aem 4501 s15 lecture notes torsion(1)

56

4 Torsion of Solid Sections: General Cross-Sections

Recall torsion of a circular section

T d T z

yuθ ur

θ x

assumed that planesections remained plane andjust rotated about z

So, we get

uθ = Θr and ur = 0 where Θ is the angle of twist

Then

u = ux = −uθ sin θ

v = uy = uθ cos θ

Thus

u = −Θr sin θ = −Θy

v = Θr cos θ = Θx

and w = uz = 0

Now, we guess that this is almost right for non-circular sections and leave rota-tion alone, but let w(x, y, z) = w(x, y) displacement, the warping of the section,be unknown but independent of z.

4.1 Prandtl’s stress function solution

Based on these assumptions about the displacemnts, and the isotropic natureof the material, we can make an educated guess for the stress fields in the barand then use the equations of 3D elasticity to see if our guess can be made towork.

03/18/15 — For individual use only. Do not distribute without permission.Copyright 2014–2015, Ryan S. Elliott

57

A reasonable guess for the stress field is the following:

σxx(x, y, z) = σyy(x, y, z) = σzz(x, y, z) = 0,

σxy(x, y, z) = 0,

σxz(x, y, z) = σxz(x, y) = 0,

σyz(x, y, z) = σyz(x, y) = 0.

That is, all stress components are zero, except the out-of-plane shear stresseswhich are the same for all cross-sections (independent of z).

Under this assumption for the stress fields, the equilibrium equations reduceto

∂σzx

∂x+

∂σzy

∂y= 0

Now, define

σzx ≡∂φ

∂yand σzy = −

∂φ

∂x

for some function φ(x, y).

Q: Why make this choice?

A: Equilibrium is identically satisfied for any φ. Check:

∂

∂x

!∂φ

∂y

"

+∂

∂y

!

−∂φ

∂x

"

=∂2φ

∂x∂y−

∂2φ

∂y∂x= 0

Since we start with stresses (not displacements), we need to make sure compat-ibility is satisfied. Six equations reduce to two:

∂

∂x

!

−∂ezy∂x

+∂ezx∂y

"

= 0

∂

∂y

!∂ezy∂x−

∂ezx∂y

"

= 0

Plug in stresses from above (recall ezx = σzx

2G , etc.) and we find

1

2G

∂

∂x

!∂2φ

∂x2+

∂2φ

∂y2

"

=1

2G

∂

∂x

#

∇2φ$

% &' (

Laplacian of φ

= 0

and


58

−1

2G

∂

∂y

!∂2φ

∂x2+

∂2φ

∂y2

"

= −1

2G

∂

∂y

#

∇2φ$

= 0

These imply that ∇2φ = F is a constant. Boundary conditions on the laterialsurfaces are [t] = [σ][n] = [0] or

σzxnx + σzyny = 0 for traction-free lateral side

Substituting for the shears:

∂φ

∂ynx −

∂φ

∂xny = 0

y

ds n

ny

nxdx

dy

x

Consider S as arc-length, then ds ⊥ n

ds

ds 1

n

ny

nx(−dx)

dyθ

θ


59

so

dy

ds=

nx

1and −

dx

ds= ny

Substituting into B.C.

∂φ

∂y

dy

ds−

∂φ

∂x

!

−dx

ds

"

= 0 or

∂φ

∂x

dx

ds+

∂φ

∂y

dy

ds= 0

This is just the chain rule for

dφ

ds= 0

Thus, φ is a constant on the boundary. Without loss of generality, we can takeφ = 0 on the boundary. Now, compute the torque – we have

n

σzy

σzx

x

y

so, T =

##

A

$

−σzxy + σzyx%

dx dy

=

##

A

$

−∂φ

∂yy −

∂φ

∂xx%

dx dy

Consider each term in turn:

##

A

y∂φ

∂ydA integration by parts ⇒

#

∂A

yφ ds−##

A

φ dA

Similarly,##

A

x∂φ

∂xdA =

#

∂A

xφ ds −##

A

φ dA


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so (noting that φ is zero on the boundary) the line integrals disappear, resultingin

T =

!!

A

φ dA+

!!

A

φ dA = 2

!!

A

φ dA

We still have not involved the twist per unit length dΘdz , so let’s look at dis-

placements. From our original assumptions for the displacements (u(x, y, z) =−Θ(z)y, v(x, y, z) = Θ(z)x and w(x, y, z) = w(x, y), we have that the out-of-plane shear strains are:

exz =1

2(∂u

∂z+

∂w

∂x) =

1

2(−

dΘ

dzy +

∂w

∂x) =

1

2G

∂φ

∂y,

eyz =1

2(∂v

∂z+

∂w

∂y) =

1

2(dΘ

dzx+

∂w

∂y) = −

1

2G

∂φ

∂x.

Take ∂/∂x of exz and ∂/∂y of eyz, then subtract to obtain:

−2GdΘ

dz= ∇2φ = constant

Thus,

dΘ

dz=∇2φ

−2G

and we usually define

T = GJdΘ

dzwhere GJ is the torsional rigidity.

