Introduction to Timoshenko Beam Theory Aamer Haque Abstract Timoshenko beam theory includes the effect of shear deformation which is ignored in Euler-Bernoulli beam theory. An elementary derivation is provided for Timoshenko beam theory. Energy principles, the stiffness matrix, and Green’s functions are formulated. Solutions are provided for some common beam problems. 1. Formulation 1.1. Introduction The goal of beam theory is to simplify the equations of solid mechanics to beams. All loads on a beam act parallel to an axis that is transverse to its longitudinal axis. The length of a beam is much larger than its width and depth. Beam theory provides equations for the deflection and internal forces of the beam. Figure 1 shows a typical deflection of a beam that is loaded from above. We assume that the initial unloaded beam’s longitudinal axis coincides with the x axis. The sign convention for the deflection u is that positive deflections have positive y values. The deflection u and angle of deflection θ are assumed to be small. Applied loads are considered positive if they act in the direction of the positive y axis. Euler-Bernoulli beam theory is extensively applied to structural analysis and design. Most design guides and manuals will exclusively use Euler-Bernoulli beams. The derivation of Euler-Bernoulli beam theory is given in: Hibbeler [3, 4], Hjelmstad [5], Kassimali [7], Krenk [9]. The problem with Euler-Bernoulli beam theory is that it is inaccurate for deep beams. Deep beams are ones in which the depth is not negligible compared to the length. Either complete solid mechanics or a more accurate beam theory is required for deep beams. Timoshenko beam theory is a simple extension to Euler-Bernoulli beam theory. Shear deformations, which are absent in Euler-Bernoulli beam theory, are included in Timoshenko beam theory. The basic assumptions for Timoshenko beam theory are: • The longitudinal axis of the unloaded beam is straight. • All applied loads act transverse to the longitudinal axis. • All deformations and strains are small. • Hooke’s law can be used to relate stresses and strains. • Plane cross sections, which are initially normal to the longitudinal axis, will remain plane after deformation. The last requirement differs from Euler-Bernoulli theory which requires that the plane cross sections also remain normal to the beam axis after deformation. For Timoshenko beams, plane cross sections will rotate due to shear forces. Figure 1: Deflection of a beam Preprint submitted to Elsevier October 26, 2016
Transcript
Introduction to Timoshenko Beam Theory
Aamer Haque
Abstract
Timoshenko beam theory includes the effect of shear deformation which is ignored in Euler-Bernoulli beam theory. Anelementary derivation is provided for Timoshenko beam theory. Energy principles, the stiffness matrix, and Green’s functionsare formulated. Solutions are provided for some common beam problems.
1. Formulation
1.1. Introduction
The goal of beam theory is to simplify the equations of solid mechanics to beams. All loads on a beam act parallel toan axis that is transverse to its longitudinal axis. The length of a beam is much larger than its width and depth. Beamtheory provides equations for the deflection and internal forces of the beam. Figure 1 shows a typical deflection of a beamthat is loaded from above. We assume that the initial unloaded beam’s longitudinal axis coincides with the x axis. The signconvention for the deflection u is that positive deflections have positive y values. The deflection u and angle of deflection θare assumed to be small. Applied loads are considered positive if they act in the direction of the positive y axis.
Euler-Bernoulli beam theory is extensively applied to structural analysis and design. Most design guides and manuals willexclusively use Euler-Bernoulli beams. The derivation of Euler-Bernoulli beam theory is given in: Hibbeler [3, 4], Hjelmstad[5], Kassimali [7], Krenk [9].
The problem with Euler-Bernoulli beam theory is that it is inaccurate for deep beams. Deep beams are ones in which thedepth is not negligible compared to the length. Either complete solid mechanics or a more accurate beam theory is requiredfor deep beams. Timoshenko beam theory is a simple extension to Euler-Bernoulli beam theory. Shear deformations, whichare absent in Euler-Bernoulli beam theory, are included in Timoshenko beam theory.
The basic assumptions for Timoshenko beam theory are:
• The longitudinal axis of the unloaded beam is straight.
• All applied loads act transverse to the longitudinal axis.
• All deformations and strains are small.
• Hooke’s law can be used to relate stresses and strains.
• Plane cross sections, which are initially normal to the longitudinal axis, will remain plane after deformation.
The last requirement differs from Euler-Bernoulli theory which requires that the plane cross sections also remain normal tothe beam axis after deformation. For Timoshenko beams, plane cross sections will rotate due to shear forces.
