[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_7,Ó Springer Science+Business Media Dordrecht 2013

Chapter 7

Plane, Straight BeamsIn this chapter we derive the kinematic and static equations for plane,straight beams. The reason why we do not treat beams in three-dimen-sional space is that this is not a book on beam theory—but rather a bookon continuum mechanics which also shows how to derive theories for spe-cialized continua such as beams and plates. Basically, the principles for thederivations remain the same independent of the dimension of the space.7.1

We shall first consider straight beams. There are at least two reasonsfor this, namely that straight beams are simpler than curved ones and thatmost beams which are used as structural members are not curved.

Fig. 7.1: A curved three-dimensional beam.

As mentioned in Chapter 6, there are (at least) three different ways todefine the term beam. We shall only consider one-dimensional beams, butbegin by mentioning that we could take the three-dimensional continuumas our point of departure and define a beam as a three-dimensional body Beams as

three-dimensional

bodies

characterized by the fact that one of its dimensions, the length, is muchlarger than the other two, the cross-sectional dimensions, see Fig. 7.1. This,of course, is a realistic way of defining the term beam.

However, it is the third, more abstract definition, which we shall employhere. According to this definition, a beam is a one-dimensional“body”whichwe supply with certain properties. This, in itself, is not very helpful. But, it Here: Beams as

one-dimensional

bodies

proves to be a much easier way to arrive at our destination because the mostfundamental decisions we must make are only concerned with the degree

7.1 In all fairness it must be mentioned that in deriving beam theories in three-dimensional space there are many more issues to deal with, making the process moredifficult.

91

Plane, Straight Beams

Fig. 7.2: A curved one-dimensional beam.

of nonlinearity and the kind of generalized strain measures we considernecessary to include. By application of the principle of virtual displacementswe then get the equilibrium equations which define the generalized stresses.

The last choice, which often proves to be a difficult one, is concerned withthe constitutive relation connecting the generalized stresses and generalizedstrains, and here we may have to apply a (simplified) three-dimensionalanalysis of a short section of the beam, see Part III. This is usually mucheasier than to perform a three-dimensional analysis of the entire beam.Although I strongly recommend the above procedure I must admit thatthere is a very important drawback to this method in that there is no surefireway to choose the strain measures correctly—rather in the most convenientway. But, if we make our decision based on experiences from the three-dimensional, real world we may get something useful. I postpone furtherdiscussion of this subject to the particular cases below.

7.1 Beam Deformation ModesIn order to establish a sound foundation for picking good strain measures fora one-dimensional beam first we discuss the most fundamental deformationmodes of a straight beam, namely bar and beam deformation, respectively.

7.1.1 Axial Deformation

The first type of deformation we consider is bar deformation, i.e. extensionBar deformation:

Change of length or shortening of the beam, see Fig. 7.3.

Undeformed beam

Axially deformed beam

Fig. 7.3: The one-dimensional body, the ruler, in undeformedand stretched state.

To be quite honest, it takes more than usual human strength to extend

Esben Byskov Continuum Mechanics for Everyone August 14, 2012

92


even a plastic ruler as much as shown in the figure.

In this case the only strain in the bar is an axial one, e.g. γ11, see (2.11)and (2.20)–(2.25) for kinematic nonlinearity, and for kinematic linearity, theaxial strain may be ε11, see (4.5).

7.1.2 Shear Deformation

The second basic kind of beam deformation, namely shear deformation, Shear deformation:

Not easy to realizesee Fig. 7.4, is not as easy to realize as the first ones—in a beam it isactually impossible to impose this deformation mode without introducingsome bending.

Undeformed beam

Shear deformation of beam

Fig. 7.4: The one-dimensional body, the ruler, in undeformedand after shear deformations.

In Fig. 7.4 there must be an ingredient of bending at the hands. Overthe rest of the beam between the hands shearing is the only deformationmode. The shear strain, which we shall call ϕ, is closely connected to theone found in three-dimensional theory, e.g. γ12 for kinematic nonlinearityor ε12 for kinematic linearity, see (2.34) and (4.30), respectively.

7.1.3 Bending Deformation

The third, very basic type of deformation bodies is beam bending, which Bending

deformation:

Change of curvature

we intend to demonstrate by the plastic ruler mentioned above. Imaginethat you stretch out your arms in front of you and try to bend the ruler bytwisting your arms, see Fig. 7.5. In this case it does not take superhumanstrength to achieve a deformation like the one shown in the figure.

Unlike in the previous cases, here we cannot appeal directly to the three-dimensional theory and find a relevant kinematic measure of bending,7.2

7.2 The first idea, which comes to mind, but proves to be wrong for curved beams, isthe change in geometric curvature. The correct one is the change in angle per length ofthe undeformed beam, see (8.25) and (8.27), but in the present case of a straight beamchange in geometric curvature is a valid measure of bending.

August 14, 2012 Continuum Mechanics for Everyone Esben Byskov

93


Bent beam

Undeformed beam

Fig. 7.5: The one-dimensional body, the ruler, in undeformedand bent state.

which we denote κ and call curvature strain or bending strain.Curvature strain κ

7.1.4 The Three Fundamental Beam Strains

To sum up, in Fig. 7.6 the three most fundamental beam strains are visual-ized for the case of an initially straight beam.

ε ϕ κ

Fig. 7.6: The three fundamental beam strains.

7.1.5 Choice of Deformation Modes

Before we may set up a theory for the plane beam shown in Fig. 7.2 we mustdecide which strain measures we consider the most imperative to include inour theory. In order to describe the elongation of the beam we need a mea-sure of the axial strain ε, which, as mentioned above, is the counterpartAxial strain ε

of e.g. γ11 of the three-dimensional continuum theory. The other impor-tant strain measure is the curvature strain κ, which, as mentioned above,Curvature strain κ

represents the change in curvature of the beam, i.e. it is an expression ofhow much the beam has been bent. The curvature strain does not enterthe three-dimensional continuum theories presented in this book and musttherefore be considered separately. These are considered the two most fun-damental strain measures of a beam because they appear in deformation ofbeams, whether they are long or short. Beams that are assumed to entailBernoulli-Euler

Beams only axial and bending deformation are called Bernoulli-Euler Beams.


94

Full Kinematic Nonlinearity

Experience shows that the third (above, the second) strain, namely theshear strain ϕ, which is the equivalent of e.g. γ12, see (2.19), of the three-dimensional theory is only important for short beams. Shear strain ϕ

Only important for

short beamsAnalyses of three-dimensional bodies and experiments show that when

the length of a homogeneous beam made of an isotropic, linear elastic ma-terial is less than about four to five times the depth of the beam, this strainbecomes significant and must be included in the analysis. Beams that areanalyzed according to a theory which involves all of the above three strainsare called Timoshenko Beams. Timoshenko Beams

Some other types of beam theories take account of more complicatedstrain patterns, but I do not intend to cover such theories here.

Kinematically, there are three cases of interest, namely the fully non-linear, the kinematically moderately nonlinear, and the linear case, respec-tively.

7.2 Fully Nonlinear Beam TheoryExcept for the famous Elastica, see Example Ex 8-2, I do not intend to derivea fully nonlinear beam theory, but limit myself to considering kinematic re-lations of various degrees of kinematic nonlinearity, see below. On the otherhand, here, we derive some kinematically full nonlinear relations becausewe exploit the results both in the section on the kinematically moderatelynonlinear straight beams, Section 7.3, and in the section on kinematicallylinear straight beams, Section 7.5.

7.2.1 Kinematics

The only kinematic relations we wish to study here are associated with the Kinematics:

Axial and bending

deformation

axial and bending deformation of the beam.

w

x, u

w + dw

(u+ du)ds0

uds

w

Fig. 7.7: Straight beam element.

