Timoshenko Beam Theory: Exact Solution for First-Order

TIMOSHENKO BEAM THEORY EXACT SOLUTION

Article

Timoshenko Beam Theory: Exact Solution for First-Order Analysis, Second-Order

Analysis, and Stability

Valentin Fogang

Civil Engineer, C/o BUNS Sarl, P.O Box 1130, Yaounde, Cameroon; [email protected]

ORCID iD https://orcid.org/0000-0003-1256-9862

Abstract: This paper presents an exact solution to the Timoshenko beam theory (TBT) for first-order analysis, second-

order analysis, and stability. The TBT covers cases associated with small deflections based on shear deformation

considerations, whereas the Euler–Bernoulli beam theory (EBBT) neglects shear deformations. Thus, the Euler–

Bernoulli beam is a special case of the Timoshenko beam. The momentcurvature relationship is one of the governing

equations of the EBBT, and closed-form expressions of efforts and deformations are available in the literature.

However, neither an equivalent to the momentcurvature relationship of EBBT nor closed-form expressions of efforts

and deformations can be found in the TBT. In this paper, a momentshear forcecurvature relationship, the equivalent

in TBT of the momentcurvature relationship of EBBT, was presented. Based on this relationship, first-order and

second-order analyses were conducted, and closed-form expressions of efforts and deformations were derived for

various load cases. Furthermore, beam stability was analyzed and buckling loads were calculated. Finally, first-order

and second-order element stiffness matrices were determined.

Keywords: Timoshenko beam; momentshear forcecurvature relationship; closed-form solutions; non-uniform

heating; stability; second-order element stiffness matrix; elastic Winkler foundation

1. Introduction

The Timoshenko beam theory (TBT) was developed by Stephen Timoshenko (1921) early in the 20th century. The

model accounts for shear deformations, making it suitable for describing the behavior of thick beams, whereas the

Euler–Bernoulli beam theory (EBBT) neglects them. If the shear modulus of the beam material approaches infinity, the

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beam becomes rigid in shear and the TBT converges towards EBBT. Thus, the Euler–Bernoulli beam is a special case of

the Timoshenko beam. First-order analysis of the Timoshenko beam is routinely performed; the principle of

virtual work yields accurate results and is easy to apply. Unfortunately, second-order analysis of the Timoshenko beam

cannot be modeled with the principle of virtual work. Various studies have focused on the analysis of Timoshenko

beams, most of which using numerical methods. Sang-Ho et al. (2019) presented a nonlinear finite element analysis

formulation for shear in reinforced concrete Beams; that formulation utilizes an equilibrium-driven shear stress function.

Abbas and Mohammad (2013) suggested a two-node finite element for analyzing the stability and free vibration of

Timoshenko beam; interpolation functions for displacement field and beam rotation were exactly calculated by

employing total beam energy and its stationing to shear strain. Hayrullah and Mustapha (2017) performed a buckling

analysis of a nano sized beam by using Timoshenko beam theory and Eringen’s nonlocal elasticity theory; the vertical

displacement function and the rotation function are chosen as Fourier series. Chen et al. (2018) used the variational

iteration method to analyze the flexural vibration of rotating Timoshenko beams; accurate natural frequencies and mode

shapes under various rotation speeds and rotary inertia were obtained. Onyia and Rowland-Lato (2018) presented a

finite element formulation for the determination of the critical buckling load of unified beam element that is free from

shear locking using the energy method; the proposed technique provides a unified approach for the stability analysis of

beams with any end conditions. Pavlovic and Pavlović (2018) investigated the dynamic stability problem of a

Timoshenko beam supported by a generalized Pasternak-type viscoelastic foundation subjected to compressive axial

loading, where rotary inertia is neglected; the direct Liapunov method was used. In stability analysis Timoshenko and

Gere (1961) proposed formulas to account for shear stiffness by means of calculation of buckling loads of the associated

Euler–Bernoulli beams. Hu et al. (2019) used matrix structural analysis to derive a closed-form solution of the second-

order element stiffness matrix; the buckling loads of single-span beams were also determined.

The momentcurvature relationship, one of the governing equations of the EBBT, has no equivalent in the TBT

literature. Furthermore, closed-form expressions of efforts and deformations are not common. In this paper a

momentshear forcecurvature relationship (MSFCR), the equivalent in TBT of the momentcurvature relationship of

EBBT, was presented. The relationship between the curvature, the bending moment, the bending stiffness, the shear

force, and the shear stiffness was then described. Based on MSFCR, closed-form expressions of efforts and



def

con

Th

Sp

dis

In

Acc

the r

In th

mod

Eq

formations w

nducted, and

2. Mater

he sign conve

Fig

ecifically, M

stributed load

first-order an

ording to the

rotation (pos

hese equation

dulus, and A

quation (4) ca

were derived,

d the bucklin

rials and Me

ention adopte

g 1 Sign c

M(x) is the be

d in the posit

nalysis the e

e Timoshenk

itive in clock

ns E is the e

is the cross s

2.1.2

an be formul

( )V x

( )M x

TIMO

, as well as f

g lengths of

ethods

2.1 Gov

2

ed for the loa

convention f

ending mom

tive downwa

quations of s

ko beam theo

kwise) of cro

elastic modu

section area.

Material an

lated as follo

()

dM x

dx

)d

EId

OSHENKO B

first-order an

single-span

verning equ

2.1.1 Static

ads, bending

for loads, ben

ment in the se

ard direction.

static equilib

(1)

ory, the bend

oss section

lus, I is the

Equations (

nd geometri

ws

)x

(dw

dx

( )x

dx

(3)

BEAM THE

nd second-ord

systems wer

uations

cs

moments, sh

nding momen

ection, V(x)

.

brium on an i

ding moment

(x) as follow

second mom

1), (3), and (

ic equations

2d

)( )

xx

x

)

EORY EXA

der element s

re determined

hear forces, a

nts, shear for

is the shear

infinitesimal

and the shea

ws:

ment of area,

(4) also apply

s

2

2

( )M x

dx

( )V x

GA

( )V x G

ACT SOLU

stiffness mat

d.

and displacem

rces, and disp

force, w(x)

element are

(2)

ar force are r

, is the she

y in second-o

( )q x

(dwGA

dx

UTION

trices. Stabili

ements is illu

placements

is the deflec

as follows:

)

related to the

ear correctio

order analysi

( )( )

xx

x

ity analysis w

ustrated in Fig

ction, and q(x

e deflection w

on factor, G i

is.

(5

)

(4)

was

gure 1.

x) is the

w(x) and

is the shear

5)




Differentiating both sides of Equation (5) with respect to x yields the following equation:

(6)

Substituting Equation (3) into Equation (6) yields the following:

(7)

Equation (7) yields the following momentshear forcecurvature relationship (MSFCR) combining bending and shear:

For a uniform beam along segments, substituting Equation (1) into Equation (8) yields,

(9)

For a tapered beam, substituting Equation (1) into Equation (8) yields,

(10)

In the case of non-uniform heating, Equation (9) becomes:

(11)

where T is the coefficient of thermal expansion, T = Tbb - Ttb is the difference between the temperature

changes at the beam’s bottom fibers (Tbb) and top fibers (Ttb), and d is the height of the beam.

Combining Equations (1) and (5) yields the rotation angle as follows:

Equations (5) to (12) apply as well in first-order analysis and in second-order analysis.

2.1.3 Summary of Timoshenko and Euler-Bernoulli beam equations

Table 1 summarizes the fundamental Timoshenko beam equations and compares them to the Euler–Bernoulli beam

equations.

2

2

( ) ( ) ( )d w x d x d V x

dx dx dx GA

2

2

( ) ( ) ( )d w x M x d V x

dx EI dx GA

2

2

( ) ( ) ( )0

d w x M x d V x

dx EI dx GA

2 2

2 2

( ) ( ) 1 ( )0

d w x M x d M x

dx EI GA dx

( ) 1 ( )( )

dw x dM xx

dx GA dx

2 2

2 2 2

( ) ( ) 1 ( ) 1 ( ) ( )0

( ) ( ) ( ( ))

d w x M x d M x d GA x dM x

dx EI x GA x dx GA x dx dx

2 2

2 2

( ) ( ) 1 ( )0,T Td w x M x d M x

dx EI GA dx d

(8)

(12)




Table 1. Summary of equations for Timoshenko and Euler–Bernoulli beams

Timoshenko beam Euler–Bernoulli beam

Statics

Geometry

Momentshear

forcecurvature

equation

The bending shear factor is defined as follows:

(13)

2.2 First-order analysis of uniform Timoshenko beams

2.2.1 Governing equations

The application of Equation (2) yields the following formulation of the bending moment:

(14)

Substituting Equation (2) into the MSFCR (Equation (9)) for a uniform beam and integrating twice yields:

(15)

The shear forces and rotation angles are determined using Equations (1) and (14), and Equations (12), (14), and (15),

respectively.

