bmbk

Notes on the Linear Analysis of Thin-walledBeams

Walter D. Pilkey and Levent Kiti̧sDepartment of Mechanical Engineering

University of VirginiaCharlottesville, Virginia

c� August 1996

Contents

Chapter I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS 11.1. Fundamental Equations of Saint-Venant Torsion 11.2. Saint-Venant’s Warping Function 61.3. Thin-walled Open Sections 91.4. Thin-walled Closed Sections 11

Chapter II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION 152.1. Geometry of Deformation 152.2. Properties of the Warping Function 182.3. Stress-Strain Relations 302.4. Equations of Equilibrium 312.5. Stress Resultants 332.6. Shear Center 352.7. Calculation of the Angle of Twist 362.8. Stress Analysis 39

Chapter III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION55

3.1. Geometry of Deformation 553.2. Equations of Equilibrium 593.3. A Multicell Analysis Example 623.4. Cross Sections with Open and Closed Parts 66

i

ii CONTENTS

CHAPTER I

SAINT-VENANT TORSION OF THIN-WALLEDBEAMS

1.1. Fundamental Equations of Saint-Venant Torsion

In this section, the fundamental equations of pure torsion are derived, startingfrom Prandtl’s assumptions about the stresses, for a prismatic beam of arbitrarilyshaped cross section, made of isotropic, homogeneous material for which Hooke’slaw is valid. The beam is subjected to end torques T as shown in Figure 1.1. Thex coordinate axis is chosen to lie along the beam axis, and the y, z coordinates inthe plane of the cross section. The coordinate origin is the centroid C of one of theend sections.

y

z

CxT T

FIGURE 1.1 Prismatic beam in torsion

When a beam is in this state of uniform torque, it is found that only the shearstresses �xy and �xz are nonzero

�x� �

y� �

z� �

yz� �

In the absence of body forces, the equations of equilibrium become

��xy

�y��xz�z

� �

��xy

�x� �

��xz�x

� �

2 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS

These equations show that the stresses are independent of x, which means that theshear stress distribution is the same over all cross sections.

By Hooke’s law of linear elasticity, only the shear strains �xy

and �xz

arenonzero

�x � �y � �z � �yz � �

The nonzero shear stresses are related to the stresses by

�xy

��xy

G�xz

��xzG

where G is the shear modulus.Only the following two of the six compatibility equations are not trivially sat-

isfied for this state of strain

�

�y

��xz�y

��

xy

�z

��

�

�z

��

xy

�z��xz�y

��

In view of Hooke’s law, these compatibility conditions can be written in terms ofthe shear stresses as

�

�y

��xz�y

��

xy

�z

��

�

�z

��

xz

�y��xy�z

��

Since neither stress depends on x, the parenthesized quantity in the precedingequation is independent of x, y, and z. Consequently

��xy

�z��

xz

�y� �C (1.1)

where C is a constant. This equation and the first of the equations of equilibriumform a set of two first-order partial differential equations to be solved, with theapplicable boundary conditions, for the stresses.

Prandtl’s stress function ��y� z� is defined by

��

�z� �

xy

��

�y� ��

xz

Stresses calculated from the stress function � satisfy the equations of equilibrium,and Eq. (1.1) becomes

��

�y��

�z�� C (1.2)

Because no forces are applied to the surface of the beam, the equality of theshear stress at the surface and the shear stress component perpendicular the bound-ary line of the cross section implies that this component is zero at all points of thecross-sectional boundary. Let the curvilinear coordinate s trace the boundary asshown in Figure 1.2. The s axis is tangential to the boundary in the direction ofincreasing s. The positive direction of the normal n to the boundary is chosen to

1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION 3

s

y

z

C

�xy

�xz

sn

�

FIGURE 1.2 Boundary condition for shear stress

make n, s, and x a right-handed orthogonal system of axes. The boundary condi-tion for the shear stress is

�xn � �xy cos�� xz sin� � �

where � is the angle from the positive y axis to the positive n axis. Since

dy

ds� � sin�

dz

ds� cos�

the boundary condition can be rewritten as

�xy

dz

ds� �

xz

dy

ds� �

which, in terms of the stress function, becomes

��

�z

dz

ds��

�y

dy

ds�

��

�s� �

This shows that the value of the stress function on the boundary remains con-stant. When the boundary of the cross section is a single closed curve, the stressfunction assumes a single constant value on it, and this value may be set equal tozero. When the boundary contains several closed curves, however, an arbitraryvalue can be assigned to the stress function only on one of these curves. On theremaining boundary curves, the stress function assumes different values.

The stress resultants over the cross section are the two transverse shear forcesVy, V

zand the torque T , which are calculated from the shear stress distribution


over the cross-sectional area A

Vy �

Z�xydA �

Z��

�zdA � �

Vz�

Z�xzdA � �

Z��

�ydA � �

T �

Z�y�

xz� z�

xy�dA � �

Z �y��

�y� z

��

�z

�dA

The first two integrals are zero by Green’s theorem, which transforms them toline integrals over the boundary curves, where � is constant. The third integralis evaluated as follows, by another application of Green’s theorem, assuming thatthe boundary value of � has been set equal to zero

T �

Z ��

��y��

�y��z��

�z

�dA � �

Z�dA

Let u, v, and w be the displacements of a point of the cross section in the x, y,and z directions, respectively. The linear strain-displacement relations give

�u

�x��v

�y�

�w

�z� � (1.3)

because all normal strains are zero, and the assumption of zero shear strain �yzmeans that

�v

�z��w

�y� � (1.4)

The functional dependence of the displacements on the coordinates x, y, z deter-mined by Eq. (1.3) is

u � u�y� z� v � v�x� z� w � w�x� y�

According to Eq. (1.4), the partial derivative of v with respect to z has no z depen-dence, and the partial derivative of w with respect to y has no y dependence. Thisimplies that v is linear function of z and w is a linear function of y. Because theshear stresses �xy and �xz are independent of x, so are the strains �xy and �xz . Thestrain-displacement relations

�xy

��u

�y��v

�x

�xz

��w

�x��u

�z

imply that the dependence of v and w on x is linear.The longitudinal displacement u�y� z� is called the warping displacement. The

warping displacement in Saint-Venant torsion has the same value for all cross sec-tions. For this theory to be applicable, the beam must be unrestrained in the lon-gitudinal direction. A cantilever beam, for instance, has a fixed end, which is notfree to undergo the same warping displacement as the other sections. If externaltorque is applied to the free end of such a beam, normal warping stresses �

xare

developed, and Saint-Venant’s solution is not applicable.Because the in-plane shear strain �yz and all normal strains are zero, the com-

ponents of the displacement in the plane of the cross section are those of a planerigid body moving in the yz plane. It will be assumed that the axis of twist is a

1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION 5

ey

ez

�

�x

P

Q

Q�

FIGURE 1.3 Displacement of a point of the cross section

line parallel to the beam axis and passes through the pointP whose coordinates inthe centroidal Cyz system are y

Pand z

P. It will also be assumed that the section

at x � � is restrained against rotation. The in-plane displacement of a point Q ofthe cross section is as shown in Figure 1.3. The point Q moves to Q� by a rotationabout P

rQ�P � rQP � vey �wez

where rQP denotes the position vector of Q measured from P , and ey, ez are unitvectors in the y, z directions. For small angles of twist �

x, the displacement v is

calculated as follows

v � rQ�P

� ey� r

QP� e

y� r cos��

x�� r cos �

x

� r�cos � cos �x� sin � sin �

x� cos �

x�

� �r sin � �x� ��z � z

P��x

where r is the length of rQP . A similar calculation gives w, and the displacementsin the yz plane are determined to be

v�x� z� � ��x�x��z � zP � w�x� y� � �x�x��y � yP � (1.5)

As mentioned earlier, the dependence of v and w on x is linear. Therefore

�x�x� � Kx

for some constant K.


The constantC appearing in Eq. (1.1) can be evaluated in terms of the angle oftwist

C � ��

xy

�z��xz�y

� �G�

�z

��u

�y��v

�x

�� G

�

�y

��u

�z��w

�x

�

� �G��x

If Eq. (1.2) is solved for C � �G��x � � and the solution is �, the solution � corre-sponding to a rate of twist of � �x is

� � �G��x�

and the torque is given by

T � �

Z�dA � �G��

x

Z�dA

The torsional constant J of the beam is defined as

J � �

Z�dA

The torsional constant, which is obtained by solving Eq. (1.2) with right hand sideequal to unity and boundary conditions that depend only on the cross-sectionalshape and dimensions, is a geometrical property of the cross section. The relation-ship between the applied torque and the angle of twist is, therefore,

T � GJ��x

(1.6)

1.2. Saint-Venant’s Warping Function

An alternative to the stress-function approach of the preceding section is Saint-Venant’s classical solution, which starts from hypotheses about the displacementfield. Saint-Venant made the assumption that the cross sections rotate about theaxis of twist, and even though the cross section warps out of its original plane, theprojection of the deformed cross section on the yz plane retains its original shapeand dimensions. The same conclusion, expressed by Eq. (1.5), was reached by thestress-function approach. If the axis of twist is taken to be the beam axis, Eq. (1.5)becomes

v�x� z� � ��x�x�z w�x� y� � �

x�x�y

If the rate of change ��x

of the angle of twist is assumed constant, and the endof the beam at x � � is assumed to be restrained against rotation, then Saint-Venant’s displacement hypothesis about the v and w displacement componentscan be expressed by

v�x� z� � ��xxz w�x� y� � ��

xxy (1.7)

The axial displacement u is assumed to be the same for all cross sections, so thatit is a function of y, z only. It is also assumed that u is directly proportional to therate of twist

u�y� z� � ��x�y� z� (1.8)

where �y� z� is an unknown function called the warping function.

