Notes on Advanced Mechanics of Solids

Notes on Advanced Mechanics of Solids

Edmundo CoronaDepartment of Aerospace and Mechanical Engineering

University of Notre DameNotre Dame, IN 46556

[email protected]

November 24, 2003

Contents

1 Elementary Topics in Mechanics of Solids 5

2 Energy Methods in Structural Analysis 23

3 Beams on Elastic Foundations 31

4 Beams with Axial Load 35

5 Shear Deformations of Beams 41

6 Unsymmetric Bending 49

7 Plastic Problems 55

8 Forces and Stresses in Solids 63

9 Kinematics of Solids in Cartesian Coordinates 73

10 Bending of Rectangular Plates 79

3

4 CONTENTS

Chapter 1

Elementary Topics in Mechanics ofSolids

Introduction

This section discusses the basics of the topics which are usually covered in elementary under-graduate courses in Mechanics of Solids or Strength of Materials. Many good textbooks existin this subject. If you no longer have your textbook, it would be a great idea to purchaseone to keep in your personal library.

The elementary topics in the area of Mechanics of Solids include the fundamentalaspects of the theories of axially loaded members, torsion of circular shafts and bending ofbeams. The basics of each case will be discussed from the perspective of a second course inmechanics of solids. Before proceeding, however, it is convenient to also start by assumingthat the structural members considered are made of homogeneous, isotropic, linearly elasticmaterial— the simplest material model.

Homogeneous, Isotropic, Linearly Elastic Materials

Consider a solid of general shape in space. The material that makes up the solid has certainmaterial properties such as modulus of elasticity, yield stress, etc. The material is homoge-neous if it has the same macroscopic properties independently of position in the solid. Thiswould exclude, for example, a plate made of alternating layers of two materials. The materialis isotropic if the properties are independent of the direction in which they are measured.This excludes materials such as fiber reinforced composites where properties in directionsalong the fibers are different from those transverse to the fibers.

The concept of linear elasticity is best understood by considering a bar made of suchmaterial as shown in Fig. 1.1(a). When unloaded, the bar has length L, diameter D andcross-sectional area A(= πD/4). If a tensile axial load P is applied uniformly over the cross-section as shown in Fig. 1.1(b) the bar will stretch by an amount ∆L and its cross-sectionalarea will shrink to A′. The engineering axial stress in the bar is defined by the familiar

5

6 CHAPTER 1. ELEMENTARY TOPICS IN MECHANICS OF SOLIDS

formula

σx =P

A. (1.1)

In this case, the axial stress is uniformly distributed over the area as shown in Fig. 1.1(c).The engineering axial strain is defined by:

εx =∆L

L. (1.2)

The lateral contraction of the bar can be characterized by the change in diameter∆D. A transverse strain can then be defined as

εy =∆D

D(1.3)

The relations between σx, εx and εy are independent of the size of the specimen and dependonly on the material properties of the bar.

The relation between σx and εx is shown in Fig. 1.1(d) for a linear elastic material.The double arrows indicate that loading and unloading occur along the same line. Theconstant of proportionality between σx and εx is the Young’s modulus of the material andis denoted by the symbol E. Hence,

σx = Eεx. (1.4)

The relation between εy and εx is also linear as shown in Fig. 1.1(e) and given by

εy = −νεx (1.5)

where ν is the Poisson’s ratio of the material.

Axially Loaded Members

Consider a axially loaded member of length L in equilibrium as shown in Fig. 1.2(a). Thelength L is significantly larger than the dimensions of the bar in other directions, which canvary with x. The bar is loaded by forces P and Q applied at each end and a distributedaxial force q with units of force per unit length. The governing equations for the behaviorof the bar can be obtained from the consideration of three fundamental items:

1. The description of the motion of material points in the bar (kinematics)

2. The equilibrium of the forces acting in the bar (equilibrium)

3. The stress-strain relation of the material (constitutive behavior).

These three items are essential in the solution of problems in mechanics of solids and willre-appear often in later sections.

7

z

x

a

a’P

y

z

a-a

a’-a’

(a)

(b)

a

a’P

P

(c)

(d)

σx

σx

εx

E

1

εy

εx

1ν

(e)

∆L

D

D-∆D

A

L

A’

Figure 1.1: Tension test on a homogeneous, isotropic material.


x dxu

x dx

x

z

* *

x

z

P Qq

dx

a b

cdm n

(a)

(b)

(c)

L

N N + dNqdx

dxdA

m n

Figure 1.2: Axially loaded member

9

Kinematics

The objective of the kinematic analysis is to determine the relationship between the displace-ment of points in the bar and the resulting strain. Since the bar carries only axial loads, theelementary theory concentrates on the description of the deformation of the bar in the axialdirection. To describe the kinematics, consider the line segment mn. This line element isalso shown in Fig. 1.2(b). Before deformation, the length of the segment is dx and its leftend is located at an axial distance x from the origin. After deformation, the left tip of theelement moves a distance u in the axial direction, so that it is now located at a distance x∗

from the origin. In addition, the line mn has a new length dx∗. It is clear from the figurethat

x∗ = x + u (1.6)

An important assumption of the theory is that u is a function of x only. Differentiation withrespect to x gives

dx∗

dx= 1 + u′ (1.7)

where ( )′ = d( )dx

. For small strain, εx can be approximated by

εx =dx∗ − dx

dx= u′ (1.8)

Equilibrium

Now let us consider the equilibrium of the axial element of the bar abcd, also of length dx.The free-body diagram of the element is shown in fig. 1.2(c). The quantity N is the resultantof the axial stress in the bar. It is given by

N =∫

AσxdA (1.9)

where dA is an infinitesimal element of cross-sectional area, shown in Fig. 1.2(c). In thepresent case, σx is constant over the cross-section, so

N = σxA. (1.10)

Note that N , however, changes value to N + dN over the length dx. Equilibrium of forcesin the axial direction gives

∑Fx = −N + N + dN + qdx = 0 (1.11)

or

N ′ = −q. (1.12)

This is the equilibrium equation of the axially loaded bar.


Constitutive Relations

The constitutive relation between σx and εx is given by (1.4) if the material is assumed tobe linearly elastic.

Final Form of the Governing Equation

Consideration of the three fundamental items above yielded a system of four equations (1.4),(1.8), (1.10), (1.12) for the four unknowns u, εx, σx, N . These equations can be reduced toa single differential equation as follows.

Substituting (1.4) into (1.9) and using (1.8) yields

N = EAu′ (1.13)

substituting this into (1.12) yields

(EAu′)′ = −q (1.14)

which is the governing differential equation for an axially loaded bar in equilibrium.Let’s suppose that q = 0 and the we need to calculate the elongation of the bar in

Fig. 1.2(a). Equation (1.14) reduces to

(EAu′)′ = 0, (1.15)

indicating that P = Q in Fig. 1.2. Integrating this differential equation yields

u − u(0) =∫ x

0

C1

EAdx (1.16)

where C1 is a constant of integration to be evaluated from a boundary condition and u(0) isthe displacement of the left end of the bar. The boundary condition is that at x = 0, N = Pso that

P = [EAu′]0 = C1 (1.17)

and

u − u(0) =P

E

∫ x

0

dx

A. (1.18)

The stretch of the bar is δ = u(L) − u(0). It is given by

δ =P

E

∫ L

0

dx

A. (1.19)

If the cross-sectional area A does not change with x, then we obtain the familiarformula

δ =PL

EA. (1.20)

11

Torsion of Circular Bars

Consider a bar of length L subjected to two end torques T1 and T2 and a distributed torqueq as shown in Fig. 1.3(a). The elementary theory of torsion is restricted to the case ofbars with circular cross-section. Analysis of other cross-sections is more complicated and istreated in elasticity books1. Since the discussion will be restricted to the case of bars withcircular cross-section, a cylindrical coordinate system is most appropriate. The coordinatedirections x, ρ, θ are in the axial, radial and circumferential directions respectively. As inthe case of axial load, L is significantly larger than the other dimensions of the bar.

Kinematics

The main kinematic characteristic in the theory is that the cross-sections of the bar rotateas rigid bodies about their center. So, for example, point m in Fig. 1.3(b), which is locatedat a distance ρ from the center of the cross-section moves to location m′ after deformation.The angle of rotation is denoted by φ. For small rotations, the displacement of m can beassumed to be equal to the tangential component v. Then

v = ρφ. (1.21)

Now consider the line element mn of length dx in Figs. 1.3(a) and (c). This lineelement is originally parallel to the x axis. Point m is located at x, ρ, θ while point nis located at x + dx, ρ, θ. After deformation, mn moves to position m′n′. The line m′n′

makes an angle γ with the x axis, where γ is the shear strain at point m. Note that thecircumferential coordinate of m′ is now θ∗ = θ + v/ρ.

Differentiation with respect to x gives

ρ(dθ∗)dx

= γ = v′ (1.22)

Note also that differentiating (1.21) with respect to x and substituting in (1.22) yields

γ = ρφ′ (1.23)

Equilibrium

Now consider the equilibrium of the section abcd, also of length dx in Fig. 1.3(a). Its freebody diagram is shown in Fig. 1.3(d). The quantity T is the torque resultant of the shearstresses τ acting on a cross-section as shown in Fig. 1.3(e). It is given by

T =∫

AτρdA. (1.24)

1See, for example: Timoshenko, S. P. and Goodier, J. N.,Theory of Elasticity, Third Edition, McGraw-Hill, 1970.


T T + dTqdx

a

ρm

m’

v

ρ

τ

(a)

(b) (c)

(d) (e)

b

cd

1T 2T

x

n

m’n’

dx

v γ

m

m n

dx

q

φ

dx

L

θ

θ

x

θ

θ*

Figure 1.3: Torsion of a circular bar.

13

Considering the equilibrium of the free body in Fig.1.3(d) by summing moments about thex axis yields ∑

Mx = −T + T + dT + qdx = 0 (1.25)

orT ′ = −q. (1.26)


The shear stress-strain relation for a linear elastic, isotropic material is

τ = Gγ (1.27)

where G is the shear modulus of the material. It can be expressed in terms of E and ν as

G =E

2(1 + ν). (1.28)

Final form of the Governing Equation


T = GJφ′ (1.29)

whereJ =

∫A

ρ2dA (1.30)

is the polar moment of inertia of the cross-section. Substituting (1.29) into (1.26) yields thedifferential equation for φ

(GJφ′)′ = −q. (1.31)

Note that it has exactly the same form as the differential equation (1.14) for u in the axiallyloaded bar.

The shear stress can be calculated based on T by substituting (1.23) into (1.29) andusing (1.27). This yields

τ =Tρ

J. (1.32)

Note that the stress varies linearly with ρ, so it is maximum on the surface of the bar.

Bending of Beams

A beam is a structural element which resists loads applied transverse to its axis as shown inFig. 1.4(a). As in the previous cases discussed, the length of the beam is much larger than itsdimensions in other directions. The elementary theory is restricted to straight beams whosecross-sections have one plane of symmetry. All transverse forces have lines of action that


are in the plane of symmetry. This plane is called the plane of bending. Bending couplesapplied to the beam must be perpendicular to the plane of bending. If the conditions aboveare satisfied, the deflection of the beam will occur exclusively in the plane of bending. Inthe discussion that follows, the x axis passes through the centroid of the cross-sections ofthe beam, while the z axis is in the plane of bending (centroidal coordinate system).

