(s+ a) − a(1 +

√2) ≤ s ≤ −a

ω(−a) = 0 ω(−2.414a) = −a2


For the line segment from to s = −a(1+2√

2) = −3.828a to s = −a(1+√

2) = −2.414a thearea swept is counterclockwise thus we have

p(s) =a√2

ω(−a(1 +√

2)) − ω(s) = +

∫ −a(1+√

2)

s

a√2ds

ω(s) = ω(−a(1 +√s)) − as√

2

∣

∣

∣

−a(1+√

2)

s= −a2 − as√

2+ a2(

1√2

+ 1)

ω(s) =a√2(a+ s) − a(1 + 2

√2) ≤ s ≤ −a(1 +

√2)

ω(−2.414a) = −a2 ω(−3.828a) = −2a2

Plotting the sectorial areas we obtain

ω=0

2 −a2

−a2 −a2

−2a2

−2a2

−a

Calculating the constant ω0 we obtain

ω0 =1

A

∫

C0

ω(s)t(s) ds

A = t(4(a√

2 + 2a)) = 2at(1 + 2√

2)∫

C0

ω(s)t(s) ds = t[1

2(a√

2)(−a2) +1

2(−a2 − 2a2)(a

√2)

+1

2(a√

2)(−a2) +1

2(−a2 − 2a2)(a

√2)

]

= −a3t4√

2

ω0 =−a3t4

√2

2at(1 + 2√

2)= −0.7388a2

Therefore the principal sectorial areas are obtained from

ωP (s) = ω(s) − ω0 = ω(s) + 0.7388a2

Plotting the principal sectorial areas we obtain

7.8. WARPING OF THIN WALL CROSS SECTIONS IN PURE TORSION 177

0.7388a2

−0.2612a2 −0.2612a2

−1.2612a2

0.7388a2

−0.2612a2−0.2612a2

−1.2612a2

0.7388a2ω=

Calculating the warping constant Jω by using the integration aid on page 287 of the textbookwe obtain

Jω =

∫

C0

ω2P (s)t(s) ds

= t[

2a(0.7388a2)2 +a√

2

3

(0.7388a2)2 + (0.7388a2)(−0.2612a2) + (−0.2612a2)2

+a√

2

3

(−0.2612a2)2 + (−0.2612a2)(−1.2612a2) + (−1.2612a2)2

+a√

2

3

(0.7388a2)2 + (0.7388a2)(−0.2612a2) + (−0.2612a2)2

+a√

2

3

(−0.2612a2)2 + (−0.2612a2)(−1.2612a2) + (−1.2612a2)2

]

Jω = 3.363a5t

7.8 Warping of Thin Wall Cross Sections in Pure Tor-

sion

dx3 τ

ττ

τ

d(p )dx3

θ

x

x

x

s

1

3

2

T

dx3ds

ds

duds

3


Consider a thin wall cross section under pure torsion. The shear center of the cross sectionis chosen as the origin for the x1-x2 coordinate system. The midline of the cross sectionis chosen as a reference line and parametrized using s the distance along the midline. p(s)denotes the perpendicular distance of line segment ds from the shear center. If one considersthe straining of an infinitesimal rectangular piece with side lengths ds and dx3 that lies onthe lateral surface created by the midline of the bar, then due to the shear stress caused bytorsion the rectangular piece will experience a shear strain γ given by

γ =du3

ds+d(pθ)

dx3

Here u3 is the warping displacement in the x3 direction (out of the plane of the cross section),θ is the twist of the cross section caused by torsion, and p = p(s) as defined above. Sincethe perpendicular distance p of the line segment ds from the shear center is only a functionof s and not x3 we obtain for the shearing strain

γ =τ

µ=du3(s)

ds+ p(s)β

where β is the rate of twist and is equal to dθdx3

. Rewritting in terms of du3

dswe obtain

du3(s)

ds=τ

µ− p(s)β =

q

µt− p(s)β (7.19)

Integrating Equation (7.19) with respect to s, one obtains

u3(s) = u3(s = 0) +

∫ s

0

q

µtds−

∫ s

0

p(s)β ds

u3(s) = u3(0) +q

µ

∫ s

0

ds

t− β

∫ s

0

p(s) ds

For a closed thin wall cross section the shear flow q is given by

q =T

2A0

where T is the applied torsion and A0 is the area enclosed by the midline. Also the sectorialarea ω(s) is defined as

