Flutter Instability of Timoshenko Cantilever Beam Carrying ...
Timoshenko Beam Theory: Exact Solution for First-Order ...
Transcript of 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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
© 2021 by the author(s). Distributed under a Creative Commons CC BY license.
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
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.”
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
(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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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,”
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
“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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Uniformly distributed load p
(B4)
Triangular distributed load (zero at x = 0, and p at x = l)
(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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Fixed–pinned
Uniformly distributed load p
(B7)
Triangular distributed load (zero at x = 0, and p at x = l)
(B8)
Non-uniform heating
(B9)
Fixed–fixed beam
Uniformly distributed load p
(B10)
Triangular distributed load (zero at x = 0, and p at x = l)
(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
p pl plEI w x x A x B x Apl x
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
p pl plEI w x x A x B x Apl x
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Pinned–pinned beam
Uniformly distributed load
(E3)
For k = -1, using Equations (40) and (47) yields the bending moment M(x) = pl²/k = -pl².
Triangular distributed load (zero at x = 0, and p at x = l)
(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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Fixed–pinned beam
Uniformly distributed load p
(E6)
Triangular distributed load (zero at x = 0, and p at x = l)
(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
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
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
pl x pl x pl pl plEI w x A B x B x A
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
pl x pl x pl pl plEI w x A B x B x A
k l k l k k kk k k k
A B
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Non-uniform heating
(E8)
Fixed–free beam
Uniformly distributed load p
(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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Triangular distributed load (zero at x = 0, and p at x = l)
(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
x x plM x A pl B pl x
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
A Bk
x x plM x A pl B pl x
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Fixed–fixed beam
Uniformly distributed load p
(E12)
Triangular distributed load (zero at x = 0, and p at x = l)
(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
pl x pl x pl pl plEI w x A B x B x A
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
pl x pl x pl pl plEI w x A B x B x A
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
x x plM x A pl B pl x
l l k
pl x pl x pl pl plEI w x A B x B x A
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
pl x pl x pl pl plEI w x A B x B x A
k l k l k k k
A B
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Fixed–fixed beam
Case of a compressive force with k -1
Case of a tensile force or a compressive force with k -1
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
Fixed–free beam
Case of a compressive force with k -1
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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:
Case of a compressive force with k -1
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)
Case of a tensile force or a compressive force with k -1
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)
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3
TIMOSHENKO BEAM THEORY EXACT SOLUTION
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
Chen, Y., Dong, S., Zang, Z., Zhang, Q., Zhang, J., Liu, X., Zhang, H., Lou, F. and Ao, C. 2018. Variational Iteration
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
Timoshenko, S.P. 1921. On the correction for shear of the differential equation for transverse vibrations of prismatic
bar. Philos Mag 1921; 41: 744–746.
Timoshenko, S.P. and Gere, J.M. 1961. Theory of Elastic Stability; McGraw-Hill: New York, USA, 1961.
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 July 2021 doi:10.20944/preprints202011.0457.v3