TECNICA ITALIANA-Italian Journal of Engineering Science

Vol. 63, No. 1, March, 2019, pp. 34-45

Journal homepage: http://iieta.org/Journals/TI-IJES

Timoshenko Beam Theory for the Flexural Analysis of Moderately Thick Beams – Variational
Formulation, and Closed Form Solution
Charles Chinwuba Ike

Department of Civil Engineering, Enugu State University of Science & Technology, Enugu State, Nigeria

Corresponding Author Email: ikecc2007@yahoo.com

https://doi.org/10.18280/ti-ijes.630105 ABSTRACT

Received: 22 January 2019 In this study, the Timoshenko first order shear deformation beam theory for the flexural
Accepted: 19 March 2019 behaviour of moderately thick beams of rectangular cross-section is formulated from
vartiational principles, and applied to obtain closed form solutions to the flexural problem
Keywords: of moderately thick rectangular beams. The total potential energy functional for the
Timoshenko beam theory, moderately thick moderately thick beam flexure problem was formulated by considering the contribution of
beams, total potential energy functional, shearing deformation to the strain energy. Euler-Lagrange conditions were then applied to
Euler-Lagrange differential equations, obtain the system of two coupled ordinary differential equations of equilibrium. The
differential equations of equilibrium, shear problem of moderately thick beam with simply supported ends subject to uniformly
deformation distributed transverse load on the entire span was solved in closed form to obtain the
transverse deflection as the sum of the flexural and shear components. Another problem of
moderately thick cantilever beam under point load at the free end was solved in closed form
to illustrate the solution of the governing equations. The transverse deflection was similarly
obtained as the sum of shear and bending components. The bending component of
deflection was found to be identical with the Euler-Bernoulli results while the shear
component was found to be dependent on the square of the ratio of the beam thickness, t, to
the span, l. It was found that as t/l<0.02, the contribution of shear to the overall deflection
is insignificant; but becomes significant for t/l>0.10. The findings are in excellent
agreement with the technical literature.

1. INTRODUCTION The need to overcome the limitations of the EBT and the
Timoshenko first order shear deformation theory motivated
The classical theory of beam flexure, also called the Euler- research in the formulation of first, second, third order and
Bernoulli beam theory (EBT) neglects the effects of the higher order shear deformation theories for beams by
transverse shear strains and deformation, and stress Levinson, Reddy, Mindlin, Vlasov and Leontiev, Stein,
concentration [1-5]. The theory is built on the Euler-Bernoulli Touratier, Shimpi, Ghugal, Sayyad etc. [11-14]. Levinson [15],
hypothesis which assumes that the transverse normal to the Krishna Murty [16], Baluch et al. [17], Bhimaraddi and
neutral axis remains normal (orthogonal) to the vertical axis Chandrashekhara [18], Brickford [19] presented parabolic
during and after flexural deformation; and this implies that shear deformation theories of beams by assuming a higher
there are no transverse shear stresses. The disregarding of variation of axial (longitudinal) displacement field in terms of
transverse shear deformation in the formulation of the the thickness coordinate variable, z. Their parabolic shear
governing equations make EBT suitable for thin (slender) deformation beam theories were formulated to apriori satisfy
beams, but unsuitable for moderately thick (deep) beams and the shear stress free boundary conditions of the top and bottom
thick beams where shear deformation plays a significant role surfaces of the beam, and thus remove the necessity of shear
in the flexural behaviour [6-8]. EBT thus underestimates the correction/modification factor. Trigonometric shear
deflections of moderately thick and thick beams where shear deformation theories of beams were formulated and developed
deformation effects significantly contribute to their behaviours. by Touratier [20], Vlasov and Leontiev [21], Stein [22] for
Timoshenko [9] presented a first order shear deformation thick beams. Their formulations however violated the shear
theory for the flexural behaviour of moderately thick and thick stress free boundary conditions at the top and bottom surfaces
beams that accounts for the effects of shear deformation and of the beam. Ghugal and Kapdis et al. [23] and Ghugal [24]
rotatory inertia. The major drawback of his theory is the improved the earlier formulations of trigonometric shear
assumption of the transverse shear strain distribution to be deformation beam theories by presenting shear deformation
uniform across the beam cross-section at any point, thus beam theories that satisfy the shear stress free boundary
requiring problem dependent shear modification factors to conditions at the top and bottom surfaces of the beam.
accurately represent the strain energy of deformation. In the present study, the Timoshenko first order shear
Cowper [10] presented mathematical expressions for the deformation beam theory is derived from fundamental
shear correction/modification factors of beams in terms of the principles, using the method of variational calculus. The
Poisson’s ratio and geometrical properties of different cross- governing system of two ordinary differential equations of
sectional shapes such as inner and outer radii. equilibrium obtained are then solved in closed form for the