∫

p(s) ds and the rate of twist is given by

β =T

µJ=

T

4A20µ

∮

ds

t

Therefore the equation for the warping displacement u3 of a closed thin wall cross sectionbecomes

u3(s) = u3(0) +T

2A0µ

∫ s

0

ds

t− Tω(s)

4A20µ

∮

ds

t(7.20)

7.8. WARPING OF THIN WALL CROSS SECTIONS IN PURE TORSION 179

For an open thin wall cross section we have q=0 and

β =T

µJ=

3T

µ∫ b

0t3(s) ds

Therefore the equation for the warping displacement u3 of an open thin wall cross sectionbecomes

u3(s) = u3(0) − 3Tω(s)

µ∫ b

0t3(s) ds

(7.21)

7.8.1 Examples

a

t

at

For the case of a closed thin wall circular cross section with constant thickness t where ais the radius of the midline we obtain the following values

A0 = πa2

∮

ds

t=

2πa

tω(s) = as

u3(s) = u3(s = 0) +T

2πa2µ

s

t− Tas2πa

4π2a4µt

u3(s) − u3(s = 0) = 0 ∀ s∴ u3(s) = 0 ∀ s no warping occurs


For the case of an open thin wall circular cross section with constant thickness t and radiusof the midline a we obtain the following values

∫ 2π

0

t3a dθ = 2πat3

u3(s) = u3(0) − 3Tω(s)

µ∫ 2π

0t3a dθ

= u3(0) − 3Tas

2πµat3= u3(0) − 3Ts

2πµt3

u3(s) − u3(0) = − 3Ts

2πµt3

u3(2πa) − u3(0) = −3Ta

µt3

Plotting the nondimensional warping displacement u3µt3

Tavs (x1

a,x2

a) we obtain

-1 -0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1 -1-0.8

-0.6-0.4

-0.2 0

0.2 0.4

0.6 0.8

1

-2.5-2

-1.5-1

-0.5 0

0.5 1

x1

x2

u3µt3

Tau3µt3

Ta

Now consider the case of a rectangular closed thin wall section as shown below. The distancealong the midline s is measured positive counterclockwise and starts at 0 midway betweencorners 1 and 4.

7.9. AXIAL STRESS IN THIN WALL SECTIONS DUE TO WARPING 181

ta

tbtb

ta

3

2 1

4

2a

2bs=0

s=bs=b+2a

s=3b+2a

s

s

A0 = 4ab∮

C0

=ds

t=

2a

ta+

2b

tb+

2a

ta+

2b

tb

= 4(a

ta+b

tb)

ω(b) = 21

2ab = ab

u3(b) = u3(0) +T

2A0µ

[

b

tb− ω(b)

2A0

(4)(a

ta+b

tb)

]

u3(b) = u3(0) +T

4A0µ

[ b

tb− a

ta

]

ω(b+ a) = ω(b) + 21

2(a)(b) = 2ab

u3(a+ b) = u3(b) +T

2A0µ

[

b

tb+a

ta− ω(b+ a)

2A0

(4)(a

ta+b

tb)

u3(a+ b) = u3(b) +T

2A0µ

[

b

tb+a

ta− b

tb− a

ta

]

u3(a+ b) = u3(b)

Therefore provided that ata

= btb

no warping will occur in the cross section.

7.9 Axial Stress in Thin Wall Sections due to Warping

So far we have only considered cases of pure twist with no restraint on warping. This leadsto a constant rate of twist β. If the torque applied is not constant and/or if warping isrestrained then the rate of twist will not be constant. Furthermore additional stresses willdevelop to maintain equilibrium in the body. These stresses will contain both axial stressand secondary shear stresses.

