torsion module 6 with solutions

7/25/2019 Torsion Module 6 With Solutions

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Module 6

Torsion

Learning Objectives

6.1 Formulation of the basic equations of torsion ofprismatic bars (St. Venant)

Readings: Sadd 9.3, Timoshenko Chapter 11

e2

e1

e3

Figure 6.1: Torsion of a prismatic bar

We will employ the semi-inverse method, that is, we will make assumptions as to the

125


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126 MODULE 6. TORSION

deformation of the twisted bar, enforce the governing equations of the theory of elasticity andfrom them derive simplified equations on a reduced set of variables. Due to the uniquenessof solutions, we can be sure that the assumptions made and the solutions found are correctfor the torsion problem.

The assumptions about the deformation resulting from the applied torque M3= T are:

Each x3 = constant plane section rotates as a rigid body about the central axis,although it is allowed to warp in the x3 direction

The rotation angle of each sectionis a linear function ofx3, i.e. (x3) =x3, where is the constant rate of twist or angle of twist per unit length.

O

e1

e2

Figure 6.2: Rigid in-plane rotation displacements for the torsion problem

Concept Question 6.1.1. Based on these assumptions and the schematic of the figure,derive the displacements corresponding to the rotation of the cross section at x3

Solution: The displacements corresponding to the rotation of the cross section atx3is:

u1 = r(x3)sin ; u2= r(x3)cos

Observing that x1 = r cos , x2= sin and replacing the angle of twist at x3, (x3) =x3:

u1= x3x2, u2 = x3x1

The out-of-plane warping displacement is assumed to be independent of x3. Our dis-placement assumption is thus reflected in the following expressions: function u3 is assumed


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6.2. STRESS FORMULATION 127

to be

u1= x3x2

u2 = x3x1

u3= u3(x1, x2)

(6.1)

6.2 Stress formulation

We start by looking at the implications of our kinematic assumptions in our strain-displacementrelations:

Concept Question 6.2.1. Use the strain-displacement relationsij = 12

(ui,j+ uj,i) and thekinematic assumptions in equations (6.1) to derive the general form of the strains in thetorsion problem. Solution:

11= u1,1= 0

22= u2,2= 0

33= u3,3= 0

23 =1

2(u2,3+u3,2) =

1

2(x1+u3,2)

31=1

2(u3,1+u1,3) =

1

2(x2+u3,1)

12 =1

2(u1,2+u2,1) = 0

(6.2)

Next, we need to consider our constitutive relations (isotropic material assumed).

Concept Question 6.2.2. Use Hookes law ij = 1E

(1 +)ij kkij

and the specific

form of the strains resulting from the assumptions of torsion theory to derive the followingrelations between the stresses and the assumed displacements:

11 = 22 = 33= 0 (6.3)

23 = G(x1+u3,2) (6.4)

31 = G(x2+u3,1) (6.5)

12 = 0 (6.6)Solution: For the normal strains, Hookes law gives

0 =11 = 1

E

11

22+33

0 =22 =

1

E

22

11+33

0 =33 =

1

E

33

11+22


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The only solution to this homogeneous systems is

11 = 22 = 33= 0

For the shear strains:

23= 12

(x1+u3,2) =1 +E

23

31 =1

2(x2+u3,1) =

1 +

E 31

12= 0 =1 +

E 12

If we follow the stress formulation, at this point we would apply the strain compatibilityrelations, but it is more direct to derive a special compatibility relation for the torsionproblem. To this end, differentiate equation (6.4) with respect to x

1

23,1 = G(+u3,21)

, equation (6.5) with respect to x2

31,2= G(+u3,12)

and subtract to obtain:31,2 32,1= 2G (6.7)

As we have done for plane stress problems, we will seek a scalar function that automati-cally satisfies the equilibrium equations. Lets see what the stress equilibrium equations looklike for the torsion problem:

Concept Question 6.2.3. Specialize the general equations of stress equilibrium: ij,j = 0(no body forces) to the torsion problem (no need to express them in terms of the strains ordisplacement assumptions as we will use a stress function)

Solution: The only non-trivial equation is the third:

31,1+32,2= 0 (6.8)

Now we can choose a stress function that will automatically satisfy equation (6.8):

31 = ,2, 32= ,1 (6.9)

It can be readily verified that this choice does indeed satisfy the equilibrium equations.To obtain the final governing equation for the stress function, we need to combine equa-

tions (6.9) with the compatibility equation (6.7)


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6.2. STRESS FORMULATION 129

Concept Question 6.2.4. Do it! Solution:

31,2 32,1= (,2),2 (,1),1= 2G

From where we get,11+,22 = 2G (6.10)

This is the final governing equation we will use in the description of torsion based onthe stress formulation. The type of equation (Laplacian equal to constant) is known as thePoisson equation.