Figure 1: Deflection of a beam
Preprint submitted to Elsevier October 26, 2016
1.2. Differential Equations
Figure 2: Forces on a beam segment
This section briefly presents an elementary derivation of the equations of Timoshenko beam theory. A complete derivationfrom the equations of solid mechanics is provided in Hjelmstad [5] and Wang et al. [16]. Figure 2 shows a small segment ofthe beam. The shear force V and bending moment M are displayed as positive in the figure according to the usual beamconvention. The load w is positive when applied upwards. We assume that the applied load is approximately constant forthe small beam segment shown in figure 2. Vertical equilibrium requires:
V + dV = V + w dxdV
dx= w (1.1)
For purposes of static equilibrium, the constant load w can be assumed to act as a point load in the middle of the segment.Equilibrium of moments results in:
M + dM = M + V dx+w
2(dx)2
dM
dx= V (1.2)
Figure 3: Pure bending deformation
The next step in deriving Timoshenko beam theory is to relate the internal forces V and M to deformation. Considerthe pure bending of the beam segment shown in figure 3. Pure bending means that the segment is bent only by constantend moments M . The y coordinate is measured from the neutral axis of the beam and is positive upwards. The neutral axisdoes not change length during the deformation of the beam. The original length of the beam segment is ds. Let dθ be theangle between the planar cross sections at the ends of the beam segment. The curvature κ is defined through the equation:
κ =1
R=
dθ
ds
The elongation of a longitudinal fiber in the beam is given by:
ds′ = (R − y) dθ = (R− y)κ ds
Since strains are assumed to be small, we shall use the engineering strain:
ε =ds′ − ds
ds= −κy
2
Notice that the strain is zero at the neutral axis. Hooke’s law gives the stress-strain relation:
σ = Eε = −κEy
where σ is the normal stress at the end planes of the segment and E is Young’s modulus. The bending moment M is relatedto normal stress σ through:
M = −
ˆ
σy dA = κE
ˆ
y2 dA = κEI
For beam theory, we need to relate the bending moment M to the deflection u. This is accomplished by substituting thecalculus definition of curvature:
M
EI= κ =
d2udx2
[
1 +(
dudx
)2]3/2
≈d2u
dx2(1.3)
The approximation is valid because dudx is assumed to be small and hence its square is smaller.
Figure 4: Pure shear deformation
Now consider the case of pure shear shown in figure 4. The constitutive relation between shear V and angle of shear γ is:
V = GAsγ (1.4)
where G is the shear modulus and As is the shear area. As is usually not equal to the area of the cross section because thedistribution of shear stress is not constant along the cross section.
Notice that shear deformation does not result in longitudinal strain. Also remember that bending deformation did notcause any shearing strain. Thus bending and shearing deformations are separable and the total angle of deformation is:
du
dx= θ − γ (1.5)
Using equations (1.1)-(1.5), we arrive at the equations for Timoshenko beam theory:
V ′(x) = w(x) (1.6)
M ′(x) = V (x) (1.7)
θ′(x) =M(x)
EI(x)(1.8)
u′(x) = θ(x) −V (x)
GAs(x)(1.9)
These equations can be reduced to the pair of differential equations for 0 < x < L:
d
dx
(
EIdθ
dx
)
= GAs
(
θ −du
dx
)
(1.10)
d
dx
[
GAs
(
θ −du
dx
)]
= w (1.11)
These equations imply that θ and u are two unknown functions whose solutions are to be determined. This is in contrast toEuler-Bernoulli beam theory where a single 4th order differential equation can be written for the displacement u. Equations(1.10) and (1.11) are used to formulate variational and energy principles in the next section.
3
Figure 5: Beam deflection using dimensionless distance ξ
Figure 6: Internal beam forces using dimensionless distance ξ
The equations (1.6)-(1.9) can be integrated to yield the following solutions:
V (x) =
ˆ x
0
w(s) ds+ V0 (1.12)
M(x) =
ˆ x
0
V (s) ds+M0 (1.13)
θ(x) =
ˆ x
0
M(s)
EI(s)ds+ θ0 (1.14)
u(x) =
ˆ x
0
θ(s) ds−
ˆ x
0
V (s)
GAs(s)ds+ u0 (1.15)
The constants of integration V0, M0, θ0, and u0 are determined using the boundary conditions for the particular problembeing solved.
The solution to the beam equations is more conveniently written using the dimensionless spatial variable ξ = x/L. Wethen reformulate functions f(x) as f(ξ). Since this involves reformulation of the function instead of mere replacement of thedependent variable, we notate this process as: f(x) → f(ξ). Integrals of f(x) are transformed in the following manner:
τ =s
L, L dτ = ds
ˆ x
0
f(s) ds → L
ˆ ξ
0
f(τ) dτ (1.16)
Using (1.16), the solutions (1.12)-(1.15) are now written as:
V (ξ) = L
ˆ ξ
0
w(τ) dτ + V0 (1.17)
M(ξ) = L
ˆ ξ
0
V (τ) dτ +M0 (1.18)
θ(ξ) = L
ˆ ξ
0
M(τ)
EI(τ)dτ + θ0 (1.19)
u(ξ) = L
ˆ ξ
0
θ(τ) dτ − L
ˆ ξ
0
V (τ)
GAs(τ)dτ + u0 (1.20)
Henceforth we shall assume prismatic beams with constant EI and GAs.