Kinematics

Consider the line segment mn, of length dx which is originally aligned with the x axis asshown in Fig. 1.4(a). The left end of the line, point m, is located by the coordinates (x,z) asshown in Fig. 1.4(b). After deformation, the left tip of the segment moves to (x∗,z∗) and itslength becomes ds∗. In addition, the line segment makes an angle β with the x axis. Fromthe figure it is clear that

x∗ = x + u (1.33)

z∗ = z + w (1.34)

where u and w are the x and z components of displacement respectively. Differentiatingwith respect to x yields

dx∗

dx= 1 + u′ (1.35)

dz∗

dx= w′ (1.36)

andds∗ =

√dx∗2 + dz∗2 =

√(1 + u′)2 + w′2 dx. (1.37)

Expanding (1.37), neglecting the term u2 compared with u and then expanding usinga binomial series yields

ds∗ = (1 + u′ +1

2w′2)dx. (1.38)

Hence the expression for the axial strain is

εx =ds∗ − dx

dx= u′ +

1

2w′2 (1.39)

Note that, for relatively small deflections, the angle β is given by

β = w′. (1.40)

The next step is to relate the displacement components of any point in the beam(u,w) to the displacement components of its centroidal axis (u,w). This can be done byusing the commonly adopted assumption that, for slender beams, plane sections originallyperpendicular to the centroidal axis remain plane and perpendicular to the centroidal axisafter deformation. This assumption is explained in Fig. 1.5. Figure 1.5(a) shows the segment

15

q

L

a b

cd

x

z

βds*

x*x

z u z*dx

y

z

y

zz

x

(a)

(b)

(c) (d)

dA

dxm nl

l

l-l

M M+dMV

V+dV

dx

w

qdx

Figure 1.4: Bending of beams


a b

cdm

x

x

z

βz

u

u

z

a

d

m

a’

d’

m’

(a)

(b)

w

w

e f

e’f ’

e f

Figure 1.5: The plane sections remain plane assumption.

abcd of Fig. 1.4(a) with one such plane section ad and a segment of the centroidal axis efhighlighted. The highlighted lines are also shown in Fig. 1.5(b). The point m represents anarbitrary point in the cross-section. The location of all lines and points after deformationaccording to the plane sections assumption are identified by the primed symbols. From thefigure it is clear that, if the plane sections assumption holds, the displacement componentsof point m are given by

u = u − βz (1.41)

w = w. (1.42)

where u and w are the displacement components of point e which is in the centroidal axis.Differentiating (1.42) and (1.40) with respect to x yields

εx = u′ − zw′′ +1

2w′2 (1.43)

which can be re-written asεx = εo

x + κz (1.44)

where εox = u′ + 1/2w′2 is called the membrane strain and κ = −w′′ the curvature. The

product κz is the bending strain.

17

Equilibrium

Now let us consider the equilibrium of the element abcd in Fig. 1.4(c). The quantities Mand V are the moment resultant of the axial stress and the force resultant of the shear stressrespectively. They are given by

M =∫

Aσxz dA (1.45)

and

V =∫

Aτxz dA (1.46)

where the element of area dA is shown in Fig. 1.4(d). Since the beam in Fig. 1.4(a) does notcarry axial loads, the corresponding stress resultant N must be zero, that is

N =∫

Aσx dA = 0. (1.47)

Summing forces in the z direction yields

∑Fz = −V + V + dV + q dx = 0 (1.48)

while sum of moments in the y direction about point c yields

∑My = −M + M + dM − V dx + q dx

dx

2= 0. (1.49)

Simplifying the previous two equations and letting dx → 0 yields the equilibrium equations

V ′ = −q (1.50)

M ′ = V (1.51)


For linear elastic materials, the axial stress-strain relation is given by Hooke’s law (1.4).

Final Form of the Governing Equation

Substituting (1.4) into (1.47) and using(1.44) yields

N = E∫

A(εo

x + κz)dA = 0. (1.52)

The second term is zero because the z axis has its origin at the centroid. Hence (1.52)indicates that, in the absence of axial load,

εox = u′ +

1

2w′2 = 0 (1.53)


which provides a relation between the axial and transverse displacement components of thebeam. Note that (1.44) reduces to

εx = κz. (1.54)


M = E∫

A(εo

xz + κz2)dA. (1.55)

Due to the centroidal nature of the z axis the first term is zero while the second term yields

M = EIκ (1.56)

whereI =

∫A

z2 dA (1.57)

is the second moment (moment of inertia) of the cross-sectional area about the y axis.Differentiating (1.51) with respect to x, substituting into (1.50) and using (1.56) and

κ = −w′′ gives the differential equation for the deflection of the beam

(EIw′′)′′ = q (1.58)

This is a fourth order differential equation. Its solution requires that four boundary condi-tions be specified to solve for the four integration constants.

The axial stress σx can be calculated in terms of M by substituting (1.54) into (1.56)and using (1.4). The final expression is

σx =Mz

I. (1.59)

Note that the stress varies linearly with z and therefore is maximum at the point in thecross-section which is furthest from the centroid in the z direction.

If a beam contains discontinuities in geometric or material properties, or it is subjectedto point loads as shown at points A through D in Fig. 1.6(a), then (1.58) is valid only betweenthe discontinuities or point loads. In these cases, matching conditions must be specifiedbetween the various ‘segments’ of the beam. The kinematic matching conditions are always

wR = wL

w′R = w′

L (1.60)

where the subscripts L and R indicate the points immediately to the left and to the rightof the discontinuity or point load. These conditions state that the beam does not break nordevelops kinks. The behavior of the shear force and bending moment can be establishedwith the aid of Fig. 1.6(b). Considering the equilibrium of the element and letting ∆x → 0the following relations are obtained:

MR − ML = −Mo

VR − VL = −Po. (1.61)

19

AB C D

PM

oo

(a)

(b)

E1 E2

O

Po

VL

ML

VR

MR

Mo

∆x

Figure 1.6: (a) Beam with geometric and material discontinuities and point loads.

Substituting from (1.51), (1.56) and recalling that κ = −w′′ we can write (1.61) in terms ofw as follows:

(EIw′′)R − (EIw′′)L = Mo

(EIw′′)′R − (EIw′′)′L = Po. (1.62)

Summarizing, (1.60) and (1.62) provide four equations that must be satisfied at allpoints where geometric or material discontinuities, or point loads are present. These equa-tions are used to adjust four of the eight constants of integration obtained from the beamsegment to the left and right of the discontinuity or point load.

Strain Energy

Every time an elastic body is loaded it stores energy that can be recovered or converted intoa different form upon removal of the loads. Consider a bow and arrow. As the archer pullsback on the string, the frame of the bow bends and stores energy (equal to the work doneby the archer). When the archer lets go of the string, the load in the bow is released andmost of the energy is transfered to the arrow in the form of kinetic energy.


The energy stored in an elastic body by virtue of its deformation is called strainenergy, is given the symbol U and is equal to the work performed during deformation froma stress-free state. In the problems considered in this chapter, only one stress component ispresent, namely a normal stress in the case of axial and bending deformations and a shearstress in the case of torsion. It is important to be able to calculate the strain energy storedin these cases. To do so we first recall the definition of work.

Definition of Work

Consider a particle P which is moving from a point A to a point B along a given path asshown in Fig. 1.7. The work done by the force F on the particle P in going from A to B isdefined by

W =∫ B

AF · ds. (1.63)

That is, only the component of the force which is parallel to the instantaneous motion of Pdoes work.

A

B

F P

s

ds

Figure 1.7: Particle moving under the influence of a force.

Strain Energy under Uniaxial Stress

Consider first the cases of bars under axial or bending loads. In these cases the only stresscomponent is the axial stress σx. If we consider a small cube of material in the bar asshown in Fig. 1.8(a) with dimensions dx, dy and dz when unloaded and subject it to a stressσx, the cube will stretch in the x direction by an amount εxdx and contract in the other

21

dxdy

dz

dx(1+ε )x

dy(1+ε )y

dz(1+ε )z

σxσx

(a) (b)

Figure 1.8: Small material element under uniaxial stress.

two directions due to the Poisson effect. From (1.63) the strain energy stored in this smallvolume, dU , which is equal to the work done by σx, will be given by

dU =∫ εx

0(σxdydz)(dεxdx) =

∫ εx

0σxdεxdV (1.64)

where dV is the volume of the cube. The strain energy density, or strain energy per unitvolume is clearly given by

uo =∫ εx

0σxdεx (1.65)

so that for the complete body, the total strain energy is given by the volume integral of uo

U =∫

VuodV. (1.66)

For a linearly elastic material the constitutive law is given by (1.4) so that

uo = E∫ εx

0εxdεx =

Eε2x

2. (1.67)

Therefore, the total strain energy is given by

U =E

2

∫V

ε2xdV (1.68)

Axially Loaded Bar

In this case, the axial strain is a function of the axial coordinate x only (see Fig. 1.2),therefore

U =E

2

∫ L

0Aε2

xdx (1.69)

If the bar is prismatic (the area is constant over the length), then

U =EA

2

∫ L

0ε2

xdx. (1.70)


Finally, if the bar is also a two-force member, εx = δ/L where δ is the stretch of the bar and

U =1

2

EA

Lδ2. (1.71)

Furthermore, in this latest case the relation between δ and the load P is given by (1.20) andthe strain energy can be written in terms of P instead as

U =P 2L

2EA. (1.72)

Bending of Beams

The plane sections-remain-plane assumption of beam theory yields a strain distribution thatis linear through the height of the beam as given by (1.54). Therefore, the expression forstrain energy is given by

U =E

2

∫ L

0

∫A(κz)2dAdx =

E

2

∫ L

oκ2dx

∫A

z2dA =E

2

∫ L

0Iκ2dx. (1.73)

If the beam is prismatic, then I is a constant and

U =EI

2

∫ L

0κ2dx. (1.74)

The moment-curvature relation for a beam is given by (1.56), so it is possible to writethe strain energy in terms of the moment as

U =1

2EI

∫ L

0M2dx. (1.75)

Being able to calculate the strain energy stored in a body may be useful in its ownright. The main applications, however, concern the use of energy methods, a sometimes veryuseful alternative to the equilibrium methods presented in this chapter.

Chapter 2

Energy Methods in StructuralAnalysis

Castigliano’s First Theorem

Consider a body subjected to N point loads. These loads can be either forces of moments asshown in Fig. 2.1. Let the loads be P1, P2, P3, . . . , PN . These loads cause the body to deformfrom the shape indicated by the solid line into the shape indicated by the dashed outline.Let the quantities δ1, δ2, . . . , δN be the displacements at the points where the correspondingloads are applied and in the same direction as the loads. Note the that if the load is a force,the displacement will be a translation but if the load is a moment, the displacement will bea rotation. Since the body has deformed, it has stored strain energy U , which can be writtenas a function of the displacements. That is

U = U(δ1, δ2, . . . , δN). (2.1)

Now suppose that one of the displacements, say δi increases by a small amount dδi,then the strain energy will increase by an amount dU equal to the work done by the forcePi. That is

dU = Pidδi. (2.2)

From equation (2.1) dU is given by

dU =∂U

∂δidδi (2.3)

From equations (2.2) and (2.3) we obtain Castigliano’s first theorem, first postulated in 1879,

Pi =∂U

∂δi. (2.4)

Note that the strain energy must be expressed in terms of the displacements.

23

24 CHAPTER 2. ENERGY METHODS IN STRUCTURAL ANALYSIS

P1

P2

P3

Pi

PN

δ1

δ2

δ3

δi

δN

Figure 2.1: Forces and corresponding displacement definitions used to derive Castigliano’stheorems.

Potential Energy

The potential energy of any conservative mechanical system in some actual configuration isdefined as the work which will be done by all of the acting forces if the system is movedfrom the actual configuration to a reference configuration. For example, consider the case ofa mass m near the surface of the earth, which is located at a height h above a reference lineas shown in Fig. 2.2. The force acting on the mass is mg, and if the mass is released andfalls to the reference, the work done by gravity will be mgh. Hence, the potential energy ofthe mass in the actual configuration (distance h above the reference) is

Π = mgh.