It requires the provision of adequate boundary conditions. As we know, stress formula-tions are useful when we can provide traction boundary conditions

Concept Question 6.2.5. Specialize the general traction boundary conditions ijnj = ti

to the torsion problem (Hint consider the loading on the (lateral) cylindrical surface of thebar and focus on a specific cross-section) Solution: The main observation is that thebar is unloaded on the sides, soti= 0. On the boundary we then have:

11n1+12n2+13n3= 0, 0 = 0

21n1+22n2+23n3= 0, 0 = 0

31n1+32n2+33n3= 0

Replacing the stresses as a function of and observing that the tangent vector on the

boundary is s = s1, s2= n2, n1, we obtain:

,2s2+ (,1)(s1) = 0, ,1s1+,2s2= 0

s= 0, or

s = 0 (6.11)

that is, the gradient of the stress function is orthogonal to the tangent or parallel to thenormal at the boundary of the cross section, which in turn implies that is constant on theboundary of the cross section. The value of the constant is really immaterial, as adding aconstant to will not affect the stresses. For convenience, we will assume this value to bezero. (We will see that in the case of multiply-connected sections this has to be relaxed).

To summarize, the torsion problem for simply-connected cross sections is reduced to thefollowing boundary value problem:

,11+,22 = 2G inside the area of the cross section (6.12)

= 0 on the boundary (6.13)

It remains to relate the function to the external torque M3 =Tapplied and to verifythat all other stress resultants are zero at the end of the bar.


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Figure 6.3: Force and moment balance at bar ends

At the bar end (x3 = 0, L), the internal stresses need to balance the external forces.

Ignoring the details of how the external torque is applied and invoking St. Venants principle,we can state, see figure:

F1 =

A

13dx2dx1=

x1

x2

,x2dx2dx1=

x1

[]xtop2

xbottom2

dx1, F1= 0

whereA is the area of the cross section. Similarly,

F2 =

A

23dx2dx1= 0

For the applied torque M3= Twe need to make sure that:

T =

A

(23x1 13x2)dA=

A

(,1x1 ,2x2)dA=

A

,ixidA

Integrating by parts by using ((xi),i= ,ixi+xi,i) and xi,i = x1,1+x2,2= 2:

T =

A

(xi),idA 2dA

The first term can be converted into a boundary line integral by using the divergence theoremon the plane

A()i,idA = A

()inids, where ni are the components of the normal to the

boundary and ds is the arc length:

T =

A

xinids+

A

2dA

The first term vanishes since = 0 at the boundary and we obtain:

T = 2

A

dA (6.14)


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6.3. SOLUTION APPROACH 131

6.3 Solution approach

The following box summarizes the overall solution procedure:

1. Compute the stress function by solving Poisson equation and associated boundary

condition (6.12)

2. Obtain torque - rate of twist relation T =T() from equation (6.14)

3. Compute stresses23, 31 from equations (6.9), strains 23, 31 follow directly from theconstitutive law

4. Compute warping functionu3(x1, x2) by integrating equations (6.5) and (6.4)

A powerful method of approaching the solution of the Poisson equation for the torsion prob-lem is based on the observation that vanishes at the boundary of the cross section. There-fore, if we have an implicit description of the boundary of our cross section f(x1, x2) = 0,

we could use the inverse method where we assume a functional dependence of(x1, x2) ofthe form (x1, x2) = Kf(x1, x2), where K is a constant to be determined. Although thisdoes not provide a general solution to the Poisson equation, it is useful in several problemsof interest. Once again as we have done when using the inverse method, if our assumedfunctional form satisfies the governing equations and the boundary conditions, uniquenessguarantees that we have found the one and only solution to our problem of interest.

