[Solid Mechanics and Its Applications] Elementary Continuum Mechanics for Everyone Volume 194 ||...

E. Byskov, Elementary Continuum Mechanics for Everyone,Solid Mechanics and Its Applications 194, DOI: 10.1007/978-94-007-5766-0_11,Ó Springer Science+Business Media Dordrecht 2013

Chapter 11

Bending and AxialDeformation of LinearElastic BeamCross-Sections11.1 Linear Elastic MaterialIn Sections 7.7 and 7.8, which are concerned with plane, straight linearlyelastic beams, the first with Bernoulli-Euler and the second with Timo-shenko beams, respectively, we assumed the following constitutive relations,see (7.86) and (7.91), respectively

Hooke’s “Law”

[N

M

]=

[EA 0

0 EI

][ε

κ

]and

N

V

M

=

EA 0 0

0 GAe 0

0 0 EI

ε

ϕ

κ

(11.1)

but, at that point we did not discuss the physical meaning of EA, GAe

and EI, except that they connected the generalized stresses and generalizedstrains. In this Chapter we shall attempt to determine A and I for a numberof different beam cross-sections, and in Chapter 12 we do the same for Ae.

11.1.1 PurposeIn Part II the meaning of the generalized quantities was clear, but when weconsider two- or three-dimensional bodies this is not necessarily the case, aswe shall see.

11.1.2 Beam Cross-Section and Beam FibersIf we were to do things completely right, we should base our derivations on Cut corners and

apply simplified

continuum

mechanics

continuum mechanics as it was introduced in Part I and treat all beams asthree-dimensional bodies subject to some kinematic or static constraints,possibly using the finite element method with three-dimensional elements.We shall, however, choose a much easier possibility, namely one which con-

August 14, 2012 Continuum Mechanics for Everyone Esben Byskov

199

200 Beams with Cross-Sections

siders the beam as consisting of “fibers,” see Fig. 11.1, where one of thesefibers is shown. When we do this, we do not account for strains perpendic-ular to the beam axis, making the derivations far simpler than in a a fullthree-dimensional analysis. Thus, we seek an engineering solution which isnot the correct one, but one that, hopefully, is sufficiently accurate.

y

z

x

σdA

Fig. 11.1: Part of a beam with a “fiber.”

In the present connection we must not take the word “fiber” on its faceBeam“fiber”

value, e.g. as a wood fiber, but rather imagine that the beam consists of in-finitely many, infinitely thin threads which cannot become longer or shorterwithout deformation of their neighbors. Before we may exploit the idea ofbeam fibers we must assume their constitutive behavior which we simplytake as a linear relation between their axial strain ε and axial stress σ, seeFig. 11.2, where the force acting on a fiber with area dA is dP .

σ ≡ limA→0

P

A=dP

dA(11.2)

Now we assume that ε

σ(x) = E(x)ε(x) (11.3)

where it is indicated that Young’s Modulus E may vary along the lengthof the beam. For simplicity we shall assume that E does not vary over thecross-section. Letting E vary would merely complicate the derivations andresults below, but it is quite easy to account for varying Young’s Moduli.In general the deformation of the fibers varies along the length

σ(x, y, z) = E(x, y, z)ε(x, y, z) (11.4)

We shall, however, limit ourselves to the case of a constant Young’sModulusAssume

E = const.

⇒ Reinforced

concrete beams

not covered

σ(x, y, z) = Eε(x, y, z) , E = const. (11.5)

thus excluding for example reinforced concrete beams since the reinforce-ment bars and the concrete have very different Young’s Moduli.

Esben Byskov Continuum Mechanics for Everyone August 14, 2012

Bending and Axial Deformation of Linear Elastic Cross-Sections 201

You may have observed that the coordinates here are denoted x, y and zinstead of xj , j = 1, 2, 3 which is the habit in the major part of this book.The main reason for abandoning the latter notation which for most purposesis the more systematic one is that, usually the coordinate along the beamaxis is called x. The advantage of using x1 and x2 instead of y and z,respectively, to indicate the axes perpendicular to the beam axis wouldbe minor, as you may see from the derivations below, since we are onlyconsidering bending about the z-axis.11.1 Then, the two coordinates in thecross-section play different roles which makes the xα-notation less obvious.Finally, the notation used here is the traditional one—not a very strongargument in my opinion.

dy

y

z h

dx εbdx σb

ytyf

εtdx σt

yb

b(y)

Fig. 11.2: Part of a beam with a “fiber” stretching over thewidth of the beam.