In the case of solids, we will always take the reference configuration to be the shapeof the unloaded structure, which has no strain energy; for example, the solid outline in Fig.2.1. In this case, the potential energy of the internal forces in the solid will be equal to thestrain energy U . The potential energy of the external forces will be

−N∑

i=1

Piδi

.Note that the potential energy of the load Pi is not equal to the work done by Pi

during loading of the structure.The total potential energy of the system is

Π = U −N∑

i=1

Piδi. (2.5)

25

m

h

Π=0

Π=mghg

Figure 2.2: Gravitational potential energy.

Principle of Stationary Potential Energy

Note that, form equation (2.5),∂Π

∂δi

=∂U

∂δi

− Pi

and, from equation (2.4) it is clear that, at equilibrium

∂Π

∂δi

= 0 (2.6)

that is, the potential energy is stationary.A useful analogy is to consider a ball which can roll in a surface under the influence

of gravity. If the surface is concave, as shown in Fig. 2.3(a) then the equilibrium position isat the bottom and corresponds to a minimum in the potential energy. Note that if the ballis pushed to the side a little it will want to return to its equilibrium position. This meansthat the equilibrium is stable. If the surface is convex, as shown in Fig. 2.3(b) then theequilibrium position is at the top of the surface, where the potential energy is a maximum.If the ball is pushed to the side then it will tend to move away from the equilibrium position.This means that the equilibrium is unstable. If the surface is flat, as shown in Fig. 2.3(c)then any position is an equilibrium position. If the ball is pushed to the side in will tendto neither return nor depart from its original equilibrium position. This is a condition ofneutral equilibrium.

The Rayleigh-Ritz Method

In many instances the exact solution of structural problems (within the applicable theory)may be very difficult or impossible to find. In these situations we use methods which canprovide an approximate solution to the problems. For example, the finite element methodhas become widely used in engineering. The Rayleigh-Ritz method is an older relative of thefinite element method and can be used to find good approximations to problems in structural


g

(a) (b) (c)

Figure 2.3: A ball resting on a surface in a gravitational field.(a) Stable equilibrium, (b)Unstable equilibrium, (c) Neutral equilibrium.

and solid mechanics. The following discussion will concentrate on the use of the method tosolve beam problems, but it can also be used in other situations.

Consider a beam subjected to a combination of arbitrary point loads, point momentsand distributed loads as shown in Fig. 2.4. The potential energy of a beam like this can bewritten as follows:

Π =1

2

∫ L

0EI(w′′)2dx −

∫ L

0qw dx −

M∑k=1

Pkw(xk) −Q∑

l=1

Mlw′(xl) (2.7)

where the point force Pk is applied at xk and the point moment Ml is applied at xl.

P1P2

M1 M2

q

Figure 2.4: Beam subjected to transverse forces and moments.

The core of the method is to guess what the deflected shape of the beam will be inorder to come up with an expression for w. In general we approximate the deflection of thebeam by choosing w in the form

w =N∑

i=1

aiφi (2.8)

where φi are the so-called shape functions, and ai are unknown coefficients. The choice of φi

must be such that they satisfy the kinematic boundary conditions. Substituting (2.8) into

27

(2.7) yields

Π =1

2

∫ L

0EI(

N∑i=1

aiφ′′i )

2dx−∫ L

0q

N∑i=1

aiφi dx−M∑

k=1

Pk

N∑i=1

aiφi(xk)−Q∑

l=1

Ml

N∑i=1

aiφ′i(xl). (2.9)

The next step is to minimize the potential energy, which is obtained by setting∂Π/∂aj = 0 for j = 1, 2, . . .N . Hence,

∂Π

∂aj

=∫ L

0EI(

N∑i=1

aiφ′′i )φ

′′jdx −

∫ L

0qφj dx −

M∑k=1

Pkφj(xk) −Q∑

l=1

Mlφ′j(xl) = 0. (2.10)

This operation yields a system of N algebraic equations for the N unknown coefficients ai.Since the expression for Π was based on linear beam theory, the algebraic equations are alsolinear and can be written in matrix form as

Ka = b (2.11)

where

Kij =∫ L

0EIφ′′

i φ′′jdx

bj =∫ L

0qφj dx +

M∑k=1

Pkφj(xk) +Q∑

l=1

Mlφ′l(xl) (2.12)

and the ais are unknowns.For example, in the case in which no point loads are applied and N = 2, (2.11)

reduces to [ ∫ L0 EIφ′′

1φ′′1dx

∫ L0 EIφ′′

2φ′′1dx∫ L

0 EIφ′′1φ

′′2dx

∫ L0 EIφ′′

2φ′′2dx

]{a1

a2

}=

{ ∫ L0 qφ1 dx∫ L0 qφ2 dx

}. (2.13)

Complementary Energy

Let us introduce the concept of a “new” type of energy1 by considering a simple, geometricallylinear, axially loaded bar as shown in Fig. 2.5(a). The material in the bar will be assumedto be nonlinear elastic, with a stress-strain response like that shown in Fig. 2.5(b). Underthese conditions, the load-displacement (P -δ) response of the bar will have a similar shapeas that shown in Fig. 2.5(c). The work done by the force P in deforming the bar is given by

W =∫ δ1

oP dδ

where δ1 is the final displacement of the bar. The strain energy stored in the bar is equal toW (U = W ). This strain energy is represented in Fig. 2.5(c) by the area between the curve

1Most of the material in this section comes from the book by Gere and Timoshenko “Mechanics ofMaterials.”


and the δ axis. The strain energy can also be calculated by integrating the strain energydensity uo over the volume of the bar, that is

U =∫

VuodV =

∫V

∫ ε1

0σ dε dV

P

δ

★

PP

dPU

σ

ε δ

σ

ε

★ 1

1

1

1

δ

σ

ε

U

u

u

d d

d

(a) (b) (c)

Figure 2.5: (a) Bar with nonlinear material behavior, (b) Stress-strain response, (c) Load-Deflection response.

Now let us define the complementary work W ? as follows:

W ? =∫ P1

0δ dP (2.14)

where P1 is the final value of the load P . The complementary work is represented in Fig.2.5(c) by the are between the P -δ curve and the P axis. Note that, although, W ? does nothave a clear physical meaning,

W + W ? = P1δ1 (2.15)

which is the area of a rectangle of sides P −1 and δ1, so in some sense W ? is the complementof W . The complementary energy U? is defined to be equal to the complementary work doneby the applied loads, so

U? = W ?. (2.16)

Note then that

U? =∫ P1

0δ dP =

∫ σ1

0

δ

Ld(

P

A

)AL =

∫V

∫ σ1

oε dσ dV =

∫V

u?dV (2.17)

where u? is the complementary energy density, or complementary energy per unit volume.If, in addition to being geometrically linear, the bar is made of a linearly elastic

material, then the relation between P and δ will also be linear as shown in Fig. 2.6 and

U? = U. (2.18)

29

P

δ

U

U

★

Figure 2.6: Strain and complementary energy for a linear structure.

Although the two quantities are numerically equal it is important to remember that theyare conceptually different quantities.

Now consider a geometrically linearly solid with strain energy density

uo =∫ ε

oσxdεx + σydεy + σzdεz + τxydγxy + τyzdγyz + τxzdγxz

and, by analogy, define the complementary strain energy to be

u? =∫ σ

oεxdσx + εydσy + εzdσz + γxydτxy + γyzdτyz + γxzdτxz. (2.19)

Note that whereas the strain energy is naturally written in terms of strains (or displace-ments), the complementary energy is naturally written in terms of stresses (or loads).

If a structure is made up of multiple members (such as a truss or frame) we knowthat, from energy conservation, the sum of the strain energy in each member must equalthe work done by the external forces and the strain energy of the whole structure. This isindependent of whether the structure is linear or nonlinear. A similar statement concerningcomplementary energy is true if and only if the structure is geometrically linear.

The Crotti-Engesser Theorem and Castigliano’s Second

Theorem

Consider the body subjected to N point loads in Fig. 2.1. The loads P1, P2, P3, . . . , PN causethe body to deform from the shape indicated by the solid line into the shape indicated bythe dashed outline. The quantities δ1, δ2, . . . , δN are the displacements at the points wherethe corresponding loads are applied and in the same direction as the loads. The body willhave complementary energy U?, which can be written as a function of the forces. That is

U? = U?(P1, P2, . . . , PN). (2.20)


Now suppose that one of the loads, say Pi increases by a small amount dPi, then thecomplementary energy will increase by an amount dU? equal to the additional complementarywork done. That is

dU? = δidPi (2.21)

as shown in Fig. 2.7.

P

dP

δ

i

i

idU = δ dP + HOT i i

★

Figure 2.7: Increment in complementary energy caused by an increase in dPi.

From equation (2.20) dU? is given by

dU? =∂U?

∂PidPi. (2.22)

From equations (2.21) and (2.22) we obtain the Crotti-Engesser theorem,

δi =∂U?

∂Pi. (2.23)

Note that the complementary energy must be expressed in terms of the loads.If the structure is linear then U = U? and

δi =∂U

∂Pi

(2.24)

provided the strain energy is written in terms of the loads, that is U = U(P1, P2, . . . , PN).This is Castigliano’s second theorem.

Chapter 3

Beams on Elastic Foundations

Introduction

The model of a beam on an elastic foundation was first developed by Winkler in 1867 in orderto analyze railroad track. The model, shown in Fig. 3.1, considers a beam of bending rigidityEI under some kind of transverse loading (distributed load or point loads) represented by thevariable q. The beam lies on a continuous, linearly elastic support which can be visualizedas a series of closely packed independent springs, that is, the deflection of one spring doesnot affect its neighbors. The stiffness of the foundation is characterized by the parameter k,which is the spring constant of the foundation per unit length along the beam. It has unitsof force over length square.

centerline

EI

q(x)

k

Figure 3.1: Beam on an elastic foundation.

In order to keep the analysis simple, we will only consider foundations which areattached to the beam so that the foundation and the beam are always in contact. The casein which the foundation can detach from the beam is considerably harder to solve.

The statement of the problem of beams on elastic foundations is as follows: Given abeam of bending rigidity EI resting on a foundation of stiffness k and loaded by specifiedtransverse loads, find the deflection of the beam. Once the deflection of the beam is known,the strain and stress in the beam can be calculated.

31

32 CHAPTER 3. BEAMS ON ELASTIC FOUNDATIONS

In addition to railroad tracks, the beam on an elastic foundation model has beenapplied to analyze buried pipelines, networks of beams in floor systems, structures thatfloat, etc.

Development of the Model

In this section we will develop the equations necessary to analyze problems of beams onelastic foundations.

Kinematics

Only beams loaded in the plane of symmetry of the cross-section will be considered inthis discussion. Hence, all deflections will also occur in the plane of symmetry. We adoptkinematics consistent with small deflections of the beam and the plane sections remainplane assumption. Furthermore the beam is not subjected to axial loading. Under thesecircumstances the axial strain in the beam is a function of x and z and is given by

εx = κz (3.1)

whereκ = −w′′. (3.2)

Equilibrium1

Consider the free-body diagram of an element of length dx of the beam as shown in Fig.3.2. The coordinate axes are taken as indicated. The stress resultants on the left side of theelement are the shear force V and the bending moment M . Both of these quantities changeby an amount dV and dM over the length of the element. The deflection of the element is w.The foundation provides a reaction of magnitude kw dx and the distributed load contributesq dx as shown in the figure.