4
2. Energy Principles
2.1. Principle of Virtual Work
2.1.1. Formulation
The pair of differential equations (1.10) and (1.11) are the starting point for developing energy principles for Timoshenkobeams. We first multiply (1.10) by a function φ and integrate by parts:
ˆ L
0
(EIθ′)′
φdx =
ˆ L
0
GAs (θ − u′)φdx
[EIθ′φ]|L0−
ˆ L
0
EIθ′φ′ dx =
ˆ L
0
GAs (θ − u′)φdx
Remember that M = EIθ′. Moving all terms to the left hand side of the equation we have:
ˆ L
0
EIθ′φ′ dx+
ˆ L
0
GAs (θ − u′)φdx − [Mφ]|L0= 0 (2.1)
φ is interpreted as a virtual angle of rotation of the plane cross section of the beam.We now multiply (1.11) by a function v and integrate by parts:
ˆ L
0
[GAs (θ − u′)]′
v dx =
ˆ L
0
wv dx
[GAs (θ − u′) v]|L0−
ˆ L
0
GAs (θ − u′) v′ dx =
ˆ L
0
wv dx
Notice that V = GAs (θ − u′). Again moving all terms to the left hand side:
ˆ L
0
GAs (θ − u′) v′ dx+
ˆ L
0
wv dx− [V v]|L0= 0 (2.2)
For this equation, v represents a virtual vertical displacement of the beam.Setting equations (2.1) and (2.2) equal to each-other and rearranging:
ˆ L
0
EIθ′φ′ dx+
ˆ L
0
GAs (θ − u′) (φ− v′) dx =
ˆ L
0
wv dx+ [Mφ]|L0− [V v]|
L0
(2.3)
The left hand side of this equation is internal virtual work and the right hand side is external virtual work. Now let
u =[
θ u]T
and v =[
φ v]T
be the actual and virtual generalized displacements. Also define the generalized force
vector fT =[
M −V]
. Then we can write the Principle of Virtual Work:
δWint(u,v) = δWext(v) (2.4)
Internal virtual work δWint and external virtual work δWext are defined as:
δWint(u,v) =
ˆ L
0
uT
[
EI 00 GAs
]
v dx (2.5)
δWext(v) =
ˆ L
0
[
0 w]
v dx + fT (L)v(L)− fT (0)v(0) (2.6)
where uT =[
θ′ θ − u′]
and v =[
φ′ φ− v′]T
.
2.1.2. Total Potential Energy
The total potential energy for Timoshenko beams is defined as:
Π(u) =1
2δWint(u,u)− δWext(u) (2.7)
The mathematical theory of transforming from virtual work equations to total potential energy is given in Hjelmstad [5],Johnson [6].
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2.1.3. Boundary Conditions
Essential Natural
u = value, V =? or u =?, V = value
θ = value, M =? or θ =?, M = value
Table 1: Essential vs natural boundary conditions
The boundary terms in equation (2.3) provide the essential and natural boundary conditions. Either M or θ can bespecified at the end points but both cannot be specified simultaneously. The same argument applies to V and u. In thecase that θ or u is specified, the corresponding virtual displacement is not permitted to vary. The possibilities for boundaryconditions are summarized in table 2.
Location Type Enforced Value Virtual Displacement
x = 0 Essential u(0) = u0 v(0) = 0x = 0 Natural V (0) = V0 v(0) is freex = 0 Essential θ(0) = θ0 φ(0) = 0x = 0 Natural M(0) = M0 φ(0) is freex = L Essential u(L) = uL v(L) = 0x = L Natural V (L) = VL v(L) is freex = L Essential θ(L) = θL φ(L) = 0x = L Natural M(L) = ML φ(L) is free
Table 2: Essential and natural boundary conditions
It should be noted that some combinations of boundary conditions are not permissible since they would violate theexistence and uniqueness principles of the differential equations. These issues will not be discussed here but are found inmathematical texts on boundary value problems.
2.1.4. Function Spaces
Solution and variational spaces must be specified so that the integrals in equation (2.3) exist and the boundary conditionsare satisfied. If the actual displacement u coincides with the virtual displacement v, then we see that both the displacementfunction and its derivative are required to be square-integrable. The set of square-integrable vector-valued functions v(x) =[
φ(x) v(x)]T
on [0, L] is denoted as:
L2 =
{
v : [0, L] → R2
∣
∣
∣
∣
∣
ˆ L
0
[φ(x)]2dx < ∞ and
ˆ L
0
[v(x)]2dx < ∞
}
The set of functions v(x) in L2 with first derivatives v′(x) =[
φ′(x) v′(x)]T
also in L2 is called:
H1 ={
v : [0, L] → R2∣
∣v ∈ L2 and v′ ∈ L2}
The subset of H1 which satisfies the essential boundary conditions is notated as H1bc. The actual displacements u =
[
θ u]T
come from H1bc. The virtual displacements v must be zero at essential boundary conditions. The function space
for virtual displacements is denoted as H1bc0. Note that H1
bc and H1bc0 will be depend on the boundary conditions of the
particular beam problem. We always have the following set relations: H1bc0 ⊆ H1
bc ⊆ H1.For the example of a beam clamped at x = 0 and prescribed displacement u(L) = uL at x = L, we have:
H1bc =
{
u : [0, L] → R∣
∣u ∈ H1, θ(0) = 0, u(0) = 0, u(L) = uL
}
H1bc0 =
{
v : [0, L] → R∣
∣v ∈ H1, φ(0) = 0, v(0) = 0, v(L) = 0}
6
2.1.5. Variational Problem
The Principle of Virtual Work states that the solution u to the Timoshenko beam problem has the property that internalvirtual work equals external virtual work for all admissible virtual displacements.