For an open thin wall cross section we have the following:

u3(s) = u3(0) − βω(s)

u3,3 = −β,3ω(s) = ε33


where u3 is the warping displacement. If the warping displacement is constant with respectto x = x3 then the axial strain will be equal to zero. However if the cross section is restrainedagainst warping and/or if β 6= constant then axial stresses will develop given by

σ33 = Eyε33 since σ11 = σ22 = 0

σ33(s, x) = −Eyd2θ

dx2ω(s) = −Ey

dβ

dxω(s)

For a closed thin wall cross section we have the following:

u3(s) = u3(s) +q

µ

∫ s

0

ds

t− βω(s)

u3,3 = ε33 =1

µ

dq

dx3

∫ s

0

ds

t− dβ

dx3

ω(s)

σ33(s, x) =Eyµ

dq

dx

∫ s

0

ds

t− Ey

d2θ

dx2ω(s)

7.9.1 Secondary Shear Stresses

dx3

dσ33

dx3

dτ s

1

2

σ33 σ +33

τ s

sτ +

Due to the axial stress that may develop from restraint of warping and/or unconstant ap-plied torsion there results additional shear stresses called secondary shear stresses. Considera small rectangular isolated piece of the thin wall with sides of length dx and ds. Forequilibrium in the x = x3 direction we require that the following equation hold:

(σ33 +∂σ33

∂xdx)tds− σ33tds+ (τs +

∂τs∂s

ds)t2dx− τst1dx = 0

∂σ33

∂xtds+

∂τs∂s

dsdxt2 + τs(t2 − t1)dx = 0

where t1 is the thickness at s, t2 is the thickness at s + ds, t is the average thickness froms to s + ds, and τs is the secondary shear stress. Dividing the above equation by dxds oneobtains

∂σ33

∂xt+

∂τs∂s

t2 + τs

(

t2 − t1ds

)

= 0

Taking the limit as ds → 0 we get

∂σ33

∂xt(s) +

∂τs∂s

t(s) + τsdt(s)

ds= 0

7.9. AXIAL STRESS IN THIN WALL SECTIONS DUE TO WARPING 183

or∂σ33

∂xt(s) +

∂qs∂s

= 0

where qs = τst is the secondary shear flow.For an open thin wall cross section the secondary shear flow is given by

∂qs∂s

= − ∂

∂x

[

− Eyd2θ

dx2ω(s)

]

t(s)

= Eyd3θ

dx3ω(s)t(s)

qs(s, x) = Eyd3θ

dx3

∫ s

0

ω(s)t(s) ds

In the above equation a constant of integration has been omitted. This is permitted as longas the principal sectorial areas are used for ω(s) and qs(s = 0) = 0.

For a closed thin wall cross section the secondary shear flow is given by

∂qs∂s

= − ∂

∂x

[Eyµ

dq

dx

∫ s

0

ds

t− Ey

d2θ

dx2ω(s)

]

t(s)

= −Eyµ

d2q

dx2t(s)

∫ s

0

ds

t+ Ey

d3θ

dx3ω(s)t(s)

qs(s, x) = Eyd3θ

dx3

∫ s

0

ω(s)t(s) ds− Eyµ

d2q

dx2

∫ s

0

t(s)

∫ ξ

0

dξ

t(ξ)

ds

For the primary shear flow q of a closed thin wall cross section we have

q(x) =2A0µβ∮

dst

thusd2q

dx2=

2A0µd2βdx2

∮

dst

The equation for the secondary shear flow in a closed thin wall cross section then becomes

qs(s, x) = Eyd3θ

dx3B(s) (7.22)

where

B(s) =

∫ s

0

ω(s)t(s) ds− 2A0∮

dst

∫ s

0

(

∫ ξ

0

dξ

t(ξ)

)

t(s) ds

Once again the principal sectorial areas should be used to ensure no net shear force resultsfrom the secondary shear stresses.


7.10 Torsion with Warping Restrained in Open Sec-

tions

For the case where warping is constrained in an open thin wall cross section the total appliedtorque will be resisted by two internal torques, the primary torque (Tp) predicted by ele-mentary torsional theory and the secondary torque (Ts) resulting from the secondary shearstresses, i.e.

T = Tp + Ts

For an open thin wall cross section the secondary torque is given by

Ts =

∫ b

0

qsp(s) ds =

∫ s=b

s=0

qs dω

Using integration by parts

Ts = qsω∣

∣

∣

s=b

s=0−

∫ s=b

s=0

ω(s)dqsds

ds

but qs is equal to 0 at the ends of the open cross section thus

Ts = −∫ s=b

s=0

ω(s)[

Eyd3θ

dx3ω(s)t(s)

]

ds

= −Eyd3θ

dx3

∫ s=b

s=0

ω(s)st(s) ds

= −Eyd3θ

dx3

∫

A

ω2(s) dA

Ts = −Eyd3θ

dx3Jω

where Jω is the warping constant of the cross section. Therefore the total applied torque inan open thin wall cross section can be written as

T = Tp + Ts = µβJ − Eyd2β

dx2Jω (7.23)

7.10.1 Example

Consider a prismatic cantilever beam with a constant applied torque T . At the fixed endwarping is restrained and thus the boundary condition requires that β = 0. At the free endwe require the axial stresses to be equal to zero thus dβ

dxmust be equal to zero.