Behind the assumption that the beam does not deflect in the directionof the z-axis lies the requirements that the beam cross-section is symmetricwith respect to the y-axis and that the loading exhibits the same symmetry.

11.1.3 Pure Axial Strain

Assume that all beam fibers are subjected to the same strain ε = const. Axial strain ε

Axial force Nand that E = const. everywhere, then the fiber stress σ is independent ofall coordinates. Therefore, it is easy to integrate the fiber stress. We expectto find N = EAε, see (11.1), and consequently we call the integral N

N =

∫

A

σdA =

∫

A

EεdA = ε

∫

A

EdA = εE

∫

A

dA

N = EAε , EA = const.

(11.6)

Thus, we have found the relationship between the axial strain and theaxial force of the beam. Now, we turn to the more interesting case ofcurvature (bending) strain as well as axial strain.

11.1 You might ask why I don’t cover bending about two axes. The reason is that I wantto address the most basic issues and Part II is only concerned with in-plane bending.



11.1.4 Both Axial and Curvature Strain in Bernoulli-Euler Beams

In Section 7.5 we developed the kinematically linear theory for straightAxial strain εCurvature strain κ

Axial force N

Bending moment

M

Bernoulli-Euler beams. This theory entails the axial strain ε and the curva-ture strain κ as its generalized strains. The axial force N and the bendingmomentM are their work conjugate generalized stresses, see e.g. Section 7.5and Section 7.3, which covers the more general case of moderate kinematicnonlinearity. We shall therefore seek expressions connecting κ andM in ad-dition to the relation between ε and N . As shown in Fig. 11.2 and Fig. 11.3

= +

dxdx12 (εb + εt)dx

12 (εb − εt)dx

h

εbdx

− 12 (εb − εt)dxεtdx

dx

Fig. 11.3: Beam strain divided into axial and bending con-tributions.

we assume that a plane cross-section remains plane after deformation. It isnot shown in those figures that we insist that the cross-sections are perpen-dicular to the beam axis, both before and after deformation, but this is aconsequence of the restriction of the Bernoulli-Euler hypothesis of vanishingshear strains, see Sections 7.4 and 7.6.

Traditionally, it is presumed that the cross-sections do not shrink orexpand, but this is not relevant when we base our derivations on the ideaof fibers. It is, of course, important when continuum mechanics is used,see also Example Ex 12-1 where it is discussed how a Bernoulli-Euler beammight be loaded.

When we consider small deformations and observe the above limitationsas regards the deformation of the cross-sections we may find that the axialdisplacement of the beam fibers u can be described by a bilinear expression

Linear

displacementu(x, y) = cx(x)(c0 + cyy) (11.7)

where we note that the axial displacement does not vary with z. Then,

Linear strain εf (x, y) =du(x, y)

dx=dcx(x)

dx(c0 + cyy) (11.8)

where εf denotes the axial fiber strain.



At this point we need to relate the generalized strains ε and κ to εf (x, y).In Fig. 11.3 the fiber strain is divided into two parts, namely one which isconstant and one which covers the rest. It should come as no surprise thatwe intend to identify the constant part as the axial beam strain ε, while wehope that the rest is closely related to the beam curvature strain κ. Basedon Fig. 11.3 we may find

ε = 12 (εb + εt) (11.9)

As regards the curvature strain we must go through a somewhat morecomplicated path. Originally, we defined the curvature strain κ as thederivative of the change in angle of the beam axis, see Section 7.2, (7.9)

Nonlinear

curvature measureκ ≡ dω

dx(11.10)

and in the case of a straight, linearly or moderately linear, see Section 7.3,(7.14), beam we found it to be