Find u ∈ H1bc such that δWint(u,v) = δWext(v) ∀v ∈ H1
bc0 (2.8)
2.1.6. Minimization Problem
The Principle of Minimization of Total Potential Energy states that the solution u to the Timoshenko beam problem isthe one that minimizes the total potential energy as defined by equation (2.7).
Find u ∈ H1bc such that Π(u) ≤ Π(v) ∀v ∈ H1
bc (2.9)
2.1.7. Concentrated Applied Moments and Forces
Suppose that N concentrated moments Mi and point loads pi are applied at xi ∈ (0, L). We allow for the possibility thatthe applied moment or point load could be zero at xi. The interval [0, L] is partitioned into subintervals:
[0, L] =
N+1⋃
i=1
[xi−1, xi]
where x0 = 0 and xN+1 = L. We included the end points of the beam in order to simplify the book-keeping. The Principleof Virtual Work (2.3) for each segment of the beam is:
ˆ xi
xi−1
EIθ′φ′ dx+
ˆ xi
xi−1
GAs (θ − u′) (φ− v′) dx =
ˆ xi
xi−1
wv dx+ [Mφ]|xi
xi−1− [V v]|xi
xi−1
Summing over all segments, we have:
ˆ L
0
EIθ′φ′ dx+
ˆ L
0
GAs (θ − u′) (φ− v′) dx =
ˆ L
0
wv dx+M(L)φ(L)−M(0)φ(0)− V (L) v(L) + V (0) v(0)
−
N∑
i=1
[
M(x+
i )φ(x+
i )−M(x−
i )φ(x−
i )]
+
N∑
i=1
[
V (x+
i ) v(x+
i )− V (x−
i ) v(x−
i )]
(2.10)
where the plus and minus superscripts indicate right and left hand limits respectively. We associate concentrated momentswith jumps in moment: Mi = −
[
M(x+
i )−M(x−
i )]
. Similarly, we equate point loads with jumps in shear: pi = V (x+
i ) −
V (x−
i ). The sign convention for applied concentrated forces is as follows: Mi is positive counter-clockwise and pi is positiveupwards. Equation (2.10) allows for discontinuities in internal forces and virtual displacements.
If all virtual displacements are continuous, then φ(xi) = φ(x+
i ) = φ(x−
i ) and v(xi) = v(x+
i ) = v(x−
i ). The Principle ofVirtual Work simplifies to:
ˆ L
0
EIθ′φ′ dx+
ˆ L
0
GAs (θ − u′) (φ− v′) dx =
ˆ L
0
wv dx+M(L)φ(L)−M(0)φ(0)− V (L) v(L) + V (0) v(0)
+
N∑
i=1
Mi φ(xi) +
N∑
i=1
pi v(xi) (2.11)
7
2.2. Principle of Complementary Virtual Work
2.2.1. Formulation
Let δM(x) be a virtual moment distribution and δV (x) be a virtual shear distribution for the beam. We restrict thesedistributions to be ones that are in static equilibrium with the virtual applied load δw(x). Multiply both sides of equation(1.8) by δM and integrate by parts:
ˆ L
0
θ′ δM dx =
ˆ L
0
M δM
EIdx
[δM θ]|L0−
ˆ L
0
θ δM ′ dx =
ˆ L
0
M δM
EIdx
ˆ L
0
M δM
EIdx − [δM θ]|
L0+
ˆ L
0
θ δM ′ dx = 0 (2.12)
Now multiply equation (1.9) by δV , integrate by parts, and use the fact that δV = δM ′ and δV ′ = δw to get:
ˆ L
0
(θ − u′)δV dx =
ˆ L
0
V δV
GAsdx
ˆ L
0
θ δV dx−
ˆ L
0
u′ δV dx =
ˆ L
0
V δV
GAsdx
ˆ L
0
θ δM ′ dx−
ˆ L
0
u′ δV dx =
ˆ L
0
V δV
GAsdx
ˆ L
0
θ δM ′ dx− [δV u]|L0+
ˆ L
0
u δV ′ dx =
ˆ L
0
V δV
GAsdx
ˆ L
0
θ δM ′ dx − [δV u]|L0+
ˆ L
0
δw u dx =
ˆ L
0
V δV
GAsdx
ˆ L
0
θ δM ′ dx− [δV u]|L0+
ˆ L
0
δw u dx−
ˆ L
0
V δV
GAsdx = 0 (2.13)
Set equation (2.12) equal to equation (2.13) and rearrange terms:
ˆ L
0
M δM
EIdx+
ˆ L
0
V δV
GAsdx =
ˆ L
0
δw u dx+ [δM θ]|L0− [δV u]|L
0(2.14)
The left hand side is complementary internal virtual work and the right hand side is complementary external virtual work.
Equation (2.14) can be written as using matrix and vector notation. Let p =[
M V]T
and q =[
δM δV]T
. Thenthe Principle of Complementary Virtual Work is:
δW∗
int(p,q) = δW∗
ext(q) (2.15)
where complementary virtual internal work W∗
int(p,q) and complementary virtual external work W∗
ext(q) are:
W∗
int(p,q) =
ˆ L
0
pT
[
1
EI 00 1
GAs
]
q dx
W∗
ext(q) =
ˆ L
0
[
0 δw]
u dx+ qT (L)u(L)− qT (0)u(0)
where u =[
θ u]T
is the actual displacement field. Remember that p is related to u through the differential equations(1.6)-(1.9).
