UNIVERSITY OF CALIFORNIA, SAN DIEGO Dept. Of Structural Engineering SE 201 – Fall 2004 Class Notes Instructor: J. P. Conte

The class notes© for this course were developed jointly by Prof. J. P. Conte at the University of California, San Diego

and Prof. E. Spacone at the University “G. D’Annunzio”, Pescara, Italy

1 BEAM THEORY - STRONG FORM

A building is made of several/different structural elements such as beams, nodes, different forms of infills, bracings, walls, etc. These elements are not necessarily all present in the same building. As for the frame skeleton itself, there are regions where classical beam theories apply, typically away from beam-column joints, and regions where classical beam theories do not apply (typically beam-column joints, corbels, etc.). The former regions are sometimes called B-regions, the latter D-regions (Schlaich et al.1). This chapter deals with the B-regions and reviews the classical Euler-Bernoulli and Timoshenko beam theories.

D B DD = DiscontinuityB = Beam

Figure 1-1 A frame and its main components

The beam theories are reviewed assuming small deformations and small displacements, thus equilibrium is enforced in the undeformed configuration. Axial, bending and torsional behaviors are studied separately.

1 Schlaich, J., Schäfer, K., and Jennewein, M. (1987). “Toward a Consistent Design of Structural Concrete.” J. PCI, 32(3), May-June, pp. 74-150.

1-2

1.1 BAR ELEMENT (AXIAL LOAD)

N

wx(x)

N+dN

dx

dx

x

Figure 1-2 Infinitesimal beam length with axial forces

Equilibrium: ( ) 0

( )

x

x

N N dN w x dxdN w xdx

− + + + =

⇒ = −

Constitutive Law: ( )N EA x ε=

Compatibility: dudx

ε =

Problem Differential Equation:

( ) ( )xd duEA x w xdx dx

⎡ ⎤ = −⎢ ⎥⎣ ⎦+ boundary conditions (essential/natural)

(1.1)

2

2 ( )xd uEA const EA w xdx

= ⇒ = −

Solution2 ( ) ( ) ( )h pu x u x u x= +

Boundary conditions must be added in order to solve the problem (i.e., determine the integration constants). Boundary solutions can be Essential or Natural. Essential b.c. are geometric restraints on the

2 Reminder: Given the general linear differential equation (DE) of order n

1

1( , , , , , ) ( )n n

n n

d d dD u c f xdx dx dx

−

− = , where ( )u x is the unknown function, c is a constant or a series of

constants, ( )f x is a known function, the solution is ( ) ( ) ( )h pu x u x u x= + , where:

( )hu x = general solution of the homogeneous DE 1

1( , , , , , ) 0n n

n n

d d dD u cdx dx dx

−

− =

( )pu x = particular solution of the non-homogeneous DE 1

1( , , , , , ) ( )n n

n n

d d dD u c f xdx dx dx

−

− =

1-3

beam ends, while Natural b.c. are force restraints on the beam ends. At a given degree of freedom, we have either one of the two conditions. In the case of a bar we have

(0) ou u= or (0) oN N= and

( ) Lu L u= or ( ) LN L N=

In civil structures, we cannot have two Natural b.c., otherwise we have a mechanism!!!! In aerospace engineering, the problem is different …

Compatibility

( )u x ( )xw x

ε N

Equilibrium

Constitutive Law

( )N EA x= ε

= du dxε ( )xdN dx w x= −Compatibility

( )u x ( )xw x

ε N

Equilibrium

Constitutive Law

( )N EA x= ε

= du dxε ( )xdN dx w x= −

Figure 1-3 Tonti’s diagram for bar problem3

3 Prof. Tonti’s webpage is www.dic.univ.trieste.it/perspage/tonti

1-4

Compatibility

( )u x ( )xw x

ε N

Equilibrium

Constitutive Law

( )N EA x= ε

= du dxε ( )xdN dx w x= −

Essential b.c. on Γu

Natural b.c. on Γt

Compatibility

( )u x ( )xw x

ε N

Equilibrium

Constitutive Law

( )N EA x= ε

= du dxε ( )xdN dx w x= −

Essential b.c. on Γu

Natural b.c. on Γt

Figure 1-4 Tonti’s diagram for bar problem + boundary conditions

Ω

Γu

Γt

2D body 1D body

Γu

Figure 1-5 Illustrations of main body Ω, essential boundary conditions on Γu and natural boundary conditions on Γt ( u tΓ = Γ ∪Γ )