So GJ =TdΘdz

=−2G∇2φ

"

2

!!

A

φ dA

#

=−4G∇2φ

!!

A

φ dA

or

J =−4∇2φ

!!

A

φ dA

So if we find φ such that


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∇2φ = −2GdΘ

dzand φ = 0 on ∂A

Then this equation tells us the torsional rigidity of the section.

4.2 Torsion of an ellipse and the Membrane Analogy

The equation of an ellipse is

x2

a2+

y2

b2= 1

ba

Recall: φ – Prandtl stress function

• equilibrium is automatic

• compatibility: ∇2φ = −2GdΘdz

• boundary condition: φ = 0 on ∂A

Stresses are

σzx =∂φ

∂yand σzy = −

∂φ

∂x

• torque is: T = 2!!

A

φ dA

So, we need φ = 0 on ∂A, i.e., for x2

a2 + y2

b2 = 1

Try

φ = c

"x2

a2+

y2

b2− 1

#

then

∇2φ = c

"2

a2+

2

b2

#

= 2c

"1

a2+

1

b2

#

= −2GdΘ

dz

So


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c = −(

1

a2+

1

b2

)−1

GdΘ

dz

(

recall1

a2+

1

b2=

a2 + b2

a2b2

)

giving

c = −a2b2

a2 + b2GdΘ

dz

Finally,

φ = −a2b2

a2 + b2GdΘ

dz

(x2

a2+

y2

b2− 1

)

Then the torque is

T = 2

∫∫

A

φ dA =−2a2b2

a2 + b2GdΘ

dz

⎡

⎣1

a2

∫∫

A

x2 dA+1

b2

∫∫

A

y2 dA−∫∫

A

dA

⎤

⎦

Now,∫∫

A

x2 dA = Iy =πa3b

4

∫∫

A

y2 dA = Ix =πb3a

4

∫∫

A

dA = A = πab

so

T = −2a2b2

a2 + b2GdΘ

dz

[1

a2πa3b

4+

1

b2πb3a

4− πab

]

= GdΘ

dz

πa3b3

(a2 + b2)


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So

GJ =TdΘdz

= Gπa3b3

(a2 + b2)

if a = b = R, we have a circle and we get

T = GdΘ

dz

!πR6

2R2

"

= GdΘ

dz

πR4

2

recall J = πR4

2 for a circle.

Plot φ over the cross-section

z φ(x, y)

x

It looks like a soap bubble, which is a membrane. The equation for a membraneis ∇2w = − P

N .

For

NN

P

w(x, y)

w – height above z = 0

Compare to

∇2φ = −2GdΘ

dz


64

The equations have the same form, thus, we can picture the stress function as asoap bubble. This is called the membrane analogy for Prandtl’s stress function.The shear stress in the plane is

σ = σzx i+ σyz j =∂φ

∂yi−

∂φ

∂xj

We can draw level curves of φ

φ = c2 φ = c1

boundary φ = 0

The normal to a level curve is

∇φ =∂φ

∂xi+

∂φ

∂yj

NOTE: let τ = σxz i+σyz j be the shear stress vector (component of the tractionvector parallel to the x− y plane). Then,

τ ·∇φ =∂φ

∂x

∂φ

∂y−

∂φ

∂y

∂φ

∂x= 0

So, τ is ⊥ to n. Thus the tangents to the level curves point in the direction ofthe shear stress.

τ

φ = c


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This allows us to sketch a soap bubble and see how the stresses are distributed.Also,

|τ | =

!"∂φ

∂x

#2

+

"∂φ

∂y

#2

= |∇φ|

so the steeper the slope of the membrane, the larger the shear stress. For theellipse

y

x

z

yz

x

b a

largest slope ⇒ largest shear |τ |

The largest shear stress is generally at the edge closest to the centroid. That is,where the membrane has the largest slope.

4.3 Torsion of Narrow Cross-Sections

Consider

ℓ

t


66

Draw a membrane for this

z τmax

x

z

y

except near the ends, the stress function only depends on x, and is independent

of y. So, we guess that φ = c!

x2 − t2

4

"

is an approximate stress function. This

is good except near the ends. NOTE: φ = 0 on x = ±t2 . Now

∇2φ = 2c = −2GdΘ

dzso c = −G

dΘ

dz

or

φ = −GdΘ

dz

#

x2 −t2

4

$

Shear stresses are

σzx =∂φ

∂y= 0

σzy = −∂φ

∂x= 2G

dΘ

dzx

maximum occurs at x = ±t

2

σyz

So the torque is

T = 2

%%

A

φ dx dy = 2

#

−GdΘ

dz

$ ℓ%

0

t2%

−t2

#

x2 −t2

4

$

dx dy

= 2ℓ

#

−GdΘ

dz

$#x3

3−

t2

4x

$&&&&

t2

−t2


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= 2ℓ

!

−GdΘ

dz

"!