(15a)

(15b)

The integration constants Ci (i = 1, 2, 3, and 4) are determined using the boundary conditions and continuity equations and

combining the deflections, the rotation angles, the bending moments, and the shear forces.

1 2( ) ( )M x q x dxdx C x C

3 4( ) ( ) ( )EI

EI w x q x M x dxdx C x CGA

2

EI

l GA

1

1 3

( ) ( )

( ) ( )

V x q x dx C

EIEI x M x dx C C

GA

( )( )

dM xV x

dx

( ) 1 ( )( )

dw x dM xx

dx GA dx

( )

( )dw x

xdx

2

2

( ) ( ) ( )0

d w x M x d V x

dx EI dx GA

2

2

( ) ( )0

d w x M x

dx EI

( )( )

dM xV x

dx



Un

Ap

No

Eq

Eq

Th

det

Fig

Uniformly d

pplying Equa

on-uniform

quation (11) i

quation (15c)

he sign conv

termining the

g 2 Sign con

element

( )

(

M x

EI w

EI

( )M x

EI w

EI

distributed

ations (14) to

m heating:

is applied ins

) becomes

2.2.3

vention for

e element sti

nvention for

stiffness ma

2

2

( )24

( )6

qx

qx

qx x

1)

( )

( )

C x C

Cw x

Cx

TIMO

2

d load q

o (15b) yield

:

stead of Equ

3 Element s

bending mo

iffness matrix

bending mom

trix

1

4 1

3 21

6

2

C x C

Cx x

Cx x

2

31

21

,

1

6 2

2

C

Cx

Cx C

OSHENKO B

2.2.2 Analy

s

ation (9). Th

stiffness ma

oments, shea

x in local co

ments, shear

2

32

2

,

1

2

C

x C

C x

2

2

( )

T

V x C

C EI

C EI

BEAM THE

ysis of some

he thermal c

atrix

ar forces, d

ordinates is i

r forces, disp

22

21

( )V x

l q

l C C

1

2

3

T

T

C

x C

x C

EORY EXA

load cases

curvature str

displacement

illustrated in

lacements, a

1

23

3

qx C

x C x

C

3 4

21

C x C

l C

ACT SOLU

rain is

s, and rotat

n Figure 2.

nd rotation a

4x C

UTION

tion angles

angles for fir

TT

T

d

(1

(1

adopted for

rst-order 4

.T

15c)

15d)

r use in

4




Let us define following vectors,

(16) (17)

The element stiffness matrix in local coordinates of the Timoshenko beam is denoted by KTbl. The relationship between

the aforementioned vectors is as follows:

(18)

Applying Equation (15c) with the distributed load q = 0, yields the following:

(19)

(20)

(20a)

Considering the sign conventions adopted for bending moments and shear forces in general (see Figure 1) and for

bending moments and shear forces in the element stiffness matrix (see Figure 2), we can set following static

compatibility boundary conditions using Equation (19):

(21)

(22)

(23)

(24)

Considering the sign conventions adopted for the displacements and rotations in general (see Figure 1) and for

displacements and rotations in the member stiffness matrix (see Figure 2), we can set following geometric compatibility

boundary conditions using Equations (20) and (20a):

(25)

(26)

(27)

(28)

; ; ; ; ; ;T T

i i k k i i k kS V M V M V w w

TblS K V

1 2 1

3 21 23 4

2 212 1 3

( ) ( )

( )6 2

( )2

M x C x C V x C

C CEI w x x x C x C

CEI x x C x l C C

1

2

1

1 2

( 0)

( 0)

( )

( )

i

i

k

k

V V x C

M M x C

V V x l C

M M x l C l C

4

21 3

3 21 2 3 4

21 2 3

( 0)

( 0)

( ) /6 /2

( ) ( 1/2 )

i i

i i

k k

k k

w x w EI w C

x EI l C C

w x l w EI w l C l C lC C

x l EI l C lC C



Th

the

Th

ma

wh

As

dis

he combinatio

e Timoshenk

he developme

atrix of the T

here

ssuming the

splacements,

Fig 3 Sign

TblK

on of Equati

ko beam:

ent of Equat

Timoshenko b

presence o

and rotation

conventions

TblK

0

1EI

TIMO

ions (16) to

tion (28a) yi

beam:

of a hinge

ns is illustrate

for bending

12

1

E

l

sym

2

1

l

1 0 0

0 1 0

1 0 0

1 0l

OSHENKO B

(18) and Equ

elds the follo

at the right

ed in Figure

moments, sh

3

6

1

4

1

EI E

l

2

12EI

GA

0 0

0 0

0 0

0 0 (

BEAM THE

quations (21)

owing widel

ht end, the

3.

hear forces, d

element stiff

2

12

1

6

1

12

1

EI

l

EI

l

2

3

0

/ 6

1/ 2

l

l

EORY EXA

to (28) yiel

ly known for

sign conven

displacemen

fness matrix

3

2

3

6

1

2

1

6

1

4

1

EI

l

EI

l

EI

l

2

2

0

0

/

)

l

l l

ACT SOLU

ds the first-o

rmulation of

ntion for be

ts, and rotati

2

2

6

EI

l

EI

l

EI

l

EI

l

0 1

1 0

/ 2 1

1 0

l

UTION

order elemen

f the first-ord

ending mom

ion angles fo

1

nt stiffness m

(2

der element

(2

(3

ments, shear

or first-order

matrix of

28a)

stiffness

29)

30)

r forces,

3 3




The vectors of Equations (16) and (17) become

The element stiffness matrix becomes:

(33)

where

(34)

2.2.4 Beam resting on an elastic Winkler foundation

For a beam resting on a Winkler foundation with stiffness Kw, Equation (2) of static equilibrium becomes,

(35) (35a)

Differentiating Equation (35) twice with respect to x, combined with the MSFCR (Equation (9)), yields:

(36)

The solution of Equation (36) yields the formulation of M(x) with four integration constants (A1, B1, C1, D1). The shape

of M(x) depends on the parameter kw - 4/2 as follows:

(36a)

(36b)

(36c)

; ; ; ;T T

i i k i i kS V M V V w w

3 2 3

2

3

3 3 3

1 ' 1 ' 1 '

3 3

1 ' 1 '

3

1 '

Tbl

EI EI EI

l l l

EI EIK

l l

EIsym

l

2

2 4

( )( ) ( )w w w

d M x EIK w x q x K k

dx l

4 2 2

4 2 2

( ) ( ) ( )( )w wK Kd M x d M x d q x

M xdx GA dx EI dx

2

3'

EI

l GA

(31) (32)

1 1 1 1 1 2 1 2

1 1 1 1 1 1 1 1 1 1

1 1 2 1 1 2

1 1 2 1 1 2

( ) cosh sinh cosh sinh ( ),

( ) cosh cosh sinh sinh ( ),

( ) cosh cos cosh sin

sinh cos sinh sin

p

p

x x x xM x A B C D M x

l l l lx x x x x x

M x A B C D M xl l l l l lx x x x

M x A Bl l l lx x x x

C D Ml l l l

( ),p x




for kw - 4/2 > 0, kw - 4/2 = 0, and kw - 4/2 < 0, respectively . Combining M(x) with Equation (35) yields the

following equations:

(37) (38)

The expression of M(x) being known, applying Equation (1) yields the shear force, and the combination of M(x) with

Equations (12) and (38) yields the rotations of the cross section.

2.3 Second-order analysis of uniform Timoshenko beams

The analysis conducted in this section holds for compressive forces smaller (absolute values) than the buckling load.