34
 𝑡
specific cases of moderately thick rectangular beams, subject and no traction on the lower surface 𝑧 = violates the
2
to uniformly distributed transverse load and moderately thick
theory of elasticity solutions of the problem. The
rectangular cantilever beam with a point load at the free end.
assumption of constant transverse shear strain is however
used to minimize computational and analytical rigours
Research aim and objectives
and any resulting errors introduced are accounted for by
the introduction of shear correction factors, k.
The general aim of this study is to present a general
(vii) The beam material is linear elastic, homogeneous and
formulation of the Timoshenko beam theory and then apply it
isotropic. Hence, the generalised Hooke’s stress-strain
to the flexural analysis of moderately thick beams. The
laws are valid. Thus:
specific objectives include:
(i) to formulate the governing partial differential equations of
1
equilibrium of Timoshenko beams using variational  xx =
E
(  xx −  yy − zz ) (2)
calculus methods.
(ii) to solve the governing system of ordinary differential
equations of the Timoshenko beam for the specific cases 1
of: (a) simply supported moderately thick beams subject to
 yy =
E
(  yy −  xx − zz ) (3)
uniformly distributed transverse load p0 over the entire
beam span; (b) moderately thick cantilever beams subject 1
to point load P at the free end.  zz =
E
( zz −  xx −  yy ) (4)

2. THEORETICAL FRAMEWORK
 xy
 xy = (5)
G
The study considered a thick beam of width b, thickness, t
and span l with a longitudinal axis coincident with the x-  yz
coordinate direction. The origin of the three dimentional (3D)  yz = (6)
G
Cartesian coordinate axis used is defined at the left end of the
beam. The beam thus occupies the 3D region defined as: 0 ≤
𝑏 𝑏 𝑡 𝑡  zx
𝑥 ≤ 𝑙, − ≤ 𝑦 ≤ , − ≤ 𝑧 ≤  zx = (7)
2 2 2 2 G
Variational formulation of the Timoshenko beam theory
where 𝜎𝑥𝑥 , 𝜎𝑦𝑦 , 𝜎𝑧𝑧 are normal stresses 𝜏𝑥𝑧 , 𝜏𝑦𝑧 and 𝜏𝑥𝑦 are
The assumptions of the formulation are: shear stresses, 𝜀𝑥𝑥 , 𝜀𝑦𝑦 , 𝜀𝑧𝑧 are normal strains, while 𝛾𝑥𝑧 , 𝛾𝑦𝑧
(i) The longitudinal axis of the unloaded undeformed beam and 𝛾𝑥𝑦 are shear strains, E is the Young’s modulus of
is straight. elasticity, G is the shear modulus, 𝜇 is the Poisson’s ratio.
(ii) All loads applied to the beam act transverse to the
longitudinal axis. E
(iii) Line elements that are normal to the middle line of the
G= (8)
2(1 + )
beam in the undeformed configuration will deform only
in the vertical direction, and will also remain vertical (vii) The deformations and strains are considered so small,
during deformation. Plane cross-sections of the beam and the strain-displacement equations of infinitesimal
which are initially orthogonal to the longitudinal axis will elasticity are used.
remain plane after deformation. Thus, the kinematic equations are:
(iv) Line elements that are tangential to the middle line (center
line) will undergo a rotation 𝜑(𝑥) which would
u
correspond to the shear angle 𝛾𝑥𝑧 at an arbitrary point  xx = (9)
along the centerline. x
(v) The total slope of the centerline results from the effects of v
 yy = (10)
bending deformation and shear deformation and can be y
expressed as the sum of the rotations due to shear
w
deformation 𝜑(𝑥) and the rotation due to bending  zz = (11)
deformation 𝛼(𝑥). Thus z
u v
 xy = + (12)
w y x
= ( x ) + ( x ) (1)
x u w
 xz = + (13)
z x
(vi) The transverse shear strain 𝛾𝑥𝑧 is constant at all points
v w
over a given cross-section of the beam. This implies the  yz = + (14)
shear stress distribution is constant or uniform over the z y
cross-section at any point on the longitudinal axis. A
constant distribution of shear stress over the cross- Displacement field
section at any point on the longitudinal axis for applied
𝑡
transverse load distribution on the upper surface 𝑧 = − , The displacement field components about the x, y, and z
2

35
coordinate axis given respectively by u(x, y, z), v(x, y, z) and  yz = 0 (30)
w(x, y, z) are given by:
 xy = 0 (31)
 dw 
u( x, y, z) = − z( x ) = − z  − ( x )  (15)
 dx  Introduction of the shear correction (modification) factor, k
yields:
v( x, y, z) = 0 (16)
cxz = kG( x) (32)
w( x, y, z) = w( x, y = 0, z = 0) = w( x) (17)
where the superscript c refers to corrected (modified) shear
It is noted that the displacement component in the stress distribution.
longitudinal direction (x-coordinate) is caused solely by
bending deformation. Internal stress resultants