7.10. TORSION WITH WARPING RESTRAINED IN OPEN SECTIONS 185

From Equation (7.23) for β we obtain

d2β

dx2− κ2β = −β0κ

2

where

κ2 =µJ

EyJωβ0 =

T

µJ

One can think of β0 as the constant rate of twist that would result if warping was notrestrained. Solving the above equation by assuming a solution of the form β(x) = e±κx weobtain

β(x) = C1 sinh(κx) + C2 cosh(κx) + β0

where C1 and C2 are constants of integration. Applying the boundary conditions

β(x = 0) = 0dβ

dx

∣

∣

x=L= 0

one obtains for β

β(x) = β0

[

1 − cosh(κ(x− L))

coshκL

]

Plotting ββ0

versus xκL

for various values of κL one obtains

β(x)β0

vs xL

for cantilever beam

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1xL

β(x

)β0 κL=1

κL=5

κL=10

κL=30


As the value of κ increases the effects of restraint of warping die out very quickly as onemoves away from the fixed end. For closed thin wall cross sections κ has a large value andnot much error is introduced in the stresses calculated away from the fixed end when warpingrestraint is neglected. For an open thin wall cross section the value of κ can be very smallthus leading to inaccurate stress calculations if warping restraint is neglected.

If the cross section is an open thin wall cross section the resulting axial stresses due torestraint of warping are given by

σ33 = −Eyω(s)dβ

dx= Eyω(s)β0κ

(

sinh(κ(x− L))

cosh(κL)

)

The largest axial stresses will occur at the fixed end where x = 0

(σ33)max = −Eyω(s)Tκ

µJtanh(κL)

7.11 I-Beam under Torque

t

t

b

aw

f

f

T

Consider an I-beam under torque with one end restrained against warping. Due to restraintof warping axial stresses will develop in the flanges of the I-beam. These axial stresses areself equilibrating and die out quickly as one moves away from the fixed end. The principalsectorial diagram for the I-Beam is shown below.

b af

4

b af

4

4

−b af

4

−b af

M

M

f

f

The corresponding axial stresses for an open thin wall cross section are proportionalto the principal sectorial areas. The axial stresses will result in a pair of Bi-moments Mf

7.11. I-BEAM UNDER TORQUE 187

developing in the top and bottom flanges. Because of the bi-moments the top and bottomflanges will be subjected to bending about the weak axis. Denoting by vf the transversedisplacement of the flanges we have

EyIfd2vfdx2

= Mf (7.24)

where If is the moment of inertia of the flanges about the weak axis and is given by If =112b3f tf . Since the cross section has two axes of symmetry the center of twist, the shear center,

and the centroid of the cross section all coincide. The transverse displacement of the flangesvf must satisfy the following equation

vf = −θa2

(7.25)

where θ is the twist produced by the applied torsion and a2

is the perpendicular distanceof the flange centroid from the center of twist for the cross section. Substituting Equation(7.25) into Equation (7.24) gives

Mf = −EyIfa

2

d2θ

dx2

Equilibrium requires that a Bi-Shear force Vf develop in the flanges when the bi-moment isnot constant, i.e

V

Vf

fVf =

dMf

dx= −EyIf

a

2

d3θ

dx3

The bi-shears are also self equilibrating and die out quickly away from the fixed end. Thebi-shear is the resultant force of the secondary shear stresses that develop in the top andbottom flanges. When a torque T is applied, a primary torque Tp and a secondary torqueTs develops. Thus

T = Tp + Ts = µJβ + aVf

= µJβ − aEyIfa

2

d2β

dx2

= µJβ − Eya2b3f tf

24

d2β

dx2


Therefore the warping constant for an I-Beam is given by

Jω =1

2Ifa

2 =a2b3f tf

24

Since the effective torsional constant is given by

J ≈ 1

3

(

2bf t3f + at3w

)

we have for the constant κ

κ =2

abf

√

2µ(2bf t3f + at3w)

Ey(bf tf )

notes6

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