2.3.1 Governing equations

A uniform beam is analyzed, an elastic Winkler foundation not being considered. The axial force N (positive in tension)

is assumed to be constant in segments of the beam. The equation of static equilibrium is as follows:

(39)

Combining Equation (39) with the MSFCR (Equation (9)) yields the following:

(40)

The solution of Equation (40) yields M(x), which contains the integration constants C1 and C2. The combination of M(x)

with Equation (39) yields following equations:

(41)

(42)

The transverse force T(x) is determined as follows:

(43)

Substituting Equations (1) and (41) into Equation (43) yields the transverse force T(x) as follows:

2 2

2 2

( ) ( )( )

d M x d w xN q x

dx dx

2 3

2 3

( ) ( ) ( ) ( )( ) ( )w w

d M x dw x d M x dq xK w x q x K

dx dx dx dx

2

2

( )(1 ) ( ) ( )

N d M x NM x q x

GA dx EI

3

3 4

( ) ( )( )

( ) ( ) ( )

dw x dM xN q x dx C

dx dx

Nw x M x q x dxdx C x C

( )( ) ( )

dw xT x V x N

dx

3( ) ( )T x q x dx C (44)



Th

Th

det

Fig

Le

we

Th

dis

Ca

Th

he combinatio

he sign conv

termining the

g 4 Sign con

4 4 ele

t us define fo

e set

he combinati

stributed load

ase 1: Comp

he solution of

S

(M x

on of M(x), E

ventions for

e element sti

nvention for

ement stiffne

ollowing vec

ion of Equa

d yields the f

pressive forc

f Equation (4

;i iT M

1) cosx A

TIMO

Equations (1

2

r bending m

iffness matrix

bending mom

ess matrix

ctors,

ation (40), th

following:

ce with k

48) is as follo

(1 )k

N k

; ;k kT M

1sx

Bl

( )N x

OSHENKO B

2) and (41) y

2.3.2 Eleme

moments, tra

x in local co

ments, transv

(45)

he bending s

-1

ows, with the

2

2

( ))

d M x

dx

2

EI

l

T

1 1sinx

Bl

) (1

BEAM THE

yields the rot

ent stiffness

ansverse for

ordinates is i

verse forces,

shear factor

e parameter

(49)

2(

kM x

l

V

)N dM

GA dx

EORY EXA

tation of the

matrix

rces, displac

illustrated in

displacemen

(Equation (

1 defined a

) 0x

;iV w

1

( )(

M xq

dx

ACT SOLU

cross section

cements, and

n Figure 4.

nts, and rotat

(13)), Equat

s shown:

; ;i k kw

1

k

k

3( )x dx C

UTION

n as follows:

d rotations

tion angles f

(46)

tion (47), an

(50)

T

3C

:

(4

adopted for

for second-or

(4

nd the absen

(4

44a)

r use in

rder

47)

nce of a

48)




Combining Equations (39) and (49) yields the following:

(51)

(52)

The combination of Equation (12) for the rotation of the cross section with Equations (49) and (51) yields the following,

with the parameter 2 defined as shown:

By applying Equations (1), (43), (49), and (51), the transverse force T(x) yields:

(55)

Considering the static compatibility boundary conditions (Equations (21) to (34)), whereby the shear forces are replaced

by the transverse forces, the geometric compatibility boundary conditions, and Equations (49) to (55), the following

equations are obtained:

(56)

The combination of Equations (18), (45), (46), and (56) yields the element stiffness matrix as follows:

(57)

1 1 1 1 1 1 1

1 1 1 1 1 1

( )sin cos

( ) cos sin

dw x x xNl A B NlC

dx l lx x

Nw x A B NC x NDl l

1 2 1 1 2 1 1 2( ) sin cos (1 )x x

Nl x A B NlC k kl l

1

2

21 1

1 1 2 21 1

1 0 0 10 0 1 0

0 1 01 0 0 0

0 0 1 0 cos sin 1

cos sin 0 0sin cos 1 0

Tbl

k lK EIll

l l

1( )T x NC

1

1

1

1 1 1 1

1 1

1 2 1

1 1 1 1 1 1

1 2 1 1 2

( 0)

( 0)

( )

( ) cos sin

( 0)

( 0) /

( ) cos sin

( ) / sin / cos

i i

i

k k

k

i i

i i

k k

k k

T T x T NC

M M x A

T T x l T NC

M M x l A B

w x w N w A ND

x N B l NC

w x l w N w A B NlC ND

x l N A l B l

1 1NC

(53) (54)




The development of Equation (57) yields the following formulation:

(58)

Accordingly, the element stiffness matrix of the beam having a hinge is as follows:

(59)

Case 2: Tensile force or compressive force with k -1

The solution of Equation (48) is as follows, with the parameter 3 defined as shown:

Combining Equations (39) and (60) and integrating with respect to x yields the following:

(62)

(63)

Combination of Equation (12) for the rotation of the cross section, with Equations (60) and (62) yields the following:

The parameter 4 (Equation (65)) has a positive value in the case of tension and a negative value in the case of

compression with k -1. Applying Equations (1), (43), (60), and (62), the transverse force T(x) yields the following:

(66)

K11 = -k2sin1/ K12 = -k(1 - cos1)/ K22 = k(cos1 -1/2sin1)/ K24 = k(-1 + 1/2sin1)/ = 2 - 2cos1 - 2sin1

K11 = -k2cos1/ K12 = -ksin1/ = sin1 - 2cos1

11 12 11 123 2 3 2

22 12 242

11 123 2

22

Tbl

EI EI EI EIK K K K

l l l lEI EI EI

K K Kl l lK

EI EIK K

l lEI

sym Kl

11 12 113 2 3

12 12 2

11 3

Tbl

EI EI EIK K K

l l lEI EI

K K Kl l

EIsym K

l

2 3 2 3 3( ) cosh sinh1

x x kM x A B

l l k

2 3 3 2 3 3 2

2 3 2 3 2 2

( )sinh cosh

( ) cosh sinh

dw x x xNl A B NlC

dx l lx x

Nw x A B NC x NDl l

2 4 3 2 4 3 2 4( ) sinh cosh (1 )x x

Nl x A B NlC k kl l

(64) (65)

2( )T x NC

(61) (60)




Considering the static compatibility boundary conditions (Equations (21) to (24)), whereby the shear forces are replaced

by the transverse forces, and the geometric compatibility boundary conditions, and Equations (60) to (66), the element

stiffness matrix can be expressed as follows:

(67)

The development of Equation (67) yields the following formulation:

(68)

Accordingly, the element stiffness matrix of the beam having a hinge is as follows:

(69a)

(69b)

(70)

2.3.3 Analysis of some load cases

Uniformly distributed load p.

Equation (48) is modified as follows:

(71a)

K11 = -k4cosh3/ K12 = -ksinh3/ = sinh3 - 4cosh3

K11 = k

4sinh

3/

K12 = -k(1 - cosh

3)/

K22 = k(cosh

3 ‐ 1/

4sinh

3)/

K24 = k(-1 + 1/

4sinh

3)/

= 2 ‐ 2cosh3+

4sinh

3

1

4

21 1

1 1 4 41 1

1 0 0 10 0 1 0

0 1 01 0 0 0

0 0 1 0 cosh sinh 1

cosh sinh 0 0sinh cosh 1 0

Tbl

k lK EIll

l l

11 12 113 2 3

12 12 2

11 3

Tbl

EI EI EIK K K

l l lEI EI

K K Kl l

EIsym K

l

11 12 11 123 2 3 2

22 12 242

11 123 2

22

Tbl

EI EI EI EIK K K K

l l l lEI EI EI

K K Kl l lK

EI EIK K

l lEI

sym Kl

2

2 2

( )(1 ) ( )

d M x kk M x p

dx l




Case 1: Compressive force with k -1. The analysis, conducted similarly to Section 2.3.2, yields

(71b)

Case 2: Tensile force or compressive force with k -1. The analysis yields

(71c)

Non-uniform heating

Combining Equation (11) and Equation (39) with q(x) = 0, yields the modification of Equation (48) as follows:

(71d)

Case 1: Compressive force with k -1. The analysis, conducted similarly to Section 2.3.2, yields

(71e)

Case 2: Tensile force or compressive force with k -1. The analysis yields

(71f)

2

1 1 1 1 1

21 1 1 1 1 1

1 2 1 1 2 1 1

( ) cos sin , ( )