Strain fields b t
2 2

Applying the strain-displacement relations, the strain fields

M=    xx z dzdy (33)
−b − t
are obtained as: 2 2

b t t
   dw   2 2 2
 xx =  −z  − ( x )   = ( − z ( x ) ) (18) M= b dy t  xx z dz = t  xx bzdz (34)
x   dx   x
− − −
2 2 2

 d 2w d  d ( x )
 xx = − z  2 −  = −z (19) t
2
d 2
 
t
dx dx dx M= −E bz dz (35)
−
dx
2
 xx = −z ( w( x) − ( x)) (20)
t
2
d
 yy =
v
=0 (21)
M = −E t bz 2 dz
dx
(36)
y −
2

w  bz 3  d d
t/2
 zz = =0 (22) M = −E  = − EI (37)
z 
 3  −t / 2 dx dx

u v
 xy = + =0 (23) b t
y x
2 2
Q( x ) =    xz dzdy (38)
−b − t
2 2
v w
 yz = + =0 (24)
z y b t t
2 2 2
Q=  dy   xz dz = b  dz  xz (39)
   dw   w −b −t −t
 xz =  −z  − ( x )   + 2 2 2
z   dx   x
 dw  w Q = bt  xz = A xz = GA( x )
 xz = −  − ( x )  + = ( x ) (25)
 dx  x QT ( x) = kQ = kGA( x ) (40)

Stress fields Total potential energy functional 

The stress fields are obtained from the stress-strain laws as: The total potential energy functional  for the moderately
thick beam is given by the sum of the strain energy U and the
 d 2w d  d ( x ) potential of the external load V as:
 xx = E  xx = − Ez  2 −  = − Ez (26)
 dx dx  dx
 = U +V (41)
 yy = 0 (27)
1
2 
U= ( xx  xx +  yy  yy +  zz  zz
zz = 0 (28) 3 R
+  xy  xy +  yz  yz +  xz  xz )dxdydz (42)
 xz = G  xz = G( x ) (29)

36
where R3 is 0  x  l, − b 2  y  b 2 , − t 2  z  t l
 =  F ( x, w( x ), ( x ), w ( x ))dx
2
(55)
0
l
V = −  p( x ) w( x )dx (43)
1
0
F ( x, w( x ), ( x ), w ( x )) = EI (( x ))2
2
t b
1
1 2 2 l
+ GAk ( w ( x ) − ( x ))2 − p( x )w( x ) (56)
U=  b  ( xx  xx +  xz  xz )dxdydz (44) 2
2 −t − 0
2 2
The integrand F is seen to be a function of two unknown
t b functions, ( x ) and w(x).
2 l
1 2
2 −t b  ( E xx + kG  xz )dxdydz
U= 2 2
(45)
2
−
2
0
F=
1
2
EI (( x ))2 +
GAk
2
(
( w ( x ))2 − 2w ( x )( x )

1 1 )
+ (( x ))2 − p( x )w( x ) (57)
U= 
2 3
E 2xx dxdydz +  kG  2xz dxdydz
2 3
(46)
R R
Differential Equations of Equilibrium
b t
2
 d 
l
1 2 2
The differential equations of equilibrium are the equations
U =  dy   E  − z  dxdz
2 −b  dx  corresponding to the minimization of the total potential energy
−t 0
2 2 functional  and are obtained by applying the Euler-Lagrange
b t
l conditions for the extremum of . Thus, the Euler-Lagrange
1 2 2
+ k  dy   G  (( x ))2 dxdz = U b + U s (47) conditions are the system of differential equations:
2 −b −t 0
2 2
F d  F 
−  =0 (58)
t
2 w dx  w 
 d 
2 l
1
U b = bE  z 2 dz    dx (48)
0
2 −t dx  F d  F 
2 −  =0 (59)
 dx   
2
 d 
l
1
2 0  dx 
Ub = EI   dx (49) F GAk
= ( 2( x ) − 2w ( x ) )
 2
F
1 2
t
l = GAk ( ( x ) − w ( x ) ) (60)
U s = kbG  dz  (( x ))2 dx (50) 
2 −t 0
2
F 1
= EI  2 ( x ) = EI  ( x ) (61)
l  2
1
U s = bGt  (( x ))2 dx (51)
2 0 F
= − p( x ) (62)
w
2
 d 
l
1
= EI    dx
2 0  dx  F GAk
= ( 2w( x ) − 2( x ) ) = GAk ( w( x ) − ( x ) ) (63)
l l
w 2
1
+ GA k (( x ))2 dx −  p( x ) w( x )dx (52)
2 0 0 Thus,

d
GAk ( ( x ) − w ( x ) ) − EI  ( x ) = 0
2
 d 
l
1 (64)
 = EI    dx dx
2 0  dx 

GAk ( ( x) − w( x) ) − EI ( x) = 0

l 2 l
1  dw  (65)
+ GA k  − ( x )  dx −  p( x ) w( x )dx (53)
0  
2 dx 0
−GAk ( w( x) − ( x)) − EI ( x) = 0 (66)
 1  d  2
l
 =   EI  
0  2  dx  d 2  dw 
EI + GAk  − ( x )  = 0 (67)
 dx 
2
2
 dx
1  dw  
+ GAk  − ( x )  − p( x ) w( x ) dx (54)
2  dx  