−t3

6

"

= GdΘ

dz

!t3ℓ

3

"

# $% &

J

This is only good for ℓ ≫ t, and is exact as ℓt → ∞. For finite values of ℓ

t wecan write

(GJ) = G · cf ·t3ℓ

3

cf = correction factor that can be found experimentally. Recall, the section alsowarps:

σzx =∂φ

∂y= 0 = G

!

−dΘ

dzy +

∂w

∂x

"

σzy = −∂φ

∂x= 2G

dΘ

dzx = G

!dΘ

dzx+

∂w

∂y

"

so

∂w

∂y=

dΘ

dzx,

∂w

∂x=

dΘ

dzy

⇒ w(x, y) =Θ

zxy (ignore constant of integration)

Then we find w ≡ dΘdz xy

w

y

x

z

x

z

y


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4.4 Shafts with Multiple Thin Sections

ℓ1

ℓ2

t2

2

1

t1

yx

Just as we ignored the contribution of the shear stress at the ends, we will ignorewhat happens at the joint. Assume

φ1 = c1

!

x2 −t214

"

φ2 = c2

!

y2 −t224

"

just as for individual sections. Then use φ1 or φ2 depending on which area weare considering. We still need

∇2φ1 = 2c1 = −2GdΘ

dz

∇2φ2 = 2c2 = −2GdΘ

dz

⇒ c2 = c2 = −GdΘ

dz

so

φ1 = −GdΘ

dz

!

x2 −t214

"

φ2 = −GdΘ

dz

!

y2 −t224

"


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Then the torque is

T = 2

∫∫

A1+A2

φ dx dy = 2

∫∫

A1

φ1 dx dy + 2

∫∫

A2

φ2 dx dy

= 2

0∫

−ℓ1

t12∫

−t12

(

−Gdθ

dz

)(

x2 −t214

)

dx dy + 2

t22∫

−t22

ℓ2∫

0

(

−Gdθ

dz

)(

y2 −t224

)

dx dy

= −2Gdθ

dz

⎡

⎣ℓ1

(x3

3−

t214x

)∣∣∣∣

t12

−t12

+ ℓ2

(y3

3−

t224y

)∣∣∣∣

t22

−t22

⎤

⎦

so

T = −2GdΘ

dz

[

2ℓ1

(

−2(t12

)3

3

)

− 2ℓ2

(

2(t22

)3

3

)]

= GdΘ

dz

[t31ℓ13

+t32ℓ23

]

(same as we get if we consider each segment separately)

So, in general

T = GdΘ

dz

N∑

k=1

(t3kℓk3

)

for N sections.

NOTE: These sections cannot form any closed regions. Ex: does not workfor this:


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How can we deal with sections that have holes? Return to our membraneanalogy:

nn

Recall we found that if dφds = 0 on the boundaries of the area, then there were

no tractions on the sides of the shaft. Thus we chose φ = 0 on the outside. Allof this applies to the sides in the hole too, but we don’t know what the value ofφ is there (can’t pick φ = 0). So

φ|hole = c = constant

So we can draw the membrane as

z constant over the hole

x

Think of there being no pressure on the membrane over the hole. So, for aclosed square section we have a picture like

z

x


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From the top

Shear stresses are all in the same direction.

4.5 Comparison of Open and Closed Tubes in Torsion

Closed

t

τR

Side view of membrane

use φ = −G

2

dΘ

dz(x2 + y2 −R2)

for

R− t ≤ r ≤ R and

φ = c1 for r < R− t

Find c1 by matching

c1 = φ(R − t) = −G

2

dΘ

dz

!

(R− t)2 −R2"

= −G

2

dΘ

dz(−2Rt+ t2) (t2 is small, so ignore)


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c1 = GdΘ

dzRt

Then the torque is

T = 2

!!

A

φ dA (ignore area over section; only take area over cut-out)

= 2c1A ≈ 2c1πR2

T ≈ GdΘ

dz(Rt) · πR2 = G

dΘ

dz(2πR3t)

Thus, J = 2πR3t for a thin-walled closed tube.

Open tube

Effect of the cut: In a closed tube, vertical (axial) planes carry a shear

z σyz

σzy

cut-away view of closed tube


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In open tubes

cutσ = 0

connected

free surface; no shear tractions

Thus σ = 0 on the top face too, near the cut edge. This is simply another wayof saying the membrane has to touch the edge at the cut. Treat this tube as arectangular section with ℓ = 2πR, t. From before,

T = GdΘ

dz

ℓt3

3= G

dΘ

dz

(2πR)t3

3= G

dΘ

dz

2

3πRt3

or GJ = G

!2π

3Rt3"

for open tubes

Now compare

(GJ)closed(GJ)open

=2GπR3t

G#23πRt3

$ =3R2

t2

For example, for R = 0.5 in, t = 0.022 in#

tR = 0.044

$

. We have

(GJ)c(GJ)o

= 3(0.5)2

(0.022)2≈ 1550 times stiffer.


aem 4501 s15 lecture notes torsion(1)

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