Figure 7.7 shows a beam element whose length in the undeformed con-figuration is ds0 and in the deformed configuration ds. The axial componentof the displacement vector u is u and the transverse component is w. Then,


95


by application of the Theorem of Pythagoras we getTheorem of

Pythagoras(ds)2 =

(ds0 + (u+ du)− u

)2+ ((w + dw) − w)

2(7.1)

When we utilize the undeformed abscissa x as the axial coordinate andintroduce the notation

( )′ ≡ d( )

dx(7.2)

we may get

(ds)2 =((1 + u′)2 + (w′)2

) (ds0)2

(7.3)

By analogy with the expression (2.12) relating the Lagrange strain tothe change in the length of the line element we define the Lagrange axialstrain γ by

(ds)2 −(ds0)2

= 2γ(ds0)2

(7.4)

and thus

γ = u′ + 12 (u

′)2 + 12 (w

′)2 (7.5)

As a measure of the bending deformations of the beam it is tempting totake the geometric curvature k which from differential geometry is knownto be

Geometriccurvature k:

Not a good

curvature strain

measure

k ≡ dω

ds=

du

dt

d2w

dt2− dw

dt

d2u

dt2((

du

dt

)2+

(dw

dt

)2)32

(7.6)

where t is a parameter that increases monotonically with s. If we make theparticular choice t = x, thenGeometric

curvature k:

Not a good

curvature strain

measure

k =u′w′′ − w′u′′

((u′)2 + (w′)2

) 32

(7.7)

For straight beams the angle before deformation is ω0 = 0 and afterdeformation it is ω, where

ω = sin−1

(dw

ds

)= sin−1

(w′

√1 + 2u′ + (u′)2 + (w′)2

)(7.8)

experience, however,7.3 shows that a better curvature strain is defined as the


96

Moderately Kinematically Nonlinear Bernoulli-Euler Beams 97

derivative of the change in angle of the beam axis.7.4 Then, the curvaturestrain κ is defined as

Correct curvature

strain κκ ≡ dω

dx(7.9)

At this point we leave the fully nonlinear straight beams in favor of thefully nonlinear curved beams in Section 8.1. We shall, however, utilize theabove formulas as a basis for the next sections which cover kinematicallymoderately nonlinear and linear straight beams.

7.3 Kinematically Moderately Nonlinear

Straight Bernoulli-Euler BeamsBy the term kinematically moderately nonlinear we imply the same restric- Kinematically

moderately

nonlinear theory

tion on the displacement field as in Chapter 3, namely that the strains—herethe axial and curvature strains—are small, but that the rotations are mod-erately small, see below and (2.42).

Here, we intend to establish a theory which excludes the shear strain, Bernoulli-Euler

beam: Shear strain

ϕ assumed ≡ 0

i.e. we are dealing with a Bernoulli-Euler beam theory.

7.3.1 Kinematics

Here, we assume that the beam is slender and omit the shear strain, asmentioned above. Thus, we only take the axial strain ε and the curvature Only strains: axial ε

curvature κstrain κ into account. Provided that the beam axis does not experiencesignificant stretching we may assume that

Little stretching|u′| ≪ 1 (7.10)

Further, since we assume that the rotations are moderate, we take

Moderate

rotations(w′)2 ≪ 1 (7.11)

Therefore, the expression (7.5) for the axial strain, which we denote themoderately nonlinear axial strain ε, becomes

Axial strain:

ε = u′ + 12(w′)2

ε = u′ + 12 (w

′)2 (7.12)

This is our axial strain measure in the kinematically moderately nonlin- Consistency in

strain measures is

important

ear theory. It is important that we make the same assumptions regardingthe order of |u′| and (w′)2 in the expressions for both the axial and thecurvature strain. Otherwise, we derive a theory that is internally inconsis-

7.3 The experience comes from observations of curved beams with finite displacementsand finite strains, see the discussion Section 8.1.3.2.

7.4 Actually, it is the derivative of the change in angle of the beam cross-section, butfor beams, which do not experience shear strains, there is no distinction.



tent with respect to order of terms, which means that basically it is useless.Thus, in view of (7.10) and (7.11) the expression for the rotation ω (7.8)simplifies to

Rotation of beam

axis: ω = w′ω = w′ (7.13)

Then, the curvature strain κ is given by

Curvature strain:

κ = w′′κ = w′′ (7.14)

In this theory, (7.6) yields the same result as (7.14) because of (7.11),but it is for a wrong reason because (7.6) is not correct. Note that oneof the strain measures, namely the curvature strain κ, is given in termsof a second derivative of the displacements, not just the gradients, i.e. thefirst derivatives. This is a common feature of many theories of specializedcontinua, such as beams, plates, and shells. Further, note that the onlynonlinear term in the strains is 1

2 (w′)2 in (7.12).

For the sake of completeness we give an interpretation of the generalizeddisplacements u and the operators l1, l2, and l11, which enter the expressionfor the generalized strains ε, see Chapter 33

u ∼[u

w

]and ε ∼

[ε

κ

]= l1(u) +

12 l2(u) (7.15)

with

l1(u) ∼[u′

w′′

], l2(u) ∼

[(w′)2

0

]and l11(u

a,ub) ∼[w′

aw′b

0

](7.16)

where (subscripts) a and b denote two different displacement fields.

Ex 7-1 Rigid Rotation of a BeamLet us consider a very simple example of a beam “deformation,” seeFig. Ex. 7-1.1. The beam is subjected to a pure rotation θ about theorigin, as shown in the figure, and should therefore not experience anystrains. The displacements are exactly given by

[u

w

]=

[x cos(θ)− x

x sin(θ)

]

⇒[u′

w′

]=

[cos(θ)− 1

sin(θ)

] (Ex. 7-1.1)

Thus, the fully nonlinear axial strain measure given by (7.5) provides

γ = (cos(θ)− 1) + 12(cos(θ)− 1)2 + 1

2sin2(θ) = 0 (Ex. 7-1.2)


98


x(1 − cos(θ))x

x, u

x sin(θ) θ

w

Fig. Ex. 7-1.1: A rigid rotated beam.

while the moderately nonlinear axial strain measure, see (7.12), pre-dicts

ε= (cos(θ)− 1) + 12sin2(θ)

=(1− 1

2θ2 − 1 +O(θ4)

)+(12θ2 +O(θ4)

)

= O(θ4) 6≡ 0

(Ex. 7-1.3)

The question is, of course, does this mean that the strain measure ε Is ε by (7.12) valid?given by (7.12) is invalid? The answer to this is that it is indeed agood strain measure provided that you only employ it for cases whichcomply with the requirement that θ is “small enough.” This exampleindicates that if θ4 ≪ 1 then the strain measure is sufficiently accurate.We must, however, not draw too wide conclusions from just one verysimple example.

Note that here the curvature strain vanishes because w′′ = 0 along thelength of the beam.

The two assumptions (7.10) and (7.11) imply that, for the theory to be Consistency is very

importantconsistent with respect to order of terms, the rotations ω must be small inthe sense that

ω2 ≪ 1 (7.17)

Unfortunately, we may not conclude that we have employed a valid the-ory7.5 if a computation based on the kinematically moderately nonlineartheory gives results that obey (7.17). The reason is that for the particularpurpose omission of terms of order ω2 erroneously might cause the predic-tions to comply with (7.17). Basically, if you linearize a problem you cannotcatch nonlinear effects, i.e. if you truncate a series expansion after degree n,

7.5 In this case: a theory which is consistent with respect to order of terms.


100 Plane, Straight Beams

then you may not expect to describe effects of order n+1 and higher. Or, inKilling all nonlineareffects

⇒ no nonlinear

effect are retained

plain English, if you kill all nonlinear effects you will not be able to see anynonlinearity. Therefore, as is the case in many applications of structuralmechanics, there is no substitute for engineering experience.