37
 d  d 4w  d2 p
− p( x ) − GAk ( w ( x ) − ( x ) ) = 0 (68) kGA  p( x ) − EI 4  = EI 2 (82)
dx
 dx  dx

d  dw 
− p( x ) − GAk  − ( x )  = 0 (69)  d 4w  EI d 2 p
dx  dx  p( x ) − EI  4  = 2
(83)
 dx  kGA dx
 d  dw 
GAk   − ( x )   + p( x ) = 0 (70) d 4w EI d 2 p
 dx  dx  EI = p( x ) − (84)
dx 4 kGA dx 2
or,
For the case of simply supported Timoshenko beam subject
to a uniformly distributed transverse load of intensity p0, the
d  dw 
GAk  ( x ) − = p (71) differential equations of equilibrium simplify to the following
dx  dx  system of two equations in ( x ) and w(x):

 d d 2w  d 3( x )
GAk  − 2 = p (72) EI = p0 (85)
 dx dx  dx 3

Decoupling the differential equations of equilibrium d 4w EI d 2 p0

EI = p0 − = p0 (86)
dx 4 kGA dx 2
From,
Summary
dM
= Q( x ) (73)
dx The equations of Timoshenko beam theory are:

dQ( x ) dQ ( x )
= − p( x ) (74) = Q ( x ) = w ( x ) (87)
dx dx

Thus, dM ( x )
= M ( x ) = Q( x ) (88)
dx
d  d  d  2

 − EI  = Q( x ) = − EI 2 (75)
dx  dx  dx d ( x ) M ( x)
= ( x ) = (89)
dx EI
Differentiating again,
du( x ) Q( x )
d  d 2  = u ( x ) = ( x ) − (90)
 − EI 2  = − p( x ) (76) dx kGA
dx  dx 

d 3 3. RESULTS
− EI = − p( x ) (77)
dx 3 Illustrative Problem

d 3 The work considered as a specific application, a moderately

EI = p( x ) (78)
dx 3 thick beam simply supported at the ends x=0 and x=l modelled
as a Timoshenko beam, carrying a uniformly distributed
Differentiating Equation (72) twice, we have transverse load of intensity p0 over the entire span as shown in
Figure 1.
d 2  d d 2w  d 2 p
kGA  − = (79)
dx 2  dx dx 2  dx 2

 d 3 d 4 w  d 2 p
kGA  3 − 4  = 2 (80)
 dx dx  dx
Figure 1. Moderately thick rectangular beam under
uniformly distributed transverse load
 p( x ) d 4 w  d 2 p
kGA  − 4 = 2 (81)
 EI dx  dx The boundary conditions are:

Multiplying by EI, M ( x = 0) = 0 (91)

38
 d 1  p0  l  
( )
2
or, ( x = 0) = 0 (92) d
For x= l =    + c2  = 0 (107)
dx dx 2 EI  2  2  

M( x = l ) = 0 (93) 𝑝0 𝑙2
+ 𝑐2 = 0 (108)
8
d
or, ( x = l) = 0 (94) 𝑝0 𝑙2
dx 𝑐2 = − (109)
8

The origin is chosen at the center of the simply supported Hence

beam to take advantage of the symmetrical nature of the beam
and the loading. The boundary conditions become: 1 𝑝0 𝑥 3 𝑝0 𝑙2
𝛼(𝑥) = ( − 𝑥) (110)
𝐸𝐼 6 8

M x= l( 2 )=0 (95)
Then

𝑑 𝑑𝑤
or,
d
dx
(
x= l =0
2 ) (96)
𝐺𝐴𝑘
𝑑𝑥
(𝛼(𝑥) −
𝑑𝑥
) = 𝑝(𝑥) = 𝑝0 (111)

Integration yields
M x=−l( 2 )=0 (97)
 dw 
GAk  ( x ) −  = p0 x (112)
 dx 
or,
d
dx
(
x=−l =0
2 ) (98)
p x
dw
( x ) − = 0 (113)
dx kGA
Integration of Equation (85) yields:
dw p x
d 2 = ( x ) − 0 (114)
EI = p0 x + c1 (99) dx kGA
dx 2

Integration of Equation (90) yields: dw 1  p0 x 3 p0 l 2 x  p0 x

=  − − (115)
dx EI  6 8  kGA
 d   p0 x
2
EI  = + c1 x + c2 (100)
Integration yields
 dx  2

Integration of Equation (100) yields: dw

w( x ) =  dx dx
x2 x2 
1 p x
3
p l 2 x  p0 x  
EI ( x ) = p0 + c1 + c2 x + c3 (101) =    0 − 0  −  dx (116)
6 2  
EI 6 8 kGA 
  
where c1, c2, and c3 are constants of integration.
The symmetrical nature of the problem demands (requires) 1  p0 x 4 p0 l 2 x 2  p0 x 2
w( x ) =  −  − + c4 (117)
that ( x ) should be an odd function. Thus EI  24 16  2 kGA

( − x ) = −( x ) (102) Applying the boundary conditions,

For 𝛼(𝑥) to be an odd function,

(
w x= l
2 )=0 (118)
c1 = 0 (103)

c3 = 0 (104) (
w x=−l
2 )=0 (119)

and We have,

p0 x 3  
( 2)
4 2
EI ( x ) = + c2 x (105) 1 p l p0 l 2  l 
w x= l =0=    0
−   
6 EI  24  2   16  2  

2
d 1  p0 x 2  p0  l 
=  + c2  (106) −   + c4 (120)
dx EI  2 2 kGA  2 


39
1  p0 l 4 p0 l 4  p0 l 2 1+ 
 − − + c4 = 0 (121) k= (131)
EI  384 64  8kGA 1.305 + 1.273