Linear curvature

measureκ =

d2w

dx2=

d

dx

(dw

dx

)(11.11)

which is a linear expression for the curvature strain as the change in angleper length of the beam. The Bernoulli-Euler conditions force the angle ofthe cross-section to be the same as the angle of the beam axis and thus wemay find the curvature strain as the change in angle of the cross-section perlength of the beam. Then from Fig. 11.3

Curvature strain

in terms of fiber

strains

κzz =(εb − εt)dx

dx

1

h=

(εb − εt)

h(11.12)

where we have indicated that the curvature is about the z-axis.

In (11.9)–(11.12) it is not explicitly stated that the strains may vary withthe axial coordinate x. In order to determine the “fiber stress”σf (x, y), andbecause of the later computation of the cross-sectional constants A, S andI, we wish to express the fiber strain εf (x, y) in terms of the generalizedbeam strains ε(x) and κ(x)

Fiber strain in

terms of axial and

bending strain

εf (x, y) = ε(x) − yκzz(x) (11.13)

which is seen to be of the same form as (11.8). The minus sign in front of thesecond term is due to the fact that a positive value of κ results in a negativecontribution to the fiber strain in the upper part of the cross-section.

11.1.5 Axial Force, Zeroth- and First-Order Moments

Our present assumption of linear elastic behavior implies linear relationsbetween the fiber strain εf and the fiber stress σf

Hooke’s “Law”σf (x, y, z) = E(x, y, z)εf (x, y) (11.14)



but symmetry about the y-axis demands that

E(x, y, z) = E(x, y,−z) (11.15)

For the time being we limit ourselves to the case of a Young’s ModulusE which is independent of the coordinates y and z

σf (x, y) = E(x)εf (x, y) (11.16)

Integrate the fiber stress σf over the cross-section to get

N(x) =

∫

A

σf (x, y)dA =

∫

A

E(x)εf (x, y)dA

= E(x)

∫

A

εf (x, y)dA = E(x)

∫

A

(ε(x)− yκzz(x)

)dA

(11.17)

and thus

Axial force N N(x) = E(x)(A(x)ε(x) − Sz(x)κzz(x)

)(11.18)

where

Cross-sectional

area ACross-sectional area : A(x) ≡

∫

A(x)

dA (11.19)

and

Static moment Sz Static moment about the z-axis : Sz(x) ≡∫

A(x)

ydA (11.20)

For the sake of completeness it may be worth mentioning that the areaZeroth-ordermoment = Area

First-order moment

= Static moment

A is the zeroth-order moment of the cross-section, while Sy is its first-ordermoment about the z-axis. By the way, some authors prefer the name firstmoment or static moment instead of first-order moment.

As (11.18) shows there are two possibilities of getting a state of pureaxial force. Either κ = 0 or Sz = 0 (or both, of course), where the first wasinvestigated in Section 11.1.3. The other possibility puts demands on thelocation of the x-axis. We shall return to that possibility in Section 11.1.8.

11.1.6 Bending Moment and Second-Order Moments

Probably it is not surprising that in the general case the stresses, givenby the fiber strain in (11.13), result in a bending moment. Therefore, wecompute their moment Mz(x) about the z-axis

Mz(x) = −∫

A

yσf (x, y)dA = −∫

A

yEεf (x, y)dA

= −E∫

A

yεf(x, y)dA = −E∫

A

(yε(x)− y2κzz(x)

)dA

= −ε(x)E∫

A

ydA+ κzz(x)E

∫

A

y2dA

(11.21)



Then,

Mz(x) = −ESzε(x) + EIzzκzz(x) (11.22)

whereMoment of InertiaIzz= Second-order

moment

Moment of inertia about the z-axis: Izz ≡∫

A

y2dA (11.23)

and where we observe that the minus sign on the first term on the left-handside of (11.22) is caused by the fact that a positive value of σf for y > 0results in a negative contribution to the bending moment Mz. The quan-tity Izz is the second-order moment about the z-axis of the cross-sectionalarea.11.2

As seen from (11.18) and (11.22) the stress-strain relations are not of thesame form as postulated in (11.1a), originally (7.86), in that the terms withSz do not appear there. The reason is that when we introduced (11.1a)and (7.86) the beam did not have depth and therefore the only naturalchoice of beam axis was the position of the one-dimensional beam itself. InSection 11.1.8 we look into the issue of the position of the beam axis.

