22311--Chapter06-v2011-v1

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Chapter 6

Torsion

• Will consider torsion of members with circular

and noncircular cross sections

• Use standard strengths of materials approach

for circular cross sections

• Use semi-inverse method of Saint-Venant for

noncircular sections in combination with thePrandtl elastic membrane (soap-film) analogy

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6.1 Torsion of a Prismatic Bar of a Circular CrossSection

• Consider

– Solid cylinder

– Cross section A

– Length L

– Torque T

– Because of radial

symmetry and view must

be same from either

end, plane sections

remain plane

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• The rotation b is a function of the axial distance z

where b = q z (6.1), and q is the angle of twist per unit length

• The displacement components are

(6.2)

(6.3)

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• To satisfy the BCs, Eqs. 6.7 must yield

– No forces on the lateral surface of the bar

– Must yield stresses s.t. the net moment is equal to T

– Net resultant force vanishes

• Direction cosines are (l,m,0)

– Stresses sPx and sPy are automatically zero

(2.10)

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• On the ends, the stresses must be distributed s.t. the net moment

is T

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• The stresses szx and szy are independent of z

• The stress vector t for any point P in a cross section is given by

(6.13)

• The stress vector t lies in the plane of the cross section and is

perpendicular to the radius vector r

• By Eq. 6.13, the magnitude of t is

(6.14)

• Therefore, t is a max at r=b• Substitution of Eq. 6.12 into Eq. 6.14 gives

(6.15)

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6.1.1 Design of Transmission Shafts

• Torsional shafts are used frequently to transmit power.

• For example, an electric motor to drive a pump

• By dynamics, the power P is

P =Tw (a)

• Where T is Torque and w is angular velocity

w = 2p f (b)

• Combining (a) and (b) T = P/2p f (c)

• If P and f are given, then T can be found using Eq. (c)

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6.2 Saint-Venant’s Semi-inverse Method• The analysis for the torsion of noncircular cross sections proceeds

in a manner similar to that for circular cross sections.

• However, a function w(x,y) is assumed to describe the warping of

the noncircular cross section.

• Consider a torsion member with a uniform cross section of a

general shape

• Any one of a number of potential shear stress distributions on the

end sections can produce a net torque T.

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• According to Saint-Venant’s Principle, the stress distribution on the

sections sufficiently far removed from the ends depends mainly on

the magnitude of T and not on the stress distribution on the ends.

For sufficiently long members, the end load distribution is not

significant.

• For S-V semi-inverse method, the starting point is to approximate

the displacement components resulting from the torque T

– This approximation is based on the observed geometric

changes in the deformed torsion member.

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6.2.1 Geometry of Deformation• S-V assumed that every straight member has an axis of twist about

which each cross section rotates as a rigid body

• Let OA and OB be line segments in the cross section for z=0 and

which coincide with the x and yaxes, respectively.

• After deformation,

– Translate the new position 0, i.e. 0*, back to coincide with 0

– Align the axis of twist along the z axis

– For small deformations, exy

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• The distortion of the cross section is called warping.

• P P* and motion is described by u, vand w

• The cross section rotates an angle b

• This rotation is the principal source of the u, v displacements

• S-V assumed b = q z

(6.16)

where y(x,y) is the warping function

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• Assuming continuous displacement components u, vand w, the

small displacement compatibility equations are automatically

satisfied

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• The state of strain at a point in the torsion member is given by the

substitution of Eqs. 6.16 into Eqs. 2.81

(6.16)

(6.17)

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• Differentiate the equation for gzx w.r.t. y

• Differentiate the equation for gzy w.r.t. x

• Subtract the 2nd equation from the 1st equation

(6.17)

(6.18)

• Eq. 6.18 is a geometrical condition (compatibility) to be satisfied for

the torsion problem.

q g

=

y

zx

q g

=

x

zy

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6.2.2 Stresses at a Point and Equations of Equilibrium

• For torsion members made of isotropic materials, stress-strain

relations for either elastic or inelastic conditions indicate that

(6.19)

• The stresses szx and szy are nonzero

• If body forces and acceleration terms are neglected, then upon

substitution of these stresses into Eqs. 2.45

(2.45)

• Gives

(6.20)

(6.21)

(6.22)

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• Eqs. 6.20 and 6.21 indicate that szx and szy are independent of z.

• The stresses szx and szy must satisfy Eq. 6.22

– Which expresses a necessary and sufficient condition for the

existence of a stress function f(x,y) (aka the Prandtl stress function)

s.t.

(6.20)

(6.21)

(6.22)

(6.23)

• Thus, the torsion problem is transformed into the determination

of the stress function f(x,y).

• BCs put restrictions on f(x,y).

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6.2.3 Boundary Conditions

• Because the lateral surface of a torsion member is free of applied

stress, the resultant shear stress t on the surface S of the cross

section must be directed tangent to the surface

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• The two shear stress components szx

and szy

may be written in

terms of t

(6.24)

(6.25)

• According to the figure,

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• The component of t in the direction of the normal n to the surface

S is zero.

• Projections of szx and szy in the normal direction give

(6.26)

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• Upon substituting Eqs. 6.23 into Eq. 6.26

(6.26)(6.23)

(6.27)

• Because the stresses are given by partial derivatives of f , it is

permissible to take this constant to be zero

(6.28)

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• The preceding argument can be used to show that the shear

stress

at any point in the cross section is directed tangent to the contour

f = constant the point.

(6.29)

• The stress distributions of szx and szy must satisfy

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(6.28)(6.28)

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6.3 Linear Elastic Solution• The stress-strain relations for linear elastic behavior of an isotropic

material are given by Hooke’s law (Eqs. 3.32)

• By Eqs. 3.32 and 6.23

(6.37)

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(6.18)

(6.37)

• Substitution of Eqs. 6.37 into Eq. 6.18

• Gives

(6.38)

• If the unit angle of twist q is specified for a given torsion member

and f satisfies the BC indicated by Eq. 6.28, then Eq. 6.38

uniquely determines the stress function f(x,y).

(6.28)

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• The elasticity solution of the torsion problem for many practical

cross sections requires special methods for finding the function

f(x,y) and is beyond the scope of the text.

• An indirect method may be used to obtain solutions for certain

types of cross sections, although it is not a general method.

• Let the boundary of the cross section for a given torsion member

be specified by the relation

F(x,y) =0 (6.39)

• Let the torsion member be subjected to a specified unit angle of

twist and define the stress function to be

f =B F(x,y) (6.40)where B is a constant.

• This function is a solution of the torsion problem provided F(x,y)=0on the lateral surface of the bar and

(6.41)

• B may be found by substituting Eq. 6.40 into Eq. 6.38.

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6.3.1 Elliptical Cross Section

• Consider an ellipse

• The stress function may be written as

(6.41)

• F(x,y) =x2/h2 +y2/b2 -1 =0on the

boundary

• Substituting Eq. 6.41 into Eq. 6.38

gives

(6.42)

in terms of the geometric parameters h and b, shear modulus Gand the unit angle of twist q.

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• The function f can be used to find the stresses

(6.43)

(6.44)

• The maximum shear stress tmax occurs at the boundary nearest

the centroid of the cross section

(6.45)

• The torque for the cross section can be found by substituting

Eq. 6.41 into Eq. 6.36

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6.3.2 Equilateral Triangle Cross Section• Consider an equilateral triangle.

• The stress function may be written as

(6.48)

• Proceeding as was done for the elliptical

cross section

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6.4 The Prandtl Elastic-Membrane (Soap-Film) Analogy

• Proposed by Prandtl (1903)

• Based on similarity of

– Membrane subjected to lateral pressure equilibrium equation

– The torsion (stress function) equation

• Consider an opening in the x-yplane that has the same shape as

the cross section of the torsion bar to be investigated.

• Cover the opening with a homogeneous membrane.

• The pressure causes the membrane to bulge out of plane.

• If the pressure is small, the slope of the membrane will be small.

• The lateral displacement z(x,y) of the membrane and the Prandtl

torsion stress function f(x,y) satisfy the same equation in (x,y)

• Require the same for both problems

– BCs – Boundary shapes

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• Prandtl Torsion Equation

(6.50)

• Elastic Membrane Equation

where zdenotes the lateral displacement due to apressure pand an initial tension S (force per unit length)

q f f

G y x

22

2

2

2

=

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• Consider an element ABCD of dimensions dxand dyof the elastic

membrane.

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• The net vertical force resulting from the tension S acting along edge AD of the membrane is

• Similarly the net vertical force resulting from the tension S acting onthe edge is

and likewise for edges AB and DC

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• The summation of the force in the vertical direction yields for the

equilibrium of the membrane element with area dx dy

(6.51)

(6.50)

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• The membrane displacement z is proportional to the Prandtl stressfunction f.

• Because the shear stresses s zx and s zy are equal to the derivatives

of f w.r.t. x and y.

the stress components are proportional to the derivatives of the

membrane displacement zw.r.t. x and y

and stresses are proportional to the slope of the membrane

• The membrane gives a visual image of the stress distribution.

• Twisting moment T is proportional to the volume enclosed by the

membrane and x-yplane

• Can be used to make valuable deductions

– e.g., for cross sections with equal area, one can deduce that a longnarrow rectangular cross section has the least stiffness and a circular

section has the greatest stiffness

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• Important conclusions may also be drawn with regard to the

magnitude of the shear stress and hence to the cross section for

minimum shear stress.

• Consider the angle section in Fig. 6.12a – At the external corners A, B, C, E and F, the membrane has zero slope.

the shear stress is zero not a design problem

– However, at the reentrant corner D, the corresponding membrane would

have an infinite slope

infinite shear stress a problem add a fillet

Fig. 6.12

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• Consider plastic vs. brittle material.

• For Geometry (a) and for static loads, stress redistributes in corner

D for torsion load in a ductile material.

• However, if the material is brittle and/or load is cyclic (fatigue), then

the load carrying capability of the section is compromised.

• Stress may be reduced by adding by drilling a hole as shown in

Geometry (b).

• Preferably redesign section as shown in Geometry (c).

Fig. 6.12

6.4.1 Remark on Reentrant Corners

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• The cross sections of many members of structures are made of

narrow rectangular parts.• Typically for primary loading of

– Tension

– Compression

– Bending

• Secondary for torsion

• For simplicity, may use membrane analogy for such sections.

6.5 Narrow Rectangular Cross Section

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• Consider a bar subject to torsion

• Assume b >>h

• Except for the region near x = ±b,

the membrane deflection is

independent of x.

• Assuming the membrane

deflection is independent of x and

parabolic w.r.t. y, then the

displacement of the membrane is:

(6.54)

where z0 is the max deflection.

• Eq. 6.54 satisfies z=0on the boundary

• If p/S is constant in Eq. 6.61, then z0

may be selected so that Eq. 6.54 is an

approximate solution of the membrane

displacement

(6.55)

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• Consider a composite section made up of joined long narrow

rectangles

• The torsional constant J can be defined

6.5.1 Cross Sections Made Up of Long Narrow Rectangles

(6.62)

where C is a correction coefficient.

• If bi > 10hi , then C =1.

• If bi < 10hi for any one or more sections, then C =0.91.

• When n=1 and b> 10h then C =1 and Eq. 6.62 is identical toEq. 6.60

• For n>1, Eqs. 6.61 take the form

where hmax is the maximum value of hi

(6.63)

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Took effective length at centerline of section

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• Consider the rectangular cross section 2b x 2h,

but discard the restriction that h << b.

• The associated membrane requires the torsion

stress function f to be even in x and y.

where

6.6 Torsion of Rectangular Cross Section Members

(a)

q f f

G y x

22

2

2

2

=

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• Slightly more complicated to analyze than simply-connected crosssection.

6.7 Hollow Thin-Wall Torsion Members and Multiply

Connected Cross Sections

• Complexity of solution is shown

graphically in Fig. 6.16.

• No shear stresses act on the lateral

surfaces of the hollow region of the

torsion member

implies zero slopeover the hole

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Solution of thin-wall torsion members is based on the assumptions:

• If the wall thickness is small

compared to the other dimensions

of the cross section, then sections

through the membrane made by

planes parallel to the zaxis and

perpendicular to the outer

boundary of the cross section are

approximately straight lines.

• Because the shear stress is given

by the slope of the membrane, this

simplifying assumption requires the

shear stress is constant through

the thickness.

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(6.53)

(6.52)

(6.65)

• Note: The shear stress around the boundary is not constant, unless

the thickness t is constant.

• The shear stress t=df/dn

where n is normal to a membrane contour curve z=constant.• Hence, by Eqs. 6.53

and Fig. 6.17b,

• By Eq. 6.52

• The quantity q=t twith dimensions [F/L] is referred to as the shear flow.• The shear flow is constant around the cross section of a thin wall

hollow torsion member and is equal to f .

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• Because f is proportional to z (Eq. 6.52), by Eq. 6.36,the torque is proportional to the volume under the membrane.

with z1 =cf1and A is the area enclosed by the mean perimeter of

the cross section

(6.66)

• A relation among t, G, q, and the dimensions of the cross section may

derived using equilibrium

and by Eqs. 6.65 and 6.52 (see previous slide)

where l is the length of the mean perimeter of the cross section

and S is the tensile force per unit length of the membrane.

(6.67)

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• Eqs. 6.66 and 6.67 are based on the simplifying assumption that the

wall thickness is sufficiently small so that the shear stress may be

assumed to be constant through the wall thickness.

• The resulting error is negligible when the wall thickness is less than1/10 of the minimum cross section dimension.

• With q=t t constant, Eq. 6.67 can be written

where t is a point-wise function of l.

• For a constant thickness cross section where l is the circumferential

length

(a)

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• For multiple constant thickness segments

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• Thin-wall hollow torsion members may have two or more

compartments.

• The plateau over each compartment is assumed to have a different

elevation zi.

• If N compartments, there are N+1unknowns TBD.

6.7.1 Hollow Thin-Wall Torsion Member having SeveralComponents

• For T specified, the unknowns are the Nvalues for the shear flow qi and the unit

angle of twist qi which is assumed to be the

same for each compartment.

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• By Eq. 6.66 the N+1unknowns are given by

and the N additional equations similar to Eq. 6.67

(6.68)

(6.69)

where

Ai is the area bounded by the mean perimeter for the ith compartment

q’ is the shear flow for the compartment adjacent to the ith compartment

t is the thickness along dl

li is the length of the mean perimeter for the ith compartment

• Note: q’ is zero at the outer boundary (no adjacent compartment).

• Then, the max shear stress occurs where the membrane has the

greatest slope, i.e. where (qi-q’)/t is the max.

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