Concept Question 6.3.1. Torsion of an elliptical barConsider a bar with elliptical cross section as shown in the figure subject to a torque T

at its ends. The boundary is described by the implicit equation

f(x1, x2) =x21a2

+x22b2 1 = 0

ab

x1

x2

Figure 6.4: Elliptical cross-section.

1. Propose a functional form for the stress function : Solution: We seek afunction of the form:

= K f(x1, x2),

whereKis an arbitrary constant and f(x1, x2) is the implicit equation describing theboundary.


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Then, our candidate is:

= K

x21a2

+x22b2 1

.

2. Use the governing equation 2= ,11+ ,22 = 2G to determine the value of theconstant Kand the final expression for . Solution: Replacing our proposedstress function into this equation we get:

K

2

a2+

2

b2

= 2G

K = G a2b2

a2 +b2,

and

= G a2b2

a2 +b2x21

a2 +x22b2 1

3. Determine the relationship between the applied torque Tand the rate of twist byusing the torque- relation (6.14) T = 2

A dA. Interpret this important relation.

Solution: The torque is determined by

T = 2

A

dA,

whereA is the elliptical cross-sectional area.

In this problem

T = 2

A

dA = 2K

A

x21a2

+x22b2 1

dA

= 2Gb2

a2 +b2

A

x21dA2Ga2

a2 +b2

A

x22dA+2Ga2b2

a2 +b2

A

dA

= 2Gb2

a2 +b2Ix2

2Ga2

a2 +b2Ix1+

2Ga2b2

a2 +b2 A,

whereIx1 = 4 ab3 andIx2 = 4a3bare the moment of inertia and A = abis the area ofthe ellipse. Replacing, we obtain:

T =G a3b3

a2 +b2

Interpretation: this states that there is a linear relation between the external torqueTand the rate of twist , the proportionality constant is the torsional stiffnesswhich


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6.3. SOLUTION APPROACH 133

has a material component (the shear modulus) and a geometric component. Note alsothat

= T

A

x21a2

+x22b2 1

(6.15)

4. Use the relations (6.9): 31 = ,2, 32 = ,1 to compute the shear stresses 13 and23 as a function of the torque T. Solution:

31 = 2G a2

a2 +b2x2 =

2T

Ab2x2

23 = 2G b2

a2 +b2x1 =

2T

Aa2x1

5. Determine the stress resultant defined by the relation= 231+223. Solution:=

231+

223 =

2T

A

x21a4

+x22b4

.

6. Determine the maximum stress resultant max and its location in the cross section.Assume a > b. What happens to the individual stress components at that point?Solution: When a > b: the maximum occurs when x1 = 0 and x2 = b, whichimplies

max = 2

AbT.

7. An elliptical bar has dimensions L = 1m, a = 2cm,b = 1cm and is made of a mate-rial with shear modulus G = 40GP a and yield stress 0 = 100MP a. Compute themaximum twist angle before the material yields plastically and the value of the torqueT at that point. Assume a yield criterion based on the maximum shear stress (alsoknown as Tresca yield criterion), i.e. the material yields plastically when max = 0.

Solution: The material yields plastically whenmax= 0. We know thatmax =

2Ab

T, which leads to

max = 02

AbT = 2G

a2b

a2 +b2 = 0

= 0

2G

a2 +b2

a2b = 0.156m1.


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8. Calculate the warping displacement u3 as a function ofT. Solution: We knowthat

31= G (u3,1 x2) , 23 = G (u3,2+x1) .

By using the expressions

= a2 +b2

Ga3b3T, T =G a

3b3a2 +b2

, 31 = 2x2

ab3T, 23 = 2x

1

a3bT,

we can write

u3x1

=

2T

Gab3+

x2 =

b2 a2

Ga3b3T x2 u

A3 =

b2 a2

Ga3b3T x2x1+g(x2)

u3x2

=

2T

Ga3b

x1 =

b2 a2

Ga3b3T x1 u

B3 =

b2 a2

Ga3b3T x1x2+f(x1).

As u3 = uA3 = u

B3, the functions f(x1) and g(x2) vanish, which leads to the following

displacement fieldu3=

b2 a2

Ga3b3T x1x2.