11.1.7 Summary of Linear Elastic Stress-Strain Rela-tions

In view of the fact that we have confined ourselves to plane beams weintroduce a simpler notation for the bending moment, the static momentand for the moment of inertia. The reason for using a more complicatednotation was that we wished to make it clear about which axes we weretaking the moments.

Therefore, introduce

Summary ofcross-sectionalpropertiesand of

linear elastic

stress-strain

relations

Cross-sectional area : A ≡∫

A

dA

Static moment about the z-axis : S = Sz ≡∫

A

ydA

Moment of inertia about the z-axis : I = Izz ≡∫

A

y2dA

Axial force : N ≡∫

A

σfdA

Bending moment about the z-axis : M =Mzz ≡∫

A

σfydA

(11.24)

11.2 It ought to be obvious that the cross-section is associated with one more second-order moment similar to Izz, namely the second-order moment Iyy ≡

∫Az2dA about the

y-axis. In addition to this there exists a mixed second-order Iyz ≡∫AyzdA. Since we

are only concerned with bending that is symmetric about the y-axis we do not need theseother second-order moments.



With this notation we get the following constitutive conditions accordingto Hooke’s “Law”

Hooke’s “Law”

[N

M

]=

[EA −ES−ES EI

][ε

κ

](11.25)

Invert this relationship and get

Hooke’s “Law”

[ε

κ

]=

1

E(AI − S2)

[I S

S A

][N

M

](11.26)

which we intend to utilize in Section 11.1.8 when we discuss the placementof the x-axis.

11.1.8 Cross-Sectional Axes—Beam Axis and Centerof Gravity

In the above derivations we chose the position of the x-axis completelyAxes of the

cross-section arbitrarily. If we can choose the axis such that the first-order moment Svanishes, then Hooke’s Law simplifies, see (11.25) and (11.26). More oftenthan not, this proves to be a fairly good idea,11.3 in part because it appearsa reasonable request that it should be possible to apply an axial loadingwithout causing bending of the beam and, in the same spirit, to applya moment load without stretching of the beam axis. These assumptionslie behind the derivations of the strictly one-dimensional beam theory, seeSections 7.5, 8.1 and 8.2 where we, without any kind of discussion, identifiedthe beam axis as the physical configuration of the one-dimensional beam.As we may see from (11.26) we can get both of the above wishes fulfilledif the static moment S vanishes. This condition may be used to fix theposition of the beam axis with the implication that for pure bending thefiber stress σf of the beam axis equals zero. Then the beam axis and the so-called neutral axis coincide. By the same token this means that the neutralNeutral axis

axis is the one which does not experience axial strain under pure momentloading.

11.1.9 The Beam Axis at the Neutral Axis

In the following we shall discuss some examples of determining the momentsof various order of some types of cross-sections. In all these cases we choosethe beam axis and the neutral axis to coincide. Then,

Hooke’s “Law” for

beam axis =

neutral axis

[N

M

]=

[EA 0

0 EI

][ε

κ

](11.27)

11.3 Saying that it is a good idea to let the beam axis be the same as the neutral axismay be a too strong statement because in the case of interconnected beams it may indeedbe better to choose a common axis which may not cause S to vanish in any of the beams.



The inverse relationship is

Hooke’s “Law” for

beam axis =

neutral axis

[ε

κ

]=

[1/(EA) 0

0 1/(EI)

] [N

M

](11.28)

At this point we may remark that it is quite easy to find the cross-sectional properties associated with other axes than the neutral axis oncewe have determined the ones related to that axis, see (11.32).

11.1.10 Independence of Results of Choice of BeamAxis

I gather that everyone would consider it unfortunate if the results as re- Strains and stresses

must not depend on

choice of beam axis

gards the stress and strain distribution over the cross-section depended onthe choice of beam axis. If that were the case, then the physical resultswould depend on the choice of coordinate system which must be deemedunacceptable. In that case, the description would not be objective and thesituation would be similar if the speed of a car depended on the coordinatesystem. Our results must be frame indifferent. We shall see that our descrip- Results must be

frame indifferenttion is objective and that our solutions are frame indifferent. In Fig. 11.4

y, y

η

a

P

z

z x

x

Fig. 11.4: Beam axes.

the beam axis, the x-axis lies at η > 0 below the neutral axis, the x-axis. Asindicated, the load consists of a force P at y = a > 0. This means that wecover loadings that consist of a force P at the center of gravity in additionto a moment, a couple C. Of course, a situation with pure bending deservesanother treatment, but, as you might observe, we can describe that situationby letting aP = −C and at the same time let P = 0 in the formulas below.Applying P at y = a along with −P at y = −a produces a a case with noresulting axial force, but with a prescribed couple C = −2P a. Thus, wemay reuse the formulas below by simple addition.



From Fig. 11.4

y = y + η (11.29)

where tilde (˜) indicates that the quantity is referred to the (x, y, z)-coor-dinate system.11.4

When we refer the axial force and bending moment to the two axes wemay see from Fig. 11.4 that

N = P , M = −aP

N = P , M = −(a+ η)P(11.30)

Utilize (11.24) to find the cross-sectional properties that refer to thez-axis

A =

∫

A

dA

S =

∫

A

ydA = 0

I = Izz =

∫

A

y2dA

(11.31)

In the same spirit the properties associated with the z-axis

Cross-sectional

properties referred

to the y- and

y-axes

A =

∫

A

dA = A

S = Sz =

∫

A

ydA =

∫

A

(y + η)dA = S + ηA

= +ηA

I = Izz =

∫

A

y2dA =

∫

A

(y2 + 2yη + η2)da = I + 2ηS + η2A

= I + η2A

(11.32)

which means that the static moment S is equal to the distance η from theneutral axis multiplied by the area A and that the moment of inertia I isequal to the moment of inertia I about the neutral axis plus the the squareof the distance η from the neutral axis multiplied by the area A. Later theseresults prove to be very valuable. From (11.32) we observe that the momentof inertia I about the neutral axis is the smallest one possible.

11.4 In this connection the meaning of the tilde ought to be clear and must not beconfused with other cases where I have used it to indicate an approximate solution.



Once we have found A, S and I we may determine η and I

η =S

a

I = I +

(S

A

)2

A ⇒ I = I − S2

A

(11.33)

Let us insert (11.30a) in (11.28) and get

[ε

κ

]=

[1/(EA) 0

0 1/(EI)

][P

−aP

]=

[P /(EA)

−aP/(EI)

](11.34)

Utilize (11.13) and (11.34) and obtain the following expression for thefiber strain εf

εf = ε− yκ =P

EA+ y

aP

EI(11.35)

In a similar fashion use (11.26) and (11.32) to get

[ε

κ

]=

1

EAI

[(I + η2A) +ηA

+ηA A

][P

−(a+ η)P

](11.36)

where we have exploited the fact that

AI − S2 = AI (11.37)

After some trivial manipulations we may get

Relation between

(ε, κ) and (ε, κ)

ε

κ

=

P

EA+ η

aP

EI

−aPEI

=

ε+ ηκ

κ

(11.38)

where (11.34) has been utilized. Finally we may use (11.38) and (11.13)with (εf , ε, κzz, y) = (εf , ε, κ, y) to get

Fiber strain is

independent of

coordinate system

εf = ε− yκ = (ε+ ηκ)− (y + η)κ = ε− yκ = εf (11.39)

and thus we have shown that the choice of beam axis does not influence thephysical results. Therefore the description is objective.