2.2.2. Complementary Total Potential Energy
The complementary total potential energy for the Timoshenko beam is defined as:
Π∗(p) =1
2δW∗
int(p,p)− δW∗
ext(p) (2.16)
8
2.2.3. Function Spaces
Suppose the virtual forces coincide with the actual forces in equation (2.14). Then the integrals exist only if the forcesare square-integrable functions. However, the virtual forces must also be in equilibrium. We denote the set of virtual forcesas L2
eq. The actual forces are derived from displacements which satisfy the essential boundary conditions. The set of actualforces is notated as L2
eqbc. Note the relationship between the sets is: L2eqbc ⊆ L2
eq ⊆ L2. The function spaces are determined
by the particular problem being solved. Unfortunately, the L2eq spaces are more cumbersome to specify than the H1
bc spaces ofvirtual work methods. This fact helps to explain why complementary virtual work is seldom mentioned in the mathematicaltheory of continuum mechanics.
2.2.4. Variational Problem
The Principle of Complementary Virtual Work states that the solution u to the Timoshenko beam problem has theproperty that complementary internal virtual work equals complementary external virtual work for all admissible virtualforces.
Find p ∈ L2eqbc such that δW∗
int(p,q) = δW∗
ext(q) ∀q ∈ L2eq (2.17)
2.2.5. Minimization Problem
The Principle of Minimization of Total Complementary Potential Energy states that the solution p to the Timoshenkobeam problem is the one that minimizes the total complementary potential energy as defined by equation (2.16).
Find p ∈ L2eqbc such that Π∗(p) ≤ Π∗(q) ∀q ∈ L2
eqbc (2.18)
2.2.6. Concentrated Virtual Moments and Forces
We now allow for the possibility of N concentrated virtual moments δMi and virtual point loads δpi applied at xi ∈ (0, L).At any location, either of these virtual forces could be zero. The interval [0, L] is partitioned into subintervals:
[0, L] =
N+1⋃
i=1
[xi−1, xi]
where x0 = 0 and xN+1 = L. The Principle of Complementary Virtual Work (2.14) for each segment is:
Matrix structural analysis requires stiffness matrices for every structural member. Only the derivation of the stiffnessmatrix is described in this paper. Details of matrix structural analysis are found in: Harrison [2], Kassimali [8], McCormac[10], McGuire et al. [11]. The Timoshenko beam element is shown in figure 7. The displacements and end forces are positivein the direction of the positive y axis. End rotations and moments are positive if they are counter-clockwise. This signconvention is demonstrated in figure 7.
The stiffness matrix K relates the beam end displacements u to the applied loads f :
Ku = f (3.1)
where u = [ u0 θ0 u1 θ1 ]T and f =[
Fu0 FM0 Fu1 FM1
]T. Equation (3.1)is explicitly written as:
K11 K12 K13 K14
K21 K22 K23 K24
K31 K32 K33 K34
K41 K42 K43 K44
u0
θ0u1
θ1
=
Fu0
FM0
Fu1
FM1
(3.2)
The most direct way of computing the elements of K is to consider the meaning of the j-th column. Kij is the end force fithat is required to maintain the unit displacement uj = 1 while keeping all other displacements at zero. Thus the procedureto compute the j-th column of K is as follows:
1. Solve the beam equations (1.17)-(1.20) with boundary conditions:
ui = δij =
{
1 if i = j
0 if i 6= j
2. Compute the end shears V0, V1 and end moments M0, M1 from the solution of the beam equations.
3. Translate from the beam sign convention to the element sign convention using:
K1j = V0, K2j = −M0, K3j = −V1, K4j = M1
This procedure will be applied in the next several pages of this paper.
3.1.2. Beam equations
The general solutions of the beam equations are required in order to compute the stiffness matrix. Since w(ξ) = 0, thesolution of the beam equations (1.17)-(1.20) are:
V (ξ) = V0 (3.3)
M(ξ) = LV0ξ +M0 (3.4)
θ(ξ) =L2V0
2EIξ2 +
LM0
EIξ + θ0 (3.5)
u(ξ) =L3V0
6EIξ3 +
L2M0
2EIξ2 + L
(
θ0 −V0
GAs
)
ξ + u0 (3.6)
Boundary conditions are used to determine the values of the constants V0, M0, θ0, and u0.We also define the parameter Φ which will be used in forming the stiffness matrix.