9. Specialize the results for the elliptical cross section to the case of a circle of radiusr= a = b Solution:

The stress function is

= G

2 x21+x

22 a

2

= G

a2

2

x21a2

+x22a2 1

The torque is

T =Ga4

2 ,

which also allows us to write the stress function as

= T

A

x21a2

+x22a2 1

,

whereA = a2.

The shear stresses are

31 = Gx2 = 2T

Aa2x2

23 = Gx1 = 2T

Aa2x1.

The stress resultant is

=

231+223 =

2T

a2A

x21+x

22,


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6.4. MEMBRANE ANALOGY 135

and its maximum is

max = 2

AaT.

The warping displacement is u3= 0, independently of the Torque T.

6.4 Membrane analogy

For a number of cross sections it is not easy to find analytical solutions to the torsion problemas presented in the previous section. Prandtl (1903) introduced an analogy that has provenvery useful in the analysis of torsion problems. Consider a thin membrane subject to a

uniform pressure load pa shown in Figure 6.6.

Figure 6.5: Schematic of a membrane subject to a uniform pressure

N is the membrane force per unit length which is uniform in all the membrane and inall directions. It can be shown that the normal deflection of the membrane u3(x1, x2) isgoverned by the equation:

u3,11+u3,22 = p

N (6.16)

The boundary condition is simply u3(x1, x2) = 0.

We observe that there is a mathematical parallel or analogy between the membrane andtorsion problems:


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Membrane Torsionu3(x1, x2) (x1, x2)pN

2Gu3,1 ,1= 23u3,2 ,2=31

V = A u3dA T2The analogy gives a good physical picture for which is useful as it is easy to visualize

deflections of membranes of odd shapes. It has even been used as an experimental techniqueinvolving measurements of soap films (see Timoshenkos book). Looking at contours ofu3 isparticularly useful.

By observing the table, we can see that:

the shear stresses are proportional to the slope of the membrane. This can be gleanedfrom the density of contour lines: the closer, the higher the slope and the higher thestress, the stress resultant is oriented parallel to the contour line.

if we measure the volume encompassed by the deformed shape of the membrane, wecan obtain the torque applied.

In particular, we can draw insights into the overall torsion behavior of general crosssections.

Figure 6.6: Examples of membranes subject to uniform pressure and sketch of deflectioncontour lines


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Concept Question 6.4.1. Consider the torsion of a rectangular bar of sides a, b. Ananalytical solution can be obtained by using Fourier series (outside scope of this class).However, the membrane analogy gives us important insights about the stress field. Sketchcontours of and make comments about the characteristics of the stress field. Solution:The two main conclusions are that: 1) the stresses are larger in the shorter side as the

contours are more bunched up on this side, 2) the stresses near the corners must be reallylow, as the contours are more spaced out toward the boundary.

Concept Question 6.4.2. Consider the cases of cross section with corners such as thoseshown in Figure 6.7. What can we learn from the membrane analogy about the stressdistribution due to torsion near the corner in the case of

1. convex corner Solution: In this case, we can seethat the contours space out near the corner, which means that the gradient of and

thus the stresses are lower. This suggests that we cancan save some weight inour structure by rounding up the corner and eliminating some material.

2. concave corner Solution: In this case, we observe that the contours bunch uptoward the corner, which means that the gradient of and thus the stresses are larger(stress concentration). This suggests that we can reduce stress concentrationsby rounding up the corner and adding some material (fillet).

Figure 6.7: Membrane analogy: corners

Concept Question 6.4.3. Torsion of a narrow rectangular cross section: In thisquestion, we will make use of the membrane analogy to estimate the stress distribution in anarrow rectangular cross section as shown in Figure 6.8.

1. Sketch the cross section of the bar and use the membrane analogy to estimate thedeformed shape of the membrane u3 and from that the shape of the contour lines of


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x1

x2

x3

T

a

b

L

Figure 6.8: Torsion of a narrow rectangular bar

Figure 6.9: Representation for Membrane analogy


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. Comment on the dominant spatial dependence of and from there, the expectedtorsion response.

Solution:

Comments: For a narrow rectangle, the contour lines will be mostly parallel to x2

and will depend only on x1 except near the ends. As a result, the stresses23 willdominate the torsion response.