7.3.2 Equilibrium Equations

Now that we have chosen the strain measures, the generalized strains, weGeneralized stresses

given by Principle

of Virtual Work

have lost our freedom to select the stress measures, the generalized stresses,but must observe the principle of virtual displacements.

LetN andM denote the generalized stresses that are the work conjugateof ε and κ, respectively. Further, let the beam end points be located at x = aand x = b. Then, the internal virtual work σ · δε is

Internal virtual

workσ · δε =

∫ b

a

(Nδε+Mδκ)dx (7.18)

In order to proceed we need an expressions for δε and δκ. We may eitherget them through use of (7.16) or directly from (7.12) and (7.14)

Variation of

strainsδε = δu′ + w′δw′ and δκ = δw′′ (7.19)

and thus (7.18) becomes

Internal virtual

work

σ · δε=

∫ b

a

(N(δu′ + w′δw′) +Mδw′′

)dx

= [Nδu]ba + [(−M ′ +Nw′)δw]ba + [Mδw′]ba

−∫ b

a

N ′δudx+

∫ b

a

(M ′′ − (Nw′)′

)δwdx

(7.20)

Let the beam be subjected to the distributed loads pu(x) and pw(x)x ∈ ] a, b [ and at the ends to the loads Pu(a), Pw(a), C(a), Pu(b), Pw(b),C(b), see Fig. 7.2. The notation should be obvious, while the convention and

pu

Pu(a) Pu(b)

Pw(a)

C(b)

Pw(b)

C(a)

pw

Fig. 7.2: Undeformed beam with loads and end forces.

consistency for the direction of the loads at x = a may seem awkward, butis chosen for convenience. The loads at the ends may either be prescribed



or they may be reactions. The external virtual work T · δu is thus

External virtual

work

T · δu=

∫ b

a

(puδu+ pwδw)dx

− Pu(a)δu(a)− Pw(a)δw(a) − C(a)δw′(a)

+ Pu(b)δu(b) + Pw(b)δw(b) + C(b)δw′(b)

(7.21)

where we have exploited (7.13).

We proceed much in the same way as in Section 4.3 where we derivedthe equilibrium equations for the case of a kinematically linear theory forthree-dimensional bodies, the important difference being the kinematicallynonlinear terms.

By equating the internal with the external virtual work we arrive atthe following variational equation (7.22) which at the first glance may lookintimidating. However, interpreted in the right way and inspected closely,it is just a collection of equations which are rather simple by themselves, aswe shall see below

0 =

∫ b

a

(N ′ + pu)δudx−∫ b

a

(M ′′ − (Nw′)′ − pw)δwdx

−(Pu(a)−N(a))δu(a)

−(Pw(a) + (M ′(a)−N(a)w′(a))δw(a)

−(C(a)−M(a))δw′(a)

+(Pu(b)−N(b))δu(b)

+(Pw(b) + (M ′(b)−N(b)w′(b))δw(b)

+(C(b)−M(b))δw′(b)

∀(δu, δw, δu(a), δw(a), δw′(a), δu(b), δw(b), δw′(b))

(7.22)

where “∀ . . .”must be interpreted as “∀ kinematically admissible . . .”.

Note that δw′ must be derived from δw in the field, but that at theends it is permissible to vary e.g. δw′(a) independently of δw(a).7.6 Thevariations vanish at the supports, but at all other points they are arbitrary.Therefore, at the latter points, i.e. unsupported ends and the entire field, thecoefficients to the variations must equal zero in order for the the principle ofvirtual displacements to be fulfilled. Since the variations are arbitrary andnon-vanishing except at the supports all their coefficients must vanish in

7.6 Both δw(a) and δw′(a) are associated with a point. We may therefore choose avariation for which δw(a) = 0 and δw′(a) 6= 0 with δw = 0 except for an infinitesimallysmall neighborhood at the left-hand end. Then, we may conclude that we may takeδw′(a) to be independent of δw(a).



order for (7.22) to hold. Therefore, (7.22) provides the static field equations

Static field

equations

N ′ + pu = 0

M ′′ − (Nw′)′ − pw = 0

}x ∈]a, b[ (7.23)

and the possible static boundary conditions

Possible static

boundary

conditions

N(a) = Pu(a) , N(b) = Pu(b)

−M ′(a) +N(a)w′(a) = Pw(a) , −M ′(b) +N(b)w′(b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.24)

where the overbar indicates that the static boundary conditions only applywhere the end loads are prescribed, i.e. at the static boundary—which is nobig surprise.

Many, probably most, authors find it convenient to introduce a staticquantity V ,7.7 which is not a generalized stress because it has no kinematiccounterpart to do internal work with

The static

quantity V is not

generalized

V ≡ −(M ′ −Nw′) (7.25)

The primary reason for the introduction of V is that it may be inter-preted as the shear force in the beam in the same spirit as N is interpretedas the axial force, and M is the bending moment. After introduction of(7.25) the static equations (7.23)–(7.24) become

Static fieldequations

including V

N ′ + pu = 0

V ′ + pw = 0

M ′ −Nw′ + V = 0

x ∈]a, b[ (7.26)


Possible staticboundaryconditions

including V

N(a) = Pu(a) , N(b) = Pu(b)

V (a) = Pw(a) , V (b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.27)

Since we are dealing with a kinematically nonlinear theory it is not sur-prising that the displacements enter the equilibrium equations, see (7.23)–(7.27). Notice, however, that no displacement components enter the equilib-rium equations (7.23a), (7.24a–b), (7.26a) and (7.27a–b) for N which meansthat sometimes N can be determined separately. Also, note that none of thestatic equilibrium equations contain u and thus w is the only displacementcomponent appearing in the equilibrium equations.

7.7 In literature in the German tradition this quantity is usually called Q for Querkraft.



7.3.3 Interpretation of the Static Quantities

We define the static quantities N , V and M as shown in Fig. 7.3, i.e. such Interpretation of

static quantities

puds0

pwds0 (V + dV )

M

(M + dM)

V

(N + dN)

N

ds0

ds

Fig. 7.3: Beam element with loads and end forces.

that N and V are parallel with and perpendicular to the undeformed beamaxis, respectively. Then, we may immediately write two of the equilibriumequations

N ′ + pu = 0

V ′ + pw = 0

}x ∈]a, b[ (7.28)

while the third requires a little care. Moment equilibrium about the left-hand end of the beam element furnishes

0 = (M + dM)− M − (N + dN)ds sin(ω) + (V + dV )ds cos(ω) (7.29)

where ω, as before, is the rotation of the beam axis.

Without approximations this may be rewritten

0 = M ′dx−(N + N ′dx)√

1 + 2γ dx sin(ω)

+(V + V ′dx)√

1 + 2γ dx cos(ω)(7.30)

where the relation (7.4) has been utilized.

Because of the restriction on the magnitude of ω we may exploit (7.17)and introduce ε, which is small compared to 1, instead of γ to get

0 = M ′ − N√1 + 2εw′ + V

√1 + 2ε (1 +O((w′)2)) (7.31)

where we have realized that both N ′ and V ′ are finite while dx is infinites-imal. The assumptions (7.10) and (7.11) result in

0 = M ′ − Nw′ + V (7.32)

Thus, the equilibrium equations (7.26) are completely equivalent to (7.28) in



conjunction with (7.29) and we may therefore interpret N as theaxial forceConclusion:

N = N , V = V

and M =MN , V as the shear force V , which are directed along the undeformed beamaxis and perpendicular to it, respectively, and M as bending moment M .