Φ =12EI
L2GAs(3.7)
10
3.1.3. Unit displacement at left end
Figure 8: Unit displacement u0 = 1
Left BC: u(0) = 1, θ(0) = 0; Right BC: u(1) = 0, θ(1) = 0
The left end boundary conditions give: u(0) = 1 ⇒ u0 = 1 and θ(0) = 0 ⇒ θ0 = 0.Solving for M0 in (3.5) with θ(1) = 0, we have:
M0 = −LV0
2
Substituting this expression into (3.6) with u(1) = 0:
0 =L3V0
6EI−
L3V0
4EI−
LV0
GAs+ 1
0 = −
(
L3
12EI+
L
GAs
)
V0 + 1
L3
12EI
(
1 +12EI
L2GAs
)
V0 = 1
V0 =12EI
L3(1 + Φ)−1
The end moments are computed to be:
M0 = −6EI
L2(1 + Φ)−1
M1 =6EI
L2(1 + Φ)−1
Translating from beam sign convention to matrix sign convention, the first column of the stiffness matrix is:
K11
K21
K31
K41
=
V0
−M0
−V1
M1
=EI
(1 + Φ)L3
126L−126L
11
3.1.4. Unit rotation at left end
Figure 9: Unit displacement θ0 = 1
Left BC: u(0) = 0, θ(0) = 1; Right BC: u(1) = 0, θ(1) = 0
The left end boundary conditions give: u(0) = 0 ⇒ u0 = 0 and θ(0) = 1 ⇒ θ0 = 1.Solving for M0 in (3.5) with θ(1) = 0, we have:
M0 = −LV0
2−
EI
L
Substituting this expression into (3.6) with u(1) = 0:
0 =L3V0
6EI−
L3V0
4EI−
L
2+ L−
LV0
GAs
0 = −
(
L2
12EI+
1
GAs
)
V0 +1
2
L2
12EI
(
1 +12EI
L2GAs
)
V0 =1
2
V0 =6EI
L2(1 + Φ)
−1
This value is used to compute M0:
M0 = −3EI
L(1 + Φ)−1 −
EI
L
M0 = −3EI
L(1 + Φ)−1 −
EI
L(1 + Φ)(1 + Φ)−1
M0 = −EI
L(4 + Φ)(1 + Φ)−1
M1 is then calculated to be:
M1 = LV0 +M0
M1 =6EI
L(1 + Φ)−1 −
EI
L(4 + Φ)(1 + Φ)−1
M1 =EI
L(2 − Φ)(1 + Φ)−1
Translating from beam sign convention to matrix sign convention, the second column of the stiffness matrix is:
K12
K22
K32
K42
=
V0
−M0
−V1
M1
=EI
(1 + Φ)L3
6L(4 + Φ)L2
−6L(2− Φ)L2
12
3.1.5. Unit displacement at right end
Figure 10: Unit displacement u1 = 1
Left BC: u(0) = 0, θ(0) = 0; Right BC: u(1) = 1, θ(1) = 0
The left end boundary conditions give: u(0) = 0 ⇒ u0 = 0 and θ(0) = 0 ⇒ θ0 = 0.Solving for M0 in (3.5) with θ(1) = 0, we have:
M0 = −LV0
2
Substituting this expression into (3.6) with u(1) = 1:
1 =L3V0
6EI−
L3V0
4EI−
LV0
GAs
1 = −
(
L3
12EI+
L
GAs
)
V0
1 = −L3
12EI
(
1 +12EI
L2GAs
)
V0
V0 = −12EI
L3(1 + Φ)−1
The end moments are computed to be:
M0 =6EI
L2(1 + Φ)−1
M1 = −6EI
L3(1 + Φ)−1
Translating from beam sign convention to matrix sign convention, the third column of the stiffness matrix is:
K13
K23
K33
K43
=
V0
−M0
−V1
M1
=EI
(1 + Φ)L3
−12−6L12−6L
13
3.1.6. Unit rotation at right end
Figure 11: Unit displacement θ1 = 1
Left BC: u(0) = 0, θ(0) = 0; Right BC: u(1) = 0, θ(1) = 1
The left end boundary conditions give: u(0) = 0 ⇒ u0 = 0 and θ(0) = 0 ⇒ θ0 = 0.Solving for M0 in (3.5) with θ(1) = 1, we have:
M0 = −LV0
2+
EI
L
Substituting this expression into (3.6) with u(1) = 0:
0 =L3V0
6EI−
L3V0
4EI+
L
2−
LV0
GAs
0 = −
(
L2
12EI+
1
GAs
)
V0 +1
2
L2
12EI
(
1 +12EI
L2GAs
)
V0 =1
2
V0 =6EI
L2(1 + Φ)
−1
This value is used to compute M0:
M0 = −3EI
L(1 + Φ)−1 +
EI
L
M0 = −3EI
L(1 + Φ)−1 +
EI
L(1 + Φ)(1 + Φ)−1
M0 = −EI
L(2− Φ)(1 + Φ)−1
M1 is then calculated to be:
M1 = LV0 +M0
M1 =6EI
L(1 + Φ)−1 −
EI
L(2− Φ)(1 + Φ)−1
M1 =EI
L(4 + Φ)(1 + Φ)−1
Translating from beam sign convention to matrix sign convention, the fourth column of the stiffness matrix is:
K14
K24
K34
K44
=
V0
−M0
−V1
M1
=EI
(1 + Φ)L3
6L(2− Φ)L2
−6L(4 + Φ)L2
14
3.2. Matrix Form
Collecting the information from the previous pages, the stiffness matrix for the Timoshenko beam element is:
K =EI
(1 + Φ)L3
12 6L −12 6L6L (4 + Φ)L2 −6L (2− Φ)L2
−12 −6L 12 −6L6L (2− Φ)L2 −6L (4 + Φ)L2
(3.8)
The equilibrium equation is thus:
EI
(1 + Φ)L3
12 6L −12 6L6L (4 + Φ)L2 −6L (2− Φ)L2
−12 −6L 12 −6L6L (2− Φ)L2 −6L (4 + Φ)L2
u0
θ0u1
θ1
=
Fu0
FM0
Fu1
FM1
(3.9)
Remember that K is only the element stiffness matrix. The global stiffness matrix is constructed by assembling the stiffnessmatrices for each element. The right hand side contains the applied forces. Forces occurring along the beam are representedas fixed end shears and moments at the ends. These statically equivalent forces are placed in the right hand side vector.Details concerning this procedure are described in texts on matrix structural analysis such as: Kassimali [8], McCormac[10], McGuire et al. [11], Timoshenko and Young [15].