1.2. SAINT-VENANT’S WARPING FUNCTION 7

The strain-displacement relations determine the strains from the assumed dis-placement field

�x�

�u

�x� �

�y�

�v

�y� �

�z�

�w

�z� �

�yz

��v

�z��w

�y� ��

xx� ��

xx � �

�xy ��u

�y��v

�x� ��x

��

�y� z

�

�xz ��u

�z��w

�x� ��x

��

�z� y

�

The stresses are then given by Hooke’s law

�x � �

�y� �

�z� �

�yz � �

�xy � G��x

��

�y� z

�

�xz � G��x

��

�z� y

�

The ratio of the change in volume to the original volume, called the cubicaldilatation, is zero

e � �x � �y � �z � �

and all surface forces are zero, so that the displacement formulation of the equa-tions of elasticity reduces to

r�u � r�v � r�w � �

where r� is the Laplacian

r� ��

�x��

��

�y��

��

�z�

The partial differential equations for the displacement components v and w aretrivially satisfied. The equation for the warping displacement u gives

r� ��

�y��

�z�� (1.9)

The boundary condition for the shear stresses on the cylindrical surface of thebeam, shown in Figure 1.2, is

�xy cos�� xz sin� � �


which, in terms of the warping function, becomes��

�y� z

�cos��

��

�z� y

�sin� � � (1.10)

The torque at any section is

T �

Z�y�xz � z�xy�dA � G��x

Z ��

�z� y

�y �

��

�y� z

�z

�dA

The integral in the preceding equation is identified as the torsional constant J

J � Iy� I

z�

Z �y�

�z� z

�

�y

�dA

The area integral can be transformed into a line integral over the boundary byapplying Green’s theoremZ �

��y�

�z��z�

�y

�dA � �

I�zdz � ydy� � �

Ir � dr

where r denotes the position vector from the centroid to points on the boundaryof the cross section. If the cross section is multiply connected, then the boundaryintegral is the sum of the line integrals along individual parts of the boundary. Thetorsional constant is given by

J � Iy � Iz �

Ir � dr (1.11)

Cy

z

r

dr

FIGURE 1.4 Solid elliptic cross section

Closed-form solutions for the warping function are known only for simpleand regular geometric shapes, such as the solid elliptic cross section shown inFigure 1.4. The equation of the elliptical boundary is

y�

a��z�

b��

The warping function is known from the theory of elasticity

� �a� � b�

a� � b�yz

1.3. THIN-WALLED OPEN SECTIONS 9

The line integral in Eq. (1.11) can be evaluated with the parametric representationof the ellipse in terms of the angle I

r � dr � �a� � b�

a� � b�

Iyz�ydy � zdz�

� �a� � b�

a� � b�

Z ��

�

�ba cos� sin� ��a� � b��d

�ab�b� � a��

��a� � b��

The area moments of inertia are

Iy ��

��ab� Iz �

�

��ba�

and the torsional constant is

J � Iy � Iz �ab�b� � a��

��a� � b��

�a�b�

a� � b�

1.3. Thin-walled Open Sections

A thin-walled section is called open if the centerline of its walls is not a closedcurve. Equivalently, a thin-walled section whose boundary is a single piecewisecontinuous closed curve is an open section. The simplest open thin-walled sectionis the narrow rectangular strip shown in Figure 1.5, for which the wall thicknesst is less than one-tenth of the length h. An approximate value for the torsional

Cy

z

h

t

�xz

FIGURE 1.5 Narrow rectangular cross section

constant J for this section will be obtained by assuming that �xy

is negligibly smalland the shear stress �xz varies linearly across the wall thickness

�xy

� � �xz

��maxy

t


The equation to be solved is Eq. (1.1), which, with these assumptions, becomes

�G��x � C ��xz�y

��max

t

This determines the differential equation for the stress function

�d�

dy� �

xz� �G��

xy

When this equation is integrated and the stress function is set equal to zero on thelonger edges of the rectangle, the result is

��y� � G��x

�t�

�� y�

�

and the torsional constant is found to be

J � �

Z�dA � �

Z �t�

�� y�

�dA �

t�A

�� Iz �

t�h

��

t�h

��t�h

where A is the area and Iz

is the area moment of inertia about the z axis.In this solution, it is not possible to set the value of the stress function to zero

on the shorter edges of the rectangle. Consequently, the stress distribution

�xz ��Ty

J

is not valid near the shorter edges, where the boundary conditions require that theshear stress be zero. In addition, the torque due to �xz is one-half the actual torqueT . This is partially because the neglected shear stresses �xy are concentrated nearthe shorter edges and have longer moment arms than the stresses �

xz.

y

z

s�s�

s�

C

n

s

FIGURE 1.6 Horseshoe section

The approximate results obtained for a narrow rectangular strip can be ap-plied to more complicated thin-walled open sections, such as the horseshoe sec-tion shown in Figure 1.6. Saint-Venant’s approximation for the torsional constant

1.4. THIN-WALLED CLOSED SECTIONS 11

is

J ��

Zt��s�ds (1.12)

where s is the coordinate that traces the median line of the section and t�s� is thewall thickness. The shear stress distribution is

�xz ��Tn

J(1.13)

where n is the normal coordinate measured from the median line. The maximumshear stress occurs at the maximum wall thickness tmax

�max �T tmax

J(1.14)

1.4. Thin-walled Closed Sections

A thin-walled section is called closed if the centerline of its walls is a closedcurve. Equivalently, a thin-walled section whose boundary is formed by twopiecewise continuous closed curves is a closed section. The boundary of a mul-ticell closed section is made up of more than two closed curves.

�xs

�

y

z

C

s

n

FIGURE 1.7 Closed thin-walled section

A closed thin-walled cross section is shown in Figure 1.7. The tangential andnormal coordinates, s and n, are chosen so that the axes n, s, x form a right-handedtriad. The coordinate s traces the median line starting from an arbitrarily selectedorigin, and the y, z coordinates of any point on the median line are functions of s.


The normal coordinate n of any point of the median line is zero. The angle ��s�is measured from the positive y axis to the positive n axis. It will be assumed thatthe shear stress is tangent to the median line and does not vary across the wallthickness. The shear flow q due to the shear stress �

xs, defined by,

q � t�s��xs�s�

will be assumed constant. Then, at any s, the derivative of the stress function withrespect to n is

��

�n�

��

�y

dy

dn��

�z

dz

dn� ��xz cos�� xy sin�

� ��xs � �q

t�s�

The stress function is then determined by setting its value equal to zero on theouter boundary of the section

��n� s� �q

�

��

�n

t�s�

�

The derivatives of the axial displacement are

�u

�y� �xy �

�v

�x�

�xy

G� ��x�z � zP �

�u

�z� �xz �

�w

�x�

�xzG

� ��x�y � yP �

where P is a point on the axis of twist. Thus, on the median line, the derivative ofu with respect to s is

�u

�s�

�xs

G� ��

x�z � z

P�dy

ds� ��

x�y � y

P�dz

ds

Let rP�s� denote the position vector of the point at s measured from the point P

rP � �y � yP �ey � �z � zP �ez

where ey, ez are unit vectors in the positive y, z directions, respectively. The unittangent vector e

sat the point with coordinates y, z is

es ��y

�sey �

�z

�sez

The unit normal vector en is, therefore, given by

en� e

s� e

x�

�z

�sey��y

�sez

The projection of the position vector rP

onto the unit normal vector is

rP � en � �y � yP ��z

�s� �z � zP �

�y

�s

The derivative of the warping displacement with respect to s becomes

�u

�s�

�xsG

� ��xrP� e

n(1.15)

1.4. THIN-WALLED CLOSED SECTIONS 13

The line integral of this derivative over the closed path formed by the median lineof the cross section is zero

�

G

I�xsds� ��

x

IrP� e

nds � � (1.16)

y

z

P

s

ds

rP

rP � en

es

en

FIGURE 1.8 Definition of sectorial area

In the preceding equation, the integral

�

IrP� e

nds

is interpreted as twice the area enclosed by the median line. Figure 1.8 showsthat the differential quantity under the integral is twice the area of the trianglewith base length ds. As the position vector sweeps through the entire median line,the integral gives the twice the area enclosed by the median line. In terms of theconstant shear flow q, Eq. (1.16) is rewritten as

q

G

Ids

t�s�� x � �


The torque resultant is calculated as the moment due to the shear stress about thepoint P

T � ex�

ZrP� �

xstdse

s� q

ZrP� �e

s� e

x�ds

� q

ZrP � ends � q �

G��x�

H ds

t�s�

This equation gives the shear stress

�xs�s� �

T

t�s�(1.17)

and the torsional constant

J �T

G��x�

�

H ds

t�s�

(1.18)

For constant wall thickness, the torsional constant becomes

J ��t

S

where S denotes the length of the median line.

CHAPTER II

THIN-WALLED ELASTIC BEAMS OF OPEN CROSSSECTION

2.1. Geometry of Deformation

Figure 2.1 shows a prismatic thin-walled beam and its cross section. The beamaxis, which is defined as the line of centroids of the cross sections, is chosen to liealong the x axis. Points on a particular cross section are specified by defining theiry and z coordinates. The coordinate s traces the median line of the cross section.Each value of s corresponds to a well-defined point of the median line, so that thecoordinates y and z of a point on the median line are functions of s.

y

z

C

s

x

FIGURE 2.1 A thin-walled beam and its cross section

It will be assumed that the shape of the median line and its dimensions remainunchanged in the yz plane when the beam undergoes a deformation under staticloads. This means that the transverse displacements, which are defined as thedisplacement components in the plane of the undeformed cross section, of a pointon the median line are those of a point belonging to a plane rigid curve constrainedto move in its own plane. Let A and B be arbitrarily chosen points of such a planerigid body in its initial position. After the body undergoes a displacement, thepoints A and B occupy new positions in space. Let A�, B� be the projections ofthese new positions onto the yz plane, as shown in Figure 2.2. Let r

BAbe the

position vector of point B measured from point A. The vector rB�A�

is given by

rB�A�� rBA cos �x � ex � rBA sin �x (2.1)

where ex is the unit vector in the direction of the positive x axis, and �x is the anglemeasured from the vector rAB to the vector rA�B�

, �� x � �, with the vector

16 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION

rAB translated such that the points A and A� become coincident. The rotationis counterclockwise, if the cross product r

BA� r

B�A�, evaluated according to the

right-hand rule, has the same direction as ex. Since

rBA

� rB�A�

� rBA

� �ex� r

BA� sin �

x� jr

BAj� sin �

xex

the rotation is counterclockwise for sin �x � �. This establishes that the sign of �xis positive when the sense of rotation from AB to A�B� is counterclockwise. Forsmall rotations, Eq. (2.1) becomes

rB�A�� rBA � �xex � rBA (2.2)

Let vA

and wA

be the displacement components of point A along the y and zaxes, and let v

Band w

Bbe the corresponding displacement components of point

B. Let the transverse displacement vector be denoted by uA

for point A and byuB for point B

uA� r

A�AuB� r

B�B

From the vector polygon A�ABB� in Figure 2.2

rB�A�� rBA � rB�B � rA�A � uB � uA � �vB � vA�ey � �wB � wA�ez

and Eq. (2.2) can be written in terms of displacement as

uB � uA � �xex � rBA (2.3)

The scalar components of this equation are

vB � vA � �zB � zA��xwB� w

A� �y

B� y

A��x

(2.4)

z

y

A

B

A�

B�

�x

ex� r

BA

rBA

rB�A�

FIGURE 2.2 Geometry of plane rigid body motion

To apply the foregoing considerations to a thin-walled beam, let A be an arbi-trarily chosen reference point, which need not be a material point of the median

2.1. GEOMETRY OF DEFORMATION 17

line, but which is assumed to be displaced as if it were rigidly attached to the me-dian line. Let e

sbe a unit vector tangent to the median line in the direction of

inreasing s as shown in Figure 2.3. The unit normal vector en

is defined so as tomake the triad e

n, e

s, and e

xa right-handed set of orthogonal vectors

ex� e

n� e

sen� e

s� e

x

Let ��s� denote the tangential component of the displacement of the point of themedian line at the coordinate s. This component is given by Eq. (2.3)

��s� � uA� e

s� �

xes� �e

x� r

A�s�� u

A� e

s� �

x�es� e

x� � r

A�s�

where rA�s� is the position vector of the point at s measured from A. In terms ofthe angle � between the s and y axes, this equation becomes

��s� � vA cos � �wA sin � � �xrA � en

Let rnA denote the projection of the vector rA onto the unit normal

rnA � rA � en

The tangential component of the displacement of the point at s is given by

��x� s� � vA�x� cos ��s� � wA�x� sin ��s� � �x�x�rnA�s� (2.5)

y

�

s

s

n

z

rA�s�

A

FIGURE 2.3 Tangential and normal components of displacement

It will now be assumed that the shear strain �xs

is negligible in a thin-walledbeam with open cross section. This means that longitudinal fibers of the beammaterial remain orthogonal to the fibers along the median line. This assumptioncan be written as

�xs

��u

�s��

�x� �


where u is the displacement of the point at s along the x axis. This leads to�u

�s� �

��

�x� �v�

A�x� cos ��s� �w�

A�x� sin ��s� � ��

x�x�rn

A�s�

in which prime denotes differentiation with respect to x. Integration with respectto s gives

u�x� s� � �v�A�x�

Z s

�

cos ��s�ds � w�

A�x�

Z s

�

sin ��s�ds � ��x�x�

Z s

�

rnA�s�ds

� �v�A�x�y�s� �w�

A�x�z�s� � ��x�x�A�s� � u��x�

where

A�s� �

Z s

�rnA�s�ds �

Z s

�rA�s� � ends

is the sectorial area and u��x� is the longitudinal displacement of the point of themedian line at s � �.