7.4 Kinematically Moderately NonlinearStraight Timoshenko Beams

The so-called Timoshenko beam theory developed in the following is applica-Kinematically

moderately

nonlinear

Timoshenko beams

ble to relatively short beams where the shear strains may not be neglected.As a rule of thumb, for an elastic beam of an isotropic material with a rect-angular cross-section the effect of shear strains becomes important when thelength of the beam is less than about four to five times its

depth. The most important feature of Timoshenko beam theory com-pared with Bernoulli-Euler beam theory is the inclusion of the shear strain.Timoshenko beam

theory includes

shear strain

For a visualization of the shear strain, see Section 7.6.

Regarding the degree of nonlinearity we shall here make assumptionssimilar to the ones in Section 7.3, i.e. that the strains—here the axial, shear,Rotation ω of

cross-sectionuncoupled fromrotation w′ of

beam axis

and curvature strains—are small, while that the rotations are moderatelysmall. But here, the rotation ω of the beam axis is uncoupled from the firstderivative w′ of the transverse displacement component, and we assume thatw′ is of the same order as ω.

7.4.1 Kinematics

We shall employ the same generalized strains later in the kinematicallylinear case, see Section 7.6, namely the axial strain ε, the shear strain ϕand the curvature strain κ. The only difference is that the axial strain isnonlinear in the same way as in the case of the moderately kinematicallynonlinear Bernoulli-Euler beams, see (7.12).

|u′| ≪ 1 , (w′)2 ≪ 1 and ω2 ≪ 1 (7.33)

7.4.1.1 Axial Strain

Axial strain ε ε = u′ + 12 (w

′)2 (7.34)

7.4.1.2 Shear StrainShear strain

ϕ = w′ − ωϕ = w′ − ω (7.35)

7.4.1.3 Curvature Strain

Curvature strain κ κ ≡ dω

dx= ω′ (7.36)

Here, the interpretation of the operators l1, l2, and l11, which enter the


Moderately Kinematically Nonlinear Timoshenko Beams 105

expression for the generalized strains ε, see Chapter 33, are

u ∼

u

w

ω

and ε ∼

ε

ϕ

κ

= l1(u) +

1

2l2(u) (7.37)

with

l1(u) ∼

u′

(w′ − ω)

ω′

, l2(u) ∼

(w′)2

0

0

, l11(u

a,ub) ∼

w′

aw′b

0

0

(7.38)

where (subscripts) a and b denote two different displacement fields.


As usual, our choice of strain measures, the generalized strains, dictatesour stress measures, the generalized stresses, because the two sets of must Generalized strains

ε, ϕ and κ

Generalized stresses

N , V and M

be each other’s work conjugate in the principle of virtual displacements.Since we work with three generalized strains we must have three generalizedstresses which we shall denote N , V and M . They are the work conjugateof ε, ϕ and κ, respectively.

When the beam end points are located at x = a and x = b the internalvirtual work σ · δε is

Internal virtual

workσ · δε =

∫ b

a

(Nδε+ V δϕ+ Mδκ)dx (7.39)

In order to proceed we need an expressions for δε, δϕ and δκ. The straindefinitions (7.34), (7.35) and (7.36) provide

δε = δu′ + w′δw′ , δϕ = (δw′ − δω) and δκ = δω′ (7.40)


Internal virtual

work

σ · δε=

∫ b

a

(N(δu′ + w′δw′) + V (δw′ − δω) + Mδω′

)dx

= [Nδu]ba + [(V + Nw′)δw]ba + [Mδω]ba

−∫ b

a

N ′δudx−∫ b

a

(V ′ + (Nw′)′

)δwdx

−∫ b

a

(M ′ + V )δωdx

(7.41)

Provided that there are no distributed moment loads, i.e. that the loads



pu

Pu(a) Pu(b)

Pw(a)

C(b)

Pw(b)

C(a)

pw


are the ones shown in Fig. 7.2, which for convenience is repeated here asFig. 7.4, the external virtual work T · δu is

External virtual

work

T · δu=

∫ b

a

(puδu+ pwδw)dx



(7.42)

which obviously is the same as (7.21).

By equating the internal and external virtual work, given by (7.41) and(7.42), respectively, we can immediately conclude that

M ′ + V = 0 , x ∈]a, b[ (7.43)

which indicates that V is not the same as the static quantity V definedby (7.25), and thus we do not recover the equilibrium equations (7.26) and(7.27) or (7.23) and (7.24). Again, by equating the internal and externalvirtual work we may get the other two static field equations

V ′ + (Nw′)′ + pw = 0 , x ∈]a, b[ (7.44)

which also does not agree with (7.26b), while

N ′ + pu = 0 , x ∈]a, b[ (7.45)

is the same as (7.26a).

Combining (7.43) with (7.44) we may get

M ′′ − (Nw′)′ − pw = 0 , x ∈]a, b[ (7.46)

and we have recovered (7.23). In order to compare with the static fieldequations (7.62) for the kinematically moderately nonlinear Bernoulli-Eulercase, we recapitulate the three static field equations (7.26) but rewrite thethird in order to compare with (7.46)

Static field

equations for

moderately

nonlinear

Bernoulli-Euler

beam

N ′ + pu = 0

V ′ + pw = 0

M ′′ − (Nw′)′ − pw = 0

x ∈]a, b[ (Bernoulli-Euler) (7.47)


Moderately Kinematically Nonlinear Timoshenko Beams 107

and collect the equilibrium equations for the present case, i.e. kinematicallymoderately nonlinear Timoshenko beam in the same way

Static field

equations for

moderately

nonlinear

Timoshenko beam

N ′ + pu = 0

V ′ + (Nw′)′ + pw = 0

M ′′ − (Nw′)′ − pw = 0

x ∈]a, b[ (Timoshenko) (7.48)

As regards the possible static boundary conditions, here we get

Possible static

boundary

conditions for

Timoshenko

beams

N(a) = Pu(a) , N(b) = Pu(b)

V (a) + N(a)w′(a) = Pw(a) , V (b) + N(b)w′(b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.49)

and also here the discrepancy between these boundary conditions and (7.27)is obvious. On the other hand, the boundary conditions for the Timoshenkobeams appear to be consistent with its differential equations.

The question is then to interpret V along with the other two staticquantities N and M , but first note that, whether we succeed in this respect,the theory developed above is a consistent one in that its stress and strain Consistent theory,

but problems with

interpretations

measures are each other’s work conjugate and therefore it is satisfactoryfrom a theoretical standpoint. Therefore, we may not necessarily need to

w′

V P

ψ

N

P

ψ

w′w′

V ∗ w′

(a) (b)beam beam

N∗

Fig. 7.5: End load and projections.Bending moments M and M∗ are not shown.

interpret the static quantities in the field x ∈]a, b[, except in order to beable to choose meaningful constitutive relations. Without mentioning it, wehave encountered a similar issue as regards the stresses and strains of thetheory for three-dimensional bodies developed in Chapter 2, in particularSection 2.6.1. There, we assumed a linear relationship between the secondPiola-Kirchhoff pseudo stresses and the Lagrange strains and, as is indicatedby “pseudo,” that stress does not measure force per present area. However,Koiter (2008) has proved that the possible error in assuming this linear



relationship is small, namely of the order of the absolute value of the largeststrain compared to 1.7.8

In order to try to interpret N , V and M consider Fig. 7.5. Let a cross-section of the beam be subjected to the load P and resolve it in two differentways, as sketched in Fig. 7.5(a) and (b). Then,

N = P sin(ψ) , V = P cos(ψ)

N∗ = P sin(ψ + w′) , V ∗ = P cos(ψ + w′)(7.50)

where we have utilized the fact that we assume that the rotation of thebeam axis is small, i.e. (w′)2 ≪ 1, in accordance with our basic assumption(7.33b). Then,

N∗ = P(sin(ψ) cos(w′) + cos(ψ) sin(w′)

)≈ P

(sin(ψ) + cos(ψ)w′

)

V ∗ = P(cos(ψ) cos(w′)− sin(ψ) sin(w′)

)≈ P

(cos(ψ)− sin(ψ

)w′)

(7.51)

i.e.