Sum of forces in the z direction yields the following equation:

dV

dx= kw − q, (3.3)

while sum of moments about a point on the right edge of the element yields

dM

dx= V. (3.4)

Taking one derivative of (3.4) with respect to x and substituting (3.3) gives,

d2M

dx2= kw − q. (3.5)

1This step can be replaced with an energy principle.

33

kw dx

q dx

M + dM

V + dV

MV

z

x

w

dx

Figure 3.2: Free body diagram of an infinitesimal length of the beam.

Constitutive Behavior

If the material of the beam is linearly elastic then the axial stress and the axial strain arerelated through Hooke’s law

σx = Eεx. (3.6)

This, together with (3.1) yields the moment-curvature relationship

M = EIκ = −EIw′′. (3.7)

Final Form of Governing Equation

Substituting (3.7) into (3.5) yields the equation for the deflection of the beam:

(EIw′′)′′ + kw = q (3.8)

and, if EI is constant,EIwiv + kw = q. (3.9)

The formulation of the beam on an elastic foundation results in a fourth-order linear differ-ential equation for the displacement w.

Solution

Equation (3.9) can be easily solved. First re-write it as

wiv + 4β4w =q

EI, where β =

4

√k

4EI. (3.10)

34 CHAPTER 3. BEAMS ON ELASTIC FOUNDATIONS

Assume that the solution of the homogeneous equation is of the form

wh = Aerx. (3.11)

The characteristic equation isr4 + 4β4 = 0. (3.12)

The four roots of this equation are:

r =√

2iβ,−√

2iβ,√−2iβ,−√−2iβ. (3.13)

After some algebraic manipulations, the roots can be written as

r = (1 + i)β,−(1 + i)β, (1 − i)β,−(1 − i)β. (3.14)

As a result the homogeneous solution wh can be written as

wh = eβx (C1 cos βx + C2 sin βx) + e−βx (C3 cos βx + C4 sin βx) . (3.15)

The particular solution wp depends on the load distribution q. For the case in which q isconstant, for example, wp = q/k so the final form of the solution of (3.9) is:

w = eβx (C1 cos βx + C2 sin βx) + e−βx (C3 cos βx + C4 sin βx) +q

k(3.16)

where C1, C2, C3 and C4 are constants of integration which are determined from the boundaryconditions.

Equation (3.16) is particularly useful in the case of long beans which can be consideredinfinite or semi-infinite in length. In the case of beams of finite length which exhibit symmetryor antisymmetry, (3.16) can be re-written in the more convenient form

w = A cosh βx cos βx + B cosh βx sin βx + C sinh βx sin βx + D sinh βx cos βx +q

k(3.17)

where A, B, C and D are, again, constants of integration to be determined from the boundaryconditions.

Calculation of Bending Moment and Stress

Once the deflection w is known as a function of x, the bending moment can be calculatedfrom equation (3.7).

The axial stress in the beam σx, can be calculated from the well-known formula

σ =Mz

I(3.18)

orσ = Eκz = −Ew′′z (3.19)

where z is measured from the neutral axis of the beam to the point of interest.

Chapter 4

Beams with Axial Load

Consider a beam subjected to both axial and transverse loads as shown in Fig. 4.1(a). Letthe beam be slightly curved when it is unloaded, that is, let it be imperfect as shown bythe solid line in Fig. 4.1(b). The initial initial deflection of the centroidal axis of the beamwill be denoted by wo and will be called the initial imperfection. The cross-section of thebeam is symmetric about the z axis as shown in Fig. 4.1(a). The x axis is set up so that itwould pass through the centroid of the cross-sections of the beam if it were perfect. All thetransverse forces have line of action along the z axis, and all moments have line of actionalong the y axis. Under these conditions, the beam will deflect exclusively along the z axis.

y

z(a)

q

L

a b

cd

z

x dxm nl

l

l-l

T

wo

(b)

Figure 4.1: (a) Beam with transverse and axial loads, (b) Definition of initial imperfection.

Kinematics

Let u and w be the displacement components of any point in the beam along the x and zaxes. Note that, due to the imperfection in the beam, w = wo and u = uo when the beamis unloaded, that is both u and w have initial values.

35

36 CHAPTER 4. BEAMS WITH AXIAL LOAD

In the discussion of the kinematics, all displacements are referred to the configurationof a perfect (straight) beam. This will be called the reference configuration. The kinematicsof the infinitesimal line segment mn in Fig. 4.1(a) originally aligned with the x axis and oflength dx in the reference configuration of the beam are shown in Fig. 4.2. Note that the leftend of the line has coordinates (x,z) in the reference configuration. Due to the imperfectionof the beam, however, this point takes the coordinates (xo,zo) when the beam is unloaded,where

xo = x + uo and zo = z + wo. (4.1)

Note that the line segment is now inclined at an angle βo and has length dso.

x

z

βds

xx

z u

zdx wo

o

o

o

βds*

o

o z*

x*u

w

*

Figure 4.2: Kinematics.

After the loads have been applied, the left end of the line segment moves to the point(x∗,z∗) where

x∗ = x + u and z∗ = z + w. (4.2)

Now the line segment is inclined at an angle β and has length ds∗.From equations (2.5) and (2.6), it can be seen that

ds∗2 − dx2 = 2ε∗xdx2 (4.3)

where

ε∗x = u′ +1

2

[(u′)2 + (w′)2

].

For intermediate deformations (u′ << 1), this expression can be simplified to

ε∗x = u′ +1

2(w′)2. (4.4)

Similarly,ds2

o − dx2 = 2εxodx2 (4.5)

37

where

εxo = u′o +

1

2w2

o. (4.6)

Since the beam is imperfect when unloaded, the stress will not be related not to thetotal deformation ds∗, but to the difference of the deformations ds∗ and dso. Note that

ds∗2 − ds2o = ds∗2 − dx2 − (ds2

o − dx2) (4.7)

or, letting ds∗2 − ds2o = 2εxdx2

εx = ε∗x − εxo. (4.8)

Hence

εx = u − uo +1

2

(w′2 − w′2

o

). (4.9)

Under the assumptions of intermediate deformations, the inclination of the line seg-ments shown in Fig. 4.2 are given by:

β∗ = w′ and βo = w′o, (4.10)

so the rotation of the line segment due to the loading of the beam is

β = β∗ − βo = w′ − w′o. (4.11)

The displacements u and w of an arbitrary point in the beam can be related to thedisplacements u and w of the centroidal axis using two assumptions. The first one is thatchanges in the cross-sectional shape of the beam are negligible. This implies that w = wand, hence, the rotation at an arbitrary point β is equal to the rotation of the centroidalaxis at the same location β. The second one is the plane sections remain plane assumption,which relates u to u and is explained in the handout Elementary Topics in Mechanicsof Solids.Therefore, in the imperfect, initial configuration,

uo = uo − βoz and βo = w′o (4.12)

while in the current, loaded configuration,

u = u − βz and β = w′. (4.13)

The axial strain in the beam is then given by

εx = εox + κz (4.14)

where the membrane strain εox is given by

εox = (u′ − u′

o) +1

2(w′2 − w′2

o ) (4.15)

and the change in curvature κ is given by

κ = −(w′′ − w′′o) (4.16)


Equilibrium

Figure 4.3 shows the free body diagram of a segment of the beam of length dx in the referenceconfiguration. Sum of forces in the x direction yields

−N cos β + (N + dN) cos(β + dβ) + V sin β − (V + dV ) sin(β + dβ) = 0 (4.17)

M

x

M+dM

V

V+dV

N+dN

N

q

ds*

dx*

dz*

β

β+dβ

z

Figure 4.3: Free body diagram used to obtain the nonlinear equilibrium equations.

For moderate rotations β we can approximate cosβ ≈ 1 and sin β ≈ β. Lettingdx → 0 (4.17) reduces to

N ′ − V β ′ − V ′β = 0. (4.18)

For slender beams, V is quite small so that the nonlinear interaction between shear androtations is customarily neglected. This assumption yields

N ′ = 0. (4.19)

Note that this equation says that the axial load in the beam is constant.Sum of forces in the z direction yields

−N sin β − V cos β + (V + dV ) cos(β + dβ) + (N + dN) sin(β + dβ) + qdx∗ = 0. (4.20)

Using the same assumptions as above and realizing that dx∗ = (1 + u′)dx ≈ dx for u′ << 1,(4.20) reduces to

V ′ + Nβ ′ + q = 0 (4.21)

39

Sum of moments about the right end of the element gives

−M +N cos β dz∗−N sin β dx∗−V cos β dx∗−V sin β dz∗+qdx∗dx∗

2+M +dM = 0. (4.22)

Using the same assumptions as above, and noting that dz∗ ≈ βdx yields

M ′ = V. (4.23)

Taking on derivative of (4.23) with respect to x and substituting into (4.21) gives

M ′′ + Nβ ′ + q = 0. (4.24)


The material considered is linearly elastic. This, combined with the plane sections remainplane assumption and the fact that z = 0 at the centroid of the cross-section yield the familiarrelations (see the bending section in the handout Elementary Topics in Mechanics ofSolids)

M = EIκ and N = EAεox. (4.25)

Final Form of the governing Equations

Substituting the first equation in (4.25) into (4.24) yields

(EIκ)′′ + Nβ ′ = −q (4.26)

or, substituting for β and κ from (4.13) and (4.16) gives the differential equation for thetransverse displacement w:

[EI(w′′ − w′′o)]

′′ − Nw′′ = q. (4.27)

If EI is constant, then

EI(wiv − wivo ) − Nw′′ = q. (4.28)

The equation for the axial displacements u can be obtained by substituting from thesecond equation in(4.25) into (4.19), which yields

N ′ = (EAεox)

′ = 0 (4.29)

which, for a prismatic beam, reduces to

N = EAεox. (4.30)

Note the following points:


• The axial force in the beam is constant

• As a result, (4.27) is a linear equation

• If N = 0 and wo = 0 then (4.27) reduces to the usual beam equation

(EIw′′)′′ = q

Chapter 5

Shear Deformations of Beams

Introduction

In many engineering applications the effect of shear in the calculation of beam deflections isneglected. This is a result of the customary assumption that plane sections originally planeand perpendicular to the beam axis before deformation remain plane and perpendicular tothe beam axis after deformation (this is commonly called the “plane sections remain plane”assumption). In some cases, however, shear deformations may be important. In these cases,beam theory needs to be corrected to account for shear deformations. Timoshenko suggestedan approximate correction, which we will study1

Before looking at the correction suggested by Timoshenko (sometimes called Tim-oshenko beam theory to differentiate it from the usual Bernoulli-Euler beam theory ), it ishelpful to review the calculation of shear stresses in beams 2.

Review: Shear Stresses in Beams

Recall that beams subjected to transverse loading develop shear stresses in addition to axialbending stresses. The presence of shear is usually illustrated by considering two beams,labeled Case 1 and Case 2 in Fig. 1. Beam 1 is a solid beam of height 2h while beam 2consists of two beams, each of height h, one on top of the other. If a load is applied asshown, beam 1 develops the usual axial stress distribution. Each beam which constitutesbeam 2, however, develops its “own” axial stress distribution and there is slip at the interfacebetween them. This indicates that the shear stress is zero at the interface and that eachbeam works independently. As a result,the beam in Case 1 is stiffer than that in Case 2. Ifwe nail the two constituents of beam 2 together, the nails will carry the shear stress at theinterface and the beam will become stiffer.

Figure 2 (a) shows the stress distribution in a beam with rectangular cross-section.

1Reference: Gere, J. M. and Timoshenko, S. P., Mechanics of Materials, Brooks/Cole Engineering Divi-sion, 1984.

2The presentation and figures come from Timoshenko’s book

41

42 CHAPTER 5. SHEAR DEFORMATIONS OF BEAMS

2h

Case 1 Case 2

Figure 5.1: Demonstration of the effect of shear. Note different axial stress distributionsindicated by the shaded areas.