The longitudinal strain � is calculated by differentiating the longitudinal dis-placement with respect to x

�u

�x� �x�x� s� � u��x� � v��A�x�y�s� � w��

A�x�z�s� � ��x�x�A�s� (2.6)

The first three terms of this equation are consistent with the Navier-Bernoulli hy-pothesis that plane sections remain plane. The contribution of the warping of thesection is expressed by the last term. For this reason the sectorial area A is calledthe warping function. The warping function depends on the sectorial origin, whichis the origin chosen for the coordinate s, and on the reference point A, termed thepole of the warping function.

2.2. Properties of the Warping Function

The warping function A

, with pole at point A and origin at s � s�, is definedas the integral

A�s� �

Z s

s�

rA�s� � e

n�s�ds

where rA�s� is the position vector of the point at s of the median line measured

from the pole A, as shown in Figure 2.4. The direction of en�s� is determined

from the convention that the axes �n� s� x� are right-handed, with the s axis in thedirection of increasing s. The magnitude of the differential quantity rA�s� �en�s�dsis twice the area of the triangle with base ds and height rA�s� � en�s�. The signof this quantity is positive if the projection of the position vector onto the unitnormal vector e

n�s� is positive. This sign is more conveniently determined on the

basis of the sense of rotation of the position vector as it sweeps through the areain the direction of increasing s. If this rotation is clockwise, the contribution to theintegral is negative, and if it is counterclockwise, the contribution is positive. Thisis verified for any point of the median line by writing the projection of the positionvector onto the unit normal in the form

rA� e

n� r

A� �e

s� e

x� � �r

A� e

s� � e

x

Thus, when the rotation of the vector rA

is counterclockwise as its tip moves inthe direction of es, the cross product rA� es is in the same direction as ex, makingrA � en positive.

2.2. PROPERTIES OF THE WARPING FUNCTION 19

y

z

A

s

ds

rA

rA � en

es

en

FIGURE 2.4 Definition of sectorial area

Let A and B be two arbitrarily selected poles for the warping function. Sup-pose that the origin for A is chosen to be at s � s� and the origin for B at s � s�,as shown in Figure 2.5. The relationship between A and B is found by the com-putation

A�s� �

Z s

s�

rB� e

nds �

Z s

s�

�rB� r

BA� � e

nds

�

Z s�

s�

rB� e

nds �

Z s

s�

rB� e

nds�

Z s

s�

rBA

� ends

� �B�s�� B�s� �

Z s

s�

rBA � �sin �ey � cos �ez�ds

� �B�s��

B�s� �

Z s

s�

rBA

� �dzey� dye

z�

� �B�s�� B�s� � �yB � yA��z�s� � z�� zB � zA��y�s� � y��


y

z

A

B

s

s�

s�

rA

rBA

rB

FIGURE 2.5 Poles and origins of warping functions

In this calculation, � is the angle between the s and y axes as shown in Figure 2.3,and y�, z� are the coordinates of the origin chosen for

A

y� � y�s�� z� � z�s��

The equation for finding the warping function A with origin s� from the thewarping function B with origin s� is, therefore,

A�s� �

B�s� �

B�s

�� z

A� z

B��y�s� � y

�� y

A� y

B��z�s� � z

�� (2.7)

When A and B are coincident points, but s� and s� are two distinct origins, thetransformation equation becomes

A�s� � B�s� � B�s�� (2.8)

showing that the effect of changing the origin of a warping function without chang-ing its pole is to add a constant to it. If, on the other hand, the origins s�, s� are thesame and the poles A, B are different, the transformation equation is

A�s� �

B�s� � �z

A� z

B��y�s� � y

�� y

A� y

B��z�s� � z

�� (2.9)

since, in this case,

B�s�� B�s��


If �s� is a warping function for a particular pole and origin, the area integral

Q� �

Z�s�dA

is called the first sectorial moment. The area integrals

Iy�

�

Zy�s��s�dA

Iz� �

Zz�s��s�dA

are known as the sectorial products of area. These definitions are analogous to thedefinitions of the first, second, and product moments of area

Qy�

ZzdA

Qz�

ZydA

Iy �

Zz�dA

Iz �

Zy�dA

Iyz �

ZyzdA

A pole for which the the sectorial products of area are both zero is called aprincipal pole. Let A and B be two poles for the warping function with origins ats� and s�, respectively. By multiplying both sides of Eq. (2.7) by y and integratingboth sides of the result over the cross sectional area, one obtains

Iy�A � Iy�B � B�s��Qz � �zA � zB��Iz � y�Qz�� yA � yB��Iyz � z�Qz�

Since the origin of the y, z axes is the centroid C of the cross section, the firstmomentsQy and Qz are both zero, so that

Iy�A � Iy�B � �zA � zB�Iz � �yA � yB�Iyz (2.10)

A similar calculation gives

Iz�A

� Iz�B

� �zA� z

B�Iyz� �y

A� y

B�Iy

(2.11)

The conditions for A to be a principal pole

Iy�A

� Iz�A

� �

are solved for the coordinates of the pole

yA � yB �Iz�B

Iz� I

y�BIyz

IyIz� I�

yz

(2.12)

zA � zB �Iz�B

Iyz� I

y�BIy

IyIz� I�

yz

(2.13)

These expressions do not depend on the origin chosen for the pole B, becauseif this origin is shifted, the resulting warping function B differs from B by a


constant value, say K, and

Iy�B �

Zy�B �K�dA � Iy�B �KQz � Iy�B

Hence, the sectorial products of area for the warping function B remain the samewhen the sectorial origin is changed, and the coordinates determined by Eqs. (2.12)and (2.13) are independent of this origin.

It is important to realize that the coordinates given by Eqs. (2.12) and (2.13) arealso independent of the pole B. If D is any arbitary pole, its sectorial products ofarea are related to those of the pole B by

Iy�B � Iy�D � �zB � zD�Iz � �yB � yD�Iyz

Iz�B � Iz�D � �zB � zD�Iyz � �yB � yD�Iy

Then

Iz�BIz � Iy�BIyz � Iz�DIz � Iy�DIyz � �yB � yD��IyIz � I�yz�

which, when substituted into Eq. (2.12), gives

yA � yD �Iz�D

Iz� I

y�DIyz

IyIz� I�

yz

The right side of this equation is the expression for the y coordinate of A with thepole D. Similarly, the z coordinate of A remains the same regardless of the poleused to find it. Thus, the principal pole depends only on the cross-sectional shapeand dimensions; it is a cross-sectional property.

If, for a given pole A, there is a sectorial origin s� such that

Q�A�

ZA�s�dA � �

the point s� is termed a principal origin. To determine s�, let B be a pole coincidentwith A but with a known origin s�. According to Eq. (2.8)

A�s� �

B�s� �

B�s��

so that the condition for s� to be a principal origin is

Q�A

� Q�B

� B�s��A � � (2.14)

This equation determines the principal origin s�

in terms of the arbitrarily selectedorigin s

�. The existence of at least one s

�is ensured, for most cross sectional

shapes, by the mean value theorem for integrals. There may be multiple solu-tions for s�, in which case any one solution can be selected as the principal origin.If another pole D, which is coincident with A and B but has its origin at s

�, is used

instead of B in determining the principal origin s�, then

Q�D

� Q�B

� B�s��A

and the condition for s� to be a principal origin, written in terms of D, becomes

Q�A� Q�D

� D�s��A � Q�B� B�s��A� �B�s�� B�s��A

� Q�B

� B�s

��A � �

This shows that the same principal origin is obtained regardless of where the ref-erence origin s� is placed. Hence, the principal origin is a cross-sectional property.


Let A be the principal pole for the warping function A, whose origin hasbeen selected arbitrarily. When this origin is changed to a principal origin, thesectorial products of area for the warping function remain zero, because, as men-tioned above, these products are independent of the sectorial origin as long as thecentroid is used as the origin of the coordinates y and z. Hence, for a given crosssection, it is possible to find a pole A and an origin s� such that Q�A

, Iy�A , andIz�A

are zero. A warping function satisfying these conditions is termed a principalwarping function. Principal warping functions will henceforth be written withouta subscript.

y

z

h

�

h

�

b

O C

tf

tw

O

s�

n�

s�

n�

s�

s�

n�

n�

FIGURE 2.6 Symmetric channel section

For the symmetric channel section shown in Figure 2.6, the warping functionwith pole and origin both at the point of intersectionO of the y axis and the medianline is given by

O�s

��

O�s�� h

�s�

O�s��

O�s��

h

�s�

where the signs are determined from the sense of rotation of the vector from O topoints on the median line. For instance, O�s�� is negative, because the positionvector rotates clockwise as it traces the median line of the upper flange.


Let A be the principal pole of the cross section shown in Figure 2.6. The coor-dinates of A will be found by using O both as a reference pole and as the sectorialorigin. Since z� � � and I

yz� �, the coordinates of A are given, according to

Eqs. (2.12) and (2.13), by

yA � yO �Iz�OIy

zA � �Iy�O

Iz

The sectorial product of area Iy�O

is zero by symmetry, and the sectorial productof area Iz�O is

Iz�O

�

Zz�s�

O�s�dA �

Z b

�

��h

��

h

�s��tfds

��

Z b

�

�h

��h

�s��tfds

��

b�h�tf�

The area moment of inertia Iy for this section is

Iy ��bh�t

f� h�t

w

��

The coordinates of the principal pole A are found to be

yA� y

O�

b�tf�btf � htw

zA� �

The principal origin s� for the principal pole A of the symmetric channel sec-tion is determined from the condition

Q�A� A�s��A � �

where A

has the arbitrarily selected origin O and is given by

A�s�� yA � yO�s�

A�s�� y

A� y

O�h

��h

�s�

A�s�� yA � yO�s�

A�s

�� y

A� y

O�h

��h

�s�

The first moment of sectorial area with pole A and origin at O happens to be zeroby symmetry

Q�A�

ZA�s�dA � �

so that the point O is the principal origin for the principal pole A, and A

is theprincipal warping function.