For a hollow cylinder of outer radius, r0 and inner radius, ri,

p0 l 2 1  −5 p0 l 4 
c4 = −   (122)
8kGA EI  384  6(1 + )(1 + m )2
k= (132)
(7 + 6 )(1 + m )2 + (20 + 12 )m 2
p0 l 2 5 p0 l 4
c4 = + (123)
8kGA 384 EI where m = b (133)
a
Hence,
where a = r0 (134)
p  x 4 l 2 x 2  p0 x 2 p l2 5 p0l 4
w( x ) = 0  −  − + 0 + (124) and r0 is the outer radius, and

EI  24 16  2 kGA 8 kGA 384 EI

b = ri (135)
p  x 4 l 2 x 2 5l 4  p0  l 2 2
w( x ) = 0  − +  +  − x  (125)
EI  24 16 384  2 kGA  4  ri is the inner radius.
For a thin-walled tube,

p
w( x ) = 0
 x 4 l 2 x 2 5l 4 
 − + +
p0 
 l2

( )
4
− x 2 
 (126) k=
2(1 + )
(136)
384  2 kbt  E 4+
2(1 + ) 
EI  24 16
 
Maximum deflection
p l  x  5
4 4 2
3 x
w ( x ) = 0   −   +  The maximum deflection occurs at the center, and is:
24 EI  l  2  l  16 

( )  
 5   t  2(1 + )  1  
2 2
p0 l 2  x −1  p l4 
 l 4  w(0) = 0   +     (137)
−
 
(127) 24 EI  16   l 
 k  4  
2k E bt
 2(1 +  ) 
p0 l 4 
 5  t   1 +  
2

w(0) =  +     (138)
p l4 
  x 
4
3 x
2
5 24 EI 16   
l 2 k 

w( x ) = 0 − 
     + 
  
24 EI  l 2  l  16 
For  = 0.25,
 t  2(1 + )   x  1 
2 2

−     − 
l  
k  l  4 
 
p0 l 4  5  t 
2 
w (0) = + 
1.25  (139)
w ( x ) = w f ( x ) + ws ( x ) (128)
(
24 EI  16  l  2 5.08475
 6 ) 

For a rectangular cross-section, k = 5
6 for  = 0.
p0 l 4  5 t 
2
w(0) =  + 0.7375    (140)
10(1 + ) 24 EI  16  l  
k= (129)
12 + 11
2
5 p0 l 4 p l4  t 
k = 0.847457627 w(0) = + 3.0729166  10−2 0   (141)
5.08475 384 EI EI  l 
k for  = 0.25
6
5.098 wT (0) = w f (0) + ws (0) (142)
k= for  = 0.30
6
5 p0 l 4
w f (0) = (143)
For a circular cross-section, 384 EI

6(1 + ) 2
t p l
4
k= (130)
7 + 6 ws (0) = 3.0721966  10 −2   0 (144)
 l  EI
For a semi-circular cross-section

40
Table 1. Variation of maximum deflection with ratios of t/l Illustrative problem
in Timoshenko beams with simply supported ends (case of
uniform loads) for 𝜇 = 0.25 A moderately thick rectangular cantilever beam of width b
and thickness t, with a span l carries a point load P at the free
t p l4 p l4 p l4 end x = 0 and is fixed at the end x = l, as shown in Figure 2.
l  0  0  0
EI EI EI
ws(0) ws(0) wT(0)
−2 −7
0.005 1.302083  10 7.68229  10 1.30216  10−2
0.01 1.302083  10−2 3.0729166  10−6 1.30239  10−2
−2 −5
0.02 1.302083  10 1.2291626  10 1.3033121  10−2
Figure 2. Moderately thick cantilever beam with a
0.05 1.302083  10−2 7.68229  10−5 1.3097652  10−2
rectangular cross-section under point load
0.08 1.302083  10−2 0.01966  10−2 1.32175  10−2
−2 −4
0.10 1.302083  10 3.0729166  10 1.332812  10−2 Considering an imaginary section x from the fee end, the
0.15 1.302083  10 −2
0.0691406  10 −2
1.3712236  10 −2 bending moment and shear force at the section is found from
−2 −2 equilibrium considerations as
0.20 1.302083  10 0.12291666  10 1.425  10−2
−2 −2
0.40 1.302083  10 0.491666  10 1.79375  10−2 M ( x ) = −Px (148)
0.60 1.302083  10−2 1.10625  10−2 2.408333  10−2
0.80 1.302083  10−2 1.96666  10−2 3.26875  10−2 Q( x) = −P (149)
1.0 1.302083  10−2 3.0729166  10−2 4.375  10−2
The differential equation of equilibrium is
For 𝜇 = 0.30
d
− EI = M ( x ) = −Px (150)
p l 4  5  t   1.3  
2 dx
w(0) = 0  +      (145)
24 EI  16  l   2  5.098
6  d
EI = Px (151)
dx
p0 l 4  5 t 
2
w(0) =  + 0.765    (146) Integrating,
24 EI  16  l  
d
p l4 t p l
4 2
 EI dx dx = EI ( x ) =  Px dx (152)
w(0) = 0 + 3.1875  10−2   0
384 EI  l  EI
w(0) = w0 ( f ) + w0 (s ) (147) Px 2
EI ( x ) = + a1 (153)
2
Table 2. Variation of maximum deflection with ratios of t/l
in Timoshenko beams with simply supported ends (case of where a1 is an integration constant.
uniformly distributed load over the entire beam span l), for Using the boundary condition at the clamped end,
𝜇 = 0.30
( x = l ) = 0 (154)
t p l4 p l4 p l4
l  0  0  0
EI EI EI We have
w0(f) w0(s) wT(0)
0.005 1.302083  10−2 7.96875  10−7 1.3021626  10−2 Pl 2
EI ( x = l ) = + a1 = 0 (155)
0.01 1.302083  10−2 3.1875  10−6 1.3024017  10−2 2
0.02 1.302083  10−2 1.275  10−5 1.303358  10−2
0.05 1.302083  10 −2
7.96875  10 −5
1.3100517  10−2
Pl 2
 a1 = − (156)
2
0.08 1.302083  10−2 2.04  10−4 1.322483  10−2
0.10 1.302083  10−2 3.1875  10−4 1.333958  10−2
Px 2 Pl 2 P ( x 2 − l 2 )
0.15 1.302083  10−2 7.171875  10−4 1.3738017  10−2 EI ( x ) = − = (157)
2 2 2
0.20 −2 −2 −2
1.302083  10 0.1275  10 1.429583  10
−2 −2
0.40 1.302083  10 0.51  10 1.812083  10−2 P( x2 − l 2 )
( x ) = (158)
0.60 1.302083  10−2 1.1475  10−2 2.449583  10−2 2 EI
0.80 1.302083  10−2 2.04  10−2 3.342083  10−2
This yields
1.0 1.302083  10−2 3.1875  10−2 4.489583  10−2