1-5

1.1.1 EXAMPLE: Cable under its own weight (constant cross section)

2 2

2 2

21 2

( )( )

( ) ( )

( )2

hp

x

u xu x

w x A material weight density

d u d uEA Adx dx E

u x C x C xE

= =

= − ⇒ = −

= + −

ρ ρ

ρρ

ρ

Boundary conditions:

2

1

(0) 0 0 ( )

( ) ( )

u C essentialP LN L P C natural

EA Eρ

= ⇒ =

= ⇒ = +

L

x

w=ρA

P

2( ) 2

( ) ' ( )

P Lu x x xEA E E

N x EAu P A L x

ρ ρ

ρ

⎛ ⎞= + −⎜ ⎟⎝ ⎠= = + −

Note the two cases in which P = 0 and w = 0.

1.1.2 EXAMPLE: Bar element axial stiffness matrix, barK

U1, N1 U2, N2

Figure 1-6 Two-node bar element: nodal displacements and nodal forces

For the case of a bar with constant EA, the bar kinematics is completely described by the end (nodal) displacements and the bar statics is completely described by the nodal forces (shown in Figure 1-6). This implies that given the nodal forces and displacements, the bar behavior is fully determined. It is important in this case to find the bar element stiffness matrix barK that relates end forces and end displacements:

bar=P K U

or

1 111 12

21 222 2

N Uk kk kN U

⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ ⎩ ⎭

(1.2)

1-6

Column 1 of barK : U1 = 1 U2 = 0

2

2 0 ( )

(0) 11 ( ) 0

d u u x Ax Bdx

u B

u L AL B AL

= ⇒ = +

= =

= = + ⇒ = −

11 21

( ) 1

( ) ' ;

xu xL

EA EA EAN x EAu k kL L L

= −

= = − ⇒ = = −

Column 2 of barK : U1 = 0 U2 = 1

(0) 01 ( ) 1

u B

u L AL AL

= =

= = ⇒ =

12 22

( )

( ) ' ;

xu xL

EA EA EAN x EAu k kL L L

=

= = ⇒ = − =

1 11 1bar

EAL

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

K (1.3)

1-7

1.2 TORSION

This problem is formally identical to that of the bar.

T T+dT

dx

mx(x)

Figure 1-7 Bar infinitesimal length

Equilibrium: ( ) 0

( )

x

x

T T dT m x dxdT m xdx

− + + + =

⇒ = −

Constitutive Law4: ( )T GJ x β=

Compatibility5: ddxθβ =

Problem Differential Equation: ( ) ( )x

d dGJ x m xdx dx

θ⎡ ⎤ = −⎢ ⎥⎣ ⎦

+ boundary conditions

(1.4)

2

2 ( )xdGJ const GJ m xdxθ

= ⇒ = −

Solution: ( ) ( ) ( )p hx x xθ θ θ= +

4 ( )GJ x = torsional stiffness. For solid cross-sections: 2( )

A

J x dA= ∫ ρ = polar moment of inertia; for

closed tubular type cross-sections: 2( ) 4 mdsJ x At

= ∫ where mA is the mean area enclosed within the

boundary of the centerline of the tube’s thickness and t is the tube’s thickness; and for open tubular type

cross-sections: ( ) 313 i i

i

J x b t= ∑ (see Gere and Timoshenko).

5 β = twist angle per unit length (rate of twist).

1-8

1.2.1 EXAMPLE: Cantilever beam subjected to uniform distributed torque.

2 2

2 2

21 2

( )( )

( )

( )2

hp

x x

xx

x

xx

m x m

md dGJ mdx dx GJ

mx C x C xGJ

θθ

θ θ

θ

=

= − ⇒ = −

= + −

mx(x) = mx

Boundary conditions:

2

1

(0) 0 0 ( )

( ) 0 ( )xx L

C essentialm LdT L GJ C natural

dx GJ

θθ

=

= ⇒ =

= = ⇒ =

( )

( ) ( )

212

x

x

mθ x = Lx xGJ

T x GJ m L xθ

⎛ ⎞−⎜ ⎟⎝ ⎠

= = −′

1.2.2 EXAMPLE: Bar element torsional stiffness matrix: torqueK

θ1, T1 θ2, T2

Figure 1-8 Two-node bar element: torsional nodal displacements and nodal forces

torque=P K U

or

1 111 12

21 222 2

T k kk kT

θθ

⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ ⎩ ⎭

(1.5)

1-9

Column 1 of torqueK : θ1 = 1 θ2 = 0

2

2 0 ( )

(0) 11 ( ) 0

d x Ax Bdx

B

L AL B AL

θ θ

θ

θ

= ⇒ = +

= =

= = + ⇒ = −

11 21

( ) 1

( ) ' ;

xxL

GJ GJ GJT x GJ k kL L L

= −

= = − ⇒ = = −

θ

θ

Column 2 of torqueK : θ1 = 0 θ2 =1

(0) 01 ( ) 1

B

L AL AL

θ

θ

= =

= = ⇒ =

12 22

( )

( ) ' ;

xxL

GJ GJ GJT x GJ k kL L L

=

= = ⇒ = − =

θ

θ

1 11 1torque

GJL

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

K (1.6)

1-10

1.3 FLEXURE

Two theories are discussed here: the Euler-Bernoulli beam theory, which considers flexural deformations only, and the Timoshenko beam theory, which considers both flexural and shear deformations. These theories are widely used to develop beam elements in finite element books and codes.

x

a

b

reference axis

b'

a'

y

u(x,y)

v(x,y) v0(x)

u0(x)

v(x,y) = v0 ( x )

Figure 1-9 Displacements of cross section points for beam

1.3.1 Euler-Bernoulli beam theory

The fundamental assumption of the Euler Bernoulli beam theory is that plane sections remain plane and normal to the longitudinal axis of the beam. This is shown in Figure 1-10. The cross section ab is normal to the longitudinal axis of the underformed beam. In the deformed configuration, the deformed cross section a’b’ is plane and normal to the longitudinal axis of the deformed beam axis. This implies that the displacements at a point at a distance y from the longitudinal axis are:

oo

o

dvu u ydx

v v

= −

=

The corresponding deformations are:

2

2

0

o oo

o o

du d vdu y ydx dx dx

dv dvdu dvdy dx dx dx

= = − = −

= + = − + =

ε ε κ

γ (1.7)

where oo

dudx

ε = is the strain at the reference axis and 2

2od v

dxκ = is the section curvature.

1-11

x

y

dvdx

0

a

bu (x) 0

v (x)0

Undeformed

Deformed b'

a'

Figure 1-10 Deformation of Euler-Bernoulli beam

yw (x)

dx

M

V V + dV

M + dM

yw (x)

dx

M

V V + dV

M + dM

Figure 1-11 Beam infinitesimal length

Equilibrium:

( )( )

2

2

00

0

⎫+ = ⎪⎪⇒ − =⎬⎪+ =⎪⎭

y

y

dV w x d Mdx w xdM dxVdx

Constitutive law: ( )M EI x κ=

Compatibility: 2

2od v

dxκ =

Differential equation: 22

2 2( ) ( )oy

d vd EI x w xdx dx

⎛ ⎞=⎜ ⎟

⎝ ⎠ (1.8)

1-12

if . EI const=4

4 ( )oy

d vEI w xdx

⇒ =

The differential equations that govern the Euler-Bernoulli beam is of the 4th order in the unknown function

( )ov x . Four boundary conditions (essential or natural) must be added to the above differential equations in order to find the answer to a given problem. At least two of these must be essential, otherwise we have a mechanism.

Compatibility

( )0v x ( )yw x

κ M

Equilibrium

Constitutive Law

( )M EI x= κ

220= d v dxκ ( )2 2 0− =yd M dx w x

Essential b.c. on Γu

Natural b.c. on Γt

Compatibility

( )0v x ( )yw x

κ M

Equilibrium

Constitutive Law

( )M EI x= κ

220= d v dxκ ( )2 2 0− =yd M dx w x

Essential b.c. on Γu

Natural b.c. on Γt

Figure 1-12 Tonti’s diagram for Euler-Bernoulli beam problem

1-13

1.3.2 EXAMPLE: Euler-Bernoulli cantilever beam under constant distributed load (constant cross section)

w

L

x

y

Figure 1-13 Cantilever beam under constant distributed load

The problem differential equation is: 4

4 ( )oy

d vEI w x wdx

= = −

Boundary conditions:

(1) ( )0 0ov = essential

(2) ( )0 0odvdx

= essential

(3) ( ) 0M L = natural

2 2

2 2

( ) ( )0 0o od v L d v LEIdx dx

= ⇒ =

(4) ( ) 0V L = natural

( ) ( ) 3 3

3 3

( ) ( ) 0 0 o odM L d v L d v LV L EIdx dx dx

= − = − = ⇒ =

Solution:

( ) ( ) ( )h po o ov x v x v x= +

Homogeneous solution ( )hov x :

4

4 0od vEIdx

=

( ) 2 31 2 3 3

hov x C C x C x C x= + + +

Particular solution ( )pov x :

4

4

pod vEI w

dx= −

1-14

4( )24

po

wv x xEI

= −

⇒ 2 3 41 2 3 4( )

24owv x C C x C x C x xEI

= + + + −

The integration constants are determined from the four boundary conditions:

(1) ( )0 0ov = 1 0C⇒ =

(2) ( )0 0odvdx

= 2 0C⇒ =

(3) 2

2

( ) 0 od v Ldx

= 23 4 2 6 0

2wC C L LEI

⇒ + − =

(4) 3

3

( ) 0 od v Ldx

= 4 41 6 0 6

w wLC L L CEI EI

⇒ − = ⇒ =

From (3), 2

3 04wLC

EI= − =

( )2

2 3 414 6 24o

w L Lv x x x xEI

⎛ ⎞= − + −⎜ ⎟⎜ ⎟

⎝ ⎠

In particular, the tip vertical displacement is:

( )4

8owLv LEI

= − 4

8wLEI

The bending moment along the beam is:

( ) ( )2

2

2 2

2 2

odv xM x EI

dxx Lw Lx

=

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

w

++ --

2

2wL

M(x)

The shear force along the beam is:

1-15

( ) ( )

( )

3

3odv xdMV x EI

dx dxw x L

= − = −

= −

wL

V(x)

w

--++

1-16

1.3.3 EXAMPLE: Euler-Bernoulli beam element stiffness matrix EB beam−K

θ1 θ2

Figure 1-14 Two-node beam element: flexural degrees of freedom only

EB beam−=P K U

or

1 11 12 1

2 21 22 2

M K KM K K

θθ

⎧ ⎫ ⎡ ⎤ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎩ ⎭ ⎣ ⎦ ⎩ ⎭ (1.9)

Column 1 of EB beam−K θ1=1θ2=0

11K 21K

42 3

1 2 3 44 0 ( )oo

d vEI v x C C x C x C xdx

= ⇒ = + + +

(1) ( )0 0ov = 1 0C⇒ =

(2) ( )0 1odvdx

= 2 1C⇒ =

(3) ( ) 0ov L = 2 33 4 L+ 0C L C L⇒ + =

(4) ( ) 0odv Ldx

= 23 4 1+2 3 0C L C L⇒ + =

3 4 2

2 1 C CL L

⇒ = − =

2

( ) 1oxv x xL

⎛ ⎞= −⎜ ⎟⎝ ⎠

(cubic Hermitian polynomial)

1-17

Bending moment6:

2

2 2

4 6( ) od vM x EI EI xdx L L

⎛ ⎞= = − +⎜ ⎟⎝ ⎠

θ1=1θ2=04EI

L2EI

L

4EIL

−

2EIL

( )M x

Column 2 of EB beam−K θ1=0

θ2=112K 22K

This case is symmetric to the case of column 1:

θ1=0θ2=1

2EIL

4EIL

2 121 2EB beam

EIL−

⎡ ⎤= ⎢ ⎥

⎣ ⎦K (1.10)

6 Note that there are two different conventions for the bending moment diagram ( )M x , shown in (a), and

for the end moments 1 2,M M , shown in (b).

++++

(a) (b)

1-18

1.3.4 Pure Shear Beam

a'

b'

b

a

x

yyw (x)

ov (x)a'

b'

b

a

x

yyw (x)

ov (x)

yw (x)

dx

M

V V + dV

M + dM

γ

γ

yw (x)

dx

M

V V + dV

M + dM

γ

γ

Figure 1-15 Beam element with shear deformations only

Equilibrium:

( ) ( ) 0

( )

y

y

V V dV w x dx

dV w xdx

− + + + =

⇒ = − (1.11)

Constitutive Law: ( ) ( ) ( )sV x GA x xγ= (1.12)

Compatibility:

( ) odvxdx

γ = (1.13)

1-19

Governing differential equation:

0( ) ( )s y

d dvGA x w xdx dx

⎡ ⎤ = −⎢ ⎥⎣ ⎦ (1.14)

If GAs = constant, then

2

0

2 ( )s yd vGA w xdx

= − (1.15)

Solution: vo(x) Note: The above governing differential equation could have been obtained by de-generating

the theory for the Timoshenko beam for the particular case.

(infinite flexural stiffness rigid in bending)EI = ∞ ⇔

0 (zero curvature zero flexural deformations)κ = ⇔

0 ) . 0oo

d x constdxακ α= = ⇒ ( = =

( ) ( )os y

dvd GA x w xdx dx

⎡ ⎤⇒ = −⎢ ⎥⎣ ⎦

( ) ( )( ) ( )

indeterminate

o os

dv dd dM xGA x V x EI xdx dx dx dx

⎛ ⎞⎜ ⎟= = − = −⎜ ⎟⎜ ⎟∞⎝ ⎠

α

κ

(1.16)

1.3.4.1 Application Example

depends on kinematic boundary conditions

1-20

1.3.5 Timoshenko beam theory

1.3.5.1 Theoretical background

The Timoshenko beam theory represents a simplification of more precise beam theories that account for shear deformations.

γ

Deformed shape according toTimoshenko theory

Deformed shape according toJourawsky theory

x

γ

Deformed shape according toTimoshenko theory

Deformed shape according toJourawsky theory

x

Figure 1-16 Cross section deformation

The fundamental assumption of the Timoshenko beam theory is that plane sections remain plane, but no longer perpendicular to the beam axis due to the shear deformation. Figure 1-16 shows the implication of such assumption. The cross section remains plane (contrarily to what happens according to Jourawsky’s theory) and the shear deformation γ is constant over the cross section. The resulting stress distribution (constant) is compared to that obtained from Jourawsky’s theory (labeled “exact”) in Figure 1-17.

SHEAR STRESSES IN RECTANGULAR SECTION

b

h

y

τ xyV

bhh y= −FHG

IKJ

643

22

"Exact" Theory Timoshenko Beam Theory

xys

VconstA

τ = =

τ xy τ xy

x

Figure 1-17 Shear stresses in rectangular cross section: “Exact” vs Timoshenko beam theory stress distributions

1-21

The essence of Timoshenko argument (Graff, Wave motion in Elastic Solids, Dover, p. 182) is as follows. The shear force V is given by the integral of the shear stresses over the cross-section. In the linear elastic case,

G=const

A A A

V dA G dA G dAτ γ γ= = =∫ ∫ ∫ (1.17)

if oγ is the shear strain at the centroidal axis, it follows that

oV G Aγ< Thus,

( )oA

V G dA G A kγ γ= =∫ where 1k <

The adjustment coefficient k is usually designated as the (Timoshenko) shear coefficient. It depends on the shape of the cross-section and must be determined, usually through stress analysis, for each cross section type. If we define sA kA= as the shear area, then

s oV GA γ=

The values of k for some simple cross sections are:

56

k = for rectangular cross section; 9

10k = for circular cross section

The vertical displacement of the beam reference axis is the sum of the flexural and shear deflections, as shown in Figure 1-18:

( ) ( ) ( )o f sv x v x v x= + (1.18)

where fv is the deflection due to flexure only, and sv is the deflection due to shear only.

( )sv x

( )fv x

( ) ( ) ( )o f sv x v x v x= +x

( )sv x

( )fv x

( ) ( ) ( )o f sv x v x v x= +x

Figure 1-18 Displacements of Timoshenko beam

1-22

It follows that

fo s

o

dvdv dvdx dx dx

α γ

= + (1.19)

This implies that the total slope of the beam axis, odv dx , is the sum of the flexural rotation fo

dvdx

α =

and of the shear rotation sdvdx

γ = . The sign convention for the flexural and the shear deformations are

shown in Figure 1-19, and they follow the sign convention of the corresponding forces shown in Figure 1-11.

combined deformation

section curvature section shear deformation

κκ

κ

+++γ

+γ

+++

Figure 1-19 Positive flexural and shear deformations

It follows that in the deformed configuration the deformed beam cross section remains plane but rotates by an angle ( )o xα , which is different from the rotation of the beam axis, o odv dx α γ= + , due to the effect of shear deformation. Euler-Bernoulli beam theory neglects the effect of shear deformation,

sV GAγ = , and therefore assumes that the beam axis and the beam cross section rotate by the same amount.