2. Based on your conclusions, obtain (x1) by simplifying and then integrating the gov-erning equation. Solution: As long as we are not close to the ends of the narrowstrip, we can assume u3,22 0 and the membrane governing equation (6.16) becomes:

u3,11 = u

3(x1) = pN

Integrating this expression twice:

u3= p

2Nx21+C1x1+C2

Applying the boundary conditions u3= 0 at x1 = a2 , we obtain:

C1= 0, C2=pa2

8N

and then:

u3(x1) = p

2N

a2

4 x21

From the membrane analogy: p

N = 2G, u3 :

(x1) =Ga2

4

x213. Obtain the torque-rate-of-twist relationTusing the expression (6.14)T = 2

A

dASolution:

T = 2

b/2b/2

a/2a/2

(x1)dx1dx2

= 2b

a/2a/2

G

a2

4 x21

dx1

= 2Gba24a

2a

21

3a

23 +1

3a

23

= 2Gb

a3

4

2

3

a3

8

T =Ga3b

3


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4. The torsional stiffness of the bar is defined as GJ, where G the shear modulus is thematerial andJ is the geometric contribution to the structural stiffness. Find Jfor thiscase and comment on the structural efficiency of this cross section: Solution:It follows immediately thatJ= ba

3

3 . This scales with a3 which is the small dimension,

i.e. the structural efficiency is terrible.

5. Find the stresses and further support your conclusions about structural efficiency:Solution: As long as we are far from the edges

23 = ,x1 = 2Gx1 = 2G T

GJx1

31 = ,x2 = 0

The stress diagram suggests that the dominant internal stresses are unable to providea large internal torque due to the small moment arm available which scales with a.

The discussion for a narrow rectangular cross section is also applicable to other narrow(open) shapes, see examples in Figure 6.10,

Figure 6.10: Other narrow open cross sections for which the solution for the rectangular caseis useful

Concept Question 6.4.4. Justify this statement and comment on the general torsionalstructural efficiency of narrow open shapes. Solution: As long as we are far fromthe ends of the narrow profile and from regions of high localized curvature such as corners,the contours of will be parallel to the edges of the narrow shape which will lead to a linearshear stress profile across the thickness, see Figure 6.11, the low moment arm and, thus, alow structural efficiency.


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Figure 6.11: Stress function contours and stress diagram for general narrow open sections

From the membrane analogy, one can observe that the volume of the deformed membranefor general narrow open cross-sections comprising several segments such as in an channel orI-beam, can be approximated by the sum of the individual volumes. The additive characterof the integral then tells us that the torque-rate-of-twist relation can be obtained by addingthe torsional stiffness of the individual components. Specifically,

Figure 6.12: Combining the membrane analogy and the solution for a rectangular thinsection to solve general open thin section torsion problems


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T =GJ = 2

A

(1 +2 +. . . )dA= 2

A1

1dA+ 2

A2

2dA+ =GJ1+GJ2+. . .

T =G

ni=1

Ji J

In the case of the channel beam, Figure 6.12, Ji = bih

3

i

3 . We observe that as we extend the

lengths of each component, the torsional stiffness only grows linearly with the total length.We will see that the situation is very different for the case of closed sections.

As for the stresses, they maximum stress will differ in each component of the thin sectionif the thickness is not uniform, since:

23 = ,1 =2T

J x1

whereJis the total geometric contribution to the stiffnessn

i=1bih

3

i

3 , so that the maximum

stress in each section is determined by its thickness:

(i)23 =

T

Jhi

The approach for thin open sections can be applied as an approximation for very slendermonolithic wing cross sections, such as shown in Figure 6.13

Figure 6.13: Use of membrane analogy for the torsion of slender monolithic wing crosssections

In this case, we can see that to a first approximation, the hypothesis for narrow sectionsapply and the same equations hold as long as we compute the torsional stiffness as:

J1

3

yTyL

h(x2)3dx2


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6.5. TORSION OF BARS WITH HOLLOW, THICK-WALL SECTIONS 143

6.5 Torsion of bars with hollow, thick-wall sections

Readings: Sadd 9.3, 9.6

Figure 6.14: Torsion of a hollow thick-wall cross section

Consider cylindrical bars subject to torsion with a cross section as shown in Figure 6.14.Just as we did for the exterior boundary, we will assume that the interior boundary orboundaries are traction free. This implies that the shear stress is parallel to the boundary

tangent, see Figure 6.15. We will call this the shear stress resultant

=s3 = 231+223.