The warping constant I�

is defined as the sectorial moment of inertia of theprincipal warping function

I��

Z��s�dA


Because the section shown in Figure 2.6 is symmetric with respect to the y axis, thewarping constant is calculated as follows

I��

Z h��

�

A�s��

�twds� � �

Z b

�

A�s��

�tfds� �

b�h�tf �btf � �htw�

��btf � htw�

C

z

yh

b�

b�

d e

t

s�

s�O

s�

FIGURE 2.7 Unsymmetric channel section

For the unsymmetric channel section shown in Figure 2.7 with the dimensions

b� � b b� � �b h � �b

and constant thickness t, the centroid C is at a horizontal distance d and a verticaldistance e from the intersection O of the lower flange and the web

d �b

�e �

�b

�

The area moments of inertia and the area product of inertia are

Iy�

��tb�

��Iz�

tb�

�Iyz

� �tb�

If the pointO is used both as the pole and the sectorial origin, the warping functionis

O�s�� O�s�� hs O�s��


To find the principal pole, the values of the sectorial products of area are needed

Iy�O

�

Zy

OdA �

Z b

�

�d� s��O�s��tds� �

Z b

�

tbs��s� � b�ds� �tb�

�

Iz�O �

ZzOdA �

Z b

�

��h � e�O�s��tds� �

Z b

�

��tb�s�

�ds� �

�tb�

�

The principal pole A has the coordinates

yA � yO �Iz�OIz � Iy�OIyz

IyIz� I�

yz

��b

��

zA � zO �Iz�OIyz � Iy�OIy

IyIz� I�

yz

��b

��

The warping function with principal pole A and origin O is

A�s�� yA � yO�s� �� b

�s�

A�s�� y

A� y

O�h� �h� e� z

A�s� �

� b�

��

��b

� s�

A�s�� e � z

A�s� �

��b

� s�

The first sectorial area moment is

Q�A�

Z h

�

A�s��tds� �

Z b�

�

A�s��tds� �

Z b�

�

A�s��tds� ��tb�

The condition for s� to be a principal origin is

Q�A

� A�s��A �

�tb�

� �tb

A�s��

or

A�s�� b�

According to Eq. (2.8), the shift of the origin to s� gives the principal warpingfunction

�s� � A�s� �

A�s

��

A�s� �

b�

which, when written out for the web and the flanges, yields

�s��

A�s

��

b�

�

� b

�s� �

b�

�s�� A�s��b�

� �

��b

� s� �

�b�

�

�s��

A�s

��

b�

�

��b

� s��b�

The principal warping function �s� shown sketched to scale in Figure 2.8. Thefunction is zero at three distinct points of the median line of the section. Any oneof these points can be regarded as the principal sectorial origin s�.

If a cross section has a symmetry axis, say the y axis, this is a principal axis, sothat I

yz� �. In calculating the coordinates of the principal poleA, using Eqs. (2.12)

and (2.13), the choice of the reference pole B is arbitrary. When B is chosen to


� b�

�

� b�

�

��b�

��

��b�

��

�b�

��

��b�

��

�

�

�

�

�

�

FIGURE 2.8 Warping function for the unsymmetric channel section

be any point on the symmetry axis y, it is readily verified that Iy�B � �. Thecoordinate zA is then

zA � zB �Iz�BIyz � Iy�BIy

IyIz� I�

yz

� �

This shows that the principal pole lies on the symmetry axis. In addition, the pointof intersection of the y axis with the median line is a principal sectorial origin.For a doubly symmetric cross section, the point of intersection of the two axes ofsymmetry is the principal pole, the principal origin, and the centroid.


A formula for the warping constant in terms of the warping function B ,whose pole and origin are arbitrary, can be derived starting from the transfor-mation equation Eq. (2.7)

A�

B�

B�s

�� z

A� z

B��y � y

�� y

A� y

B��z � z

�� (2.15)

Since the origin s� for A

is a principal origin, the area integral of Eq. (2.15) is

� � Q�A

� Q�B

� B�s��A� y��zA � z

B�A� z��yA � y

B�A (2.16)

The first moments of area Qy

and Qz

have been set equal to zero in the calculationof the right side of this equation because y and z are centroidal axes. The result ofEq. (2.16) allows Eq. (2.15) to be rewritten as

A � B �Q�B

A� �zA � zB�y � �yA � yB�z (2.17)

The conditions for A to be a principal pole are then

Iy�A

� Iy�B

� �zA� z

B�Iz� �y

A� y

B�Iyz

� � (2.18)

Iz�A

� Iz�B

� �zA� z

B�Iyz� �y

A� y

B�Iy� � (2.19)

The warping constant is given by

I� � I�B � ��zA � zB�Iy�B � ��yA � yB�Iz�B �Q��B

A

� �yA � yB��Iy � �zA � zB�

�Iz � ��yA � yB��zA � zB�Iyz

which is simplified by using Eqs. (2.18) and (2.19) to

I�� I

�B�Q��B

A� �y

A� y

B��I

y� ��y

A� y

B��z

A� z

B�Iyz� �z

A� z

B��I

z

(2.20)

The channel section with double flanges shown in Figure 2.9 has uniformthickness t, and it is symmetric with respect to the y axis. The point O, whichis the point of intersection of the median line and the y axis, is chosen as a conve-nient pole and origin for the warping function

O�s�� s� �

h��

O�s�� h�

�s� � � s� � b

O�s

��

h��s�

� � s�� b

The area moment of inertia Iy

is calculated using the centerline dimensions

Iy�

th��

�tbh��

�tbh��

and the sectorial product of area Iz�O

is found, taking advantage of symmetry,

Iz�O � �

Z b

��h��O�s��tds� � �

Z b

��h��O�s��tds�

�tb�

��h�� h��


A Cy

z

h� h�

b

s�

s�

s�

O

FIGURE 2.9 Channel section with double flanges

The principal pole A is on the symmetry axis y

yA� y

O�Iz�O

Iy� y

O�

b��h�� h��

h�� b�h�� h��

The warping constant is given by Eq. (2.20)

I� � I�O � �yA � yO��Iy �

tb��h�� h��h�� b�h�� h��

��h�� b�h�� h��

Since point O is the principal origin, the principal warping function can be writtenas

�s�� y

A� y

O�s

�

�s�� y

A� y

O�h��h��s�

�s�� yA � yO�h��h��s�

from which the value found for I� from Eq. (2.20) can be verified by evaluating

I� � �

Z h��

�

�s��tds� � �

Z b

�

�s��tds� � �

Z b

�

�s��tds�


2.3. Stress-Strain Relations

The only significant stresses will be assumed to be the normal stress�x

and theshear stress �

xs. Although the strain �

xswas taken to be negligible in Section 2.1,

the corresponding stress �xs is not assumed to be zero, and it will be noticed thatthis contradicts Hooke’s law. The distribution of normal stress across the wallthickness will be assumed to be uniform. The shear stress due to unrestrained, orSaint-Venant, torsion will be assumed to be linear, with a zero value at the me-dian line. All other shear stresses will be assumed to be constant across the wallthickness.

According to the kinematic assumption that the median line of the cross sec-tion is inextensible, the strain �s is zero

�s ��

E��s � �x� � �

where E is the modulus of elasticity and is Poisson’s ratio. The longitudinalstrain is

�x�

�

E��x� �

s� �

��

E�x

The normal stress �x is written, using the kinematical expression for �x given inEq. (2.6), as

�x� �E

�u��x�� v��

A�x�y�s� �w��

A�x�z�s� � ��

x�x�

A�s��

(2.21)

in which the material constant �E is

�E �E

��

ds

dx

��x ��

x

�xdx�tds�

xtds

�q ��q

�sds�dx

qdx

FIGURE 2.10 Forces on a wall element

The shear stress is determined from the x component of the force equilibriumequation for the wall element shown in Figure 2.10. If it is assumed that there is

2.4. EQUATIONS OF EQUILIBRIUM 31

no longitudinal applied load on the surface

�q

�s� t�s�

��x�x

� �

where t is the thickness of the wall, and the shear flow q is defined by

q�x� s� � �xs�x� s�t�s�

The shear flow is found by integrating the equilibrium equation with respectto s

q�x� s� � q��x��

Z s

�

��x

�xt�s�ds (2.22)

where q��x� is the shear flow at s � �. By substituting the expression

��x

�x� �E

�u��x�� v��

A�x�y�s� �w��

A�x�z�s� � ��

x�x�

A�s��

(2.23)

into Eq. (2.22) and writing dA � t�s�ds for the element of cross-sectional area, theshear flow can be written as

q�x� s� � q��x� � �E

�v��A�x�Q

z�s� � w��

A�x�Q

y�s� � ��

x�x�Q

�A�s� � u��x�A�s�

�(2.24)

where

A�s� �

Z s

�

t�s�ds �

Z s

�

dA

Qy�s� �

Z s

�

z�s�dA

Qz�s� �

Z s

�

y�s�dA

Q�A

�s� �

Z s

�

A�s�dA

2.4. Equations of Equilibrium

The equations of equilibrium will be written for a differential element of lengthdx of the beam. It will be assumed that the coordinate axes y, z are centroidal, andthat only the principal the warping function is being used. Let px be appliedforce per unit length of the beam in the longitudinal direction. The normal stressresultant on the differential element isZ �

�x��x�x

dx�dA�

Z�xdA � dx

Z��x�x

dA

and the equilibrium of forces in the x direction gives

dx

Z��

x

�xdA� pxdx � �

The first term is evaluated by integrating the expression on the right side of Eq. (2.23)over the cross sectional area. The result is

�E�u��x�A � v��

A�x�Q

z� w��

A�x�Q

y� ��

x�x�Q

�

�� p

x� �

which simplifies to�Eu��x�A� px � �


Let py denote the applied force in the y direction per unit length of the beam.The direct shear force in the y direction balances the applied force in this directionZ

�q

�xcos �dsdx � pydx � �

where � is the angle between the y and s axes, and q is the direct shear flow. Sincedy � ds cos �,Z