The shear force acting in the cross-section shown is V . Figure 2 (b) shows the shear stressesin the element m-n. Recall that shear stresses come in pairs acting in perpendicular planes,so in this case we have a shear stress τxz in the y − z plane and an equal stress τzx in thex − y plane (since τxz = τzx we will refer to both by τ). The assumptions about the shearstress distribution are:

• τ is parallel to V .

• τ distribution is uniform across the width of the cross-section.

Figure 3 (a) shows a side view of an element of beam of length dx, while Fig. 3 (b)shows the front view of the cross-section. The bending stress distribution is given by

σx =Mz

I(5.1)

where M is the bending moment, and I is the moment of inertia of the cross-section. Noticethat a shear stress τ acts along the segment p-p1.

Considering the equilibrium of the segment p-p1-n-n1, we get that

∑Fx = −

∫ h2

z1

Mz

Ib dz +

∫ h2

z1

(M + dM)z

Ib dz − τb dx = 0, (5.2)

sodM

dx

1

I

∫ h2

z1

z dz − τ = 0 (5.3)

butdM

dx= V (5.4)

43

Figure 5.2: Shear stress in a beam cross-section.

Figure 5.3: Beam element under shear stress


then

τ =V

I

∫ h2

z1

z dz (5.5)

or

τ =V Q

Ib(5.6)

where

Q =∫ h

2

z1

bz dz, (5.7)

the first moment of the shaded area about the y axis.Note that for a rectangular cross-section,

τ =V

2I

(h2

4− z2

1

)(5.8)

so the shear stress distribution is parabolic. The shear stress is zero at z = −h2

and z = h2,

and it is maximum at z = 0.Equations (7) and (8) can be used with some other cross-sectional shapes provided

that the appropriate expression for Q is adopted.

The Timoshenko Approximation

Since the shear stress τ varies parabolically over the cross-section, the shear strain, given by

γ =τ

G, (5.9)

where G is the shear modulus of the material, must vary in the same manner. This has thefollowing implications:

• Cross-sections become warped as shown in Fig. 4.

• The maximum shear strain occurs at the neutral axis.

• The shear strain is zero at points such as m1, p1, n1 and q1, so sections there are normalto the beam surface.

Warping of the cross-sections can be explained with the help of Fig. 5. An undeformedbeam segment is shown in Fig. 5(a). It has been divided into five equal blocks as shown.Upon the application of pure shear stress, each block deforms as shown in Fig. 5(b). Thefarther a block is from the neutral axis, the smaller its shear strain, so the net result, shownin Fig. 5 (c) is that γ1 > γ2 > γ3 where γ3 = 0 since it is the shear strain at the edge of thecross-section.

If we let the height of the blocks go to zero and the number of blocks go to infinity,the profile of the section under pure shear becomes smooth as shown in Fig. 6 (a) (the

45

Figure 5.4: Cross-section warping due to shear stress.

τ

τ

γγ

γ

1

2

3

cγ

=0

(a) (b) (c)

Figure 5.5: (a) Division of longitudinal beam segment into finite blocks (b) Center blockunder pure shear stress. (c) Visualization of shear deformations based on the finite blocks


undeformed shape of the section is shown by the dashed rectangle). The contribution ofshear to the deflection can now be visualized by rotating the element clockwise until thedeformed section is vertical at point m so that it coincides with the undeformed sectionthere. This is like enforcing the plane sections remain plane at point m. It can be senthat the shear deformation will contribute to the deflection of the beam. The slope of thecontribution of shear to the deflection of the axis of the beam is equal to the maximum shearstrain γc, which is also the shear strain at the neutral axis. Therefore,

dws

dx= γc (5.10)

where ws is the shear contribution to the deflection.

(a) (b)

γc

m n γc

mn

Figure 5.6: (a) Warping of the cross-section due to shear (b) Visualization of the contributionof shear to beam deflection (deformations are exaggerated).

Timoshenko proposed thatdws

dx=

αsV

GA(5.11)

where V is the shear force, A is the cross-sectional area of the beam and G is the shearmodulus of the material. Therefore, V

Ais the average shear stress on the cross-section of the

beam. Finally, “αs is a numerical factor (or shear coefficient) by which the average shearstress must be multiplied to obtain the shear stress at the centroid of the cross-section.” Theshear coefficient depends on the shape of the cross-section. For example, for a rectangularcross-section, αs = 3

2.

Taking one derivative of (11) we get

d2ws

dx2=

αs

GA

dV

dx= −αsq

GA, (5.12)

47

where q is the distributed load on the beam.Finally, the total deflection of the beam is taken to be

w = wb + ws (5.13)

where wb is the deflection due to bending alone. Therefore,

d2w

dx2=

d2wb

dx2+

d2ws

dx2(5.14)

ord2w

dx2= − M

EI− αsq

GA(5.15)

which is commonly written as

d2w

dx2= − 1

EI

(M +

αsEI

GAq)

. (5.16)

So, given a particular problem, this equation can be integrated twice to evaluate the trans-verse displacement of a beam considering the effect of shear deformations. From (16) wecan see that shear deformations can be important for materials with low shear modulus(compared to Young’s modulus). This is often true in the case of composite materials.

Chapter 6

Unsymmetric Bending

Introduction

So far we have considered beams that have cross-sections with at least one plane of symmetry,which we have required to coincide with the plane of bending. In addition, the transverseloading in the beam has always been required to act in the plane of bending. Thereforethe bending moment is perpendicular to the plane of bending. Under these conditions, thebeams deflect exclusively in the plane of bending. Some examples of the cases treated so farare shown in Fig. 1.

Plane ofBending

Figure 6.1: Examples of cross-sectional shapes with one plane of symmetry.

We will now consider beams which bend in two planes simultaneously1. We will startwith the simplest case which is that of a beam with doubly symmetric cross-section, butwhich is loaded by a transverse load which is not in the plane of bending. One such exampleis shown in Fig. 2 (a).

1Reference: Gere, J. M. and Timoshenko, S. P., Mechanics of Materials, Brooks/Cole Engineering Divi-sion, 1984

49

50 CHAPTER 6. UNSYMMETRIC BENDING

z

y

x

x L

P

z

y

n

n

M

M

My

z

β

θ

A

(a) (b)

θ

Figure 6.2: Beam with doubly symmetrical cross-section subjected to a skew load.

Beams With Doubly Symmetric Cross-sections Under

Skew Loads

This case is relatively simple since the loading can be decomposed into components lyingin the two planes of symmetry of the cross-section as shown in Fig. 2 (b). Since the beamequation derived previously is linear as long as the deflections are small and the materialremains linear, the deflection in each of the y and z axes can be calculated separately.

For bending in the x − z plane we can use the familiar equation

−EIyd2w

dx2= My (6.1)

where w is the deflection along the z axis. In the case of bending in the x− y plane we have

EIzd2v

dx2= Mz (6.2)

where v is the deflection along the y axis (why is the sign different in this case?). Finally,the total beam deflection is given by the vector sum of wy and wz.

Note that the stress produced at point A is given by

σx =Myz

Iy− Mzy

Iz. (6.3)

The neutral plane (plane of zero stress) can be found by setting σx equal to zero. So, in thecase considered,

M cos θz

Iy− M sin θy

Iz= 0 (6.4)

which yields

tanβ =z

y=

Iy

Iz

tan θ (6.5)

51

General Theory of Pure Bending

The general theory of pure bending refers to the treatment of beams with general cross-sectional shape (symmetric and unsymmetric). Figure 3 shows some possible unsymmetriccross-sectional shapes.

Figure 6.3: Examples of cross-sectional shapes with no planes of symmetry.

Figure 4 shows a general shape cross-section. The origin of the y-z axes is locatedat the centroid of the area. The sign conventions for the curvatures κy and κz in the x-yand the x-z planes respectively, are shown in Figure 5. The convention is such that positivemoments yield positive curvatures. From this figure, it is easy to see that

κy = d2v/dx2 and κz = −d2w/dx2. (6.6)

y

z

A

Mz

My

Figure 6.4: Unsymmetric cross-section with non-principal centroidal axes.

Using the plane sections remain plane assumption, the axial strain at point A is givenby

εx = κzz − κyy (6.7)


+x

y

+ x

z

yz

Figure 6.5: Curvature sign convention.

so the axial stress isσx = E (κzz − κyy) (6.8)

The normal force can now be evaluated and, recalling that it must be zero for purebending, yields

N =∫

Aσx dA = 0 or κz

∫A

z dA − κy

∫A

y dA = 0 (6.9)

which is automatically satisfied since the origin of the y-z axes is located at the centroid.The moment about the y axis, My is given by

My =∫

Aσxz dA = Eκz

∫A

z2 dA − Eκy

∫A

yz dA (6.10)

orMy = EIyκz − EIyzκy (6.11)

where Iy is the centroidal moment of inertia of the area about the y axis and Iyz is theproduct of inertia of the area.

Similarly,

Mz = −∫

Aσxy dA = Eκz

∫A

yz dA − Eκy

∫A

y2 dA (6.12)

orMz = −EIyzκz + EIzκy. (6.13)

Solving for κy and κz we get

κy =MzIy + MyIyz

E(IyIz − I2

yz

) (6.14)

and

κz =MyIz + MzIyz

E(IyIz − I2

yz

) . (6.15)

Substituting(14) and (15) into (8) gives the expression for the stress σx

σx =(MyIz + MzIyz)z − (MzIy + MyIyz)y

IyIz − I2yz

. (6.16)

53

Shear Stresses in Thin-Walled Beams Bent About Non-

principal Axes

In this section we consider the shear stress distribution in thin-walled beams when subjectedto bending about non-principal axes as shown in Fig. 6.6(a). The shear forces applied are Vy

and Vz. In the treatment of thin-walled sections it is customary to neglect the shear stresseswhich act in the through-thickness direction and to assume that the shear stresses act alongthe mid-surface as shown in Fig. 6.6 (b). As customary, the origin of the y and z axes is atthe centroid of the cross-section. Note that the thickness of the section is a function of s.

y

zx

ab

cd

Vy

Vz

s

a b

cd

τ

dx

S

(a)

(b)

F1

F2

A

dA

t(s)

Figure 6.6: Shear stresses in a beam of thin-walled open cross-section

Consider the free-body diagram of the element abcd. The faces of the element whichface the x axis have area A. Equilibrium yields∑

Fx = F2 − F1 + τt dx = 0 (6.17)

where t is the thickness at point c and

F1 =∫

Aσx dA and F2 =

∫A

(σx +

dσx

dxdx

)dA (6.18)

so

τ = −1

t

∫A

dσx

dxdA. (6.19)


From equation (6.16),

dσx

dx=

(dMy

dxIz + dMz

dxIyz)z − (dMz

dxIy + dMy

dxIyz)y

IyIz − I2yz

, (6.20)

but, according to the sign convention used as shown in Fig. 6.7,

dMy

dx= Vz and

dMz

dx= −Vy (6.21)

so, finally we obtain

τ = − 1

t(IyIz − I2yz)

[Vz

(Iz

∫A

z dA − Iyz

∫A

y dA)

+ Vy

(Iy

∫A

y dA − Iyz

∫A

z dA)]

, (6.22)

which is the final expression for the shear stress.