( ) cos sin2

( ) sin cos

x x plM x A B T x px NC

l l kx x p

Nw x A B x NC x NDl lx x

Nl x A B plx NlCl l

2

2 3 2 3 2

22 3 2 3 2 2

2 4 3 2 4 3 2

( ) cosh sinh , ( )

( ) cosh sinh2

( ) sinh cosh

x x plM x A B T x px NC

l l kx x p

Nw x A B x NC x NDl l

x xNl x A B plx NlC

l l

1 1 1 1 1

1 1 1 1 1 1

1 2 1 1 2 1 1

( ) cos sin , ( )

( ) cos sin

( ) sin cos

T

x xM x A B EI T x NC

l lx x

Nw x A B NC x NDl lx x

Nl x A B NlCl l

2 3 2 3 2

2 3 2 3 2 2

2 4 3 2 4 3 2

( ) cosh sinh , ( )

( ) cosh sinh

( ) sinh cosh

T

x xM x A B EI T x NC

l lx x

Nw x A B NC x NDl l

x xNl x A B NlC

l l

2

2 2 2

( )(1 ) ( ) T

d M x k kk M x EI

dx l l




2.3.4 Beam resting on an elastic Winkler foundation with an axial load

In the equation of static equilibrium, the axial force N (the value of which is positive in tension) is assumed to be

constant in segments of the beam. The stiffness of the Winkler foundation is denoted by Kw.

(72)

The combination of Equation (72) with the MSFCR (Equation (9)) yields the following:

(73)

Differentiating twice both sides of Equation (73) with respect to x and combining the result with the MSFCR (Equation

(9)) yields the following:

(74)

The solution of Equation (74) yields the formulation of M(x) with four integration constants. From M(x), combined with

Equations (1) and (73), the shear force V(x) and the deflection w(x) can be deduced. The application of Equations (12)

and (43) yields the transverse forces T(x) and the rotations of the cross section (x).

2.3.5 Stability of the Timoshenko beam

The formulations of bending moment M(x), deflection w(x), rotation of the cross section (x), and transverse force T(x)

are used to determine the buckling load.

For k -1, Equations (49) to (55) are considered to satisfy the boundary conditions and continuity conditions.

For k -1, Equations (60) to (66) are considered to satisfy the boundary conditions and continuity conditions.

The resulting eigenvalue problem is solved to determine the buckling loads.

2 2

2 2

( ) ( )( ) ( )w

d M x d w xN K w x q x

dx dx

2

2

( )( ) 1 ( ) ( )w

N d M x NK w x M x q x

GA dx EI

4 2 2

4 2 2

( ) ( ) ( )1 ( )w wK KN d M x N d M x d q x

M xGA dx EI GA dx EI dx



Le

Th

5/6

res

of

Mo

Ap

Th

3. Resul

3.1

t us calculate

he characteris

6. The calcu

sults are pres

virtual work

oreover, clos

ppendix B.

T

he results of t

lts and Discu

First-orde

3.1.1 Unif

e the respons

stics are as fo

ulation of the

sented in the

k.” Table 2 li

sed-form exp

Table 2. De

this study are

TIMO

ussion

r analysis o

form Timos

ses of a pinn

Fig 5

follows: p =

e deflections

Supplement

sts the result

pressions of

eflections of

Node

position

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

e exact.

OSHENKO B

f Timoshenk

henko beam

edpinned b

Beam subje

10 kN/m, L

s with the pr

tary Material

ts obtained w

single-span

f the beam: P

n

Pr

wo

BEAM THE

ko beams

m subjected

beam subjecte

ected to a un

= 10 m, b =

rinciple of v

l “Deflection

with the princ

systems for

Principle of v

rinciple of v

ork (exact r

0.00000

0.00382

0.00722

0.00988

0.01157

0.01215

0.01157

0.00988

0.00722

0.00382

0.00000

EORY EXA

to uniforml

ed to a unifo

niformly distr

= 0.3 m, H =

virtual work

n calculation

ciple of virtu

various supp

virtual work,

virtual

esults)

0

2

2

7

7

2

2

0

ACT SOLU

y distribute

ormly distribu

ributed load

0.5m, E = 34

is presented

of a pinned

ual work and

port conditio

and present

Presen

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

UTION

ed load

uted load as

4.5106 kN/

d in Append

pinned beam

those obtain

ons and loadi

study

nt study

00000

00382

00722

00988

01157

01215

01157

00988

00722

00382

00000

shown in Fig

/m², = 0.3,

dix A. Detail

m using the p

ned in this stu

ings are pres

gure 5.

and =

ls of the

principle

udy.

sented in



Le

De

fix

(M

of

Th

We

the

“B

fix

t us calculate

etails of the

xedpinned b

MuL) for differ

virtual work

Table 3.

= E

M

M

M

M

he results of t

3.2

e determined

e bending she

Buckling anal

xedfree beam

3.1.2 Bea

e the respons

analysis an

beam under c

rent values o

k (exact value

. Moments i

I/GAl²=

MFEM =

MuL =

MFEM =

MuL =

this study are

Second-or

3.2.1 Stab

d the bucklin

ear factor. D

lysis of a pi

m,”and “Buc

TIMO

ms subjecte

ses of a unifo

F

nd results a

concentrated

of the bendin

es).

in beam subj

0.0000

-12.89

13.92

Calculat

-12.89

13.92

e exact.

rder analysis

bility of beam

ng loads of u

Details of the

innedpinned

ckling analys

OSHENKO B

d to concent

orm fixedpi

Fig 6 Beam s

re listed in

d load.” Table

ng shear facto

ected to a co

0.0250

Calcu

-11.99

14.25

tions accordi

-11.99

14.25

s of Timoshe

ms

uniform singl

analysis and

d beam,” “B

sis of a fixed

BEAM THE

trated load

inned beam s

subjected to

Appendix

e 3 displays

or, calculated

oncentrated lo

0.0500

ulations as d

-11.21

14.55

ing to the pri

-11.21

14.55

enko beams

le-span beam

d results are l

Buckling ana

dfixed beam

EORY EXA

subjected to

a concentrate

C and in th

the moment

d as describe

oad

0 0.075

escribed in t

-10.5

14.8

inciple of vir

-10.5

14.8

s

ms with vario

listed in App

alysis of a fi

m.”

ACT SOLU

a concentrat

ed load

he Suppleme

ts at the fixed

d in this pap

50 0.1

his paper

52 -9

80 15

rtual work (e

52 -9

80 15

ous support c

pendix D and

xedpinned

UTION

ted load, as s

entary Mate

d end (MFEM

per and accor

000 0

.92 -

5.03 1

exact values)

.92 -

5.03 1

conditions fo

d in the Supp

beam,” “Bu

shown in Fig

erial “Analy

M) and under

rding to the p

.1250

-9.38

15.23

-9.38

15.23

or different v

plementary M

uckling analy

ure 6.

sis of a

the load

principle

0.1500

-8.89

15.42

-8.89

15.42

values of

Materials

ysis of a




The buckling load Ncr is defined as follows:

(75)

Values of the buckling factor are listed in Table 4. Closed-form expressions of the matrices expressing the boundary

conditions are presented in Appendix D. To determine the buckling loads the determinants of those matrices are set to

zero. Closed-form expressions of the buckling factors for a pinned–pinned beam, a fixed–free beam, and a fixed–fixed

beam are also presented in Appendix D.

Table 4. Buckling factors for Timoshenko beam with various support conditions

= EI/GAl² = 0.000 0.025 0.050 0.075 0.100 0.1250 0.150

Pinnedpinned beam (SS–SS)

= 1.0000 1.1163 1.2220 1.3192 1.4096 1.4946 1.5750

Fixedpinned beam (F–SS)

= 0.6992 0.8716 1.0146 1.1392 1.2510 1.3530 1.4474

Fixedfree beam (F–FR)

= 2.0000 2.0608 2.1198 2.1772 2.2332 2.2877 2.3410

Fixedfixed beam (F–F)

= 0.5000 0.7048 0.8623 0.9951 1.1122 1.2181 1.3155

We recall that the exact values of the buckling factors for = 0.0 (corresponding to the Euler–Bernoulli beam) for the

support conditions SS–SS, F–SS, F–FR and F–F are 1.00, 0.700, 2.00, and 0.500, respectively.

3.2.2 Beams subjected to uniformly distributed load and an axial force

Let us calculate the responses of a uniform pinnedpinned beam subjected to a uniformly distributed load and an axial

force, as shown in Figure 7.