3.3. Meaning of Φ
The parameter Φ, defined by equation (3.7), appears in every term of the element stiffness matrix. Φ measures the ratioof the bending stiffness to shear stiffness. For a very large shear stiffness, Φ is close to zero. In this case, the solutions forthe Timoshenko beam match the solutions to Euler-Bernoulli beam theory. This is expected since shear deformations areignored for Euler-Bernoulli beams.
The relative importance of shear deformation can be explored further by examining the material and shape constants inΦ. We will ignore all numerical constants contained in Φ and focus on dimensional factors instead. For rectangular sections,the moment of inertia and shear area scale according to the following dimensions:
I ∼ wd3, As ∼ wd
where w is the width and d is the depth of the section. Noting that E and G are usually the same order of magnitude, wehave:
Φ ∼
(
d
L
)2
Thus, Φ ∼ 0 when L ≫ d. For I-sections, the analysis is slightly different:
I ∼ wf tfd2, As ∼ twd
where wf is the flange width, tf is the flange thickness, and tw is the web thickness. For most sections, tf and tw are muchnearer in magnitude than they are to d. Also, wf is usually smaller than d. Thus we can write:
Φ ∼wfd
L2<
(
d
L
)2
We again observe the justification of Euler-Bernoulli beam theory when L ≫ d.
15
4. Green’s Functions
4.1. Formulation
Figure 12: Download unit point load at ξ = µ
The Green’s functions are the solutions to the beam equations (1.17)-(1.20) for a downward unit point load acting atlocation x = a. The point load is represented as a delta function: w(x) = −δ(x− a). In terms of the dimensionless variableξ, we set µ = a/L. The delta function obeys the scaling property:
δ(x− a) = δ(L(ξ − µ)) =1
Lδ(ξ − µ) (4.1)
Thus we write: w(ξ) = −δ(ξ−µ)/L. We also require the fact that the integral of the delta function is the Heaviside function:
ˆ ξ
0
δ(τ − µ) dµ = H(ξ − µ) =
{
1 if ξ ≥ µ
0 if ξ < µ(4.2)
Detailed information on Green’s functions and the delta function are found in: Richards and Youn [12], Richtmyer [13],Stakgold [14]. Successive integrals of the Heaviside function are given by the formula:
ˆ ξ
0
(τ − µ)n H(τ − µ) dτ =(ξ − µ)n+1
n+ 1H(ξ − µ), n ≥ 0 (4.3)
These properties allow explicit integration of the solution (1.17)-(1.20).The shear force is computed to be:
V (ξ) = L
ˆ ξ
0
w(τ) dτ + V0
V (ξ) = L
ˆ ξ
0
−1
Lδ(τ − µ) dτ + V0
V (ξ) = −H(ξ − µ) + V0 (4.4)
The bending moment is given as:
M(ξ) = L
ˆ ξ
0
V (τ) dτ +M0
M(ξ) = L
ˆ ξ
0
[−H(τ − µ) + V0] dτ +M0
M(ξ) = L [−(ξ − µ)H(ξ − µ) + V0ξ] +M0 (4.5)
The angle of deflection can be simplified:
θ(ξ) = L
ˆ ξ
0
M(τ)
EIdτ + θ0
θ(ξ) =L2
EI
ˆ ξ
0
[
−(τ − µ)H(τ − µ) + V0τ +1
LM0
]
dτ + θ0
θ(ξ) =L2
EI
[
−1
2(ξ − µ)2 H(ξ − µ) +
1
2V0ξ
2 +1
LM0ξ
]
+ θ0
θ(ξ) =L2
2EI
[
−(ξ − µ)2 H(ξ − µ) + V0ξ2 +
2
LM0ξ
]
+ θ0 (4.6)
16
Finally, the deflection is calculated in the form:
u(ξ) = L
ˆ ξ
0
θ(τ) dτ −L
GAs
ˆ ξ
0
V (τ) dτ + u0
u(ξ) =L3
2EI
ˆ ξ
0
[
−(τ − µ)2 H(τ − µ) + V0τ2 +
2
LM0τ +
2EI
L2θ0
]
dτ −L
GAs
ˆ ξ
0
[−H(ξ − µ) + V0] dτ + u0
u(ξ) =L3
2EI
[
−1
3(ξ − µ)3 H(ξ − µ) +
1
3V0ξ
3 +1
LM0ξ
2 +2EI
L2θ0ξ
]
+L
GAs[(ξ − µ)H(ξ − µ)− V0ξ] + u0
u(ξ) =L3
6EI
[
−(ξ − µ)3 H(ξ − µ) + V0ξ3 +
3
LM0ξ
2
]
+L
GAs[(ξ − µ)H(ξ − µ)− V0ξ] + Lθ0ξ + u0 (4.7)
Expressions (4.4)-(4.7) are the Green’s functions for the Euler-Bernoulli beam. Let Gq(ξ, µ) denote the Green’s functionfor the value of quantity q at location ξ due to the unit downward load at location µ. We let q ∈ {V,M, θ, u} and list all thecorresponding Green’s functions:
Notice that the constants of integration V0, M0, θ0, and u0 depend on the boundary conditions and are functions of thelocation µ of the point load.