[N∗

V ∗

]≈[

1 w′

−w′ 1

] [N

V

](7.52)

with the inverse relation

[N

V

]≈ 1

1 + (w′)2

[1 −w′

w′ 1

][N∗

V ∗

]≈[1 −w′

w′ 1

] [N∗

V ∗

](7.53)

where we have exploited the condition (7.33b).

The first resolution of the forces is the one that is valid for the moderatelynonlinear Bernoulli-Euler beam, while the second is the more intuitive onewhere the axial force is directed along the deformed beam axis and theshear force is perpendicular to it. The latter is the one which applies to theElastica, see Example Ex 8-2. One might think that a combination of theseresolutions, for example considering N together with V ∗ could furnish theequilibrium equations (7.48). This is, however, not possible7.9 and we must

live with the fact that we cannot interpret N , V and M . Fortunately, aswe shall see later in Example Ex 18-2, which is concerned with determiningthe buckling load of a linearly elastic pinned column using the above theory,the theory can be applied with success.

7.8 Note that this does not preclude finite displacements and rotations.7.9 At least some very qualified people and I have not been able to do so.


Kinematic Linearity 109

7.5 Kinematically Linear Straight Bernoulli-Euler Beams

A kinematically linear theory for straight beams is easily derived from the Kinematically

Linear

Bernoulli-Euler

Beams

theory given in Section 7.3. In addition to the assumption (7.10)

|u′| ≪ 1 (7.54)

we impose the restriction

|w′| ≪ 1 (7.55)

in contrast to (7.11)

(w′)2 ≪ 1 (7.56)

Then, all kinematic nonlinearity vanishes with the result that the gen-eralized strains become

Generalized

strains ε and κε = u′ and κ = w′′ (7.57)

while the kinematically moderately nonlinear theory has the generalizedstrains given as

ε = u′ + 12 (w

′)2

and κ = w′′ (7.58)

By going through the derivations of Section 7.3.2 and omitting all kine-matically nonlinear terms we get the following static field equations

Static field

equations

N ′ + pu = 0

M ′′ − pw = 0

}x ∈ ] a, b [ (7.59)


Possible static

boundary

conditions

N(a) = Pu(a) , N(b) = Pu(b)

−M ′(a) = Pw(a) , −M ′(b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.60)

Just to be sure, we note that, as in the case of moderate kinematicnonlinearity, N and M are the generalized stresses.

Introduce the shear force V by (recall that in these theories V is not ageneralized stress)

V ≡ −M ′ (7.61)

then, the static field equations may be written in the following fashion whichis similar to and may be derived from (7.27).



The result is

Static fieldequations

including V

N ′ + pu = 0

V ′ + pw = 0

M ′ + V = 0

x ∈ ] a, b [ (7.62)

and the possible static boundary conditions become

Possible staticboundaryconditions

including V

N(a) = Pu(a) , N(b) = Pu(b)

V (a) = Pw(a) , V (b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.63)

For the interpretation of the static quantities the reader is referred to Sec-tion 7.3. In the present, kinematically linear case, the generalized stresses,i.e. the axial force and the bending moment, act on the undeformed beam,which is a fact that must be taken into consideration in the interpretationof the generalized stresses, in particular in Fig. 7.3.

7.6 Kinematically Linear Straight TimoshenkoBeams

We shall here make the same assumptions as in Section 7.5 regarding thedegree of nonlinearity, i.e. that the strains—here the axial, shear, and cur-vature strains—as well as the rotations are small.

As mentioned in Section 7.4, Timoshenko beam theories are applicable torelatively short beams where the shear strains may not be neglected. Recallthat as a rule of thumb, for an elastic beam with a rectangular cross-sectionthe effect of shear strains becomes important when the length of the beamKinematically

Linear Timoshenko

Beams

is less than about four to five times its depth. If the beam is made of, say anorthotropic material with the stiff direction along the beam axis, then shearstrains may always play an important role. A similar observation holds forsandwich beams.

The derivations below follow the same pattern as that of Section 7.4 andof Section 7.5. As mentioned in Section 7.4 we did not visualize the shearstrain there, but chose to defer it to the present case. In part the reasonfor this is that the theory developed in Section 7.4 was more complicatedthan I like—and probably not completely satisfactory. As before, the firstconsequence of considering shear strains is that we need three generalizeddisplacements to describe deformations of the beam, namely the axial dis-placement component u, the transverse displacement component w and, inaddition, to these the rotation ω of the beam “cross-section.”7.10

7.10 The reason for the quotation marks is that we are dealing with a one-dimensionalbeam theory, i.e. the beam theory does not, by itself, entail a thickness, but the meaningshould be clear from Fig. 7.6.


Kinematically Linear Timoshenko Beams 111

7.6.1 Kinematics

Here, we must define the three generalized strains, which are strains sketched Timoshenko beam:

Shear strain ϕ 6≡ 0in Fig. 7.6, namely the axial strain ε, the shear strain ϕ, and the curvaturestrain κ, to the same degree of nonlinearity as in Section 7.3.1, namely

|u′| ≪ 1 , |w′| ≪ 1 and |ω| ≪ 1 (7.64)

7.6.1.1 Axial Strain

Once more (7.57) defines the axial strain

Axial strain εε = u′ (7.65)

7.6.1.2 Shear Strain

We take the shear strain ϕ to be the same as for the moderately nonlinearversion of the Timoshenko beam theory, see (7.35).

Shear strain

ϕ = w′ − ωϕ = w′ − ω (7.66)

7.6.1.3 Curvature Strain

Because we have abandoned the assumption of kinematic coupling betweenthe transverse displacement component w and the rotation ω the curvaturestrain κ must not be defined in accordance with (7.57) but as in (7.9)

Curvature strain

κ 6= w′′κ ≡ dω

dx= ω′ (7.67)

which, by the way, holds regardless of the degree of nonlinearity.

7.6.1.4 Interpretation of Shear Strain

While the physical interpretation of the two strains ε and κ probably isclear, both from Fig 7.6 and from our intuition, the shear strain ϕ deservesspecial attention. Consider a two-dimensional element of the beam, seeFig. 7.6,7.11 and subject it to the two simple displacement fields shown.But, first we need to define two directions. One is the direction of the beamaxis, which is indicated by the dash-dot line, and the other is the normal tothe cross-section of the beam and is given by the arrow. Note that, here,the term cross-section should be taken literally because we are dealing witha two-dimensional beam.

7.11 The deformation patterns shown in the figure are not possible for an entire beamcross-section because some ingredient of S-shape is unavoidable. So, the sketches mustbe taken with a grain of salt.

You may, however, ask if such an S-shape does not violate the idea associated withextension and pure bending that the cross-sections remain plane. The answer is thatthese different kinds of displacements do not interact. A similar conclusion holds fortorsion of beams where most cross-sections experience warping.



Case (a)

Recall that all displacements are infinitesimal.

Then, the displacements may be determined as

u(a)1 = 0 and u

(a)2 = w′x1 (7.68)

w′

x2

x1

(a)

ω

x2

x1

(b)

Fig. 7.6: Infinitesimal element of Timoshenko beam. Twofundamental cases of shear strain. In order to avoid clog-ging of the figure, the x1-axis is not coinciding with thebeam axis.

The result is that the strains (4.5) become

ε(a)11 = 0 , ε

(a)12 = ε

(a)21 = 1

2w′ and ε

(a)22 = 0 (7.69)

and thus the beam shear strain ϕ(a) is

ϕ(a) = 2ε(a)12 = w′ (7.70)

Case (b)

In this case the displacements are

u(b)1 = −ωx2 and u

(b)2 = 0 (7.71)

and

ε(b)11 = 0 , ε

(b)12 = ε

(b)21 = − 1

2ω and ε(b)22 = 0 (7.72)

and thus

ϕ(b) = 2ε(b)12 = −ω (7.73)

Combination of Cases (a) and (b)

Any strain situation can be composed of cases (a) and (b) and therefore wemay take the beam shear strain ϕ to be

Shear strain

ϕ = w′ − ωϕ = w′ − ω (7.74)

This is the strain utilized in Timoshenko beam theories.


Kinematically Linear Timoshenko Beams 113

7.6.1.5 Interpretation of Operators

Here, we only need to interpret the linear operator l1 which enters theexpression for the generalized strains ε, see Chapter 33. The generalizeddisplacements are

u ∼

uwω

(7.75)

and the interpretation of the linear operator l1 is

l1(u) ∼

u′

(w′ − ω)

ω′

(7.76)

while the other operators l2, and l11 do not enter the description due to itslinearity.

7.6.2 Generalized Stresses

As mentioned above, our choice of strain measures, the generalized strains,dictates our stress measures, the generalized stresses, because these two setsmust be each other’s work conjugate in the Principle of Virtual Displace-ments. Since we work with three generalized strains we must have threegeneralized stresses which, as before, we shall denote N , V and M . Theyare the work conjugate of ε, ϕ and κ, respectively.


When the beam end points are located at x = a and x = b the internalvirtual work σ · δε is

Internal virtual

workσ · δε =

∫ b

a

(Nδε+ V δϕ+Mδκ)dx (7.77)

In order to proceed we need an expression for δε, δϕ and δκ. The straindefinitions (7.65), (7.74) and (7.67) provide

Generalized

strains ε, ϕ and κδε = δu′ , δϕ = (δw′ − δω) and δκ = δω′ (7.78)


Internal virtual

work

σ · δε=

∫ b

a

(Nδu′ + V (δw′ − δω) +Mδω′

)dx

= [Nδu]ba + [V δw]ba + [Mδω]ba

−∫ b

a

N ′δudx−∫ b

a

V ′δwdx −∫ b

a

(M ′ + V )δωdx

(7.79)

Provided that there are no distributed moment loads, i.e. that the loadsare the ones shown in Fig. 7.2, which for convenience is repeated here



pu

Pu(a) Pu(b)

Pw(a)

C(b)

Pw(b)

C(a)

pw


as Fig. 7.7, the external virtual work T · δu is

External virtual

work

T · δu=

∫ b

a

(puδu+ pwδw)dx



(7.80)

which obviously is the same as (7.21).

By equating the internal and external virtual work we can immediatelyconclude that

First static field

equationM ′ + V = 0 , x ∈]a, b[ (7.81)

which indicates that V is the same as the static quantity V introduced by(7.61). Furthermore, from (7.79) and (7.80) we may also get the other twostatic field equations. The first is

Second static field

equationV ′ + pw = 0 , x ∈]a, b[ (7.82)

which agrees with (7.62b), and the second isThird static field

equationN ′ + pu = 0 , x ∈]a, b[ (7.83)

which is the same as (7.62a). Thus we recover the equilibrium equations(7.62).

As regards the possible static boundary conditions we get

Possible static

boundary

conditions

N(a) = Pu(a) , N(b) = Pu(b)

V (a) = Pw(a) , V (b) = Pw(b)

M(a) = C(a) , M(b) = C(b)

(7.84)

which are the same as (7.63).

We may now realize that equilibrium equations of the linear Bernoulli-Kinematic linearity :

Same static

equations for B-E

and Timoshenko

Euler theory and the linear Timoshenko theory are the same, although thegeneralized strains are different,7.12 but recall that in the Bernoulli-Eulertheory the shear force V is not a generalized quantity.

7.12 One could have hoped that the same held for moderate kinematic nonlinearity.


Elastic Bernoulli-Euler Beams 115

7.7 Plane, Straight Elastic Bernoulli-EulerBeams

In applications of beam theories, be they kinematically nonlinear or linear,there are two constitutive models which dominate the picture completely.One is linear (hyper)elasticity, the other is perfect plasticity. I do not addressthe latter issue here, but devote much of Part III to determination of cross-sectional properties of linearly elastic beams. Whether the Bernoulli-Eulerbeam is kinematically nonlinear or linear the elastic constitutive model,Hooke’s “law,”7.13 is taken to be

Hooke’s “law”

[N

M

]=

[D11 0

0 D22

][ε

κ

](7.85)

where D11 signifies the axial stiffness and D22 the bending stiffness, re-spectively, and the cross-sectional properties are referred to an axis aboutwhich coupling between axial and bending actions does not occur. For ahomogeneous, isotropic cross-section (7.85) is usually written

Hooke’s “law”

[N

M

]=

[EA 0

0 EI

][ε

κ

](7.86)

where E is Young’s Modulus, A is the cross-sectional area, I is the moment Axial stiffness EA

Bending stiffness

EI

of inertia of the cross-section, and EA and EI denote the axial stiffnessand the bending stiffness of the beam, respectively. There is an importantobservation to make in connection with the structure of (7.86), namely thatthe constitutive matrix is a diagonal one and, therefore all coupling betweenaxial and transverse displacement components must come either from theboundaries or from a kinematically nonlinear effect, which is described bythe term 1

2 (w′)2 of the axial strain measure, see (7.12), and in the differential

equation (7.23b) is given by the term −(Nw′)′.

When we introduce the relation between M and κ, see (7.86), in thestatic field equation of the kinematically moderately nonlinear theory (7.23)we obtain

N ′ + pu = 0 and (EIκ)′′ − (Nw′)′ − pw = 0 (7.87)

and when the curvature-displacement relation (7.14)

κ = w′′ (7.88)

is exploited the result is

Equilibrium

equations in terms

of N and w

N ′ + pu = 0

(EIw′′)′′ − (Nw′)′ − pw = 0

}x ∈]a, b[ (7.89)

7.13 Remember, it is a model—not a law, but this is the traditional name.



These relations deserve some comments. First, using A and I in theconstitutive relation may be justified through application of the contin-uum mechanical theories developed in Part III with the comment that theSometimes N can

be determinedindependent of M

and w.

This true for

statically

determinate

structures

moment-curvature relation is less obvious than the axial force-axial strainrelation. Second, it may not seem like a good idea to utilize only the consti-tutive relation that expresses the bending momentM in terms of the curva-ture strain κ and let the axial force N remain untouched. The reason whythis is sometimes useful is that in many cases N may be determined with-out recourse to the constitutive relation because it is, statically determinatei.e. it may be found by use of the static equations, only. In some othercases, N may not be statically determinate, but may be found through aniterative process which entails that N is assumed known before each stepand corrected afterwards. This is often the case when you do a computa-tion by hand, while in more general purpose procedures such as a FiniteElement program the constitutive relation connecting N and ε is always in-voked. Third, it is important to bear in mind that the constitutive relationfor the bending moment is the second row of (7.86), i.e. M = EIκ, andnot M = EIw′′ because in order to get the latter expression the kinematicrelation κ = w′′, which only holds for kinematically linear or moderatelynonlinear Bernoulli-Euler beams, has been exploited.7.14

Ex 7-2 When Is the Linear Theory Valid?It would be comforting to have a strict criterion for the validity ofthe kinematically linear beam theory. The conditions given by (7.54)

wM

R

θ θ

CC

w

x

R

L

R

Fig. Ex. 7-2.1: Beam loaded by couples at the ends.

and (7.55) are useful but not very precise. We may concentrate on

7.14 The reason why I stress the third point is that too many people get these topicsmixed up.


Elastic Bernoulli-Euler Beams 117

the approximations inherent in the choice (7.57) of curvature strain.For this purpose, consider the beam shown in Fig. Ex. 7-2.1. Thebeam is allowed to slide on the rollers, thus keeping their distance atL during the loading history which is important for the full nonlin-ear theory, while the kinematically linear theory does not recognizekinematic shortening which implies that the curve is longer than thechord.

Ex 7-2.1 Full Nonlinear Analysis

This is one of the very few cases where the full nonlinear analysis isas easy as the kinematically linear one. It is clear that, independentof the theory, the bending moment M is constant and equal to −C.Then,

R =1

κ=EI

C(Ex. 7-2.1)

Furthermore,

wNLM = R(1− cos(θ) = RC

1−

√1−

(L

2R

)2

=EI

C

1−

√1−

(CL

2EI

)2

(Ex. 7-2.2)

This may look somewhat suspicious because it seems as if the displace-ment is inversely proportional to the applied load. A Taylor expansionabout C = 0 provides

wNLM ≈ 1

8

(CL

EI

)+

1

128

(CL

EI

)3

+ · · · (Ex. 7-2.3)

which is comforting to know.

For convenience let us non-dimensionalize in the following fashion

w ≡ w

L, C ≡ CL2

EI(Ex. 7-2.4)

to get

wNLM =

1

C

(1−

√1− 1

4C2

)≈ 1

8C + 1

128C3 + · · · (Ex. 7-2.5)

Note also that

θNL = sin−1( 12C ) (Ex. 7-2.6)

Ex 7-2.2 Linear Analysis

Using (7.89b) with N = 0 and pw = 0 we may find the solution

wL = 12Cξ(1− ξ) where ξ ≡ x

L(Ex. 7-2.7)

and thus

wLM = 1

8C (Ex. 7-2.8)



and

θL =dw

dx

∣∣∣∣x=0

= 12C (Ex. 7-2.9)

We may see that the linear approximations are equal to the first termof the Taylor expansion of the full nonlinear solution, as expected, and

LinearNonlinear

wM

C21.751.51.2510.750.50.250

0.5

0.4

0.3

0.2

0.1

0

Fig. Ex. 7-2.2: Beam with couples at the ends.Maximum normalized displacement.

seen from Figs. Ex. 7-2.2 and Ex. 7-2.3. Clearly, Fig. Ex. 7-2.2 in-dicates that the linear theory is valid for quite large loads, i.e. loadswhich cause a maximum displacement of about one tenth of the lengthof the beam which is more than is acceptable in most problems ofstructural engineering. The same load is associated with a rotation ofabout 40◦ at the supports, see Fig. Ex. 7-2.3, which is much more thanwe usually allow. Are we therefore justified in claiming that the kine-

LinearNonlinear

θ

C21.751.51.2510.750.50.250

1.75

1.5

1.25

1

0.75

0.5

0.25

0

Fig. Ex. 7-2.3: Beam with couples at the ends.Rotation at support.

matically linear theory is valid even for very large loads? The answer is


The Euler Column 119

simply: no. The reason is that even a small axial load would increasethe displacements considerably if computed by the nonlinear theory,while the results from the linear theory would remain the same. Anindication of this may be inferred from Example Ex 8-2, the Elastica.We are therefore not much closer to an estimate of the validity of thekinematically beam theory. On the other hand, the condition (7.55),which often is interpreted as |w′| . 0.1 rad ≈ 5◦ still seems to be agood estimate provided that no axial forces cause a strong increase inkinematic nonlinearity.

Ex 7-2.3 Moderately Nonlinear Analysis

You might ask why I did not apply the kinematically moderately non-linear theory to the above problem. The reason for not doing it is that(7.26c) clearly does not predict any kinematic nonlinearity unless thereis an axial force.

In Example Ex 8-2 we may see that the kinematically moderately non-linear theory does not perform well in postbuckling of the Elastica, seeFig. Ex. 8-2.2. On the other hand, in most cases, kinematically nonlin-ear analyses of frames are based, not on the full nonlinear theory, buton the kinematically moderately nonlinear theory and often with goodresults. There are several reasons for this. Among these is that bothnonlinear theories predict the same classical buckling load for moststructures. Another reason is that in postbuckling there are other ef-fects that govern than the ones related to the strain measures. A thirdreason is that in the presence of geometric imperfections the displace-ments at maximum load are often so small that the difference betweenthe postbuckling predictions according to the full and the moderatelynonlinear theories become unimportant. Finally, when plasticity en-ters then the displacements at maximum load are usually even smallerthan in the elastic cases.

We shall return to some of the above issues in Part IV, Buckling.

For our next example, the Euler Column, as for the Elastica, it would bemeaningless to try to apply the kinematically linear theory because the onlyload is an axial force and we are interested in seeing whether such a loadmight cause the column to experience transverse displacements.

Ex 7-3 The Euler Column

Probably the most prominent example of a nonlinear beam is the“EulerColumn” with pinned ends which is shown in Fig. Ex. 7-3.1. For this The Euler Columnbeam, or column, the stiffnesses are constant over the length of thebeam. The only load is the axial force λP , where λ is a scalar loadparameter, and the load is compressive.

For this structure (7.87a) simply is

N ′ = 0 (Ex. 7-3.1)



w

λP

xL

Fig. Ex. 7-3.1: The Euler Column.

which, together with the relevant static boundary condition (7.24b)

N(L) = −λP (Ex. 7-3.2)

provides

N(x) = −λP (Ex. 7-3.3)

Even if a transverse load, say pw(x), had been applied, (Ex. 7-3.3) stillholds. The static boundary conditions associated with the transverseloads follow from (7.24e) and (7.24f)

M(0) = 0 ⇒ w′′(0) = 0

M(L) = 0 ⇒ w′′(L) = 0(Ex. 7-3.4)

where the constitutive relation connecting M and κ as well as thestrain-displacement relation for κ and w′′ has been exploited. If wenormalize P such that

P = π2EI

L2(Ex. 7-3.5)

(7.89b) yieldsDifferentialequation foreigenvalueproblem

wiv + λ(πL

)2w′′ = 0 (Ex. 7-3.6)

which, together with the boundary conditions, constitutes a linear

eigenvalue problem in λ. The solution is7.15

Solutionλ(n) = n2

w(n) = ξ(n) sin(nπ

x

L

)}n ∈ [1, 2, . . . ,∞] (Ex. 7-3.7)

where λ(n) are the eigenvalues and ξ(n) designates the amplitudes ofBuckling Load= Classical Critical

Load λc

= Euler Load λE

Buckling mode w(1)

the associated eigenfunctions w(n). As in all other eigenvalue problemsthe amplitudes are undetermined. The value λ(1) = 1 is the Buckling

Load, which is also called the Classical Critical Load λc or Euler Load

λE of the column, and w(1) is usually referred to as the Buckling Mode.In most cases, only the lowest eigenvalue is of major interest, exceptpossibly when a number of other eigenvalues are close to the lowest,

7.15 If you don’t believe me you might check my claim, simply insert (Ex. 7-3.7) in(Ex. 7-3.6).


The Euler Column 121

which indeed is not the case for the Euler column in that the secondeigenvalue is four times the first. For shells it is, however, typical thatthere is a cluster of buckling modes associated with almost the samebuckling load. In any case, when the eigenfunctions are utilized as abasis for a series expansion all terms are needed, no matter how farthe higher eigenvalues lie from the lowest.

For later purposes, see Example Ex 32-5, write the potential energyassociated with the transverse displacement w for the Euler Column

Potential energyfor Euler Column

ΠP(w) =12

∫ L

0

EI(w′′)2dx+ 1

2

∫ L

0

N(w′)2dx (Ex. 7-3.8)

where it may seem strange that we may extract only one part of thepotential energy. In general, this it not admissible, but here the actionin the axial and transverse directions may be separated in this respect,as may be seen from the structure of (Ex. 7-3.1)–(Ex. 7-3.4). ThusN may be regarded as a constant in (Ex. 7-3.8). If you still haveyour doubts, you may realize that you can construct (Ex. 7-3.9) from(Ex. 7-3.6) when (Ex. 7-3.5) is reintroduced.

When we exploit the fact that EI as well as N are constants and thatN is given by (Ex. 7-3.3) we get

Potential energyfor Euler Column

ΠP(w) =12EI

∫ L

0

(w′′)2dx− 1

2λP

∫ L

0

(w′)2dx (Ex. 7-3.9)

It is easily verified that requiring that the first variation of ΠP(w)vanishes provides the correct differential equation and boundary con-ditions, see below.

0 = δΠP(w) = EI

∫ L

0

w′′δw′′dx− λP

∫ L

0

w′δw′dx (Ex. 7-3.10)

or

0 = EI [w′′δw′]l0 −EI

∫ L

0

w′′′δw′dx

− λP [w′δw]l0 + λP

∫ L

0

w′′δwdx

(Ex. 7-3.11)

and, finally

0 =EI [w′′δw′]l0 −EI [w′′′δw]l0 + EI

∫ L

0

wivδwdx

− λP [w′δw]l0 + λP

∫ L

0

w′′δwdx

(Ex. 7-3.12)

which, because of the arbitrariness of δw in the field, provides thedifferential equation (7.89).

As regards the boundary terms their verification is left as an exercisefor the reader.



7.8 Plane, Straight Elastic TimoshenkoBeams

Independent of the degree of kinematic nonlinearity the linear elastic consti-tutive model, Hooke’s “law,”7.16 for Timoshenko beams is

Hooke’s “law”

N

V

M

=

D11 0 0

0 D22 0

0 0 D33

ε

ϕ

κ

(7.90)

where D11 denotes the axial stiffness, D22 the shear stiffness, and D33 thebending stiffness, respectively. This separation between the axial and thebending deformation is only possible when the cross-sectional propertiesare referred to an axis about which coupling between axial and transverseactions does not occur.

Introduce Young’s Modulus E, the shear modulus G, the cross-sectionalAxial stiffness EAEffective shearstiffness GAe

Bending stiffness

EI

area A, the so-called effective cross-sectional area Ae, see Part III for itsdefinition, and the moment of inertia I of the cross-section. For a homo-geneous, isotropic cross-section we let EA, GAe and EI denote the axialstiffness , the effective shear stiffness, and the bending stiffness of the beam,respectively.7.17

Now, (7.90) may be written

Hooke’s “law”

N

V

M

=

EA 0 0

0 GAe 0

0 0 EI

ε

ϕ

κ

(7.91)

As was the case for the elastic Bernoulli-Euler beams the constitutivematrix is a diagonal one, see (7.91). In order to get an idea about theimportance of accounting for the shear flexibility we shall study the simpleexample of a clamped–free beam, see Example Ex 7-4 below.

Ex 7-4 A Cantilever Timoshenko BeamThe cross-sectional properties of the beam shown inFig. Ex. 7-4.1 areCantilever

Timoshenko beam assumed to be independent of the axial coordinate x. The only load isthe transverse force P .

7.16 Again, remember, it is a model—not a law, but this is the traditional name.7.17 The reason GAe is called the Effective Shear Stiffness is that the real shear stiffnessof a cross-section is not equal to GA because the shear strain is never constant over thecross-section. This has the inconvenient consequence that it is necessary to perform arather complicated analysis for every kind new of cross-section. Fortunately, over theyears, such analyses have been carried out for almost any conceivable cross-section, andthe results may be found in a large number of standard tables. In Part III, Beams withCross-Sections, we touch this subject for the rectangular cross-section.


A Cantilever Timoshenko Beam 123

x

w

P

L

Fig. Ex. 7-4.1: A Cantilever Beam.

First, let us note that the structure is statically determinate, i.e. allgeneralized stresses can be found from the static conditions, includ-ing the static boundary conditions, alone. For this structure simplecomputations yield

N = 0 , V = P andM = PL(1− ξ) , ξ ≡ x

L(Ex. 7-4.1)

Introduce the moment-curvature relation (7.90c) and the definition(7.67) of the curvature strain κ in (Ex. 7-4.1c) and get

ω′ = (1− ξ)PL

EI(Ex. 7-4.2)

with the solution

Rotation ωω = (ξ − 12ξ2)

PL2

EI(Ex. 7-4.3)

where we have exploited the kinematic boundary condition

ω(0) = 0 (Ex. 7-4.4)

Now, combine ((Ex. 7-4.1b) with the shear strain definition (7.74) andthe constitutive relation between the shear stress and the shear strain,see (7.90b)

P = GAe(w′ − ω) (Ex. 7-4.5)

After introduction of (Ex. 7-4.3) we may find

w′ =P

GAe

+(ξ − 1

2ξ2) PL2

EI(Ex. 7-4.6)

with the (kinematic) boundary condition7.18

w(0) = 0 (Ex. 7-4.7)

Thus, the solution for w is

Transversedisplacement w

w =PL3

3EI

(32ξ2 − 1

2ξ3 +

3EI

GAeL2ξ

)(Ex. 7-4.8)

As an example assume that the cross-section is rectangular with depthH and width B, then A = BH and I = 1

12BH3.

7.18 I stress that this is the only kinematic boundary condition on w. Note that unlikein the case of a Bernoulli-Euler beam w′(0) 6= 0.



When we introduce the shear modulus G given by (5.15) this becomesTransverse

displacement wfor a rectangular

cross-section

w =PL3

6EI

(3ξ2 − ξ3 + (1 + ν)

(A

Ae

)(H

L

)2ξ

)(Ex. 7-4.9)

A computation, which I do not attempt to perform here, shows that forthis kind of cross-section Ae ≈ 5

6A, but see Section 12.3 where possible

values of Ae are discussed. When we further assume that ν = 14, which

is a fairly common value, (Ex. 7-4.8) becomes

Displacement wfor a rectangular

cross-section

w =PL3

3EI

(32ξ2 − 1

2ξ3 + 3

4

(H

L

)2ξ

)(Ex. 7-4.10)

The tip deflection is

Tip deflectionw(L)

w(L) =PL3

3EI

(1 + 3

4

(H

L

)2)

(Ex. 7-4.11)

The value of w(L) for the equivalent Bernoulli-Euler beam isTip deflectionwBE(L) for a

Bernoulli-Eulerbeam

wBE(L) =PL3

3EI(Ex. 7-4.12)

and thus the second term in the parenthesis signifies the amplificationdue to the finite shear stiffness.

As you can see, if the length of the beam is 5 times its depth, which is aDisplacementsamplified in relation

to theBernoulli-Euler

solution

rather stocky beam, the increase in the prediction of the tip deflectionby using the Timoshenko beam theory is only 3%. You must, however,not take this as a universal truth in that e.g. for sandwich beams, whichconsist of a flexible core and two stiff face plates, the influence of shearflexibility must never be neglected.


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