11.1.11 Distribution of Axial Strain and Axial Stress

In Section 11.1.10 we concluded that the fiber strain εf isFiber strain is

linear over the

cross-section

εf = ε− yκ (11.40)

which means that the fiber strain ε is linear over the cross-section, seeFig. 11.2. Under the assumption of linear elasticity the fiber stress σf willalso be linearly distributed over the cross-section because the constitutivemodel (11.16) together with (11.40) gives:

Fiber stress is

linear over the

cross-section

σf = E(ε− yκ) (11.41)

If we exploit the constitutive relations (11.27) we may express the fiberstress in terms of the axial force N and the bending moment M

Navier’s formula σf =N

A− y

M

I(11.42)

In the literature this formula is known as Navier’s Formula after theFrench engineer Claude Louis Marie Henri Navier.

In order to use a simpler nomenclature we shall omit the lower index f

and talk about the axial strain ε and the axial stress σ in stead of εf andσf , respectively. Later, when we wish to refer to the strain and stress of thebeam axis we must indicate this explicitly. A commonly used notation usesa lower index 0 to signify the neutral axis. If, however, we insist on usinganother beam axis we need another indicator.

11.1.12 Examples of Moments of Inertia

Our first example is a rectangular cross-section which is important in it-self, but the results are useful in many connections, see examples Ex 11-3and Ex 11-4.

Ex 11-1 Rectangular Cross-SectionThe rectangular cross-section, see Fig. Ex. 11-1.1, probably is the sim-Rectangular

cross-section plest example in this connection. By z we denote the axis about whichthe static moment S vanishes,11.5 while z is an axis that we may placeas we see fit. The reason it lies at the bottom of the cross-section isthat that choice makes the following derivations easy.

Utilize (11.24) to get the quantities referred to the z-axis

A =

∫

A

dA = bh

S = Sz =

∫

A

ydA = 12bh2

I = Izz =

∫

A

y2dA = 13bh3

(Ex. 11-1.1)

11.5 You may have guessed that the figure is misleading in this respect.



12b

12 b

η

y, y

z

z

h

x

Fig. Ex. 11-1.1: Rectangular cross-section.

In the same fashion we may find the quantities referred to the z-axis

Area of rectangleA = bh

A =

∫

A

dA = bh

S = Sz =

∫

A

ydA = b

∫ h−η

−η

ydy = 12b(h2 − 2hη)

I = Izz =

∫

A

y2dA = 13b(h3 − 3h2η + 3hη2 − 2η3)

(Ex. 11-1.2)

If we demand that S = 0, then

S = 0 ⇒ η = 12h (Ex. 11-1.3)

which results in Moment if inertiaof rectangleI = 1

12bh3

I = 112bh3 (Ex. 11-1.4)

It is probably not surprising that in the present case the x-axis lies atthe center of the cross-section, simply because the center of gravity ofa rectangle is its mid-point.11.6

Our next example is almost as simple as Ex 11-1 and also very fundamentalin that beams with circular cross-sections are used in many structures.

Ex 11-2 Circular Cross-SectionCircular cross-sections, see Fig. Ex. 11-2.1, are often used in high- Circular

cross-sectionvoltage masts. It ought to be self-evident that the center of gravitycoincides with the center of the circle. Consequently, we know imme-diately where we should place the x, y, z-coordinate system. We needthe following simple geometric relations

y = R sin θ , dy = R cos θdθ

dA = (2R cos θ)dy = 2R2 cos2 θdθ(Ex. 11-2.1)

11.6 By the way, the result I = 112

bh3 for the rectangle is assumed known to studentswho have passed an introductory course on statics and strength of materials.



y

zθ

dθdy

y

z xxR

Fig. Ex. 11-2.1: Circular cross-section.

Once more we use (11.24) and find

Area of circleA = πR2

Moment of inertiaof circle I = π

4R4

A =

∫

A

dA =

∫ +π/2

−π/2

2R2 cos2 θdθ = πR2

S =

∫

A

ydA =

∫ +π/2

−π/2

2R3 cos2 θ sin θdθ = 0

I =

∫

A

y2dA =

∫ +π/2

−π/2

2R4 cos2 θ sin2 θdθ =π

4R4

(Ex. 11-2.2)

where the result (Ex. 11-2.2a) most likely is known in advance, and(Ex. 11-2.2b) is the expected.