2 2/( )crN EI l



Th

be

def

De

to

Th

pin

he bending m

nding shea

flection curv

etails of the r

a uniformly

Table 5

k

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

he limits of

nnedpinned

Fig

moments at m

ar factor ,

ves are prese

results are pr

distributed l

5. Moments

= 0.

k = MM

7.50 2

6.00 0

5.00 0

4.00 0

3.00 0

2.00 0

1.00 0

0.00 0

1.00 0

2.00 0

3.00 0

4.00 0

f k correspo

d beam”.

TIMO

g 7 Beam su

mid-span (po

and the res

ented in App

resented in th

oad and an a

at position L

025

MP /pl² =

2.4474

0.5278

0.3453

0.2561

0.2032

0.1683

0.1435

0.1250

0.1107

0.0993

0.0900

0.0822

onding to b

OSHENKO B

ubjected to a

osition L/2) a

sults are list

pendix E, to

he Supplemen

axial force.”

L/2 for differ

k =

-6.50

-6.00

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

buckling are

BEAM THE

a uniformly d

are calculate

ted in Table

ogether with

entary Materi

rent values o

= 0.05

MMP /pl² =

7.8536

1.3947

0.5242

0.3215

0.2313

0.1804

0.1477

0.1250

0.1083

0.0955

0.0854

0.0772

e listed in t

EORY EXA

distributed lo

ed for differe

e 5. Closed-

those relate

ial “Bending

of shear facto

= k =

-5.50

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

the Supplem

ACT SOLU

oad and an ax

ent values of

-form expres

ed to other s

analysis of a

or and axial l

= 0.075

MMP /p

4.2593

1.0825

0.4316

0.2684

0.1944

0.1522

0.1250

0.1060

0.0920

0.0812

0.0727

mentary Mat

UTION

xial force

f axial force

ssions of be

support cond

a pinnedpin

load

pl² = k =

3

5 -4.5

6 -4.0

4 -3.0

4 -2.0

2 -1.0

0 0.0

0 1.0

0 2.0

2 3.0

7

aterial “Buck

e (coefficient

ending mom

ditions and l

nned beam su

= 0.10

= MMP /p

50 1.363

00 0.655

00 0.319

00 0.210

00 0.157

00 0.125

00 0.103

00 0.088

00 0.077

kling analys

t k) and

ment and

oadings.

ubjected

pl² =

35

53

97

08

70

50

38

87

74

sis of a



Le

sho

De

fix

ben

mo

Le

k =

and

Th

t us calculat

own in Figur

etails of the a

xedpinned b

nding mome

oments (MFEM

Table 6.

= EI/MF

M

= EI/MF

M

t us calculate

= -1.5 (Equat

d (58). Deta

he calculation

3.2.2a Be

te the respon

re 8. The com

analysis and

beam subjec

ents and defl

M) and under

. Moments i

GAl² =

FEM =

MuL =

GAl² =

FEM =

MuL =

3.2.3 Seco

e the second

tion (47)),

ails of the re

n of the elem

TIMO

ams subject

nses of a uni

mpressive for

Fig 8 Beam

d results are

cted to a co

ections for o

r the load (M

in beam subj

Coefficient

0.0000

-15.65

16.73

Coefficient

0.0000

-11.04

12.03

ond-order el

-order eleme

= 0.05 (Equ

sults are pre

ment stiffness

OSHENKO B

ted to concen

iform fixed

rces are assu

m subjected

listed in App

ncentrated l

other support

MuL) for differ

ected to a co

of axial forc

0.0250

-16.99

19.72

t of axial forc

0.0250

-9.31

11.27

lement stiffn

ent stiffness m

uation (13)),

esented in th

matrix of th

BEAM THE

ntrated load

pinned beam

umed smaller

to a concent

pendix F an

load and an

t conditions

rent values o

oncentrated lo

ce k = -4.0

0.0500

-18.98

23.70

ce k = 4.0

0.0500

-7.98

10.60

ness matrix

matrix of a b

and length L

he Suppleme

his Timoshen

EORY EXA

d

m subjected

r (absolute va

trated load an

d in the Sup

axial force.

are presente

of bending sh

oad and an a

Coeffi

0.0000

-17.60

18.72

Coeffi

0.0000

-10.32

11.29

beam with th

L = 4.0 m. T

ntary Materi

nko beam yie

ACT SOLU

to a concent

alues) than th

nd an axial fo

pplementary

.” Furthermo

d in Append

hear factor an

axial force

cient of axia

0 0.025

-21.5

24.6

icient of axia

0 0.025

-8.3

10.2

e following c

The matrix is

ial “Second-

elds:

UTION

trated load a

he buckling

force

Material “B

ore, closed-f

dix F. Table

nd axial forc

al force k = -

50 0.0

58 -29

68 35

al force k =

50 0.0

9 -6

22 9.

characteristic

s calculated u

-order eleme

and an axial

load.

ending analy

form expres

6 lists the fi

e.

-6.0

500

9.28

.46

6.0

500

.98

35

cs:

using Equati

ent stiffness

load, as

ysis of a

sions of

ixed-end

ons (57)

matrix.”




(78)

Let us now calculate the element stiffness matrix of the beam with the formula presented by Hu et al. (2019):

(79) The aforementioned characteristics become P = 1.5 EI/L², = 1- P/(ksGA) = 1- 1.5 0.05 = 0.925,

and

Details of the calculations are presented in the abovementioned Supplementary Material.

(80)

The results of this study are in good agreement with those of Hu et al. (2019).

In fact, both formulas are identical since following equivalences exist between the parameters considered by Hu et al.

(2019) (, ), and those considered in the present study (k,, 1, 2 ): = 1+ k, = 1, = 2, (/L)2 = -k/L2.

The equations for the determination of the buckling loads (see Appendix D) are also identical to those of Hu et al.

(2019) (Table 1). However, the formula presented by Hu et al. (2019) only applies for compressive forces with k -1.

For this beam subjected to a tensile force (k = 1.5 (Equation (47))), the stiffness matrix becomes (Equation (67):

(81)

² / 1.5 / 0.925 1.273PL EI

10 0 ( / )² 0 1 0 0 1

( / )² 0 0 0 0 / 1 0

0 0 ( / )² 0 sin 1

( / )²cos ( / )²sin 0 0 / sin / cos 1 0

Tbl

L

L LK EI

L cos L

L L L L

0.0917 0.2303 0.0917 0.2303

0.2303 0.6759 0.2303 0.2454

0.0917 0.2303 0.0917 0.2303

0.2303 0.2454 0.2303 0.6759

TblK EI

0.0917 0.2303 0.0917 0.2303

0.2303 0.6759 0.2303 0.2454

0.0917 0.2303 0.0917 0.2303

0.2303 0.2454 0.2303 0.6759

TblK EI

0.8850 1.4784 0.8850 1.4784

1.4784 4.6884 1.4784 1.2253

0.8850 1.4784 0.8850 1.4784

1.4784 1.2253 1.4784 4.6884

TblK EI




4. Conclusion

The momentshear forcecurvature relationship presented in this study enabled the derivation of closed-form solutions

for first-order analysis, second-order analysis, and stability of Timoshenko beams. The results showed that the

calculations conducted as described in this paper are exact. Closed-form expressions of efforts and deformations for

various load cases were presented, as well as closed-form expressions of second-order element stiffness matrices (the

axial force being tensile or compressive) in local coordinates. The determination of element stiffness matrices (ESM)

enables the analysis of systems with the direct stiffness method. We showed that ESM can also be determined by the

presence of hinges.

Calculation of bending moments and shear forces:

Influence of tensile force: With increasing tensile force, bending moments decrease (in absolute values), and with

increasing bending shear factor, bending moments decrease (in absolute values).

Influence of compressive force: With increasing compressive force, bending moments increase (in absolute values), and

with increasing bending shear factor, bending moments also increase (in absolute values).

Stability of the beam: With increasing bending shear factor, the buckling load decreases.

The following aspects not addressed in this study could be examined in future research:

Analysis of linear structures, such as frames, through the transformation of element stiffness matrices from local

coordinates in global coordinates.

Second-order analysis of frames free to sidesway with consideration of P- effect.

Use of the direct stiffness method, since element stiffness matrices are presented.