41
 Pxz Pl 3 Pl
 xx = (159) w(0) = w( x = 0) = + (172)
I 3EI kGA

This result for 𝜎𝑥𝑥 is the same result obtained using the Pl 3  3E I 
Euler-Bernoulli beam theory. w(0) = 1 +  (173)
From the second differential equation of equilibrium, 3EI  kG Al 2 

Pl 3  3E 1  t  
2
d  dw  p( x )
 − ( x )  + =0 (160) w(0) = 1 +    (174)
dx  dx  kGA 3EI  kG 12  l  

Integration yields
Pl 3  
2
E t
w(0) = 1 +    (175)

3EI  4Gk  l  
dx
− ( x ) + 
p( x )dx
=0 (161) 
dx kGA
From Equation (8),
dw Q( x )
− ( x ) + =0 (162) E
dx kGA = 2(1 +  ) (176)
G
dw Q( x )
− ( x ) = − (163)
Pl 3  2(1 + )  t  
2
dx kGA
w(0) = 1 +    (177)
3EI  4k  l  

dw Q( x ) P P( x2 − l 2 )
=− + ( x ) = − + (164)
dx kGA kGA 2 EI
Pl 3  1 +   t  
2
w(0) = 1 +    (178)
Integrating, 3EI  2k  l  


dw  −P
 
P( x2 − l 2 )  5.08475
 dx dx =  dw = w( x ) =   kGA + 2 EI
 dx

(165) For  = 0.25, k =
  6
Pl 3  
2
1.25  t 
w(0) = 1 +    (179)
−Px P  x3 2  3EI  2  5.08475 l 
w( x ) = +  − l x  + a2 (166) 6 
kGA 2 EI  3 
Pl 3  
2
t
where a2 is the constant of integration. w(0) = 1 + 0.7375   
Using the boundary conditions at the fixed end, 3EI  l 

w (0) = w f (0) + ws (0) (180)
w( x = l ) = 0 (167)
Table 3. Variation of maximum deflection with ratios of t/l
We have in cantilevered Timoshenko beams under point load P at the
free end (for 𝜇 = 0.25)
−Pl P  l3 3 
w( x = l ) = 0 = +  − l  + a2 (168)
kGA 2 EI  3 
t Pl 3 Pl 3 Pl 3
l   
EI EI EI
wf(0) ws(0) w(0)
Pl P  2l 3 
a2 = − − 
0.005 0.333333 6.145833  10−6 0.33333
kGA 2 EI  3  0.01 0.333333 2.45833  10−5 0.333358
3 3 −5
Pl 2Pl Pl Pl 0.02 0.333333 9.8333  10 0.333432
a2 = + = + (169)
kGA 6 EI kGA 3EI 0.05 0.333333 −4 0.333948
6.14583  10
0.08 0.333333 1.57333  10−3 0.33491
Thus, −3
0.10 0.333333 2.45833  10 0.335792
0.15 0.333333 −3 0.338865
P  x3 5.53125  10
2  Px Pl 3 Pl
w( x ) =  − l x − + + (170)
2 EI  3
0.20 0.333333 9.8333  10−3 0.343167
 kGA 3 EI kGA
0.40 0.333333 0.039333 0.372666
0.60 0.333333 0.0885 0.421833
P  x3 l 2 l3  P 0.80 0.333333 0.157333 0.490667
w( x ) =  − x+ + (l − x )
EI  6 2 3  kGA 1.0 0.333333 0.245833 0.5791666

w ( x ) = w f ( x ) + ws ( x ) (171)