Sometimes it is convenient to express the above results in terms of thediameter d the area A of the circle

Area of circleA = π

4d2

Moment of inertiaof circle

I = 116Ad2

A =π

4d2

I =π

64d4 = 1

4AR2 = 1

16Ad2

(Ex. 11-2.3)

Our two first examples of cross-sections were characterized by theirsimplicity. During the analysis of the next two cross-sections we shallexploit the result from example Ex 11-1 in connection with (11.32c) tohandle more complicated cases.

Ex 11-3 T-Shaped Cross-SectionMany structures are designed with T-shaped beam cross-sections, seeT-shaped

cross-section Fig. Ex. 11-3.1. The reason for its popularity is that it is inexpensiveand easy to inspect for corrosion, while a circular tube, for instance,may corrode from the inside without any indication of deteriorationon the outside. As mentioned above, in the derivations below we shallutilize some of the result from Example Ex 11-1 and (11.32).

Obviously, there is no intuitively “correct” choice of the (x, y, z)-co-ordinate system, except that the y-axis must respect the symmetry.However, three possibilities seem rather obvious, namely choosing the



z

z

y, y

12bf

η

tf

hw

e

12 bf

2× 12 tw

x

Fig. Ex. 11-3.1: T-shaped cross-section.

x-axis at the top of the cross-section, or, as shown in Fig. Ex. 11-3.1,at the bottom, or at the junction between web and flange.

Referring to the figure we may easily find

Area of T-shapedcross-sectionA = twhw + tf bf

Aw = twhw

Af = tfbf

A = Aw + Af = twhw + tf bf

Sw = Aw

(12hw − η

)= −hwtw

(η − 1

2hw

)

Sf = Af

(hw + 1

2tf − η

)= bf tf

(hw + 1

2tf − η

)

S = Sw + Sf = Af

(hw + 1

2hk − η

)+ Aw

(12hw − η

)

(Ex. 11-3.1)

where index w indicates the web and index f the flange. Demand thatthe static moment S about the z-axis vanishes and get

η =hw(2Af + Aw) + tfAf

2A

=twh

2w + 2bfhwtf + bf t

2f

2(bf tf + hwtw)

(Ex. 11-3.2)

The distance e from the junction between flange and web to the cen-troid is then

e = hw − η =h2wtw − bf t

2f

2A(Ex. 11-3.3)

and the moment of inertia I becomes

I =Aw(η − 12hw)

2 + 112twh

3w

+Af (e+12tf )

2 + 112t3f bf

(Ex. 11-3.4)



or

I = Aw(e− 12hw)

2 + 112twh

3w

+Af (e+12tf )

2 + 112t3f bf

(Ex. 11-3.5)

or

Moment of inertiaof T-shaped

cross-section I

I = Aw(e− 12hw)

2 + 112Awh

2w

+Af (e+12tf )

2 + 112Af t

2f

(Ex. 11-3.6)

Regarding the T-shaped cross-section in Example Ex 11-3 we did not assumeanything about the thicknesses of the web and flange in relation to the widthand depth of the cross-section. It is not difficult to derive formulas for thin-walled T-shaped cross-sections from the above formulas by realizing that inthat case tf is small compared to hw and bf , but I shall not do it here. It isnot necessary to assume tw small, but chances are that it is, otherwise theflanges would not contribute sufficiently.

In the next example, namely Example Ex 11-4, we shall, however, assumethat the thickness of the flanges is much smaller than the depth and widthof the I-shaped profile we investigate.

Ex 11-4 Thin-Walled I-Shaped Cross-SectionNumerous structures are designed with beams with doubly symmetric,Thin-walled

I-shapedcross-section

thin-walled I-shaped cross-section because it has proved to be a ratherinexpensive way of obtaining good stiffness and strength associatedwith bending about the z-axis. Such a cross-section has much smallerstiffness and strength associated with the y-axis, but often this is ofless concern. Its torsional stiffness is also very low. Once more we uti-

z

y

tf

tf

12hw

12hw

h

2× 12 tw

12b

12b

x

Fig. Ex. 11-4.1: I-shaped cross-section.

lize some of the results we obtained in Example Ex 11-1 and (11.32).