1-23

The horizontal and vertical displacement fields are given by:

o ou u yα= −

ov v=

The corresponding deformations/strains are:

o o

oo

du ddu ydx dx dx

dvdu dvdy dx dx

αε

γ α

= = −

= + = − + (1.20)

The section generalized deformations are:

oo

o

oo

dudx

ddx

dvdx

ε

ακ

γ α

=

=

= −

(1.21)

From the constitutive law (Hooke’s law) Eσ ε= and Gτ γ= (1.22)

The section generalized forces are:

oo

A

o

A

sA

duN dA EA EAdx

dM ydA EI EIdx

V dA GA

σ ε

ασ κ

τ γ

= = =

= − = =

= =

∫

∫

∫

(1.23)

1.3.5.2 Differential equations

Equilibrium:

( ) 0

0

ydV w xdxdM Vdx

+ =

+ = (1.24)

1-24

Constitutive laws:

( ) ( ) ( )( ) ( ) ( )s

M x EI x x

V x GA x x

κ

γ

=

= (1.25)

Compatibility:

( ) ( )

( ) ( ) ( )

o

oo

d xx

dxdv x

x xdx

ακ

γ α

=

= −

(1.26)

Differential equations:

( ) 0

0

os o y

o os o

dvd GA w xdx dx

dv ddGA EIdx dx dx

α

αα

⎛ ⎞⎛ ⎞− + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞− + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(1.27)

if .EI const= and .sGA const= (uniform prismatic beam)

( )2

2

2

2

0

0

o os y

o os o

d v dGA w xdx dx

dv dGA EIdx dx

α

αα

⎛ ⎞− + =⎜ ⎟

⎝ ⎠

⎛ ⎞− + =⎜ ⎟⎝ ⎠

(1.28)

There are two coupled differential equations that govern the Timoshenko beam. They are both of 2nd order in the unknown functions ( )ov x and ( )o xα . Four boundary conditions (essential and/or natural) must be added to the above differential equations in order to find the answer to a given problem. At least two of these must be essential, otherwise we have a mechanism.

1-25

Compatibility

( )( )

0

0

v x

xα ( )yw x

,κ γ ,M V

Equilibrium

Constitutive Law

( )( )s

M EI x

V GA x

κ

γ

=

=

( )= − = −ydV dMw x Vdx dx

Essential b.c. on Γu

Natural b.c. on Γt

0 00= = −

d dvdx dxα

κ γ α

Compatibility

( )( )

0

0

v x

xα ( )yw x

,κ γ ,M V

Equilibrium

Constitutive Law

( )( )s

M EI x

V GA x

κ

γ

=

=

( )= − = −ydV dMw x Vdx dx

Essential b.c. on Γu

Natural b.c. on Γt

0 00= = −

d dvdx dxα

κ γ α

Figure 1-20 Tonti’s diagram for Timoshenko beam problem

1-26

x

a'

b'

Undeformed

Deformed

( )0 xα

a

b( )0u x

( )0v x

( )0 xα

Timoshenko beam theory:plane sections remain planebut not normal to beam axis

x

direction of

deformed section

slope of cross-section dueto bending deformation

shear deformation γ

deformed beam axisw/o shear deformation

( )0 xα

deformed beam axiswith shear deformation

0dv dx

total slope of beam axis0 :dv dx

Figure 1-21 Kinematics of Timoshenko Beam Theory

1-27

Figure 1-22 Formulas for shear coefficient ( ⋅sA = k A ); ν = Poisson’s ratio; neutral axis is shown as a chain-dotted line (Cowper, 1966)

1-28

1.3.6 EXAMPLE: Timoshenko cantilever beam under tip load (constant cross section)

P

x

y

L

E, G, I, As

P

x

y

L

E, G, I, As

Figure 1-23 Cantilever beam with load P at end

The problem differential equations are:

( )2

2

1 0o oy

s

d v d w xdx dx GA

α− = − = (1.29)

2

2 0o os o

dv dGA EIdx dx

αα⎛ ⎞− + =⎜ ⎟⎝ ⎠

(1.30)

Boundary conditions:

(1) ( )0 0ov = essential

(2) ( )0 0oα = essential (Note that ( )0

0odvdx

≠ !)

(3) ( ) 0M L = natural

( ) ( )0 0o od L d LEI

dx dxα α

= ⇒ =

(4) ( )V L P= natural

( ) ( ) ( )os s o

dvGA L GA L L Pdx

γ α⎛ ⎞= − =⎜ ⎟⎝ ⎠

From Eq. (1.30)

2

2o o

os

dv dEIdx GA dx

αα= − (1.31)

Substitute into (1.29) to obtain 3 3

3 30 0o o o os

s

d d d dEIGAdx GA dxdx dxα α α α⎛ ⎞

− − = ⇒ =⎜ ⎟⎜ ⎟⎝ ⎠

1-29

21 2 3( )o x C C x C xα = + +

From (1.31)

23 1 2 3

2o

s

dv EI C C C x C xdx GA

= − + + +

Integrating with respect to x:

2 34 3 1 2 3

2 1 12 3o

s

EIv C C C x C x C xGA

⎛ ⎞= + − + + +⎜ ⎟

⎝ ⎠

Determine the integration constants by applying the boundary conditions:

(1) ( )0 0ov = 4 0C⇒ =

(2) ( )0 0oα = 1 0C⇒ =

(3) ( ) 0od Ldx

α= 2 3 2 0C C L⇒ + =

(4) ( ) ( )os o

dvGA L L Pdx

α⎛ ⎞− =⎜ ⎟⎝ ⎠

3 3 2 2

PEIC P CEI

⇒ − = ⇒ = −

2 PLCEI

⇒ =

Solution:

( )

( )

2 3

2

2 6

2

os

o

P PL Pv x x x xGA EI EIPL Px x xEI EI

α

⎧ = + −⎪⎪⎨⎪ = −⎪⎩

In the above solution

Contribution of shear deformations: ( )ss

Pv x xGA

=

Contribution of flexural deformations: ( ) 2 3

2 6fPL Pv x x xEI EI

= −

In particular:

( )3

3os

duedue toto flexureshear

PL PLv LGA EI

= + (1.32)

( )2

2oPLLEI

α =

1-30

( ) 2

2o

sduedue toto flexureshear

dv L P PLdx GA EI

= +

(see figures on page 1-32)

Bending moment along beam axis:

( ) ( )odM x EI P L xdxα

= = −

PL

PLP

P

M(x)PL

PLP

P

M(x)

Shear force along beam axis:

( ) os o

dvV x GA Pdx

α⎛ ⎞= − =⎜ ⎟⎝ ⎠

( ) 2 3

2 6os

P PL Pv x x x xGA EI EI

= + −

P

PLP

P

V(x)P

PLP

P

V(x)

In order to better understand the relative importance of shear and flexural deformations, we study Eq. (1.32) in more detail. Consider a b h× rectangular section and express the beam depth h as a function of the beam length, h f L= , with 0 1f< < . For values of f close to 1 and larger than 1, beam theories do not apply any more. We are in this case dealing with discontinuous regions where other theories must be considered. With this notation, we have sA kbh= (with k defined in Section 1.3.5.1)

and 3 12I bh= . Thus,

1-31

( ) ( )( )3

1 23

2 1 4 ;2 1o

P Ev L GE kbf bf

c cPE f f

νν

⎛ ⎞+⎛ ⎞= + =⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

⎛ ⎞= +⎜ ⎟

⎝ ⎠

(1.33)

where ( )

1

2 1c

kbν+

= (ν = Poisson’s ratio), 24cb

= are two constants. It follows that when the beam is

very “slender”, f is very small and the shear contribution to ( )ov L (represented by the first term in Eq. (1.33)) can be neglected with respect to the flexural term (the second term in Eq. (1.33)). In this case, the solution given by Timoshenko beam theory “degenerates” into that given by Euler-Bernoulli beam theory. When the beam becomes “short”, that is when f approaches 1, the two terms in Eq. (1.33) are comparable and the use of the Timoshenko beam theory is justified.

1-32

2

2oPLEI

α =

s

Flexural deformations only, i.e., EI finite GA = ∞

Bernoulli-EulerBeam:

P

3

3oPLvEI

=

2

2odv PL

dx EI=

2

2oPLEI

α =

s

Flexural deformations only, i.e., EI finite GA = ∞

Bernoulli-EulerBeam:

P

3

3oPLvEI

=

2

2odv PL

dx EI=

Figure 1-24 Bernoulli-Euler beam with flexural deformations only

oα = 0

s

Shear deformations only, i.e., EI = GA finite

∞(Pure) ShearBeam:

s

PGA

γ = P

o

s

dv Pdx GA

=

os

PLvGA

=

oα = 0

s

Shear deformations only, i.e., EI = GA finite

∞(Pure) ShearBeam:

s

PGA

γ = P

o

s

dv Pdx GA

=

os

PLvGA

=

Figure 1-25 Pure shear beam with shear deformations only

2

2oPLEI

α =

s

Flexural and shear deformations, i.e., EI finite GA finite

TimoshenkoBeam:

P

3

3os

PL PLvEI GA

= +

2

2o

s

dv PL Pdx EI GA

= +

higher-order beam theories

2

2oPLEI

α =

s

Flexural and shear deformations, i.e., EI finite GA finite

TimoshenkoBeam:

P

3

3os

PL PLvEI GA

= +

2

2o

s

dv PL Pdx EI GA

= +

higher-order beam theories

Figure 1-26 Timoshenko beam with flexural and shear deformations

bK

sK

bK sK

1 1 1

eq b sK K K= + b

eqs

KK

K<⎧⎨<⎩

Two sources of flexibility: shear and flexural

bK

sK

bK sK

1 1 1

eq b sK K K= + b

eqs

KK

K<⎧⎨<⎩

Two sources of flexibility: shear and flexural

1-33

1.3.7 EXAMPLE: Timoshenko beam element stiffness matrix T beam−K

α1 α2

Figure 1-27 Two-node Timoshenko beam element: flexural degrees of freedom only

T beam−=P K U or

1 11 12 1

2 21 22 2

M K KM K K

αα

⎧ ⎫ ⎡ ⎤ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎩ ⎭ ⎣ ⎦ ⎩ ⎭ (1.34)

NOTE: 1α and 2α are the rotations of the end cross-sections and not of the neutral/reference axis at the

end of the beam!

Column 1 of T beam−K : α1=1α2=0

11K 21K

( )2

2

1 0o oy

s

d v d w xdx dx GA

− = − =α

2

2 0o os o

dv dGA EIdx dx

αα⎛ ⎞− + =⎜ ⎟⎝ ⎠

Solution:

( )

2 34 3 1 2 3

21 2 3

2 1 12 3o

s

o

EIv C C C x C x C xGA

x C C x C xα

⎧ ⎛ ⎞= + − + + +⎪ ⎜ ⎟

⎨ ⎝ ⎠⎪ = + +⎩

Boundary conditions:

(1) ( )0 0ov = 4 0C⇒ =

(2) ( )0 1oα = 1 1C⇒ =

(3) ( ) 0ov L = 2 33 2 3

2 1 11 02 3s

EI C L C L C LGA

⎛ ⎞⇒ − + + + =⎜ ⎟⎝ ⎠

(4) ( ) 0o Lα = 22 3 1+ 0C L C L⇒ + =

1-34

3 2

12 2

6 s

CL EI

GA

⇒ =+

2 21 2

26 s

LC

L EILGA

⇒ = − −+

22 2

11 2 2( ) 12 2

6 6

o

s s

Lx x x

L EI L EILGA GA

α

⎛ ⎞⎜ ⎟⎜ ⎟= − + +⎜ ⎟+ +⎜ ⎟⎝ ⎠

It follows that the bending moment over the beam axis is

2 2

( )( )

1 12 2 2

6 6

o

s s

d xM x EIdx

LEI x

L EI L EILGA GA

α=

⎛ ⎞⎜ ⎟⎜ ⎟= − − +⎜ ⎟+ +⎜ ⎟⎝ ⎠

thus,

11 1 2

21 2 2

1 2(0)2

6

1 2( )2 2

6 6

s

s s

LK M M EI

L EILGA

L LK M L EIL EI L EIL

GA GA

⎛ ⎞⎜ ⎟⎜ ⎟= = − = +⎜ ⎟+⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟= = − − +⎜ ⎟+ +⎜ ⎟⎝ ⎠

Upon introduction of

2

12

s

EIGA L

Φ = (1.35)

a dimensionless measure of the flexural-to-shear stiffness ratio, the above expressions simplify to

1-35

( )

( )

11

21

4(1 )

2(1 )

EIK

LEI

KL

+Φ=

+Φ

−Φ=

+Φ

Column 2 of T beam−K : α1=0

α2=112K 22K

From symmetry, it is easy to verify that

( )

( )

12

22

2(1 )

4(1 )

EIK

LEI

KL

−Φ=

+Φ

+Φ=

+Φ

It follows that the stiffness matrix is

( )

( ) ( )( ) ( )4 22 41T beam

EIL−

⎡ ⎤+Φ −Φ= ⎢ ⎥−Φ +Φ+Φ ⎣ ⎦

K (1.36)

The above expression for T beam−K allows to arrive at conclusions similar to those drawn for the example in Section 1.3.6. If the beam is very slender, the shear stiffness is very large relative to the bending stiffness, then 0Φ→ and the stiffness matrix of the Timoshenko beam approaches that of the Euler-Bernoulli beam. At the limit, as 0Φ = , the two matrices coincide.

4 20

2 4K K− −

⎡ ⎤Φ = ⇒ = = ⎢ ⎥

⎣ ⎦T beam EB beam

EIL

1-36

PP

Figure 1-28 Beam with load P at midspan: flexural deformations only

PP

Figure 1-29 Beam with load P at midspan: shear deformations only