42
 5.098 and (84). Two specific illustrative cases of the closed form
For  = 0.3, k = analytical solution of the Timoshenko beam equations were
6
considered and solved. They were:
(i) moderately thick beam of rectangular cross-section
Pl 3  
2
1.3  t 
w(0) = 1 +    (181) simply supported at x =  l and subject to a uniformly
3EI  2  5.098
6  
l 

2
distributed load of intensity p0, and
(ii) moderately thick rectangular cantilever beam of span l
Pl 3  
2
t and beam b and thickness t with a point load P, applied at
w(0) = 1 + 0.765    (182) x=0, and fixed (clamped) at x=l.
3EI  l 
 Successive integration of the governing ODE was used to
obtain the general solution for the unknown rotation 𝛼(𝑥) in
Table 4. Variation of maximum deflection with ratios of t/l terms of three integration constants as Equation (101). The
in cantilevered Timoshenko beams under point load P at the requirement that 𝛼(𝑥) be an odd function, imposed by the
free end (for 𝜇 = 0.30) symmetry of the problem was used to obtain two integration
constants c1 and c3 as Equations (103) and (104). The third
t Pl 3 Pl 3 Pl 3 unknown integration constant c2 was obtained using the
l   
EI EI EI boundary conditions at the simply supported ends as Equation
wf(0) ws(0) w(0) (109). The solution for 𝛼(𝑥) was thus fully determined as
0.005 0.33333 6.375  10−6 0.33334 Equation (110). Equation (110) was used to obtain the
0.01 0.33333 2.55  10−5 0.33336 differential equation for w(x) as Equation (115). Integration of
Equation (115) with respect to x gave the general solution for
0.02 0.33333 1.02  10−4 0.33344
w(x) as Equation (117). Application of the boundary
0.05 0.33333 −4 0.33397
6.375  10 conditions Equations (118) and (119) gave the constant of
0.08 0.33333 1.632  10−3 0.334965 integration c4 as Equation (123), yielding the solution for w(x)
0.10 0.33333 2.55  10−3 0.33588 as Equation (125). The solution for w(x) is observed to be
−3
expressed in terms of a flexural component wf(x) and a shear
0.15 0.33333 5.7375  10 0.339071
component ws(x). The maximum deflection was observed to
0.20 0.33333 0.0102 0.34353 occur at the center x = 0, and was found as Equation (138). For
0.40 0.33333 0.0408 0.374133 𝜇 = 0.25, the maximum deflection was obtained as Equation
0.60 0.33333 0.0918 0.425133
(141), which is decomposable into flexural and shear
0.80 0.33333 0.1632 0.49653
components. For 𝜇 = 0.30, the maximum deflection was
1.0 0.33333 0.255 0.58833
obtained as Equation (147) which is also expressible as shear
component and flexural components. In general, the
expression for wmax is observed to depend on the ratio of the
4. DISCUSSION
beam thickness to the beam span l. Values of wf(0), ws(0) and
w(0) for various values of t/l ranging from t/l =0.005 to t/l=1
This work has successfully presented a variational
are tabulated for Poisson ratio values 𝜇 = 0.25 and 𝜇 = 0.30,
formulation of the Timoshenko theory for the static flexural
and presented as Tables 1 and 2. Tables 1 and 2 show that for
analysis of moderately thick beams with rectangular cross-
t/l <0.02 the contribution of shear to the overall deflection is
sections.
insignificant being less than 0.1 %. At t/l =0.1 for 𝜇 = 0.25,
The formulation assumed a linear elastic, isotropic,
the contribution of shear to the total deflection increases to
homogenous beam material and accounted for the effect of
transverse shear deformation. The displacement field 2.305 %, and to 46 % for t/l =0.60 for 𝜇 = 0.25. For the
components were assumed as Equations (15)-(17), and the moderately thick cantilever beam, the unknown rotation 𝛼(𝑥)
kinematic relations for small displacement elasticity – was obtained by integration and use of boundary conditions at
Equations (9-14) – used to obtained the strain fields as the clamped end as Equation (158). The deflection w(x) was
Equations (18), (20), (21), (22), (23), (24) and (25). Hooke’s obtained using Equation (158) and the Timoshenko beam
generalised stress-strain laws given as Equations (2)–(7) were equation as the ODE Equation (164). The general solution by
used to obtain the stress fields as Equations (26)–(31). The integration was found as Equation (166). Application of
internal stress resultants were obtained from the requirement boundary conditions at the fixed end yielded the unknown
of equilibrium of internal and external forces as Equations (37) constant of integration as Equation (169). The solution for w(x)
and (40). The total potential energy functional  was was thus obtained as Equation (171), which is decomposable
obtained as the sum of the strain energy U and the potential of into flexural and shear components. The maximum deflection
the external load V as Equation (54). The total potential energy was observed to occur at the free end and is given by Equation
functional  was observed to be a function of one (178). The maximum deflection was observed to depend upon
independent variable, x, and two unknown functions w(x) and the ratio t/l. For 𝜇 = 0.25, maximum deflection expression
𝛼(𝑥) and derivatives of the two unknown functions. The was found as Equation (180); while for 𝜇 = 0.30, the
Euler-Lagrange conditions which are a system of two maximum deflection expression was obtained as Equation
differential equations were used to obtain the differential (182). Values of the maximum deflection for various values of
equations of equilibrium of the Timoshenko beam under static the ratio t/l for Poisson’s ratio 𝜇 = 0.25, and 𝜇 = 0.30 are
flexure as a system of two coupled ordinary differential presented in tabular form as Tables 3 and 4 respectively for the
equations – Equations (64) and (70). The system of two moderately thick rectangular cantilever beam problem.
coupled ordinary differential equations are expressed in
decoupled form for homogeneous beams as Equations (78)

43
5. CONCLUSIONS deformation theory. International Journal of Engineering
Research 5(3): 565-571.
The following conclusions are drawn from this study: [7] Sayyad A. (2012). Static flexure and free vibration
(i) the problem of flexure of moderately thick beams analysis of thick order shear deformation theories.
modelled and idealized using Timoshenko’s first order International Journal of Applied Mathematics and
shear deformation theory can be presented in variational Mechanics 8(14): 71-87.
form as a problem of finding the unknown displacements [8] Ghugal Y. (2006). A two dimensional exact elasticity
w(x) and 𝛼(𝑥) that minimize the total potential energy solution of thick beam. Departmental Report 1.
functional  or in differential form in terms of the Department of Applied Mechanics, Government
system of two coupled ordinary differential equations Engineering College, Aurangabad India, pp 1-96.
obtained by use of the Euler-Lagrange conditions on the [9] Timoshenko SP. (1921). On the correction for shear of
total potential energy functional. differential equation for transverse vibrations of
(ii) the Timoshenko beam flexure problem can be formulated prismatic bars. Philosophical Magazine, Series 6 41:
and solved using variational calculus methods. 742-746.
(iii) mathematical solutions obtained show that the shear [10] Cowper GR. (1966). The shear coefficients in
component of transverse deflections are expressed in Timoshenko beam theory. ASME Journal of Applied
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the span l. [11] Ghugal YM, Sharma R. (2009). Hyperbolic shear
(iv) mathematical expressions obtained for the transverse deformation theory for flexure and vibration of thick
deflection show the transverse deflection is decomposed isotropic beams. International Journal of Computational
into a shear component and a flexure component. Methods 6(4): 585-604.
(v) the analytical expression obtained for the flexure [12] Ghugal YM, Shimpi RP. (2002). A review of refined
component of the transverse deflection are exactly the shear deformation theories for isotropic and anisotropic
same as obtained by the Euler-Bernoulli theory. laminated beams. Journal of Reinforced Plastics and
(vi) as the ratio of t/l becomes very small, t/l<0.02 the Composites 21: 775-813.
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(overall) deflection is insignificant, being less than 0.1 % supported beam using trigonometric shear deformation
and the solutions obtained become very close to the theory. International Journal of Scientific and Research
solutions obtained for the Euler-Bernoulli beam theory. Publications 2(11): 1-7. 10.29322
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deflection becomes significant as t/l >0.1 and increases to using refined shear deformation theory. International
46 % for t/l=0.60 for the case of simply supported Journal of Civil and Structural Engineering 3(2): 321-
moderately thick beam for 𝜇 = 0.25. 335.
[15] Levinson M. (1981). A new rectangular beam theory.
Journal of Sound and Vibration 74: 81-87.
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analysis of thick isotropic beam using hyperbolic shear [24] Ghugal YM. (2006). A simple higher order theory for

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 beam with transverse shear and transverse normal effect. A area of cross-section
Departmental Report No 4, Applied Mechanics I moment of inertia
Department, Government College of Engineering, F integrand in the total potential energy
Aurangabad, India, pp. 1-96. functional

partial derivative with respect to w
w
NOMENCLATURE
w( x ) derivative of w(x) with respect to x
l length of beam a1 a2 
 constants of integration
b width of beam c1 c2 c3 c4 
t thickness of beam p0 intensity of uniformly distributed load
x, y, z three dimensional Cartesian coordinate P point load
3D three dimensional (three dimensions)
( x) rotation due to shear deformation  integral

 xz transverse shear angle  double integral

( x ) rotation due to bending deformation
x longitudinal coordinate axis of beam
 triple integral or volume integral

w transverse deflection d
ordinary derivative with respect to x
 xx ,  yy ,  zz normal strains dx
 xy ,  yz ,  xz shear strains
Subscripts
 xx ,  yy ,  zz normal stresses
 xy ,  yz ,  xz shear stresses f flexural
s shear
G shear modulus
b bending
E Young’s modulus of elasticity
max maximum
 Poisson’s ratio
u, v, w displacement field components in the x, y, Superscripts
and z coordinate directions respectively
k shear correction (modification) factor c corrected
M(x) bending moment T Timoshenko
Q(x) shear force
U strain energy functional Abbreviations
V potential of external load
 total potential energy functional EBT Euler-Bernoulli beam theory
p(x) applied load distribution R3 three dimensional region of integration

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