Without any computations we may see that the centroid must coincidewith the x-axis shown in the figure. This makes the following deriva-tions much easier. By “thin-walled” we understand that the thicknesstf of the flanges is very small in comparison with the depth h andwidth b of the cross-section.11.7 In mathematical terms we require

Definition of“thin-walled”

tf ≪ h ⇒ hw ≈ h ; tf ≪ b

tw ≪ b ; tw ≪ h(Ex. 11-4.1)

Referring to Fig. Ex. 11-4.1 we may find

Area A andmoment of inertiaI of thin-walledI-shapedcross-section

Aw = twhw ≈ twh

Af = tfb

A = Aw + Af = twhw + tfb ≈ twh+ tf b

I = 112twh

3w + 2Af

(12(hw + tf )

)2+ 2 1

12bt3f

(Ex. 11-4.2)

providing

Moment of inertiaI of thin-walled I

I ≈ 112h3wtw + 2h2

wbtf = 112Awh

2w + 2Afh

2w (Ex. 11-4.3)

or


I ≈(1 + 6

Af

Aw

)112Awh

2w =

(1 + 6

Af

Aw

)Iw (Ex. 11-4.4)

where Iw denotes the moment of inertia of the web. This result mayalso be given in the form


I ≈(1 +

Aw

24Af

)IF (Ex. 11-4.5)

where IF is the moment of inertia of both flanges. Both (Ex. 11-4.4)and (Ex. 11-4.5) show that it pays better to add material to the flangesthan to the web—provided that only the bending stiffness about the z-axis is of concern. If the web is too thin, then it may buckle leavingthe cross-section worthless, or it may not be able to carry the shearstress necessary to “transport” axial stresses between the upper andthe lower flange which may be just as fatal.

Our last example deals with a circular tube. In various truss structures thisprofile is used very often due to the fact that it offers a feasible alternativein terms of usage of material. It is particularly advantageous with respectto torsional stiffness, while its bending stiffness is less than that of the I-profile of the same amount of material. The latter cross-section has a verylow torsional stiffness seen in this light. It must be acknowledged that itmakes for a somewhat uneasy feeling that circular tubes are difficult toinspect for internal corrosion.

11.7 In order to compute the moment of inertia about the z-axis it is not necessary toput any restrictions on the thickness tw of the web. On the other hand, it is not likelythat one should choose a value of tw that is of the same order of magnitude as h or b.



Ex 11-5 Circular Tube—Ring-Shaped Cross-SectionIn this example we intend to determine the area and moment of inertiaCircular tube.

Ring-shapedcross-section

of a circular ring as shown in Fig. Ex. 11-5.1. Here we may exploit the

y

z

tr

xR

Fig. Ex. 11-5.1: Ring-shaped, circular cross-section.

result (Ex. 11-2.3b), in that we may subtract the results for a circlewith radius r from results for the circle with radius R.11.8

Area A andmoment of inertia

I for a circulartube

AR = πR2 , Ar = πr2 , A = π(R2 − r2)

IR =π

4R4 , Ir =

π

4r4 , I =

π

4(R4 − r4)

(Ex. 11-5.1)

When we introduce

r = R− t (Ex. 11-5.2)

where t is the wall thickness we may find

A = π(R2 − (R − t)2) and I =π

4

(R4 − (R − t)4

)(Ex. 11-5.3)

If we are dealing with thin cross-sections, i.e. when t ≪ R, we mayfind the very simple result

Area A andmoment of inertiaI for a thin-walled

circular tube

A ≈ 2πrt

I ≈ πR3t or I ≈ 12AR2

(Ex. 11-5.4)

The above examples do not cover all the cross-sections that are used instructural engineering, but if you come across one which is not covered hereyou ought to be able to “do the sums yourself,” as the British sometimessay.

11.8 We may do this because the circles have coinciding centroids.


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