Figure 6.15: Shear stress resultant

It also implies that s

= 0 (see equation (6.11)) and = const at the interior boundariesas well as on the external boundary. However, we cannot assume that the constant will bethe same. In fact, each boundary i will be allowed to have a different constant i. Wecan still assign one constant arbitrarily which we will keep setting as zero for the externalsurface, i.e. 0 = 0 on 0.


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The values of the constant for each internal boundary is obtained by imposing the con-dition that the warping displacement be continuous (single valued):

0 =

i

du3=

i

(u3,1dx1+u3,2dx2)

=i

31G

+x2

dx1+23

G x1

dx2

=

1

G

i

(31dx1+23dx2) +

i

(x2dx1 x1dx2)

The first integrand31dx1 + 23dx2= ds, whereis the resultant shear stressandds is thearc length. The second integral can be rewritten using Greens theorem:

i

(x2dx1 x1dx2) =

i

x1,1+x2,2d = 2A(i)

Summarizing,

0 = 1

G

i

ds 2A(i),

i

ds= 2 GT/J

A(i) = 2A

JT (6.17)

The value of the constant i on each internal boundaryi can be determined by applyingthis expression to each interior boundary.

Internal holes also lead to a modification of the external torque equilibrium condition atthe end of the bar equation (6.14). For the case ofNholes, we obtain:

T = 2

A

dA+Ni=1

2iA(i) (6.18)

With the exception of a few simple cases, it is generally difficult to obtain analyticalsolutions to torsion problems with holes. We will look at a few specific cases.

These results are also very important in the context of box-beams, as we shall see laterin the class.

Concept Question 6.5.1. Consider the case of aHollow elliptical sectionIt is important

that the interior boundary be an ellipse scaled from the outer boundary, that is the ellipsesemi-radii are ka, kb, where k


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6.5. TORSION OF BARS WITH HOLLOW, THICK-WALL SECTIONS 145

x1

x2

Figure 6.16: Torsion of a bar with a hollow elliptical cross section

2. Show that this implies that the interior boundary coincides with a contour line of thestress function we developed for the solid elliptical section and find the value of1= Cfor which the contour line matches the interior boundary for a given k . Solution:The stress function developed for the solid elliptical section was, see equation (6.15):

solid= T

A

x21a2

+x22b2 1

The contours ofsolid are given by:

T

A

x21a2

+x22b2 1

= C

x21a2

+x22b2

= CA

T + 1

The implicit function for the elliptical hole can be rewritten as:

x21a2

+x22b2

=k2

Comparing the last two expressions we obtain:

CA

T + 1 =k2 C=

T

A(k2 1)

So the value constant value of the stress function on the interior boundary is:

1= T

A(k2 1)

3. Evaluate the value of on the interior boundary when k = 0.5 relative to the value ofat the center of the bar in the solid section Solution: The value of at thecenter of the bar in the solid section is:

solid(0, 0) = T

A

02

a2+

02

b2 1

=

T

A


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According to the previous expression, the value of of on the interior boundary whenk= 0.5 is:

C= T

A(0.52 1) = 0.75

T

ASo for k = 0.5, the constant value of on the interior boundary is 3/4 of the value at

the center for the solid section.

4. Comment on the possibility of using the same function in both cases (solid and hol-low): does it satisfy the governing equation ,ii = 2Gand the boundary conditions?What about the warping displacement compatibility condition (6.17)? Solution:The stress function for the solid section will satisfy the governing equation at all pointsof the hollow section by virtue of the fact that it did so at those same points for thesolid section.

The external boundary condition hasnt been modified, = 0 there in the hollow caseas well.

The stress function is constant on the inner boundary and its value is 1 = TA

(k21).This satisfies the requirement that the interior boundary surface is traction free andimplies that the shear stress resultant is tangent to the interior boundary. It alsoimplies that any contour line in a solid section is traction free under torsion, i.e. it isas if the individual strips of material inbetween contour lines acted independently intheir contribution to torsion.

It can be readily verified that the warping displacement compatibility condition (6.17)is satisfied as well.