Analysis of positions of discontinuity (interior supports, springs, hinges, abrupt change of section), since closed-

form expressions of bending moments, shear or transverse forces, rotation angles, and deflection are known.

Beams resting on Pasternak foundations, the Pasternak soil parameter can be considered as a tensile force.

Supplementary Materials: The following files were uploaded during submission:

“Deflection calculation of a pinnedpinned beam using the principle of virtual work,”

“Analysis of a fixedpinned beam under concentrated load,”

“Buckling analysis of a pinnedpinned beam,”




“Buckling analysis of a fixedpinned beam,”

“Buckling analysis of a fixedfree beam,”

“Buckling analysis of a fixedfixed beam,”

“Bending analysis of a pinnedpinned beam subjected to a uniformly distributed load and an axial force,”

“Bending analysis of a fixedpinned beam subjected to a concentrated load and an axial force,” and

“Second-order element stiffness matrix.”

Funding:

Acknowledgments:

Conflicts of Interest: The author declares no conflict of interest.

Appendix A Deflection calculation at position x0 with the principle of virtual work

(A1)

M(x), V(x) are the bending moment and shear force due to the distributed load, respectively. m(x), v(x) are the bending

moment and shear force due to a virtual unit load at the point of interest x0, respectively.

M(x) = px(l-x)/2 V(x) = p(l/2-x)

m(x) = x(l-x0)/l for x x0 m(x) = x0(l-x)/l for x x0 (A2)

v(x) = 1-x0/l for x < x0 v(x) = -x0/l for x > x0

Substituting Equations (A2) into Equation (A1) yields,

(A3)

0

( ) ( ) ( ) ( )( )

M x m x V x v xx dx dx

EI GA

2 0 0 00 0 0 0

1 1( ) ( ) 1 (1 )

24 2

x l x xx pl x l x x pl

EI l l GA l




Appendix B Closed-form expressions of single-span systems for various support conditions and loadings

The equations of Sections 2.2.1 and 2.2.2 are applied, and the corresponding boundary conditions are satisfied. Closed-

form expressions of bending moment and deflection curves are as follows:

Pinned–pinned beam

Uniformly distributed load p

(B1)

Triangular distributed load (zero at x = 0, and p at x = l)

(B2)

Non-uniform heating

Fixed–Free beam


(B4)


(B5)

Non-uniform heating

(B6)

22

24 3 2 3

( )2 2

1( ) ( )

24 6 2 2

p plM x x plx

p pl plEI w x x x x pl x

3

5 3 3

( )6 6

1 7( )

120 6 6 360 6

p plM x x x

lp pl

EI w x x x pl xl

23

2 35 3 2

( )6 2 3

1( )

120 2 6 6 2

p pl plM x x x

l

p pl pl plEI w x x x x x

l

2

24 3 2 3

( )2 2

1( )

24 12 2 24 2

p plM x x x

p pl plEI w x x x x pl x

2( ) 0 ( )2 2

T TEI EIM x EI w x x lx

2( ) 0 ( )2

TEIM x EI w x x

(B3)




Fixed–pinned


(B7)


(B8)

Non-uniform heating

(B9)

Fixed–fixed beam


(B10)


(B11)

2 2

24 3 2 3

( )2

( ) ( )24 6 2

5 12 1

8(1 3 ) 8(1 3 )

pM x x Aplx Bpl

p pl plEI w x x A x B x Apl x

A B

22

2 34 3 2

( )2 2 12

1( )

24 12 12 2 2

p pl plM x x x

p pl pl plEI w x x x x x

3 2

25 3 2 3

( )6

( )120 6 2

9 20 14

40(1 3 ) 240(1 3 )

pM x x Aplx Bpl

l


l

A B

3 2

25 3 2 3

( )6

( )120 6 2

0.15 2 0.8 12

1 12 24(1 12 )

pM x x Aplx Bpl

l


l

A B

3 2 211 1 1 1

1 1

1( ) ( )

6 2

2 1/ 3 2 1/ 3

T

T T

AM x A x B EI w x x B EI x l A x

EI EIA B

l




Non-uniform heating

(B12)

Appendix C Fixedpinned Timoshenko beam subjected to a concentrated load

Equations (19), (20), and (20a) are applied on both sides of the concentrated load, x1 at the left side (x1 a) and x2 at the

right side (x2 b):

(C1)

Similar equations are applied on the right side (with the variable x2). Following boundary conditions and continuity

conditions are applied:

w(x1 = 0) = 0: D1 = 0

(x1 = 0) = 0: - l²×A1 + C1 = 0

w(x1 = a) = w(x2 = 0): -A1/6×a3 - B1/2×a2 + C1×a + D1 = D2

(x1 = a) = (x2 = 0): -A1/2×a2 - B1×a + C1 - l²×A1= C2 -l²×A2 (C2)

M (x1 = a) = M (x2 = 0): A1× a + B1 = B2

Q (x1 = a) - Q (x2 = 0) = P: A1 - A2 = P

M(x2 = b) = 0: A2× b + B2 = 0

w(x2 = b) = 0: -A2/6×b3 - B2/2×b2 + C2×b + D2 = 0

The unknowns are determined by solving the system of equations, and the moments can be calculated as follows:

Fixed-end moment MFEM = B1, and moment under the load MUL = B2.

For calculation with the principle of virtual work the fixed-end moment and the moment under the load can be

expressed as follows:

(C3)

1 1 1 1 1 1

3 21 11 1 1 1 1 1

2 211 1 1

( ) ( )

( )6 2

( )2

M x A x B V x A

A BEI w x x x C x D

AEI x x B x l A C

2

(1 / ) / // /

6 1/ 3 /FEM uL FEM

b l a l b lM Pl M b l M Pab l

EI GAl

( ) ( ) 0TM x EI EI w x




For other support conditions (pinnedpinned, fixedfree, and fixedfixed) the boundary conditions (Equations (C2)a,

(C2)b, (C2)g, and (C2)h) are modified accordingly, and the bending moment and deflection curves are deduced using

Equation (15c) with q = 0.

Appendix D Buckling loads of single-span beams with various support conditions

Compressive force with k -1: The eigenvalue problem led to following equations: sinh3 = 0,

sinh3 - 4cosh3 = 0, cosh3 = 0, and 2 – 2cosh3 + 4sinh3 = 0 for a pinned–pinned beam, a fixed–

pinned beam, a fixed–free beam, and a fixed–fixed beam, respectively. Recalling that 4 is negative, those equations

have no nontrivial solutions; consequently, no buckling occurred due to the action of a compressive force with k -1.

Compressive force with k > -1

Pinned–pinned beam: the results are presented in the Supplementary Material “Buckling analysis of a pinnedpinned

beam.” To determine the buckling load, the determinant of the matrix M expressing the boundary conditions is set to

zero. The matrix M, the determinant equation, and the buckling factor are as follows:

(D1)

Fixed–pinned beam: the results are presented in the Supplementary Material “Buckling analysis of a fixedpinned

beam,” with the matrix and the equation as follow:

(D2)

11 1

1 1

1.00 0.00 0.00 1.00

1.00 0.00 0.00 0.00sin 0, 1 ²

cos sin 0.00 0.00

cos sin 1.00 1.00

M

1 2 12

1 1 1 1 1

1 1

1.00 0.00 0.00 1.00

sin cos 00.00 1.00 0.00

cos sin 0.00 0.00 sin cos 0

cos sin 1.00 1.00

Mk




Fixed–free beam: the results are presented in the Supplementary Material “Buckling analysis of a fixedfree beam,”

with the matrix, the equation and the buckling factor as follows

(D3)

Fixed–fixed beam: the results are presented in the Supplementary Material “Buckling analysis of a fixedfixed beam,”

with the matrix and the equations as follows:

(D4)

Appendix E Timoshenko beams subjected to various load cases and an axial force

The compressive forces are assumed smaller (absolute values) than the buckling load.

Case 1: Compressive force with k -1. The bending moment and the deflection curve are denoted by Mc1(x) and

wc1(x), respectively. Let us recall the parameters 1 and 2 defined in Equations (50) and (54), respectively.

(E1)

Case 2: Tensile force or compressive force with k -1. The bending moment and the deflection curve are denoted

by Mc2(x) and wc2(x), respectively. Let us recall the parameters 3 and 4 defined in Equations (61) and (65),

respectively.