4.2. Arbitrary Loads
Consider the case of N point loads pi at location µi. By the principle of superposition, the total response for quantity qis the sum of responses due to each point load pi. Thus the solution to the Euler-Bernoulli beam equations is:
q(ξ) =
N∑
i=1
pi Gq(ξ, µi) (4.12)
where q ∈ {V,M, θ, u} and Gq(ξ, µ) is the corresponding Green’s function.For an arbitrary load distribution w(ξ) acting downwards, the total response is given by the integral:
q(ξ) = L
ˆ 1
0
w(µ)Gq(ξ, µ) dµ (4.13)
17
4.3. Uniform Loads
For a uniform load w (weight per unit length) acting downwards, the solution (4.13) becomes:
q(ξ) = wL
ˆ 1
0
Gq(ξ, µ) dµ (4.14)
In order to proceed further, we require the evaluation of certain types of definite integrals. The first of which involvesnon-negative integer powers of ξ − µ multiplied by H(ξ − µ):
ˆ 1
0
(ξ − µ)n H(ξ − µ) dµ =ξn+1
n+ 1, n ≥ 0 (4.15)
Another useful definite integral is:ˆ 1
0
(1− µ)n dµ =1
n+ 1, n ≥ 0 (4.16)
Formulas (4.15) and (4.16) are proved in Haque [1].Using the Green’s functions (4.8)-(4.11) and applying the formula (4.15) to (4.14):
V (ξ) = wL
[
−ξ +
ˆ 1
0
V0(µ) dµ
]
(4.17)
M(ξ) = wL2
[
−1
2ξ2 + ξ
ˆ 1
0
V0(µ) dµ
]
+ wL
ˆ 1
0
M0(µ) dµ (4.18)
θ(ξ) =wL3
2EI
[
−1
3ξ3 + ξ2
ˆ 1
0
V0(µ) dµ+2
Lξ
ˆ 1
0
M0(µ) dµ
]
+ wL
ˆ 1
0
θ0(µ) dµ (4.19)
u(ξ) =wL4
6EI
[
−1
4ξ4 + ξ3
ˆ 1
0
V0(µ) dµ+3
Lξ2ˆ 1
0
M0(µ) dµ
]
+wL2
GAs
[
1
2ξ2 − ξ
ˆ 1
0
V0(µ) dµ
]
+ wL2ξ
ˆ 1
0
θ0(µ) dµ+ wL
ˆ 1
0
u0(µ) dµ (4.20)
18
5. Examples
5.1. Simply-supported Beam
5.1.1. Boundary Conditions
Figure 13: Simply-supported beam
5.1.2. Green’s Functions
The left end boundary conditions give: u(0) = 0 ⇒ u0 = 0 and M(0) = 0 ⇒ M0 = 0. The right end boundary conditionsare used to determine θ0 and V0.
Using M(1) = 0 in equation (4.5) we compute the value of V0:
0 = L [−(1− µ) + V0]
V0 = 1− µ
The value of θ0 is computed using u(1) = 0 in equation (4.7):
It is seen that Green’s functions for θ and u are the sum of the corresponding Green’s functions for simple beams and a lineardisplacement field connecting the ends.
5.4.3. Uniform Load
V (ξ) = wL
(
1
2− ξ
)
M(ξ) =wL2
2ξ(1 − ξ)
θ(ξ) = −wL3
24EI(1− 6ξ2 + 4ξ3) +
1
L(u1 − u0)
u(ξ) = −wL4
24EI(ξ − 2ξ3 + ξ4)−
wL2
2GAsξ(1− ξ) + ξu1 + (1 − ξ)u0
22
5.5. End-restrained Beam
5.5.1. Boundary Conditions
Figure 17: End-restrained beam
5.5.2. End Moments
The left end displacement boundary condition gives u(0) = 0 ⇒ u0 = 0.We first use equation (4.5) to derive an expression for V0 in terms of the end moments: M0 = M(0) and M1 = M(1).
M1 = L [−(1− µ) + V0] +M0
V0 =M1 −M0
L+ (1− µ)
This expression and the end moment-angle relations are incorporated into equations (4.6) and (4.7):
−M1
β1
=L2
2EI
[
−(1− µ)2 +M1 −M0
L+ (1− µ) +
2
LM0
]
+M0
β0
0 =L3
6EI
[
−(1− µ)3 +M1 −M0
L+ (1− µ) +
3
LM0
]
−1
GAs[M1 −M0] + L
M0
β0
The solution of this system of equations for M0 and M1 is:
