Ch6 Torsion

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Advanced Mechanics of MaterialsMECH411

Chapter 6Torsion

University of Balamand, Department of Mechanical Engineering1/15/2014

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Chapter 6

Torsion

In this Chapter, we will study torsion of long prismatic bars. Only linearly elastic andisotropic materials will be considered. First, bars with circular cross-sections will be

considered and the formulas will be derived using the mechanics of materials approach.

The same problem will be re-considered using the semi-inverse method of Saint-Venant.

This will set the ground to treat, using the last approach, the more general torsion problem where the cross-section is not circular.

6.1 Torsion of a Prismatic Cylinder of Circular or Annular Cross-Section.

Mechanics of Materials Approach.

Consider the prismatic cylinder of circular cross-section shown in Figure 1 and markedwith grid lines in the longitudinal and in the transverse directions. The grid lines in the

longitudinal direction are initially lines parallel to the axis of the cylinder, they are its

generators. The others are circles in planes perpendicular to the axis of the cylinder.

Figure 1. Deformation of circular cylinders

It is noticed that after loading, each longitudinal line will deform into a helix and that the

circles will remain circles and therefore the plane of each circle will remain a plane. Also,radial lines of the circle will remain radial and they rotate from their initial directions. Fora uniform cylinder subject to end loading torque, the amount of rotation is proportional to

the applied torque and the distance of the circle from the fixed support. This rotation is

known as the angle of twist. Due to symmetry, each helix will be similar to the otherhelix.

Figure 2 shows a small piece of the cylinder of length x isolated to carry the necessarycalculations to derive the desired formulas.




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Figure 2. The distortion of a small element

Initially, the angle between a typical longitudinal line and a circle is 90 degrees (Angle at point A on the lateral surface of the cylinder). The shear strain is constant for a given

radial distance from the axis of the cylinder and is given, for small deformations, by

2

2 (6.1)

The arc length BD can be expressed in two manners and in the limit as x becomes

infinitesimal it can be written as

dxd BD (6.2)

In particular, at the circumference of the cylinder, we can write,

dxd cmax

(6.3)

Thus

max

c

(6.4)

stating that the shear strain varies linearly along the radial distance.

Since the material is assumed to be linearly elastic, G , then the shear stress

will satisfy the following relation:

max

c

(6.5)




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These shear stresses should balance the applied torque T. We therefore have,

c

J dA

cdAT

A

max

max2

(6.6)

where J represents the polar moment of inertia of the circular cross section. For a circular

cross-section of radius c it is given by 4

2c J

and for a hollow cylinder, it is

given by )(2

44

0 icc

where ci and co designate the inner and outer radii of the

annular region, respectively.

From Equation (6), we can deduce the value of cmax

, and substituting in the

expression of given in equation (5) to get

J T

(6.7)

To recapitulate the assumptions used in the derivation of the above formulae, we note thefollowing:

The material is isotropic and linearly elastic

The geometry is prismatic and axi-symmetric. The torque is along the axis ofsymmetry

A plane perpendicular to the axis of symmetry remains a plane. A circle in that plane and centered on the axis remains a circle. A radius of that circle remains a

radius and its length does not vary

A line parallel to the axis of symmetry becomes a helix

Small deformations are assumed

Before finishing this section, we will derive the expression of the angle of twist. From

equation (2), we can express the infinitesimal angle of twist, and by using equation (7),

we can write

dx JG

T dx

Gdxd

(6.8)

Thus the total angle of twist will be obtained by integration as

dx JG

T L

0

(6.9)

which can be reduced, for the case of a uniform cylinder loaded by a constant torque to

JG

TL

(6.10)




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6.2 Torsion of a Prismatic Cylinder of Circular Cross-Section. Semi-Inverse

Method.

According to this approach, we propose a displacement field; the strains are then derived

from the displacement strain relations. Thereafter, the stresses are derived from theconstitutive equations of the material. Equilibrium can be checked, tractions can be

calculated, and the displacement and traction boundary conditions have to be satisfied.

In case a cylindrical coordinate system is chosen, we let the z-axis be along the axis ofthe cylinder and the ρ and θ axes be in the perpendicular plane. In case a Cartesian

coordinate system is chosen, we consider the x3 axis along the axis of the cylinder and the

x1 and x2 axes in the plane of the cross-section. In cylindrical coordinates, thedisplacement field is assumed as follows:

z uuu

z r

and0

(6.11)

where θ represents the twist per unit length JG

T and z represents the rotation of the

cross section at a distance z from the plane z = 0. The problem can be solved in

cylindrical coordinates. However, this will require that the formulation be represented in

such coordinate system. Another alternative is to convert the displacement field to theCartesian system where we have:

0and;; 3132231 u x xu x xu (6.12)

The strains can be calculated as:

2and;2;02

1;0

1232131,22,112332211 x xuu (6.13)

The material being linear elastic, the stresses can be expressed as:

12321312332211 and;;0 xG xG (6.14)

These stresses automatically satisfy the equilibrium equations provided there are no body

forces.

In matrix form, the stress tensor can be written as

0

00

00

12

1

2

xG xG

xG

xG

(6.15)




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As for the boundary conditions, the ends and the lateral cylindrical boundary (or

boundaries in case of a hollow cylinder) still have to be checked. The components of the

outer unit normal on the lateral faces are 0,,21

c xc xn and therefore the traction

vector nt

vanishes on the surface and the boundary conditions on the lateral surfaceare satisfied. A similar conclusion can be drawn on the inner boundary in the case of a

hollow cylinder.

The traction, resultant force and moment at the end L x 3

can be calculated as

)0,,( 12 x xGnt

(6.16)

021

dA xG F

A

(6.17)

012

dA xG F

A

(6.18)

k GJ k dAr Gk dA x xGdA

x x

x x

k ji

GdAt r M A A A A

22

2

2

1

12

21 )(

0

0 (6.19)

yielding just one component representing the torque T , and thatGJ

T ,

J

Tx2

13

and, J

Tx1

23 . Thus we recover the results obtained in Section 6.1. Therefore, on one

hand, the assumptions to this torsion problem according to the mechanics of materialstheory, meet with the current proposition of the displacement field. On the other hand,

this solution satisfies all the equations of the theory of elasticity. Therefore, in this case,

the mechanics of materials solution is the solution as required by the theory of elasticity.




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Examples:

Example 1:

The solid steel shaft shown has a diameter of 20 mm, if it is subjected to the two torques,determine the reactions at the fixed supports A and B.

Solution:

Equilibrium, using the free body diagram, it is seen that the problem is statically

indeterminate since there is only one available equation of equilibrium, whereas T A and

TB are unknown.

∑

Compatibility, since the ends of the shaft are fixed, the angle of twist of the one end ofthe shaft with respect to the other must be zero, hence the compatibility equation can be

written as:

⁄

This relationship can be expressed In terms of the unknown torques by using the load-

displacement relationship, . We get:

or

Solving equations 1 and 2 yields:




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6.3 Torsion of a Prismatic Cylinder of Non Circular Cross-Section. Semi-Inverse

Method.In this case, the displacement field differs from the previous one by assuming a non-

vanishing3

u displacement. More specifically,3

u is a function of1

x and2

x and

independent of 3 x . The proposed displacement field is now:

213132231 ,and;; x xu x xu x xu (6.20)

where is named the warping function. This field indicates that the points of the cross-

section, in addition to their in-plane motion, move out of the plane of the cross-section, ingeneral, differently for every point. However, points in different cross-sections, having

the same1

x and2

x coordinates, move the same amount in the3

x direction.

Now the strains can be calculated as:

1

2

232

1

13123322112

and2

;0 x

x

x

x

(6.21)

The material being linear elastic, the stresses can be expressed as:

1

2

232

1

1312332211 and;;0 x x

G x x

G

(6.22)

The first two equilibrium equations in the 1 x

and 2 x

directions are automaticallysatisfied. The third one in the

3 x direction yields, assuming no body forces:

02

2

2

2

1

2

x x

(6.23)

This is nothing but Laplace equation in . Thus the problem consists of solving for a

harmonic function (a function that satisfies Laplace equation) and satisfying the

prescribed boundary conditions.

Now let us partially differentiate, in Equation (21), the strain component 13 with respect

to2

x and the component23

with respect to1

x and subtract to obtain:

1,232,13 (6.24)

which is a compatibility condition that should be satisfied in case the strains are

prescribed. This condition can be formulated in terms of the stress by differentiating the




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components in Equation (22) in the case of linear material, or by substituting for the

stresses in Equation (24) leading to:

G21,232,13 (6.25)

6.3.1 Solution Using Prandtl Stress FunctionThe formulation above is set to solve for the displacement field, and more specifically, to

solve for the component of the displacement in the3

x direction. An alternative

formulation consists of solving for a potential function, of the stress field defined such

that:

1

32

2

31 and; x x

(6.26)

Working with this stress potential is usually useful since, more often, the traction

boundary conditions are prescribed rather than the displacement boundary conditions.

Based on the definition of Equations (6.26), the 3 equilibrium equations are satisfied.

Actually, the first two are always automatically satisfied for the current stress tensor

while the equilibrium equation in the3

x direction reads:

0012,21,

21

2

12

2

3,332,321,31

x x x x

(6.27)

Thus, with such a choice of the stress function, equilibrium will be automatically

satisfied.

By substituting in term of the stress function defined in (26), Equation (25) becomes

G x x

22

2

2

2

1

2

22,11,

(6.28)

which is a Poisson’s equation in the function .

Now, to express the boundary conditions as a function of , notice that the traction on

the lateral surface is vanishing. The outward unit normal on the surface can be expressed

in terms of the components 1dx and 2dx of a small element ds tangent along the

boundary (see Figure 3) as jds

dxi

ds

dxn 12




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Figure 3 Tangent and normal to a curve

Therefore, the traction vector at the lateral surface is

00

0

00

00

00

1

32

2

31

1

2

3231

32

31

ds

dx

ds

dx

ds

dx

ds

dx

nt

(6.29)

By substituting for the stress function in the last component, we obtain

01

1

2

2

1

32

2

31

ds

d

ds

dx

xds

dx

xds

dx

ds

dx (6.30)

Hence is constant on the boundary and, without loss of generality, this constant can be

set to zero. This same argument can be used to show that the shear stress at any point in

the cross section is directed tangent to the contour constant through the point.

On the cross section, the stresses should add up to a zero force and to a moment

balancing the external torque, hence

0212,311 dxdxdA F

A A

(6.31)

0211,322

dxdxdA F

A A

(6.32)

A A

dA x xdA x xT 2,21,1312321 (6.33)

It can be shown that the integrals of Equations (31) and (32) actually vanish and that after

integration by parts the torque T can be written as

A

dAT 2 (6.34)




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which represents twice the volume under the function (surface) 21

, x x over the two

dimensional domain A. Note that was taken to vanish at the boundary of A.

Therefore, the problem now reduces to solving for the function in the domain A,

satisfying Poisson’s equation (28), having zero value at the boundary of A, and satisfyingthe condition (34) when a torque is prescribed. Note also that the torque T and the unit of

twist are related. This relation should depend on the cross-section rotational inertia andthe material stiffness as will be shown in the following illustrative examples.

Examples :

Example 1: Elliptical Cross-Section We consider a prismatic bar of elliptical

cross-section of major axis 2a and minoraxis 2b. The bar is subjected to a prescribed

torque T . The angle of twist is to be found.

The stress field is to be determined and, in particular, the maximum shear stress is to

be calculated.

Note that the ellipse can be represented by the equation

01,2

2

2

2

2

1

21

b

x

a

x x x F

(6.35)

For this problem, the proposed stress potential is

1,2

22

2

21

21b

x

a

x B x x

(6.36)

that vanishes at the boundary.

By su bstituting the expression in (36) into Poisson’s Equation (28) the following relationis obtained

22

22

ba

Gba B

(6.37)

linking the geometrical cross-section parameters a and b, the shear modulus, G, the unitangle of twist, , and the constant B.

Using Equation (34), the torque can be calculated as

BA I b

B I

a

BdA BdA x

b

BdA x

a

BdAT

x x

A A A A

222

222

212 22

2

22

2

12

(6.38)




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Expressing the area ( ba A ), and area moments of inertia for an ellipse in term of the

geometrical cross-section parameters ( 43

2

ba I x

), we obtain

ba BT (6.39)

Now B can be calculated in terms of the applied torque using Equation (39). The unit

angle of twist can be calculated via Equation (37) and the shear stresses can be calculatedusing Equations (26) resulting in:

ba

xT

xba

xT

b

Bx

x 3

1

1

323

2

2

2

2

31 2and;22

(6.40)

The largest shear stress occurs at the boundary nearest the centroid of the cross-sectionand it is

222

2

max

22

ba

T

ba

bGa

(6.41)

The unit twist angle is

33

22

baG

baT

(6.42)

which can be written as

J G

T

(6.43)

where J represents the torsional rigidity of the elliptical cross-section and is defined by

22

33

ba

ba J

(6.44)

Example 2: Equilateral Triangle Cross-Section We consider a prismatic bar of cross-section formed of

an equilateral triangle. The stress function is defined by

33

23

3

23

2, 1212121

h x

h x x

h x x

h

G x x

(6.45)

that vanishes at the boundary of the triangle.




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Similar calculations as in the previous case lead to

3max

2

315

h

T

(6.46)

The unit twist angle is

GJ

T

hG

T

4

315

(6.47)

where J represents the torsional rigidity of the triangular cross-section and is defined by

315

4h

J (6.48)

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