(E2)

The parameter 4 (Equation (65)) has a positive value in the case of tension and a negative value in the case of

compression with k -1.

21

1 1

1.00 0.00 0.00 1.00

0.00 1.00 0.00cos 0, 4 ²

cos sin 0.00 0.00

0.00 0.00 1.00 0.00

M

2

1 1 2 12

1 2 11 1

1 1 12 1 2 1

1

21 1

1

1.00 0.00 0.00 1.001 cos sin sin 0

0.00 1.00 0.002 2cos sin 0

cos sin 1.00 1.00

sin cos 1.00 0.00 sin 2sin cos 02 2 2

1sin 0 1 4 , 2sin cos

2 2 2

M

k

k

1 02

1 2/ 1 1k k k k

3 4/ 1 1k k k k




Pinned–pinned beam

Uniformly distributed load

(E3)

For k = -1, using Equations (40) and (47) yields the bending moment M(x) = pl²/k = -pl².


(E4)

Non-uniform heating

(E5)

2 2 21

1 1 11

4 4 2 3 421

1 1 12 2 21

2 2 23

2 3 33

4 4

2 32 2

1 cos( ) cos sin

sin

1 cos( ) cos sin

sin 2 2

1 cosh( ) cosh sinh

sinh

1 c( ) cosh

c

c

c

c

pl x pl x plM x

k l k l k

pl x pl x pl pl plEI w x x x

k l k l k k k

pl x pl x plM x

k l k l k

pl x plEI w x

k l k

2 3 423

3 23

oshsinh

sinh 2 2

x pl pl plx x

l k k k

2 2 2

1 1 1 1 1 1 1 1 1 1 1

1 11 1

2 2 2

2 2 3 2 3 2 2 3 2 3 2

2 2

( ) cos sin , ( ) cos sin

1 1

sin tan

( ) cosh sinh , ( ) cosh sinh

1

c T c

T T

c T c

T T

x x l x l x lM x A B EI EI w x A B A

l l k l k l k

A EI B EI

x x l x l x lM x A B EI EI w x A B A

l l k l k l k

A EI B EI

3 3

1

sinh tanh

2

1 11

4 33

1 121

2

2 33

4 33

1 323

( ) sinsin

1 1( ) sin

sin 6 6

( ) sinhsinh

1 1( ) sinh

sinh 6 6

c

c

c

c

pl x plM x x

k l k

pl x pl plEI w x x x

k l k k k

pl x plM x x

k l k

pl x pl plEI w x x x

k l k k k




Fixed–pinned beam


(E6)


(E7)

22 2

1 1 1 1 1

4 4 2 3 42

1 1 1 1 1 2 1 1

1 2 1 11 1

1 2 1 1 2 1

2 22 2 3 2 3

( ) cos sin

( ) cos sin2

1/ 2sin sin / 1/ 1/ 2 cos 1/

sin cos sin cos

( ) cosh sinh

c

c

c

x x plM x A pl B pl

l l k

pl x pl x pl pl plEI w x A B x B x A

k l k l k k kk k k

A B

x xM x A pl B pl

l

2

4 4 2 3 42

2 2 3 2 3 4 2 2

3 4 3 32 2

3 4 3 3 4 3

( ) cosh sinh2

1/ 2sinh sinh / 1/ 1/ 2 cosh 1/

sinh cosh sinh cosh

c

pl

l k


k l k l k k kk k k

A B

2 21 1 1 1 1

4 4 3 43

1 1 1 1 1 1 2 1

1 2 11 1

1 2 1 1 2 1

2 22 2 3 2

( ) cos sin

( ) cos sin6

1/ 6 1/ sin / 1/ 1/ 6 cos 1/

sin cos sin cos

( ) cosh sinh

c

c

c

x x plM x A pl B pl x

l l k


k l k l k k kk k k k

A B

xM x A pl B pl

l

3

4 4 3 43

2 2 3 2 3 2 4 2

3 4 32 2

3 4 3 3 4 3

( ) cosh sinh6

1/ 6 1/ sinh / 1/ 1/ 6 cosh 1/

sinh cosh sinh cosh

c

x plx

l k


k l k l k k kk k k k

A B




Non-uniform heating

(E8)

Fixed–free beam


(E9)

1 1 1 1 1

2 2 2

1 1 1 1 1 1 2 1

1 2 11 1

1 2 1 1 2 1

2 2 3 2 3

2 2

2 2 3 2

( ) cos sin

( ) cos sin

sin 1 cos

sin cos sin cos

( ) cosh sinh

( ) cosh s

c T

c

T T

c T

c

x xM x A B EI

l l

l x l x l lEI w x A B B x A

k l k l k k

A EI B EI

x xM x A B EI

l l

l x lEI w x A B

k l k

2

3 2 4 2

3 4 32 2

3 4 3 3 4 3

inh

sinh 1 cosh

sinh cosh sinh coshT T

x l lB x A

l k k

A EI B EI

22 2

1 1 1 1 1

4 4 2 3 42

1 1 1 1 1 1

11 1

2 1 2

22 2

2 2 3 2 3

4 4

2 2 3 2 3

( ) cos sin

( ) cos sin2

tan 1 1

cos

( ) cosh sinh

( ) cosh sinh

c

c

c

c

x x plM x A pl B pl

l l k

pl x pl x pl pl plEI w x A B x x A

k l k l k k k

A Bk

x x plM x A pl B pl

l l k

pl x pl xEI w x A B

k l k l

2 3 4

22

32 2

4 3 4

2tanh 1 1

cosh

pl pl plx x A

k k k

A Bk





(E10)

Non-uniform heating

(E11)

21

1 1 11 1

23

2 2 33 3

cos( ) 1 , ( ) 1 cos ,

cos cos

cosh( ) 1 , ( ) 1 cosh ,

cosh cosh

c T c T

c T c T

xl xlM x EI EI w x EI

k l

xl xlM x EI EI w x EI

k l

2 21 1 1 1 1

4 4 3 43

1 1 1 1 1 2 1 1

11 1

1 2 2

2 22 2 3 2 3

4

2 2

( ) cos sin

( ) cos sin6

1/ 2 1/ tan 1/ 2 1/1

cos

( ) cosh sinh

( ) c

c

c

c

c


l l k


k l k l k k kk k

A Bk


l l k

plEI w x A

k

4 3 43

3 2 3 4 2 2

32 2

3 4 4

osh sinh6

1/ 2 1/ tanh 1/ 2 1/1

cosh

x pl x pl pl plB x B x A

l k l k k kk k

A Bk




Fixed–fixed beam


(E12)


(E13)

22 2

1 1 1 1 1

4 4 2 3 42

1 1 1 1 1 2 1 1

1 2 1 2 1 2 11 1

1 2 1 1 2 1

22 2 3

( ) cos sin

( ) cos sin2

1 cos / 2 sin / 1 cos / 1/ 2sin

2 2cos sin 2 2cos sin

( ) cosh

c

c

c

x x plM x A pl B pl

l l k


k l k l k k k

A B

xM x A pl B

l

22

2 3

4 4 2 3 42

2 2 3 2 3 4 2 2

3 4 3 4 3 4 32 2

3 4 3 3 4 3

sinh

( ) cosh sinh2

1 cosh / 2 sinh / 1 cosh / 1/ 2sinh

2 2cosh sinh 2 2cosh sinh

c

x plpl

l k


k l k l k k k

A B

2 21 1 1 1 1

4 4 3 43

1 1 1 1 1 2 1 1

1 2 1 2 1 2 11 1

1 2 1 1 2 1

2 2

( ) cos sin

( ) cos sin6

1/ 6 1 cos 1/ 2 sin / 1/ 2 1 cos / 1/ 6 sin,

2 2cos sin 2 2cos sin

( )

c

c

c


l l k


k l k l k k k

A B

M x A

2 23 2 3

4 4 3 43

2 2 3 2 3 4 2 2

3 4 3 4 3 4 32 2

3 4 3 3 4 3

cosh sinh

( ) cosh sinh6

1/ 6 1 cosh 1/ 2 sinh / 1/ 2 1 cosh / 1/ 6 sinh,

2 2cosh sinh 2 2cosh sinh

c

x x plpl B pl x

l l k


k l k l k k k

A B




Non-uniform heating

(E14)

Appendix F Timoshenko beams subjected to a concentrated load and an axial force

The compressive forces are assumed smaller (absolute values) than the buckling load.