M M + dM

V V + dV

z

x

yy

zzz

yM + dM

V V + dV

y

x

zz

yyy

Mz

z y

Figure 6.7: Axial stress resultants in the x-z and y-z planes

Chapter 7

Plastic Problems

Introduction

When metals are stressed beyond certain limit, the relation between stress and strain becomesnonlinear, and Hooke’s law is no longer valid. For example, Figs. 7.1(a) and (b) showschematics of typical stress-strain curves for hot rolled structural steel and cold rolled steelrespectively. The limit of linear elasticity for these materials is called the proportional limit.This is the point when the stress-strain curve ceases to be linear. Since such limit is difficultto determine accurately, especially in the case in Fig. 7.1(b) the concept of a yield stresshas been introduced as a criterion to determine the end of linearly elastic behavior. In Fig.7.1(a), the yield stress is defined as the value of the stress plateau which immediately followsyield. In Fig. 7.1(b) the yield stress is defined by the so-called 0.2% strain offset criterion asindicated in the figure.

σ

ε

σ

ε

σo σo

1E

1

E

0.002(a) (b)

1

E

Figure 7.1: Examples of stress-strain curves.

Most structures are designed so that they remain linearly elastic during service. Manyinstances exist, however, when it is necessary to study the behavior of structures or solids

55

56 CHAPTER 7. PLASTIC PROBLEMS

after yielding has occurred. For example:

• Predict the behavior of the structures should an overload occur

• Analysis of manufacturing processes involving cold work

• Design of structures based on ultimate load rather than yield load.

In order to maintain simplicity, we will only consider structural elements which aresubjected to a uniaxial state of stress and undergo relatively small strains.

Approximation of Uniaxial Stress-Strain Curves

Generally, it is necessary to approximate a measured stress-strain curve to simplify analy-sis and make it more systematic. Many approximations to stress-strain curves have beenproposed, some of which are presented in Fig. 7.2.

Elastic-Perfectly Plastic Fit This is the simplest approximation and is shown in Fig.7.2(a). It contains two parameters: Young’s modulus E and a yield stress parameterσY . Note that the stress remains constant after yield.

Elastic-Linearly Hardening Fit This is a three parameter model and takes the effectof strain hardening1 into account, as shown in Fig. 7.2(b). The three parametersare: Young’s modulus E, yield stress parameter σY and the post-yield slope E ′. Ageneralization of this model consists of using a piecewise linear approximation for thewhole stress-strain curve as shown in Fig. 7.2(c). This approximation is usually adoptedin commercially available numerical analysis codes.

Ramberg-Osgood Fit This three-parameter fit belongs to the family of power-law fits andis used when a smooth fit may be desirable. It is given by the equation

ε =σ

E

[1 +

3

7

(σ

σY

)n−1].

The three parameters used are: Young’s modulus E, Ramberg-Osgood yield stress σY

and the hardening exponent n. The parameters σY and n can be found by trial anderror. A good initial guess for σY is σo. A good guess form n can be obtained byplotting log(ε − σ/E) vs log σ and fitting a straight line through the data. The slopeof the line is equal to n.

1Strain hardening simply means that the stress keeps rising with increasing strain after yield.

57

σ

ε(a)

σ

ε(c)

σ

ε(d)

σ

ε(b)

Yσ

1E Actual

Fit

Yσ

1E

1 E’

1E

σY

σ1

σ2σ3

εY

ε1

ε2ε3

Figure 7.2: Approximation of stress-strain curves: (a) Elastic-perfectly plastic, (b) Bilinear,(c) Multi-linear, (d) Ramberg-Osgood.


L

Cross-section

y

z

dA=b(z)dzdzb(z)

Figure 7.3: Beam with doubly symmetric cross-section.

Inelastic Bending

This section considers bending of a beam with doubly symmetric cross-section of height has shown in Fig. 7.3. The restriction of symmetry of the cross-section about the y axis willbecome apparent later2. The material of the beam is assumed to be elastic-perfectly plastic,with equal properties in tension and compression as shown in fig. 7.4.

Kinematics

The assumption that plane sections originally perpendicular to the beam’s axis remain planeand perpendicular to the deflected axis will be retained. Furthermore, we will require thatthe axial load in the beam be zero. Under these conditions, it has been shown that the axialstrain is given by

εx = κz (7.1)

where κ = −w′′ if the deflection of the beam w is small.


When the material is in the elastic range, the stress σx is related to εx through Hooke’s lawσx = Eεx. This, combined with (7.1) yields the well-known relation

M = EIκ. (7.2)

The expression for the stress in the beam is

σx = Eεx = Eκz =Mz

I. (7.3)

2Recall that symmetry about the z axis, combined with the requirement that all forces act in this planeof symmetry and all moments perpendicular to it guarantees that the deflection of the beam will occurexclusively along the z axis.

59

−σ

σ

σ

−ε ε ε

Y

YY

Y

1

E

Figure 7.4: Assumed elastic-perfectly plastic stress-strain curve.


ε

z h/2

-h/2

σ

z h/2

-h/2

(a) (b)

εY−εY σY−σY

Figure 7.5: Strain and stress distributions if the material is elastic everywhere.

The strain and stress distributions are shown in Figs. 7.5(a) and (b) respectively. They areboth linear.

As the loading on the beam increases so does the curvature and eventually, the max-imum stress in the beam reaches the yield stress σY , which occurs when the strain is εY asshown in Fig. 7.4, for the first time. At this time, the first yield condition is

εY = κYh

2(7.4)

where κY is the yield curvature. Upon use of Hooke’s law (7.4) can be re-written and solvedfor κY , yielding

κY =2σY

Ehor MY =

2IσY

h. (7.5)

where the quantity MY is called the yield moment. Note that in account of the symmetry ofthe cross-section and of the stress-strain response of the material, yielding occurs at z = ±h/2simultaneously.

Further increase in curvature causes the material to enter the plastic range, so thatthe post yield strain and stress distributions are as shown in Fig. 7.6(a) and (b) respectively.The assumption that plane sections remain plane still holds in the plastic range, hence thestrain varies linearly with z but the stress distribution is no longer linear. As the curvatureof the beam increases, the plastic regions move inwards from the edges of the beam towardsthe neutral axis. Note that in account of symmetry, the stress distribution is symmetricabout the y axis. This means that the neutral axis of the beam remains located at z = 0,something that would not be true if the cross-section was only symmetric about the z axis.This restriction greatly simplifies the analysis and it is the main reason the double symmetryrestriction on the cross-section was adopted. The moment is given by

M =∫

Aσxz dA (7.6)

61

ε

z h/2

-h/2

σ

z h/2

-h/2

(a) (b)

εY−εY σY−σY

h

h

Figure 7.6: Strain and stress distributions if some of the material in the beam has yielded.

where dA = b(z) dz as shown in Fig. 7.3. Hence

M = 2

[∫ h

0Eκz2b(z)dz +

∫ h2

hσY zb(z) dz

]. (7.7)

For a rectangular section b(z) = b, a constant, so (7.7) reduces to

M = 2

[Eκb

h3

3+

σY b

2

(h2

4− h2

)]. (7.8)

The height of the elastic region h can be easily determined by noticing that at z = h,εx = εY = σY /E, so that

h =σY

Eκ. (7.9)

Substituting (7.9) into (7.8) yields the expression for the moment in terms of thecurvature

M =σY bh2

4− σ3

Y b

3E2κ2(7.10)

Note that both the curvature and the moment are assumed to be positive in thisexpression, and that changing the sign of κ does not change the sign of M . To account forthe sign of the curvature, (7.10) needs to be slightly modified as follows:

M = sgn(κ)

(σY bh2

4− σ3

Y b

3E2κ2

). (7.11)

Note that as κ → ∞ the moment approaches the value

Mu =σY bh2

4(7.12)

which is called the ultimate moment of the beam and occurs when the elastic region disap-pears and the stress distribution is as shown in Fig. 7.7.


σ

z h/2

-h/2

σY

−σY

Figure 7.7: Stress distribution at the ultimate moment.

Note that, for a rectangular cross-section, the yield curvature and the yield momentare given by

κY =2σY

Ehand M =

σY bh2

6(7.13)

respectively. Using these expressions, the curvature of the beam can be written as follows

κ

κY= sgn(M)

1√3 − 2 |M |

MY

(7.14)

Example: Cantilever Beam

Chapter 8

Forces and Stresses in Solids

External Forces

Consider a body of general shape (sometimes called the solid mechanics potato) on whicha set of forces with resultants, F1 . . . F5 act, as shown in Fig. 8.1. These forces are calledexternal forces. External forces can be divided into two kinds:

1. Surface Forces These forces act on the boundary of the body. They are usually givenin terms of force per unit area. Note that concentrated forces are used for conveniencein Fig. 8.1. In actuality all forces act over a finite area. Examples include pressure,reaction forces, etc.

2. Body Forces These forces act in the inside of the body. They are usually given in termsof force per unit volume. Examples include gravity forces, centrifugal forces, etc.

F1

F2

F3F4

F5

x

y

z

Figure 8.1: External force resultants acting on an arbitrary body

If the body in Fig. 8.1 is in equilibrium, the forces F1 . . . F5 are not arbitrary. Theymust satisfy the equilibrium conditions; that is, the vector sum of all forces as well as of

63

64 CHAPTER 8. FORCES AND STRESSES IN SOLIDS

all moments acting on the body have to have zero resultant. The equilibrium equations invector form are: ∑

F = 0 and∑

M = 0, (8.1)

or if we consider the Cartesian frame shown, we can write the equations in component form:

∑Fx = 0,

∑Fy = 0,

∑Fz = 0,

∑Mx = 0,

∑My = 0,

∑Mz = 0. (8.2)

which are the familiar six equations of equilibrium.

Internal Forces

In order for a solid not to break when external forces are applied, it needs to develop internalforces that keep it together. For example, consider the situation in Fig. 8.2. Here the bodyhas been split into two parts. If the body as a whole is in equilibrium, then each part mustalso be in equilibrium. The force resultant F and the moment resultant M represent theloads which act on the surface where the body was split. When the body is in one piecethese resultants arise from the cohesive forces in the material.

The operation of splitting a body in order to study the internal force and momentresultants is very commonly done. For example, in beam analysis we often split a beam in aplane perpendicular to its axis in order to calculate the shear force and the bending moment,which are the internal loads which keep the beam together.

F

F M

M

F1

F2

F3F4

F5

x

y

z

Figure 8.2: Detail of cohesive force resultants which keep the body together.

The force and moment resultants in Fig. 8.2 are the resultants or static equivalents ofa distributed cohesive force which acts over the areas generated by splitting the body. Thisdistributed load, of course, is measured in units of force per unit area. Now consider small(infinitesimal) elements of area as shown in Fig. 8.3. Each area element has two importantcharacteristics for our purposes: Its size, given by ∆A and its orientation, given by the

65

outward unit vector n. A force ∆F acts in each area element. This force is equal to thevalue of the distributed load acting at the location of the small area times ∆A. Now comesan important concept: The traction at a point is defined by:

n

T = lim∆A→0

∆F

∆A. (8.3)

Note thatn

T is a vector with units of force per unit area. The n on top of the T indicatesthat the traction depends on the orientation of the outward normal of the area considered.

F1

F2

F5

n

F∆A∆

Figure 8.3: Definition of traction.

If the we extract a cube from the inside of the body that has its faces aligned withthe coordinate axes, then the components of the traction vector acting on each face withnormal along the positive direction of the axes are shown in Fig. 8.4. The tractions on eachof the faces are given by:

i

T = σxi + τxyj + τxzk (8.4)j

T = τyxi + σyj + τyzk (8.5)k

T = τzxi + τzyj + σzk (8.6)

where i, j and k are unit vectors along the x, y and z axes respectively.The nine quantities σx, σy, σz, τxy, τyx, τxz, τzx, τyz and τzy define the state of stress

at a point. σx, σy and σz are called the normal stresses and τxy, τyx, τxz, τzx, τyz and τzy arecalled the shear stresses.

Subscript Order and Sign Convention

The meaning of the subscript in the normal stresses is obvious from the figure. The subscriptsin the shear stresses have the following order: The first subscript indicates the orientation ofthe face of the cube on which the shear component acts while the second subscript determinesthe direction in which it points. For example, in Fig. 8.4, τxz acts on the face that has normali and points along the z axis, but τzx acts on the face that has normal k and points alongthe x axis.


x

y

z

σx

σy

σz

τxy

τyx

τxz

τzxτyz

τzy

i

j

k

Figure 8.4: Traction components

The sign convention for the stress component is as follows. A stress component ispositive if the outward normal of the face in which it acts and the direction in which it pointshave the same sign, otherwise it is negative. All components in Fig. 8.4 are positive.

Cauchy’s Formula

Now, what happens if the orientation of the area in which we are interested does not coincidewith the coordinate directions? In this case we have the situation shown in Fig. 8.5. In thisfigure we consider an arbitrarily oriented area dA. The unit normal to this area is given by:

n = nxi + nyj + nzk (8.7)

where nx = n · i, ny = n · j and nz = n · k are the direction cosines of n. Note thatn2

x +n2y +n2

z = 1. The areas of the element in Fig. 8.5 which are aligned with the coordinateaxes are given by dAx = nxdA, dAy = nydA, dAz = nzdA and the volume of the element is

dV =

√2nxnynz

3dA3/2.

The body force B and the tractionn

T are given by

B = Bxi + Byi + Bzk (8.8)n

T =nTx i+

nTy j+

nTz k. (8.9)

Summing forces in the x direction we get:

−σxdAx − τyxdAy − τzxdAz + BxdV +nTx dA = 0

67

which, letting dA → 0, yields

nTx= σxnx + τyxny + τzxnz.

Repeating for the y and z directions,

nTy= τxynx + σyny + τzynz

nTz= τxznx + τyzny + σznz

or, in matrix form,

nTxnTynTz

=

σx τyx τzx

τxy σy τzy

τxz τyz σz

nx

ny

nz

. (8.10)

This is Cauchy’s formula. Given the state of stress at a point and the orientation of a surfacepassing through that point, it calculates the traction there.

σxσy

σz

τxy

τyx

τxz

τzx

τyz

τzy

Tn

n

dA

dA

dA

dA

x

z

y

dx

dy

dz

x

y

z

B

Figure 8.5: Element used for the derivation of Cauchy’s formula.

Note that the stress matrix represents the state of stress at a point. It is a functionof position (and time for a dynamic problem). The traction vector depends on the state ofstress and on the orientation of the plane of interest.


i

j

k

τ + τzy zy,zdz

τ + τxy xy,xdx

σ + σy y,ydy

τzy

τxy

σy By

dx

dy

dz

x

y

z

Figure 8.6: Traction components in the y direction.

Equilibrium

In order for the internal forces in the body to be in equilibrium, the stress components cannotbe arbitrary, they must satisfy the equations of equilibrium. Figure 8.6 shows the stresscomponents acting on a cube of material of dimensions dx by dy by dz which correspond tothe traction components along the y axis. The traction components in the x and z directionshave been omitted for clarity. Note that the magnitude of each stress component can changegoing from one face of the cube to another. The change can be represented using Taylorseries expansions. Note that partial derivatives have been indicated by a comma followed bya subscript. For example,

τxy,x =∂τxy

∂x.

Terms of higher order than the linear have been omitted. The component of the body forcein the y direction has also been included. Summing forces in the y direction yields

∑Fy = −σydxdz − τxydydz − τzydxdy +

(σy + σy,ydy)dxdz + (τxy + τxy,xdx) dydz + (τzy + τzy.zdz) dxdy + Bydxdydz = 0

which givesτxy,x + σy,y + τzy,z + By = 0

Equilibrium of forces in the x and z directions yields two more equilibrium equationsso that we obtain a set of three force equilibrium equations

σx,x + τyx,y + τzx,z + Bx = 0

69

c

xτ + τxy xy,xdx

τxy

Byτ + τyx yx,ydy

τyx

ydy

dx

Bx

Figure 8.7: Traction components in the x-y plane.

τxy,x + σy,y + τzy,z + By = 0 (8.11)

τxz,x + τyz,y + σz,z + Bz = 0.

Three more equilibrium equations can be obtained by considering moment equilib-rium. Figure 8.7 shows the top view of the infinitesimal block in Fig. 8.6. All tractioncomponents in the x-y plane have been identified with arrows but only those which cancontribute to a moment about the z axis have been labeled. Summing moments about thecenter of the element, point c, yields

τxydydzdx

2− τyxdxdz

dy

2+ (τxy + τxy,xdx)dydz

dx

2− (τyx + τyx,ydy)dxdz

dy

2= 0.

Dividing by dxdydz and letting dx, dy, dz → 0 gives

τxy = τyx.

Considering moment equilibrium about the x and y axes yields two more identities.In summary, moment equilibrium indicates that

τxy = τyx

τyz = τzy (8.12)

τzx = τxz.

This indicates that the stress matrix in (8.10) is symmetric, so it has six independent entries.


That is, the stress matrix can be written as

σ =

σx τxy τxz

τxy σy τyz

τxz τyz σz

(8.13)

Linear Elastic Constitutive Relations

In order to generalize Hooke’s law to three dimensions, consider the blocks depicted in Fig.8.8. Since the block in Fig. 8.8(a) has only the stress σx applied to it, the strains in theblock will be

εx =σx

Eεy = εz = −νεx = − ν

Eσx. (8.14)

Similarly, for the block in Fig. 8.8(b),

εy =σy

Eεx = εz = −νεy = − ν

Eσy, (8.15)

and for the block in Fig. 8.8(c),

εz =σz

Eεx = εy = −νεz = − ν

Eσz. (8.16)

In the case of the block in Fig. 8.8(d), the state of stress is pure shear, so no normalstrains will develop and

γxy =1

Gτxy. (8.17)

Since the material considered is isotropic, the relation between the other two shear strainsand stresses will be

γyz =1

Gτyz γxz =

1

Gτxz. (8.18)

If the material is subjected to a multi-axial state of stress, the stress-strain relationwill be given by the superposition of equations (8.14)-(8.18). Thus,

εx

εy

εz

γxy

γyz

γxz

=

1E

− νE

− νE

0 0 0− ν

E1E

− νE

0 0 0− ν

E− ν

E1E

0 0 00 0 0 1

G0 0

0 0 0 0 1G

00 0 0 0 0 1

G

σx

σy

σz

τxy

τyz

τxz

. (8.19)

Note that even though the stress was previously shown to be a 3 × 3 matrix, it is oftenconvenient to represent it as a six-dimensional vector when writing the constitutive equations.

71

y

z

x

y

z

σx

σx

x

z

σyσy

x

y

zσz

σzx

yτxy

τxy

τxy

τxy

(a)

(d)(c)

(b)

Figure 8.8: Stress-strain relations for an isotropic material.

Chapter 9

Kinematics of Solids in CartesianCoordinates

Strain Components

This section discusses the kinematics of solids in Cartesian coordinates. The objective isto find the strain-displacement relations or, in other words, the expressions for the strainsdeveloped during deformation of a solid body in terms of the displacements of the points inthe body.

Consider a reference state of a body which is defined by zero displacement as shownin Fig. 9.1. Usually, this corresponds to the position of the body when it is unloaded. Twopoints m and n define an infinitesimal line segment of length ds. The coordinates of point mare (x, y, z) while those of n are (x+dx, y +dy, z +dz). After the body is loaded it moves tothe current state, also shown in Fig.9.1. The points m and n now occupy positions m′ andn′ and the line segment between them has a new length ds∗. The coordinates of location m′

are now (x∗, y∗, z∗). The displacement components of point m in the x, y and z directionsare u, v and w respectively. The figure indicates that the displacement is due to both rigidbody motion and deformation of the body.

It is clear form Fig. 9.1 that

x∗ = x + u

y∗ = y + v (9.1)

z∗ = z + w.

Also note that

ds2 = dx2 + dy2 + dz2

ds∗2 = dx∗2 + dy∗2 + dz∗2. (9.2)

From (1.1), partial differentiation1 with respect to x, y and z gives

x∗,x = 1 + u,x x∗

,y = u,y x∗,z = u,z (9.3)

1Partial differentiation will be indicated by using a comma followed by subscripts. For example, u,x = ∂u∂x

73

74 CHAPTER 9. KINEMATICS OF SOLIDS IN CARTESIAN COORDINATES

x

y

z

m

n

m’n’

uv

w

(x,y,z)ds

ds*

(x*,y*,z*)

Reference StateCurrent State

Figure 9.1: Deformation of a body in Cartesian coordinates.

and similarly for y∗ and z∗. Also from (9.3)

dx∗ = x∗,xdx + x∗

,ydy + x∗,zdz = (1 + u,x)dx + u,ydy + u,zdz. (9.4)

A similar procedure yields the expressions for dy∗ and dz∗. Substituting these expressionsinto the equation for ds∗2 in (9.2) gives the following expression

ds∗2 − ds2 = 2εxdx2 + 2εydy2 + 2εzdz2 + 2γxydxdy + 2γxzdxdz + 2γyzdydz (9.5)

where

εx = u,x +1

2

[u2

,x + v2,x + w2

,x

]

εy = v,y +1

2

[u2

,y + v2,y + w2

,y

]

εz = w,z +1

2

[u2

,z + v2,z + w2

,z

](9.6)

γxy = u,y + v,x + u,xu,y + v,xv,y + w,xw,y

γxz = u,z + w,x + u,xu,z + v,xv,z + w,xw,z

γyz = v,z + w,y + u,yu,z + v,yv,z + w,yw,z

Note that the quantity ds∗2 − ds2 does not depend on the rigid motion of the body2 and,as a result, is a measure of the deformation of the body. The quantities εx, εy, εz, γxy, γxz,and γyz are called the components of strain. Note that these expressions are valid for largedeformations of the body since no assumptions regarding the magnitude of u, v and w weremade.

2Rigid body motion has, by definition, ds∗2 − ds2 = 0.

75

Linear Strains

In many structural engineering problems the deformations of the components under con-sideration are very small. This provides an opportunity to greatly simplify the expressions(9.6). The linear or infinitesimal strain expressions are obtained by neglecting all terms ofhigher order than linear in the displacements which gives

εx = u,x

εy = v,y

εz = w,z (9.7)

γxy = u,y + v,x

γxz = u,z + w,x

γyz = v,z + w,y.

This simplification introduces two restrictions on the deformations pictured in Fig. 9.1

1. The length of ds∗ has to be almost the same as the length of ds.

2. The orientation of ds∗ cannot be very different from the orientation of ds.

Recall from elementary mechanics of solids that the linear strain of the line mn isgiven by

ε =ds∗ − ds

ds(9.8)

which, for |ε| << 1 gives

εds2 =1

2(ds∗2 − ds2) = εxdx2 + εydy2 + εzdz2 + γxydxdy + γxzdxdz + γyzdydz.

If l, m and n are the direction cosines of ds then dx = l ds, dy = m ds and dz = n ds so that

ε = εxl2 + εym

2 + εzn2 + γxylm + γxzln + γyzmn. (9.9)

is the strain of an arbitrarily oriented line of length ds.

Compatibility Conditions

Note that if u, v and w in (9.6) are given as continuous functions of x, y and z then all straincomponents can be easily calculated. The converse, however, is not true. If εx, εy, εz, γxy,γxz, γyz are given as arbitrary functions of x, y and z then we have six equations to solvefor the three displacement components. This, in general, cannot yield a valid solution for u,v and w unless the strain components are related in some way. In other words, the strainscannot be prescribed arbitrarily.


a

b c

d

a

b c

da

b c

d

a

bc

d

(a)

(c) (d)

(b)

Figure 9.2: The significance of compatibility.

For example, Fig. 9.2(a) shows a plane body on which the triangle abcd has beenscribed. Figures 9.2(b), (c) and (d) show the same body after it has been deformed. If thestrain components are not related appropriately, then the situation in Fig. 9.2(b), where thematerial has crossed over itself or the situation in Fig. 9.2(c) where a gap has appeared mayresult. The only acceptable strain field is the one which produces smooth displacement fieldssuch as the one shown in Fig. 9.2(d).

Compatibility Conditions for Linear Strains

From equations (9.7) note that

εx,yy = u,xyy εy,xx = v,yxx γxy,xy = v,xxy + u,xyy.

If the displacement field is to be continuous and smooth, then

εx,yy + εy,xx = γxy,xy.

Similar arguments yield six compatibility equations:

εx,yy + εy,xx = γxy,xy

εz,xx + εx,zz = γxz,xz

77

εy,zz + εz,yy = γyz,yz (9.10)

(−γyz,x + γxz,y + γxy,z),x = 2εx,yz

(γyz,x − γxz,y + γxy,z),y = 2εy,zx

(γyz,x + γxz,y − γxy,z),z = 2εz,xy.

At this point it seems that the strain-displacement relations are still not balancedbecause we now have 12 equations, six form (9.7) and six from (9.10) but only nine unknowns,three displacement components and six strain components. It can be shown, however thatonly three of the compatibility equations are independent so that in reality we have nineindependent equations for nine quantities.

Chapter 10

Bending of Rectangular Plates

Introduction

Plates are flat structural members which have one dimension (the thickness, t) which ismuch smaller than the in-plane dimensions. For example, in the plate shown in Fig. 10.1,t << a and t << b, but a and b can be of the same order. For simplicity, we will considerrectangular plates only since they can be analyzed using Cartesian coordinates. The problemwhich will be considered, shown in Fig. 10.1, can be stated as follows: A flat, rectangularplate of given dimensions a, b, and t, lies in the x-y plane with its edges aligned with thecoordinate axis. The support conditions at the edges are given. The plate is made of anisotropic and homogeneous linearly elastic material with Young’s modulus E and Poisson’sratio ν and is subjected to a transverse pressure p(x, y). Find the transverse deflection ofthe plate as well as the stress and strain in the plate.

p(x,y)

a b

x,u

y,v

z,w

t

Figure 10.1: A plate under transverse loading.

79

80 CHAPTER 10. BENDING OF RECTANGULAR PLATES

Kinematics

Define the mid-surface of the plate as the surface which bisects the thickness and let itlie in the x-y plane as shown in Fig. 10.1 when the plate is unloaded. The displacementcomponents of any point in the plate along the x, y and z axes are u, v and w respectively.

The simplest set of plate kinematics are based on the following assumptions, all ofwhich have been discussed with regards to beam bending:

1. Shear deformations are negligible

2. The thickness of the plate does not change during loading

3. The plate is initially perfectly flat

4. The deflection of the plate is small

5. Plane sections originally perpendicular to the middle surface remain plane and per-pendicular to the middle surface after deformation

6. Intermediate class of deformation kinematics are sufficient to describe the motion ofpoints in the plate.

Note that assumption 5 is a direct result of assumption 1 and implies that the throughthickness shear strains will be zero. That is, γxz = γyz = 0.

Let the displacement components of the middle surface of the plate be (u, v, w). Fromassumptions 2, 5 and 6 (u.v, w) can be related to (u, v, w) as follows:

u = u − w,xz

v = v − w,yz (10.1)

w = w

The strain-displacement relations for intermediate deformations are

εx = u,x +1

2w2

,x

εy = v,y +1

2w2

,y (10.2)

γxy = u,y + v,x + w,xw,y

Substituting (10.2) into (10.3) yields the following expressions for the strain compo-nents of a point in the plate:

εx = εox + κxz

εy = εoy + κyz (10.3)

γxy = γoxy + 2κxyz

81

yx

1/κx

y

x

1/κy

Figure 10.2: Normal curvatures of a plate.

where

εox = u,x +

1

2w2

,x

εoy = v,y +

1

2w2

,y

γoxy = u,y + v,x + w,xw,y

κx = −w,xx (10.4)

κy = −w,yy

κxy = −w,xy.

The quantities εox, εo

y and γoxy are the membrane strains of the plate, while κx, κy

correspond to the curvatures of the plate as shown in Figs. 10.2 and κxy represents the twistof the plate.

Stress Resultants

The stresses in the plate produce a set of force and moment intensities resultants as shownin Fig. 10.3(a) and (b) respectively. The quantities Nx, Ny and Nxy are the in-plane forceintensities and are given by

Nx =∫ t/2

−t/2σxdz, Ny =

∫ t/2

−t/2σydz, Nxy =

∫ t/2

−t/2τxydz. (10.5)


N

M

x

x

My

Mxy

Ny

Nxy

Mxy

Nxy

Qy

Qx

Figure 10.3: Stress resultants in a plate.

The quantities Qx and Qy are the through-thickness shear force intensities and are given by

Qx =∫ t/2

−t/2τxzdz, Qy =

∫ t/2

−t/2τyzdz. (10.6)

Note that the force intensities have units of force per unit length. The quantities Mx, My

and Mxy are the moment intensities and are given by:

Mx =∫ t/2

−t/2σxz dz, My =

∫ t/2

−t/2σyz dz, Mxy =

∫ t/2

−t/2τxyz dz. (10.7)

Note that the moment intensities have units of moment per unit length.


In addition to the assumptions made in the kinematics, the through-thickness normal stressσz will be assumed to be zero. The material will be assumed to be homogeneous, isotropicand linear elastic so that Hooke’s law holds. Therefore,

83

σx =E

1 − ν2(εx + νεy)

σy =E

1 − ν2(εy + νεx) (10.8)

τxy =E

2(1 + ν)γxy

The simplest plate bending theory is obtained by restricting the cases considered toplates which carry no in-plane forces, that is to cases where Nx = Ny = Nxy = 0. Substituting(10.4) into (10.9) and into (10.5) and enforcing the condition of zero in plane-forces yieldsthe following equations

Nx =Et

1 − ν2(εo

x + νεoy) = 0

Ny =Et

1 − ν2(εo

y + νεox) = 0 (10.9)

Nxy =Et

2(1 + ν)γo

xy = 0

These three equations indicate that, in the absence of in-plane forces, εox = εo

y = γoxy =

0. Therefore, the expressions for the strain components reduce to

εx = κxz, εy = κyz, γxy = 2κxyz. (10.10)

Principle of Virtual Work

Under the current set of assumptions, the principle of virtual work is given by

∫V(σxδεx + σyδεy + τxyδγxy)dV = δWext (10.11)

where the left hand side of the equation is the internal virtual work and δWext is the externalvirtual work. V represents the volume of the plate. Substituting from (10.10) and breakingup the integral over the volume yields

∫ a

0

∫ b

0

∫ t/2

−t/2(σxzδκx + σyzδκy + 2τxyzδκxy) dz dy dx =

∫ a

0

∫ b

0pδw dy dx. (10.12)

Substituting from (10.7) yields the principle of virtual work in terms of moment intensitiesand curvatures

∫ a

0

∫ b

0(Mxδκx + Myδκy + 2Mxyδκxy − pδw) dy dx = 0 (10.13)


Substituting (10.4) into (10.9) and into (10.7) yields the following expressions for themoment intensities in terms of the curvatures

Mx = D(κx + νκy)

My = D(κy + νκx) (10.14)

Mxy = D(1 − ν)κxy

where

D =Et3

12(1 − ν2). (10.15)

Substituting (10.15) into (10.13) yields

∫ a

0

∫ b

0{D [(κx + νκy)δκx + (κy + νκx)δκy + 2(1 − ν)κxyδκxy] − pδw} dy dx = 0 (10.16)

or, substituting from (10.5),

D∫ a

0

∫ b

0[(w,xx + νw,yy)δw,xx + (w,yy + νw,xx)δw,yy + 2(1 − ν)w,xyδw,xy] dy dx −

∫ a

0

∫ b

0pδw dy dx = 0. (10.17)

Integrating the first integral by parts yields

[D∫ b

0[(w,xx + νw,yy)δw,x − (w,xx + νw,yy),xδw] dy

]a

0

+

[D∫ a

0[(w,yy + νw,xx)δw,y − (w,yy + νw,xx),yδw] dx

]b0+[

D(1 − ν)∫ b

0[w,xyδw,y − w,xyyδw] dy

]a

0

+ (10.18)

[D(1 − ν)

∫ a

0[w,xyδw,x − w,xxyδw] dx

]b0+

D∫ a

0

∫ b

0[(w,xxx + νw,yyxx + w,yyyy + νw,xxyy + 2(1 − v)w,xxyy) − p]δw dy dx = 0.

The first four lines of this equation give the boundary conditions whereas the last line givesthe governing differential equation for bending of plates by noting that δw is arbitrary. Thisdifferential equation is

D(w,xxxx + 2w,xxyy + w,yyyy) = p (10.19)

or, using the double Laplacian operator

D∇4w = p (10.20)

85

The boundary terms evaluated at the sides of the plate with x = 0 and x = a yieldthe following conditions:

D∫ b

0(w,xx + νw,yy)δw,xdy = 0 →

∫ b

oMxδw,xdy = 0.

This implies that either w,x is prescribed or Mx = 0.

D∫ b

0(w,xx + νw,yy),x + (1 − ν)w,xyy)δw dy = 0 →

∫ b

0Qxδw dy = 0.

This implies that either w is prescribed of Qx = 0

D(1 − ν)∫ b

0w,xyδw,y dy = 0 →

∫ b

0Mxyδw,y dy = 0.

This implies that either w,y is prescribed or Mxy = 0.Similar reasoning for the sides with y = 0 and y = b yields: either w,y is prescribed

or My = 0, either w is prescribed or Qy = 0 and either w,x is prescribed or Mxy = 0.Some common boundary conditions, say in the sides with x = 0 and x = a are:

Simple support: w = 0 and Mx = 0 along the edge. The first condition implies thatw,y = w,yy = 0. Therefore, the boundary conditions in terms of w are w = 0 andw,xx = 0. The first condition implies that w,y = 0.

Clamped support: Both w = 0 and w,x = 0 along the edge. Note that the first conditionimplies that w,y = 0.

Free end: Mx = Qx = Mxy = 0. This leads to an inconsistency since three independentconditions appear to be valid, but only two conditions per side can be prescribed. Thisproblem can be resolved by noticing that Qx and Mxy can be combined as shown in Fig.10.4. This figure considers a segment of an edge of length 2dy. Figure 10.4(a) showsthe original system of loads. In Fig. 10.4(b) the twisting moment has been replaced bystatically equivalent couples. Figure 10.4(c) shows the resulting situation in the centerpart of the segment, which has length dy. The resultant force intensity in this centralsegment, Rx is shown in Fig. 10.4 and is given by

Rx = Qx − Mxy,y.

Hence the boundary conditions for the free edge are

Mx = 0 → w,xx + νw,yy = 0

Rx = Qx − Mxy,y = 0 → w,xxx + (2 − ν)w,xyy = 0.


dy dy

Q

Mxy

Mxy

+ M dyxy,y

dyx

dy

( (dy

M Mxy+ M dyxy,yxy

Qx

Qx

M Mxy+ M dyxy,yxy

Qx-Mxy,y

(a)

(b)

(c)

(d)

Figure 10.4: Derivation of the equivalent transverse shear force intensity used to enforceboundary conditions at a free edge.

Notes on Advanced Mechanics of Solids

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