�q

�xcos �ds �

�q

�xy

��edges

�

Zy�

�s

�q

�xds �

Zy�

�x

�t��

x

�x

�ds �

Zy��

x

�x�dA

where it has been assumed that the edges are free of shear stresses. The last termis evaluated by differentiating the expression on the right side of Eq. (2.23) withrespect to x and integrating the result over the cross sectional areaZ

y��x�x�

dA � �E�u��x�Q

z� viv

A�x�I

z�wiv

A�x�I

yz� �iv

x�x�I

y�

�� E

��viv

A�x�I

z�wiv

A�x�I

yz

�Hence, the equation of equilibrium in the y direction becomes

�EIzvivA �x� � �EIyzw

ivA �x� � py (2.25)

Let pz

denote the applied force in the z direction per unit length of the beam.The direct shear force in the z direction balances the applied force in this directionZ

�q

�xsin �dsdx� p

zdx � �

where � is the angle between the y and s axes, and q is the direct shear flow. Sincedz � ds sin �,Z

�q

�xsin �ds �

�q

�xz

��edges

�

Zz�

�s

�q

�xds �

Zz�

�x

�t��

x

�x

�ds �

Zz��

x

�x�dA

where it has been assumed that the edges are free of shear stresses. The last termis evaluated using Eq. (2.23)Z

z��x�x�

dA � �E�u�� x�Qy � vivA �x�Iyz �wivA �x�Iy � �ivx �x�Iz�

�� E

��viv

A�x�I

yz� wiv

A�x�I

y

�The equation of equilibrium in the z direction is

�EIyzvivA�x� � �EI

ywivA�x� � p

z(2.26)

The total torque at the section is the sum of two parts

T � Tt � T�

The torque Tt is due to the shear stresses resulting from pure, or unrestrained,torsion. It is related to the angle �x of rotation by

Tt � GJ��x�x�

2.5. STRESS RESULTANTS 33

The torque T� is called the warping torque. It is due to the shear flow q. For abeam element of length dx, equilibrium of the torques about the pole A gives

ex�

ZrA��q

�xdxdse

s� GJ��

xdx�m�x�dx � �

where rA

is the vector from the pole A to the point at s, and m�x� is the appliedtorsional moment per unit length. Since

�rA � es� � ex � rA � �es � ex� � �rA � en

the first term in the torque equation can be rewritten as

ex�

ZrA��q

�xdxdse

s� �

Z�q

�xdxr

A� e

nds � �

Z�q

�xdxd

An integration by parts givesZ

�q

�xd �

�q

�x

��edges

�

Z�

�s

�q

�xds �

Z��

x

�x�t�s�ds

so that the torsional equilibrium equation becomes

�

Z��x�x�

dA� GJ��x �m�x� � �

which is brought to its final form using Eq. (2.23)

�EI��ivx �GJ��x � m�x� (2.27)

2.5. Stress Resultants

The stress resultants for the normal stress �x

are the axial force N , the bendingmomentsM

yand M

z, and the bimoment M

�, which are defined by

N �

Z�xdA

My �

Zz�xdA

Mz � �

Zy�xdA

M��

Z�

xdA

The normal stress is given by Eq. (2.21)

�x� �E

�u��x�� v��

A�x�y�s� �w��

A�x�z�s� � ��

x�x��s�

�The stress resultants are evaluated, recalling that the origin of the coordinates y, zis the centroid, and that is the principal warping function

�BB�

NMy

Mz

M�

CCA � �E

�BB�A � � �� Iyz �Iy �

� Iz Iyz �

� � � �I�

CCA�BB�u��v��A

w��

A��x

CCA


Hence

�Eu��x� �N �x�

A

�Ev��A�x� �IyzMy�x� � IyMz�x�

IyIz� I�

yz

�Ew��

A�x� � �IzMy�x� � I

yzMz�x�

IyIz� I�

yz

�E��x�x� � �M��x�

I�

The normal stress is found in terms of the stress resultants by using these expres-sions in Eq. (2.21)

�x�

N

A�IyzMy � IyMz

IyIz � I�yzy �

IzMy � IyzMz

IyIz � I�yzz �

M�

I�(2.28)

In Eq. (2.24), let the point s � � be placed at the free edge so that the shearflow q��x� is zero, and suppose that there is no longitudinal external load p

xon

the beam. Then Eq. (2.4) shows that u��x� is zero, and the shear flow is given by

q�x� s� � �E�v��A �x�Qz�s� � w��

A �x�Qy�s� � ��x �x�Q��s��

(2.29)

Let the shear stress resultants Vy, V

z, and T

�be defined by

Vy �

Zq�x� s�ds cos ��s� �

Zq�x� s�dy

Vz�

Zq�x� s�ds sin ��s� �

Zq�x� s�dz

T��

Zq�x� s�d

(2.30)

The definition of Qz�s� is

Qz�s� �

Z s

�y�s�dA

Integration by parts givesZQzdz � zQz

��A

�

ZzdQz � �

ZzydA � �Iyz

Similarly,ZQzdy � yQ

z

��A

�

ZydQ

z� �

Zy�dA � �I

z

and ZQzd � Q

z

��A

�

ZydA � �I

y��

2.6. SHEAR CENTER 35

The corresponding integrals for Qy�s� areZQydy � �IyzZQydz � �IyZQyd � �

The definition of Q��s� is

Q��s� �

Z s

��s�dA

Integration by parts givesZQ�dz � zQ

�

��A

�

ZzdQ

��

ZzdA � �I

z��

Similarly, ZQ�dy � �Iy� � �

and ZQ�d � Q�

��A

�

Z�dA � �I�

The stress resultants are evaluated using Eq. (2.30) and Eq. (2.29)

Vy� � �E

�Izv��A�x� � I

yzw��

A�x�

�Vz� � �E

�Iyzv��A�x� � I

yw��

A�x�

�T�� EI

��x�x�

Substitution of these results into Eq. (2.29) gives the shear flow

q�x� s� � �IyQz�s� � IyzQy�s�

IyIz � I�yzVy�IzQy�s� � IyzQz�s�

IyIz � I�yzVz�Q��s�

I�T�

(2.31)

The total shear stress is found by adding Saint-Venant’s torsional stress to the con-tribution from q�x� s�

�xs

��T

tn

J�q

t(2.32)

where n is the coordinate measured from the median line in the normal direction.

2.6. Shear Center

A beam is said to be in pure flexure if the angle of twist �x�x� is identically

zero. The shear center S is defined as the point, in the plane of the cross section,through which the line of action of the transverse shear forces must pass for thebeam to be in pure flexure. Suppose that a beam is in pure flexure under the actionof transverse shear forces in the y direction only, as shown in Figure 2.11. The lineof action of V

ypasses through the shear center S. From Figure 2.11, the moment


y

z

es

q

en

SVy

A

rA

FIGURE 2.11 Shear center calculation

of the shear stresses about point A, which is the principal pole of the warpingfunction , is calculated by integrating

dMA� e

x� �r

A� qdse

s� � r

A� �e

s� e

x�qds � r

A� e

nqds � qd

over the cross-sectional area

MA�

Zq�x� s�d � T

�� EI

��x�x� � �

This moment is equal to the moment of Vy about A

MA� e

x� �r

SA� V

yey� � �z

A� z

S�Vy� �

which shows that the z coordinates of the shear center and the principal pole areidentical. A similar computation with the beam cross section subjected only to Vzshows that the y coordinates of A and S are also identical. The shear center andthe principal pole are, therefore, the same point.

2.7. Calculation of the Angle of Twist

The warping stresses in a thin-walled beam depend on the bimoment M�

andthe warping torsion T

�

M��x� � � �EI

��x�x�

T��x� � � �EI��

x �x�

The torque equilibrium equation 2.27, solved with the applicable boundary condi-tions, determines the angle of twist as as function of x. With the definition

c� �GJ�EI

�

2.7. CALCULATION OF THE ANGLE OF TWIST 37

Eq. (2.27) is rewritten in the form

d��x

dx�� c�

d��x

dx��

m�x��EI�

(2.33)

The most common boundary conditions on the angle of twist are those forfixed, simple, free or beam supports. At a fixed support, no twisting or warpingoccurs. These kinematical conditions are expressed by

�x� � ��

x� �

where the second condition is obtained by setting equal to zero the warping com-ponent, which is proportional to ��

x, of the longitudinal displacement u�x� s�. A

simple support does not allow twisting and is free of normal stress

�x� � ��

x� �

where the second condition expresses that the bimoment is zero

M��

Z�

xdA � �

At a free support there are two statical conditions, one expressing that there isno normal stress, and the other that the total torque is zero. The second of theseconditions is

Tt� T

�� GJ��

x� �EI

��x

� �EI��c��

x� ��

x� � �

Thus, for a free support, the boundary conditions are

��x � � c��x � ��x � �

The general solution of Eq. (2.33) is

�x�x� � C� � C�x�C� cosh cx� C� sinh cx��

cGJ

Z x

�

�c�x� �� sinh c�x� ��m��d�

(2.34)

where Ck, � � k � �, are the constants of integration, and one end of the beamis assumed to be at x � �. The bimoment is obtained from Eq. (2.34) by twodifferentiations

M��x� � �JG�C� cosh cx� C� sinh cx��

c

Z x

�

m�� sinh c�x� ��d� (2.35)

The warping torque is the derivative of M��x�

T��x� � �cGJ�C� sinh cx� C� cosh cx��

Z x

�

m�� cosh c�x� ��d� (2.36)

The pure torsion torque is

Tt�x� � GJ�C� � cC� sinh cx� cC� cosh cx��

Z x

�

m�� cosh c�x� ��d�

(2.37)

and the total torque is given by

T �x� � GJC��

Z x

�

m��d� (2.38)

A concentrated torque is applied at the unsupported end of the cantileverbeam shown in Figure 2.12. For this loading condition, the distributed torque


x

L

T�

FIGURE 2.12 Cantilever beam with end torque

m�x� is zero, because there is no external torque for the cross sections that lie be-tween the two end sections. The external torque is set equal to the total torque atx � L

T �L� � GJC�� T

�

The other boundary condition at x � L is that the cross section is free of normalstress

M��L� � �GJ�C� cosh cL� C� sinh cL� � �

At the fixed end the boundary conditions are

�x�� C� � C� � �

��x�� C� � cC� � �

The angle of twist is determined from these boundary conditions

�x�x� �T�cGJ

�cx� sinh cx� tanh cL�� cosh cx�

�

x

L

T�a

FIGURE 2.13 Cantilever beam with torque at x � a

If the torque is applied at a point x � a � L as shown in Figure 2.13, thedistributed torque is no longer zero. The concentrated torque of magnitude T� canbe expressed as a distributed torque in terms of the Dirac delta function

m�x� � T��x� a�

The angle of twist is calculated from Eq. (2.34) as

�x�x� � C� � C�x�C� cosh cx� C� sinh cx�

T�cGJ

�c�x� a�� sinh c�x� a��U �x� a�

2.8. STRESS ANALYSIS 39

where U denotes the unit step function. The boundary conditions are

�x�� C� � C� � �

��x�� C� � cC� � �

T �L� � GJC� � T� � �

M��L� � �JG�C� cosh cL �C� sinh cL��

T�c

sinh c�L � a� � �

The angle of twist for � � x � a is

�Lx�x� �

T�cGJ

�cx� sinh cx� �� cosh cx�

� sinh c�L� a�

cosh cL� tanh cL

��

and for a � x � L

�Rx �x� � �Lx �x��T�cGJ

�c�x� a�� sinh c�x� a��

2.8. Stress Analysis

As a first example, stresses in a cantilever beam of length L, with its fixed endat x � � and its free end at x � L, will be analyzed. The cross section of the beam,shown in Figure 2.14, is symmetric with respect to the y axis and is of constantthickness t. The load is a single vertical force of magnitude P applied at the freeend of the beam. The point of application of P on the cross section is the lowerend of the left flange.

The centroid is located by the dimension a

a �h��b

�� h�

��h� b�� b

��


Iy�

t

��b�� b��

Iz� tb

�a� � tb

��h � a��

t

�a� � �h� a��

Iyz � �

The warping function whose origin and pole are both chosen to be point O can bewritten from Figure 2.14 as

O�s

�� hs

�O�s

�� hs

�

The warping function O is zero on the branches s�, s�, and s�. To calculate they coordinate of the shear center using Eq. (2.12), its is necessary to evaluate thesectorial product of area

Iz�O �

Zz�s�O�s�dA � �

Z b��

�

hs��tds� �thb�

�

��

The shear center location is then given by Eq. (2.12) as

yS� y

O�Iz�O

Iy� y

O�

hb��b�� b��


P

y

z

C S

h

b�

b�

a

O

s�

s�

s�

s�

s�

FIGURE 2.14 I-beam cross section

or

yS� ��h � a� �

hb��b�� b��

� �h��b�� b�� hb�b��b

�� b��

��b� � b� � h��b�� b��

which shows that S is to the left of the centroid for b�� b

�. The warping constant

I� is found from Eq. (2.20), which for this cross section becomes

I�� I

�O� �y

S� y

O��I

y� �

Z b��

�

h�s��tds� �h�b�

�

�b�� b��

t

��b�� b��

th�

��

b��b��

b�� b��

The principal warping function, the origin of which can be taken atO, is foundby transforming

Oaccording to Eq. (2.9)

�s� � O�s� � �yS � yO�z�s�


In terms of the branch coordinates sk, � � k � �, defined in Figure 2.14, theprincipal warping function is

�s�� y

S� y

O�z�s

�� y

S� y

O�s

�

�s�� yS � yO�z�s�� yS � yO�s��s��

�s�� hs

�� y

S� y

O�z�s

�� h � y

S� y

O�s

�

�s�� hs� � �yS� y

O�z�s�� h� y

S� y

O�s�

The applied force P at the free end of the beam does not pass through theshear center. The force-couple equivalent of P at the shear center S is the force Pand the torsional moment T

�of P about S

T� � �a � jyS j�P �Phb��b�� b��

The angle of twist of the beam is, therefore, determined as for the beam of Fig-ure 2.12

�x�x� �T�cGJ

�cx� sinh cx� tanh cL�� cosh cx�

�

For the torsional constant J , Saint-Venant’s approximation can be used

J �t�

�h� b� � b��

The constant c depends on material constants and cross-sectional dimensions

c� �GJ�EI�

�Gt��h� b� � b��

�Eh��

b��b��

�

The internal forces at the clamped end are

Vz� P M

y� �PL T � T�

The torsional shear stress, which is proportional to ��x

, is zero at the clampedend. The shear stress distribution over the cross section at the fixed end x � � isgiven by Eq. (2.32) as

�xs�s� � �

PQy�s�

tIy�T�Q��s�

tI�(2.39)

In Eq. (2.39), the first moment area Qy�s� is calculated using section cuts such asthose indicated in Figure 2.15. For instance, with the section cut on the right flange

Qy�z� �

Z z

�b��

ztdz �t

�

�z� �

b��

�

�

The corresponding shear stress distribution is

��z� �P

�Iy

�b��

�� z�

�


y

z

C Sb�

b�

FIGURE 2.15 Transverse shear flow directions and section cuts

Similarly, the shear stress in the left flange is given by

��z� �P

�Iy

�b�� z�

�

and the shear stress in the web is zero.The second contribution to the shear stress in Eq. (2.39) is the warping shear

stress

��s� � �T�Q��s�

tI�

where T� is equal to the entire torque T�, since the pure torsion torque Tt is zero atthe clamped end. The first sectorial area moment Q��s� of the principal warpingfunction is calculated for the part of the cross section cut off at s, remembering thatthe integration starts at a free edge. The section cuts indicated in Figure 2.15 maybe used for this calculation. For instance, with the section cut on the right flange

Q��s��

Z s�

b��

��yS � yO�s�tds� �t�y

S� y

O�

��b��

�� s��

Similarly

Q��s��

Z s�

b��yS � yO�s�tds� � �

t�yS � yO�

��b�� s��

Q��s��

Q��s

��

Z s�

b��

�yS� y

O� h�s

�tds

��

t�h� yS� y

O�

��b�� s��

Q��s��

Z s�

b��

�h� yO � yS�s�tds� � �t�h� y

S� y

O�

��b�� s��


The shear stresses are expressed by

��s�� T�thb��

�b��

�� s��

��s�� T�thb��

�b�� s��

��s��

��s

��

�T�thb��

�b��

�� s�

��

��s�� T�thb��

�b�� s��

The sign of ��s�� is positive, which means that the shear flow is in the directionof increasing s�, hence upward. Similarly, the sign of ��s�� is negative, so theshear flow is in the direction of decreasing s�, hence upward. Thus, warping shearstresses on the right flange are directed upward, but the signs of ��s�� and ��s��show that the warping shear stresses on the left flange are directed downward.

The normal stress at the fixed end due to bending is

��z� �Myz

Iy

� �PLz

Iy

The normal warping stress is

��s� �M��s�

I�

where M� is the bimoment at the fixed end

M�� EI

��x��

T�c

tanh cL

To calculate the stresses numerically, the following dimensions will be as-sumed

b� � b h � b� � �b L � ��b t �b

��

Poisson’s ratio will be taken to be � ��. The modulus of elasticity �E and theshear modulus G are then

�E �E

��

��E

��

G �E

��

�E

�

The transverse and warping shear stress distributions at the clamped end of thebeam are sketched in Figure 2.16. The force-couple equivalent of the transverseshear stress �

�at the shear center S is a single force of magnitude V

z� P . The

warping shear stress ��

is statically equivalent to a couple. The total shear stresson the right flange is zero, so that, at the fixed end of the beam, all shear stressesare carried by the left flange.

The bending and warping normal stresses are shown in Figure 2.17. The max-imum normal warping stress exceeds the maximum bending stress. The bendingstresses are statically equivalent to the bending momentM

y� �PL. The warping


stresses are statically equivalent to zero force and zero couple. When consideredseparately for the two flanges, these stresses are equivalent to two equal and op-posite bending moments. The maximum stresses are shown in Table 2.1. Thereference stress �� is defined as

�� P

b�

��

��

��

��

��

��

FIGURE 2.16 Transverse and warping shear stresses at the fixed end

As a second example, the stress distribution in the simply supported beamshown in Figure 2.18 will be determined. The load is a vertical force of magnitudeP at midspan. The cross section, whose centroid C is also the shear center S, isshown in Figure 2.19. The wall thickness t is the same for the flanges and the web.


�

�

�

��

��

�

�

�

�

��

��

FIGURE 2.17 Bending and warping normal stresses at the fixed end

The area moments of inertia and the area product of inertia for this cross sec-tion are

Iy�

�tb�

Iz�th�

��tbh�

�Iyz

�thb�

�

The torsional constant, calculated from Saint-Venant’s approximation, is

J �ht�

�

�bt�

�t��h� �b�


MAXIMUM STRESS RIGHT FLANGE LEFT FLANGE

�� (Transverse) 6.7 1.7�� (Warping) 6.7 13.3��

�(Bending) 266.7 133.3

�� (Warping) 91.3 365.0

TABLE 2.1 Maximum stresses at the clamped end

L

�

P

L

x

FIGURE 2.18 Simply supported beam with midspan load

The warping function with pole and origin both at point O is

O�s�� O�s�� O�s�� hs�

The principal warping function is given by Eq. (2.17) as

�s� � O�s� �Q�O

A� �yS � yO�z�s�

Since

Q�O

�

Z b

�

hs�tds

��

thb�

�

and

A � t�h� �b�

the principal warping function is

�s��

hs��

�hb�

��h� �b�

�s�� hb�

��h� �b�

�s�� hs��

�hb�

��h� �b�


P

y

z

b

b

h

C

s�

s�

s�

OC

FIGURE 2.19 Cross section of the simply supported beam

The warping constant is found from Eq. (2.20)

I�� I

�O�Q��O

A� �y

S� y

O��I

y�

th�b��b� �h�

��h� �b�

The applied load at x � L�� is equivalent to a torsional couple T� and a trans-verse force P at the shear center

T� �Ph

�

The applied torque per unit length can be written in terms of the Dirac delta func-tion

m�x� � T��x�L

��

The angle of twist is calculated from Eq. (2.34) as

�x�x� � C� � C�x� C� cosh cx� C� sinh cx� ��x�

where ��x� � � for � � x � L�� and

��x� �T�cGJ

c�x�

L

�� sinh c�x�

L

��

for L�� x � L.


0 0.2 0.4 0.6 0.8 1

0

0.1

0.2

0.3

0.4

0.5

0.6

x�L

�x

FIGURE 2.20 Angle of twist for the simply supported beam

The boundary conditions at the two simple supports

�x��

x�L� � � �� L� � �

are solved for the integration constants

C�� C

��

T��GJ

C�� C

��

T� sinh cL��

cGJ sinh cL

For the left half of the beam

�x�x� �

T�

�cGJ

�cx� �

sinh cL��

sinh cLsinh cx

�

M��x� �

T�c sinh cL

sinhcL

�sinh cx

Tt�x� �

T��T� sinh cL��

sinh cLcosh cx

T��x� � T�

sinh cL��

sinh cLcosh cx

The qualitative behavior of these functions over the entire span of the beam can beseen in Figures 2.20, 2.21, and 2.22. In Figure 2.22, the torques Tt and T� are shown


0 0.2 0.4 0.6 0.8 1

0

0.1

0.2

0.3

0.4

x�L

M�

FIGURE 2.21 Bimoment for the simply supported beam

as fractions of the applied torque T�. The total torque is the sum of Tt

and T�

T �x� �T��

if x �L

�

T �x� � �T��

if x �L

�

The stresses at x � L�� at the section just to the left of the applied torque willbe calculated. The transverse shear stress at this section is

��s� � �IzQy�s� � IyzQz�s�

t�IyIz� I�

yz�

Vz

From Figure 2.19, on the right flange, the first moments of area are

Qy�s��

Z s�

b

�s�tds� �t�b� � s�

��

�

Qz�s

��

Z s�

b

�h

�tds

��

ht�b� s��

�

and

��s�� P �s� � b��bh� hs� � �bs��

�tb��h� b�


0 0.2 0.4 0.6 0.8 1

-0.4

-0.2

0

0.2

0.4

x�L

Tt

T�

FIGURE 2.22 Pure torsion and warping torques for the simply supported beam

The value of ��s�� for s� � b is negative, which means that the the stress is in thenegative s�, or the positive z, direction. For the web, the first moments of area are

Qy�s��

tb�

�

Qz�s

��

t�s�� hs� � bh�

�

The shear stress in the web is

��s�� P ��s�� hs� � h�

�tbh��h� b�

Similarly, on the left flange,

��s

��

P �b� s��bh� �bs� � hs��

�tb��h� b�

The warping shear stress at x � L�� is

��s� � �Q��s�

tI�

T�


where the warping torque T� is T��, because the pure torsion torque Tt is zero atmidspan. The expressions for the warping shear stresses are

��s�� T�h�b� s��bh� hs� � �bs��

�I��h� �b�

��s�� T�hb

��s� � h�

�I��h� �b�

��s�� T�h�b� s��bh� hs� � �bs��

�I��h� �b�

The normal stress distribution at x � L�� due to bending is

��

IyzMy

IyIz � I�yzy �

IzMy

IyIz � I�yzz

where My � PL��. The normal stress due to warping is

�� M�

I�

where

M� �T�

�ctanh

cL

�

The shear stress distribution at x � L�� is sketched in Figure 2.23. The forceresultant of the transverse shear stress �� over the two flanges is equal to the totalshear force P��. The transverse shear stresses over the web are statically equiva-lent to a zero force-couple. The warping shear stress �� is equivalent a torsionalmoment. The transverse shear stress adds to the warping shear stress over the leftflange, but subtracts from it over the right flange.

The normal stress distribution at x � L�� is sketched in Figure 2.24. Thenormal stress�� due to bending is statically equivalent to a bending moment aboutthe y axis. The warping stress �� is statically equivalent to a zero force-couple. Thebending and warping stresses are additive over the left flange.


�

�

�

�

�

��

�

�

�

�

��

FIGURE 2.23 Shear stresses for the simply supported beam


�

�

�

�

�

�

��

�

�

�

�

�

��

FIGURE 2.24 Normal stresses for the simply supported beam

CHAPTER III

THIN-WALLED ELASTIC BEAMS OF CLOSEDCROSS SECTION

3.1. Geometry of Deformation

A closed thin-walled cross section is shown in Figure 3.1. The tangential andnormal coordinates, s and n, are chosen so that the axes n, s, x form a right-handedtriad. The coordinate s traces the median line starting from an arbitrarily selectedorigin, and the y, z coordinates of any point on the median line are functions of s.The normal coordinate n of any point of the median line is zero. The angle ��s� ismeasured from the positive y axis to the positive n axis.

�xs

�

y

z

C

s

n

FIGURE 3.1 Closed thin-walled section

As in Chapter II, it will be assumed that the shape of the median line andits dimensions remain unchanged in the yz plane when the beam undergoes adeformation under static loads. This means that the transverse displacements,which are defined as the displacement components in the plane of the undeformed

56 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION

cross section, of a point on the median line are those of a point belonging to a planerigid curve constrained to move in its own plane. Let S be the shear center of thecross section shown in Figure 3.2 and let ��s� denote the tangential component ofthe point of the median line at the coordinate s. As shown in Chapter II, ��s� canbe written as

��x� s� � vS�x� cos ��s� � w

S�x� sin ��s� � �

x�x�h�s� (3.1)

where vS

, wS

are the displacements of the shear center in the y, z directions, and his the projection, onto the unit normal vector en, of the position vector r�s� of thepoint at s

h � r � en

y

�

s

s

n

z

r�s�

S

FIGURE 3.2 Tangential and normal components of displacement

It will now be assumed that the shear strain �xs

of the median line is equal toits value found in Saint-Venant torsion. This assumption can be written as

�xs

��u

�s��

�x�

�xsG

�qttG

where qt is the constant shear flow of Saint-Venant torsion

qt �Tt

and denotes twice the area enclosed by the median line

�

Ihds

3.1. GEOMETRY OF DEFORMATION 57

The derivative of u with respect to s is

�u

�s�

Tt

tG� v�S cos � �w�

S sin � � ��xh

from which the displacement of the point at s along the x axis is obtained by inte-gration

u � u� �

Z s

�

Tt

tGds� ��x

Z s

�

hds� v�Sy �w�

Sz

From Eq. (1.18)

TtG

��xH ds

t�s�

with which the axial displacement becomes

u � u� � ��x

H ds

t

Z s

�

ds

t� ��

x

Z s

�

hds� v�Sy �w�

Sz

The warping function for a closed section is defined by

�s� �

Z s

�

hds�H ds

t

Z s

�

ds

t(3.2)

The first term of the preceding equation will be recognized as the sectorial area, orthe warping function for an open section. The warping displacement can now bewritten in the same form as it was in Chapter II for open cross sections

u�x� s� � u��x�� v�

S�x�y�s� �w�

S�x�z�s� � ��

x�x��s� (3.3)

It is easily verified that the presence of the second integral in Eq. (3.2) does notchange Eq. (2.7) for changing the pole of the warping function from B to A

A�s� � B�s� � B�s�� zA � zB��y�s� � y�� yA � yB��z�s� � z�� (3.4)

The equations for finding the principal pole, or the shear center, also remain thesame as those for open sections

yS � yB �Iz�B

Iz� I

y�BIyz

IyIz� I�

yz

(3.5)

zS � zB �Iz�B

Iyz� I

y�BIy

IyIz� I�

yz

(3.6)

The principal warping function is given in terms of a warping function B, whosepole and origin are arbitrarily chosen, by Eq. (2.17)

�s� � B�s� �

Q�B

A� �z

S� z

B�y�s� � �y

S� y

B�z�s� (3.7)

Similarly, the warping constant is given by

I�� I

�B�Q��B

A� �y

S� y

B��I

y� ��y

S� y

B��z

S� z

B�Iyz� �z

S� z

B��I

z(3.8)


a

bC

y

z

s�

s�

s� O

s�

FIGURE 3.3 Thin-walled box section

As an example, consider the thin-walled rectangular cross section of uniformthickness t shown in Figure 3.3. If point O is used both as pole and origin, thewarping function is

O�s�� ab

a � bs�

O�s�� a�

a� b�s� � b�

O�s

��

b�

a� bs�

O�s�� ab

a� b�b� s��



Iy �tb��b � a�

�Iz �

ta��a� b�

�The shear center is at the centroid of the rectangle. The principal warping functionis found by an application of Eq. (3.7)

�s�� b�b� a�

��a� b��s� � a�

�s��

a�a� b�

��a� b��s

�� b�

�s�� b�b� a�

��a� b��s� � a�

�s�� a�a� b�

��a� b��s� � b�

The principal warping function is zero for a square cross section, which accordingto the theory being described here, is free of warping. The warping constant forthe rectangular box section is

I��

ta�b��b� a��

��a� b�

In the theory developed by Benscoter for closed thin-walled sections, the rateof angle of twist ��x in Eq. (3.3) is replaced by an arbitrary function � of x, so thatthe fundamental kinematical assumption for the warping displacement becomes

u�x� s� � u��x�� v�

S�x�y�s� � w�

S�x�z�s� � ��x��s� (3.9)

The normal strain �x is written as

�x� u�� v��

Sy � w��

Sz � �� (3.10)

The shear strain �xs

is

�xs ��u

�s��

�x� ��xh � �

�

�s� ��xh� �

�h�

tH ds

t

�(3.11)

3.2. Equations of Equilibrium

The normal stress �x is obtained from Hooke’s law as

�x� �E�

x� �E�u�� v��

Sy � w��

Sz � ��

The shear stress is the sum of the bending and the torsional contributions

�xs

� �b� �

t

The torsional contribution is

�t� G�

xs� G��

xh�G��h� k�

where the abbreviation

k �

tH ds

t

has been introduced. As for open cross sections, the bending shear stress has nocorresponding shear strain.


The stress resultants for the normal stress �x are the axial force N , the bendingmomentsM

yand M

z, and the bimoment M

�, which are defined by

N �

Z�xdA

My �

Zz�dA

Mz � �

Zy�dA

M� �

Z�dA

The stress resultants are evaluated, recalling that the origin of the coordinates y, zis the centroid, and that is the principal warping function�

BB�NMy

Mz

M�

CCA � �E

�BB�A � � �� I

yz�I

y�

� Iz

Iyz

�

� � � �I�

CCA�BB�u��v��Sw��

S��x

CCA

Hence

�Eu��x� �N �x�

A

�Ev��S�x� �IyzMy�x� � I

yMz�x�

IyIz� I�

yz

�Ew��

S�x� � �IzMy�x� � I

yzMz�x�

IyIz� I�

yz

�E��x�x� � �M��x�

I�

The normal stress is found in terms of the stress resultants by using these expres-sions in Eq. (2.21)

�x �N

A�IyzMy� I

yMz

IyIz� I�

yz

y �IzMy� I

yzMz

IyIz� I�

yz

z �M�

I�

(3.12)

The total torque T is

T �

Zh�

xsdA � G��

x

Zh�dA�G�

�Zh�dA�

ZhkdA

�

The last part of this expression can be evaluated asZhkdA �

ZhdA

tH ds

t

��

H ds

t

� J

A cross-sectional property, sometimes called the polar constant, is defined by

Ih�

Zh�dA

so that the torque T is

T � GIh��

x � G��Ih � J�


The torque equilibrium equation for a length dx of the beam givesdT

dx�m�x� � GIh�

��

x � G��Ih � J� �m�x� � �

where m�x� is the applied torque about the shear center S per unit length of thebeam.

As in Chapter II, the equilibrium equation in the longitudinal direction, in theabsence of applied axial load px, gives

t��x�x

��q

�s� �

The equilibrium of forces in the longitudinal direction remains the same�Eu��

��x�A� p

x� �

which gives u��x� � �. Hence

�q

�s� �t

��x�x

� t �E�v��Sy � w��

Sz � ��

The shear flow is

q�x� s� � q��x� ��E�v��S�x�Q

z�s� �w��

S�x�Q

y�s� � ��

x�x�Q

��s��

(3.13)

As in Chapter II, the shear stress resultants Vy , Vz , and T� be defined by

Vy �

Zq�x� s�dy

Vz �

Zq�x� s�dz

T��

Zq�x� s�d

The warping torque is calculated by integrating both sides of Eq. (3.13) withrespect to

T�� E��

xQ��s�d � � �E��

xI�

because, as shown Chapter II by an integration by parts,ZQ��s�d � �I

�

Since Zq�h� k�ds � G�I

h� J��

x� ��

it follows that

� �EI�� G�I

h� J��

x� ��

The equation for the angle of twist is found by eliminating � from

GIh��

x � G��Ih � J� �m�x� � �

G�Ih� J��

x� �� EI

��

From the first of these equations

�� Ih

Ih� J

��x �m

G�Ih� J�


The differential equation for the angle of twist is

�EI�Ih

Ih � J�ivx�GJ��

x� m�x� �

�EI�

G�Ih � J�m��x� (3.14)

When attempting to use Eq. (3.14), it is possible to encounter cross sections forwhich Ih and J are equal. For the rectangular box section of Figure 3.3, the polarconstant Ih is

Ih�

tab�a� b�

�

and the torsional constant J is

J ��ta�b�

a� b

For a square cross section, with a � b,

Ih � J � tb�

and Eq. (3.14) cannot be used. In general, when the polar constant is the identicalto the torsional constant, the cross section is free of warping, and the warpingfunction is everywhere zero. The differential equation for the angle of twist thenreduces to

GJ��x �m�x� � �

This is the governing equation for Saint-Venant torsion with a variable distributedmoment m�x�.

3.3. A Multicell Analysis Example

If the area enclosed by the outer wall of a cross section is subdivided into anynumber of other closed thin-walled sections, the beam is a multicell structure, anexample of which is shown in Figure 3.4. For such cross sections, the condition ob-tained in Chapter I by taking the line integral of the derivative of the longitudinaldisplacement u with respect to s around a closed contour is used

�

G

Ii

qds

t�s��

xi� � (3.15)

where the integral is taken around the contour of the ith cell, and i

is twice thearea enclosed by the contour of the ith cell.

In pure torsion the shear flow in each cell has a constant value. On a sharedwall, such as the one of length b in Figure 3.4, the shear flows are additive

q��

� q�� q

�

where q�, q� are the individual shear flows in the two cells and q�� is the shear flowin the shared part of the wall. The condition in Eq. (3.15) is applied to the two cells,assuming that the thickness t is uniform throughout the cross section,

��x��

�q��d� e�

Gt�q�b

Gt� �

��x� �

q�b

Gt�

�q��a� b�

Gt� �

3.3. A MULTICELL ANALYSIS EXAMPLE 63

C

cz

cy

S

y

z

a

b

d

e

t

FIGURE 3.4 Two-cell thin-walled cross section

The directions assumed for the shear flows q�, q� are indicated in Figure 3.5. Thetotal torque in the section is

T � q�� q

��

The torsional constant can be calculated from

J �T

G��x

�q�� q

��

G��x

The warping function with respect to any arbitrarily chosen pole O is writtenas

O�s� �

Z s

�

hds ��

G

Z s

�

qds

t

where the first integral is the sectorial area. In the second integral, q denotes theshear flow corresponding to ��x � �. For the example two-cell section, with the scoordinates defined in Figure 3.5, the warping function O, whose pole and origin


C

S

O

y

z

s�

s�

s�

s�

s�

s�

s�

s

q�

q�

FIGURE 3.5 Definitions for the two-cell thin-walled section

are both chosen to be point O, is written as

��O �s��

q�s�Gt

��O �s

��

��O �a� � as� �

�q� � q��s�Gt

��O �s

��

��O �b� � bs

��q�s�Gt

��O �s

��

��O �a� �

q�s�Gt

��O �s��

��O �a� � as� �

q�s�Gt

��O �s��

��O �d� b� � �d� b�s� �

q�s�Gt

��O �s��

��O �e� � �a� e�s

��q�s�Gt

�O �s

� �

��O �d� � bs

�q�sGt

3.3. A MULTICELL ANALYSIS EXAMPLE 65

Let the dimensions of the cross section be

d � �� mm e � �� mm b � �� mm a � � mm t � �� mm

Then, based on centerline dimensions, the location of the centroid C is defined by

cy� �� mm c

z� �� mm

The area moments of inertia and the area product of inertia, based on centerlinedimensions, are found to be

Iy� �� mm� I

z� � �� mm� I

yz� �� mm�

The sectorial areas

� � �ed � �� mm� � � �ab � �� mm�

are needed in the calculation of the torsional constant J . The shear flows are

q� ��G��x

��q� �

��G��x��

where the numerical values in the numerators are in mm�. With the shear flowsdetermined, the torsional constant can be calculated

J �q�� q��

G��x� �� mm�

and the warping function O

becomes

O�s�� s��

O�s�� s� � ��

��

O�s�� s� � ��

��

O�s��

�� s��

��

O�s��

�� s��

��

O�s

��

�� s� � ��

��

O�s

��

�� s� � ��

��

O�s� �

�� s�

��

where the dimension of the s coordinates is in mm, and the dimension of O is inmm�.


Next the section properties dependent on O are calculated

Q�O�

ZOdA � �� mm�

I�O �

Z�OdA � �� mm�

Iy�O �

ZyOdA � �� mm�

Iz�O

�

Zz

OdA � � �� mm�

The shear center coordinates are

yS� y

O�Iz�OIz � Iy�OIyz

IyIz � I�yz� �� mm

zS� y

O�Iz�OIyz � Iy�OIy

IyIz � I�yz� �� mm

The principal warping function, whose pole is the shear center S, can be ob-tained from O by the transformation

�s� � O�s� �Q�O

A� �zS � zO�y�s� � �yS � yO�z�s�

where A is the cross-sectional area

A � t��e� �d� b� �a� � �� mm�

The warping constant I� can be determined either by integrating the square of theprincipal warping function over the cross-sectional area, or by the transformationformula

I�� I

�O�Q��O

A� �y

S� y

O�Iy� �z

S� z

O�Iz� ��y

S� y

O��z

S� z

O�Iyz

� �� mm�

3.4. Cross Sections with Open and Closed Parts

Some cross sections contain both closed cells and open branches. An exampleis shown in Figure 3.7. In analyzing such cross sections, the warping functions forthe open branches are found as described in Chapter II. The warping functionsfor the closed cells are found as described in the two-cell example of the preced-ing section. The contribution of the open branches to the torsional constant J isusually negligibly small, so that J can be calculated for the closed cells of the crosssection alone.

For the cross section shown in Figure 3.7, the origin of the user coordinatesystem is placed at point O, with the y axis horizontal and pointing left, the z axisvertically downward. The node coordinates are shown in Table 3.1 below. Thewall thicknesses are in Table 3.2, each line entry of which lists two nodes and thethickness of the wall segment between them.

The properties of the cross section are listed in Table 3.3. A qualitative ideaof the distributions of normal stress and strain due to warping is provided by theprincipal warping function. Because this section has straight walls, the warping

3.4. CROSS SECTIONS WITH OPEN AND CLOSED PARTS 67

S

�

�

�

��

FIGURE 3.6 Principal warping function for the two-cell thin-walled section

CS

O ��

� �

�

�

��

FIGURE 3.7 Thin-walled section with three cells

function is piecewise linear. Table 3.4 lists numerical values of the warping func-tion, and Figure 3.8 shows how it varies along the median line.

As a final example, the cross section shown in Figure 3.9, which resemblescertain thin-walled sections found in automobile frames, will be analyzed. Theuser coordinate system for this cross section has its origin at point O, with the yaxis horizontal and directed toward the left and the z axis vertically downward.The position coordinates of the nodes of the section are listed in Table 3.5. Thethickness is uniform for the entire cross section. The principal warping functionfor the section is sketched in Figure 3.10.

The results derived above by the approximate linear theory of beams withthin-walled cross section are compared with the results calculated by the the com-puter program BEAMSTRESS in Table 3.7. When the wall thickness is very small,


Node y coordinate (mm) z coordinate (mm)� ��

� ��

��

� ��

� ��

� ��

��

� ��

� ��

��

TABLE 3.1 Nodal coordinates of the cross section shown in Figure 3.7

First Node Second Node Thickness (mm)� � ��

� ��

� ��

� � ��

� ��

� � �

� �

� �

� � �

� ��

� ��

� � ��

TABLE 3.2 Wall thicknesses of the cross section shown in Figure 3.7

the linear theory agrees well with the results obtained by this program, which pro-vides a finite-element calculation based on the elasticity formulation. As the wallthickness increases, the linear theory results become less accurate, and it is pos-sible to have very large errors in the section properties, especially in the warpingconstant. As the thickness is changed, the warping function of the linear theorydoes not change, because the median line of the section determines this function.The elasticity formulation considers the warping function as a function of y andz, and when the boundary of the section is changed, the warping function alsochanges. The large errors in I

yzare a consequence of the assumption in the lin-

ear theory that the elements of cross-sectional area are entirely concentrated at themedian line.


Cross-Sectional Area (mm�) ��

Centroid yC

(User Coordinates) (mm) �

Centroid zC (User Coordinates) (mm) ��

Shear Center yS (Centroidal Coordinates) (mm) �

Shear Center zS (Centroidal Coordinates) (mm) ��

Shear Center yS

(User Coordinates) (mm) �

Shear Center zS

(User Coordinates) (mm) � ��

Area Moment of Inertia Iy

(mm�) ��

Area Moment of Inertia Iz (mm�) ��

Area Product of Inertia Iyz (mm�) �

Polar Constant Ih

(mm�) ��

Torsional Constant J (mm�) ��

Warping Constant I�

(mm�) ��

TABLE 3.3 Properties of the cross section in Figure 3.7

Node m Node n m

(mm�) n

(mm�)

� ��

� ��

� � ��

� � ��

� ��

� ��

� ��

� � ��

� � ��

� ��

� � ��

� ��

TABLE 3.4 Values of the principal warping function for the section in Figure 3.7


S

�

�

��

�

�

�

��

FIGURE 3.8 Warping function for the section in Figure 3.7

C

S

O

� �

�

�

�

� �

FIGURE 3.9 Thin-walled cross section


Node y coordinate (mm) z coordinate (mm)� ��

� ��

��

� ��

� ��

� ��

� ��

� ��

� � ��

TABLE 3.5 Nodal coordinates of the cross section in Figure 3.9

Node m Node n m

(mm�) n

(mm�)

� � ��

� � ��

� ��

� ��

� ��

� ��

� ��

� � ��

� � ��

� � ��

TABLE 3.6 Numerical values of the principal warping function sketched in Figure 3.10


S

� �

�

�

�

� �

FIGURE 3.10 Principal warping function for the section in Figure 3.9


Thickness Property Linear Theory BEAMSTRESS Difference (%)

Area (mm�) ��

Iy

(mm�) ��

� mm Iz (mm�) ��

Iyz (mm�) ��

J (mm�) ��

I�

(mm�) ��

Area (mm�) ��

Iy

(mm�) ��

� mm Iz

(mm�) ��

Iyz (mm�) ��

J (mm�) � � ��

I�

(mm�) ��

Area (mm�) ��

Iy

(mm�) ��

�� mm Iz

(mm�) ��

Iyz (mm�) ��

J (mm�) ��

I�

(mm�) ��

Area (mm�) ��

Iy

(mm�) ��

�� mm Iz

(mm�) ��

Iyz (mm�) ��

J (mm�) ��

I�

(mm�) ��

TABLE 3.7 Properties of the cross section in Figure 3.9

bmbk

Documents

Transcript of bmbk