Equations (49) to (55) are applied in case of a compressive force with k -1, and Equations (60) to (66) are applied in

case of a tensile force or a compressive force with k -1. In the case of a compressive force with k -1, the

constants of integration left of the concentrated force are A11, B11, NC11, and ND11, and right of the concentrated force

they are A12, B12, NC12, and ND12. In the case of a tensile force or a compressive force with k -1, the constants of

integration left of the concentrated force are A21, B21, NC21, and ND21, and right of the concentrated force they are A22,

B22, NC22, and ND22.

Fixedpinned beam

Case of a compressive force with k -1: following boundary conditions and continuity conditions are applied:

w(x1 = 0) = 0: -A11 + ND11 = 0

(x1 = 0) = 0: -B112 + NC11×l = 0

w(x1 = a) = w(x2 = 0): -A11cos1a/l - B11sin1a/l + NC11×a + ND11 = -A12 + ND12

(x1 = a) = (x2 = 0): A112sin1a/l - B112cos1a/l + NC11×l = -B122 + NC12×l (F1)

M (x1 = a) = M (x2 = 0): A11cos1a/l + B11sin1a/l = A12

T (x1 = a) - T (x2 = 0) = P: NC11 – NC12 = P

M(x2 = b) = 0: A12cos1b/l + B12sin1b/l = 0

w(x2 = b) = 0: -A12cos1b/l - B12sin1b/l + NC12×b + ND12 = 0

1 1

2 2

( ) ( ) 0

( ) ( ) 0c T c

c T c

M x EI EI w x

M x EI EI w x




Solving the equation system yields the constants of integration as follows:

Case of a tensile force or a compressive force with k -1: the corresponding boundary conditions and

continuity conditions are applied. Solving the equation system yields the constants of integration as follows:

2

111 1

11

112 1 2 1 2

11

12 1 1

12

12

1 112

1 1

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cos sin 1 1 0 0 1

sin cos 0 0 0

cos sin 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cos sin 0 0

0 0 0 0 cos sin 1

l

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

ND l lb b

bl l

1

0

0

0

0

0

0

0

P

4

213 3

21

214 3 4 3 4

21

22 3 3

22

22

3 322

3

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cosh sinh 1 1 0 0 1

sinh cosh 0 0 0

cosh sinh 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cosh sinh 0 0

0 0 0 0 cosh sin

l

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

ND l lb

l

1

3

0

0

0

0

0

0

0

h 1

P

bb

l

(F2)

(F3)




Pinnedpinned beam

Case of a compressive force with k -1

Case of a tensile force or a compressive force with k -1

111 1

11

112 1 2 1 2

11

12 1 1

12

12

1 112

1 1

1 0 0 1 0 0 0 0

1 0 0 0 0 0 0 0

cos sin 1 1 0 0 1

sin cos 0 0 0

cos sin 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cos sin 0 0

0 0 0 0 cos sin 1

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

ND l lb b

bl l

1

0

0

0

0

0

0

0

P

213 3

21

214 3 4 3 4

21

22 3 3

22

22

3 322

3

1 0 0 1 0 0 0 0

1 0 0 0 0 0 0 0

cosh sinh 1 1 0 0 1

sinh cosh 0 0 0

cosh sinh 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cosh sinh 0 0

0 0 0 0 cosh sinh

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

ND l lb

l

1

3

0

0

0

0

0

0

0

1

P

bb

l

(F4)

(F5)




Fixed–fixed beam



2

111 1

11

112 1 2 1 2

11

12 1 1

12

12

2 1 2 112

1 1

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cos sin 1 1 0 0 1

sin cos 0 0 0

cos sin 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 sin cos 0

0 0 0 0 cos sin

l

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

lND l lb b

bl l

1

0

0

0

0

0

0

0

1

P

4

213 3

21

214 3 4 3 4

21

22 3 3

22

22

4 3 4 322

3

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cosh sinh 1 1 0 0 1

sinh cosh 0 0 0

cosh sinh 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 sinh cosh 0

0 0 0 0 cosh

l

A a aa

l lBa aNC l ll lND

a aA

l lB

NCb b

lND l l

1

3

0

0

0

0

0

0

0

sinh 1

P

b bb

l l

(F6)

(F7)




Fixed–free beam


Case of a tensile force or a compressive force with k -1:

2

11

1 111

11

2 1 2 1 211

121 1

12

12

121 1

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cos sin 1 1 0 0 1

sin cos 0 0 0

cos sin 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cos sin 0 0

0 0 0 0 0 0 1 0

lA

a aaB l l

NC a al l

ND l lA a a

l lB

NCb bNDl l

1

0

0

0

0

0

0

0

P

4

21

3 321

21

4 3 4 3 421

223 3

22

22

223 3

1 0 0 1 0 0 0 0

0 0 0 0 0 0

cosh sinh 1 1 0 0 1

sinh cosh 0 0 0

cosh sinh 0 0 1 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0 cosh sinh 0 0

0 0 0 0 0 0 1 0

lA

a aaB l l

NC a al l

ND l lA a a

l lB

NCb bNDl l

1

0

0

0

0

0

0

0

P

(F8)

(F9)




For a concentrated load P applied at the tip of the fixedfree beam, the moment and deflection curves are as follows:

It is worth mentioning that the results of Equation (F10) satisfy the following equations of equilibrium using the free

body diagram:

For the case of an external moment M*(in counterclockwise) applied at the beam the continuity equations are modified

as follows:


M (x1 = a) - M (x2 = 0) = M* A11cos1a/l + B11sin1a/l - A12 = M* (F12)

T (x1 = a) - T (x2 = 0) = 0: NC11 – NC12 = 0 (F13)


M (x1 = a) = M (x2 = 0): A21cosh3a/l + B21sinh3a/l - A2 = M* (F14)

T (x1 = a) - T (x2 = 0) = 0: NC21 – NC22 = 0 (F15)

11 1 1

2 2

3 3 2 31 1

1 1 12 2 2

32 3 3

4 4

3 3 2 33 3

2 3 34 4 4

tan( ) cos sin

tan tan( ) cos sin

tanh( ) cosh sinh

tanh tanh( ) cosh sinh

c

c

c

c

x Pl xM x Pl

l l

Pl x Pl x Pl PlEI w x x

k l k l k k

x Pl xM x Pl

l l

Pl x Pl x Pl PlEI w x x

k l k l k k

(F10)

1 1

2 2

( 0) ( ) 0

( 0) ( ) 0c c

c c

M x Pl Nw x l

M x Pl Nw x l

(F11)




References

Abbas, M-O. and Mohammad, K. 2013. Finite element formulation for stability and free vibration analysis of

Timoshenko beam. Advances in Acoustics and Vibration, vol 2013. https://doi.org/10.1155/2013/841215

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Approach for Flexural Vibration of Rotating Timoshenko Cantilever Beams. International Journal of Structural

Stability and Dynamics, Vol. 18, No. 12, 1850154 (2018). https://doi.org/10.1142/S0219455418501547

Hayrullah, G.K. and Mustapha, O.Y. 2017. Buckling analysis of non-local Timoshenko beams by using Fourier series.

Int J Eng Appl Sci, vol. 9(4), 89-99. https://doi.org/10.24107/ijeas.362242

Hu, Z.P., Pan, W.H. and Tong, J.Z. 2019. Exact solutions for buckling and second-order effect of shear deformable

Timoshenko beam–columns based on matrix structural analysis. Appl. Sci. 2019, 9(18), 3814.

https://doi.org/10.3390/app9183814

Onyia, M.E. and Rowland-Lato, E.O. 2018. Determination of the critical buckling load of shear deformable unified

beam. IJET, Vol 10 No 3. DOI: 10.21817/ijet/2018/v10i3/181003026

Pavlovic, R. and Pavlović, I. 2018. Dynamic stability of Timoshenko beams on Pasternak viscoelastic foundation.

Theoretical and Applied Mechanics 45(00):5-5, 2018. DOI:10.2298/TAM171103005P

Sang-Ho, K., Sun-Jin, H. and Kang, S.K. 2019. Nonlinear finite element analysis formulation for shear in reinforced

concrete beams. Appl. Sci. 2019, 9, 3503. https://doi.org/10.3390/app9173503

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Timoshenko Beam Theory: Exact Solution for First-Order