In conclusion, the function for the solid section can be used for the hollow section as

well without any modification.5. Is there anything at all that changes, e.g. stiffness, stresses and rate of twist for a given

torque, etc.? Solution: We expect the stiffness to besmaller for the hollow section, which implies that for a given torque the rate of twistwill be larger. To see this, compute the torque-rate-of-twist relation for this case:

T = 2

hollow

d = 2

solid

d 2

core

d

since is the same:

T =G

a3b3

a2 +b2

G

(ka)3(kb)3

(ka)2 + (kb)2

=G(1 k4) a3b3

a2 +b2

Also, notice that since is the same as in the solid section, for a given torque thestresses have to be the same (where there is material of course).

The final an overarching conclusion is that the solution for the hollow section can beobtained by simply removing the inner elliptical core from the solid section.


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6.6. TORSION OF BARS WITH THIN HOLLOW CROSS-SECTIONS 147

6. Compute the torsional efficiency = JA

of the hollow elliptical cross section relativeto the solid section as a function ofk. How would you optimize the cross section tomaximize stiffness relative to weight? Solution:

=(1 k4) a

3b3

a2+b2

(1 k2)ab

=1 k4

1 k2

a2b2

a2 +b2

For the solid case: = a2b2

a2+b2. The relative value is then:

relative=1 k4

1 k2

The denominator (representative of the relative area) decreases much faster than thenumerator (representative of the relative stiffness). This means that the efficiencyincreases as we take out more and more material from the core. The main conclusionis that the most efficient cross sections for torsion are thin hollow cross sections.

6.6 Torsion of bars with thin hollow cross-sections

Consider the limit case of a very thin hollow (closed) section, Figure 6.17

Figure 6.17: Thin hollow cross section

Since the inner and outer boundaries are nearly parallel, the resultant shear stress willbe nearly parallel to the median line throughout.

Also, the gradient of across the thickness and therefore the resultant stress will bealmost constant and equal to

n

10t(s)

, where t(s) is the thickness.This is in stark contrast to the thin open sections where was zero on both boundaries

across the thickness (it is in fact the same boundary) and adopted a parabolic profile


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inbetween which resulted in a linear shear stress distribution which changed signs across thethickness.

We can also make the following approximation in the computation of the warping dis-placement compatibility condition (6.17):

ds 2GA

whereA Aouter Ainner.The resisting torque provided by this cross section can then be computed as, Figure 6.18

Figure 6.18: Computation of the torque for a closed thin section

dT =h(s)(s)t(s)ds, T =

dT =

(s)t(s)h(s)ds

whereh(s) is the moment arm.We can show that the product(s)t(s) is a constant along the boundary for anys. Based

on Figure 6.6 and imposing equilibrium:

F3= 0 : AtAdx3+BtBdx3 = 0 AtA= BtB, q= (s)t(s) =constant

where we have defined qas the shear flow.Then,

T = t

h(s)ds

but hds = 2d, then:T = 2 tA()

= T

2At


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6.6. TORSION OF BARS WITH THIN HOLLOW CROSS-SECTIONS 149

which is known as Bredts formula. To find the torque-rate-of-twist relation, we replace in

T

2At

ds= 2GA

and we obtain:

T =G 4A2

dst From where we find that the stiffness is given by:

J= 4A2

dst

Concept Question 6.6.1. Compare the result of this approximation with the exact theoryfor a hollow circular bar of radius R and thickness t Solution:

For a thin-hollow (thin) circular bar of radius R and thickness t, we have that

Jthin= 4A2

dst

,

where

ds

t =

1

t

ds=2R

t

andA2 =

R2

2,


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which leads to following geometric stiffness

Jthin= 2tR3.

The geometric stiffness for a solid circular bar of radius R is J = R4

2 . If we consider

a thick-hollow (thick) bar with external radius Re = R and internal radius Ri = R

t =R(1 t/R), we can compute the geometric stiffness as

Jthick = R4e

2

R4i2

=

2

R4e R

4i

=

2R4

1

1

t

R

4

=

2R4

1

1 4

t

R+ 6

t2

R2 4

t3

R4+

t4

R4

.

If we neglect the nonlinear terms in t, the previous expression can be approximated as

Jthick 2tR3,

which is equivalent to the geometric stiffness we obtain for a thin-hollow circular bar.

torsion module 6 with solutions

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