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: [email protected]
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 푡 specific cases of moderately thick rectangular beams, subject and no traction on the lower surface 푧 = violates the to uniformly distributed transverse load and moderately thick 2 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 =( xx − yy − zz ) (2) calculus methods. E (ii) to solve the governing system of ordinary differential equations of the Timoshenko beam for the specific cases 1 yy =( yy − xx − zz ) (3) of: (a) simply supported moderately thick beams subject to E 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 =( zz − xx − yy ) (4) E
2. THEORETICAL FRAMEWORK =xy (5) xy 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 ≤ 푏 푏 푡 푡 푥 ≤ 푙, − ≤ 푦 ≤ , − ≤ 푧 ≤ =zx (7) 2 2 2 2 zx G
Variational formulation of the Timoshenko beam theory where 휎 , 휎 , 휎 are normal stresses 휏 , 휏 and 휏 are 푥푥 푦푦 푧푧 푥푧 푦푧 푥푦 The assumptions of the formulation are: shear stresses, 휀푥푥, 휀푦푦, 휀푧푧 are normal strains, while 훾푥푧, 훾푦푧 (i) The longitudinal axis of the unloaded undeformed beam and 훾푥푦 are shear strains, E is the Young’s modulus of is straight. elasticity, G is the shear modulus, 휇 is the Poisson’s ratio. (ii) All loads applied to the beam act transverse to the longitudinal axis. E G = (8) (iii) Line elements that are normal to the middle line of the 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 휑(푥) which would u correspond to the shear angle 훾푥푧 at an arbitrary point =xx (9) along the centerline. x (v) The total slope of the centerline results from the effects of v = (10) bending deformation and shear deformation and can be yy y expressed as the sum of the rotations due to shear w deformation 휑(푥) and the rotation due to bending =zz (11) deformation 훼(푥). Thus z uv = + (12) w xy yx = ()()xx + (1) x uw = + (13) xz zx (vi) The transverse shear strain 훾 is constant at all points 푥푧 vw over a given cross-section of the beam. This implies the yz = + (14) shear stress distribution is constant or uniform over the zy 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 푡 transverse load distribution on the upper surface 푧 = − , 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) c xz =kG () x (32) (17) w(,,) x y z= w (, x y = 0, z = 0) = w () x 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 bt 22 M =z dzdy (33) Applying the strain-displacement relations, the strain fields xx −−bt are obtained as: 22
b t t dw 2 2 2 xx = −z − () x =() − z () x (18) M= dy z dz = bzdz (34) x dx x xx xx −b − t − t 2 2 2
d2 w d d() x = −zz − = − (19) t xx 2 2 d dx dx dx M=− E bz2 dz (35) dx − t 2 xx = −z( w()() x − x ) (20) t 2 d v M=− E bz2 dz (36) yy = = 0 (21) dx − t y 2
w 3 t/2 = = 0 (22) bz d d zz z ME= − = −EI (37) 3 dx dx −t/2
uv = + = 0 (23) bt xy yx 22 (38) Q() x = xz dzdy −−bt vw 22 yz = + = 0 (24) zy b t t 2 2 2 (39) Q= dy xz dz = b dz xz dw w −b − t − t 2 2 2 xz = −zx − () + z dx x
dw w Q= bt xz = A xz = GA () x xz = − − ()()xx + = (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 d2 w d d() x potential of the external load V as: =E = − Ez − = −Ez (26) xx xx 2 dx dx dx =UV + (41)
=0 (27) yy 1 U =( xx xx + yy yy + zz zz 2 R3 =zz 0 (28) + xy xy + yz yz + xz xz )dxdydz (42)
xz =G xz = G () x (29)
36 3 bbtt l where R is 0,xl −22 y , −22 z = F(, x w (), x (), x w ()) x dx (55) 0 l V=− p()() x w x dx (43) 1 0 F(,(),(), x w x x w ()) x = EI (()) x 2 2 tb l 1 1 22 + GAk( w ( x )− ( x ))2 − p ( x ) w ( x ) (56) U =() xx xx + xz xz dxdydz (44) 2 2 −−tb0 22 The integrand F is seen to be a function of two unknown tb functions, ()x and w(x). 1 22l U = ()E22 + kG dxdydz (45) 2 xx xz −−tb0 1 22GAk 22 F= EI(()) x + ( (()) w x− 2()() w x x 22 11 +( (x ))2 − p ( x ) w ( x ) (57) U= E 22 dxdydz + kG dxdydz (46) ) 22xx xz RR33 Differential Equations of Equilibrium bt l 2 1 22d The differential equations of equilibrium are the equations U=− dy E z dxdz 2 dx corresponding to the minimization of the total potential energy −−bt0 22 functional and are obtained by applying the Euler-Lagrange bt l conditions for the extremum of . Thus, the Euler-Lagrange 1 22 + k dy G( ( x ))2 dxdz = U + U (47) conditions are the system of differential equations: 2 bs −−bt0 22 F d F −=0 (58) t 2 l 2 w dx w 1 2 d Ub = bE z dz dx (48) 2 dx − t 0 F d F 2 −=0 (59) dx l 2 1 d (49) Ub = EI dx F GAk 2 0 dx =(2 (x ) − 2 w ( x )) 2
t F l 1 2 =GAk( ()() x − w x ) (60) U= kbG dz( ( x ))2 dx (50) s 2 − t 0 2 F 1 =EI 2 ( x ) = EI ( x ) (61) 2 1 l U= bGt( ( x ))2 dx (51) s 2 0 F =−px() (62) w l 2 1 d = EI dx 2 dx F GAk 0 = (2w ( x )− 2 ( x ))( = GAk w ( x ) − ( x )) (63) w 2 1 ll +GAk( ( x ))2 dx − pxwxdx ( ) ( ) (52) 2 00 Thus,
l 2 d 1 d GAk(( x ) − w ( x )) − EI ( x ) = 0 (64) = EI dx dx 2 dx 0 ll2 1 dw GAk()( x ) − w ( x ) − EI ( x ) = 0 (65) + GAk− ()()() x dx − pxwxdx (53) 2 dx 00 −GAk() w( x ) − ( x ) − EI ( x ) = 0 (66) l 2 1 d = EI 2 2 dx d dw 0 EI2 + GAk − ( x ) = 0 (67) 2 dx dx 1 dw +GAk − ()()() x − p x w x dx (54) 2 dx
37 d d42 w d p −p() x − GAk( w( x )− ( x )) = 0 (68) kGA p() x−= EI EI (82) dx 42 dx dx
d dw 42 −p() x − GAk − (x ) = 0 (69) d w EI d p dx dx p() x−= EI (83) dx42kGA dx
d dw GAk − (x ) + p ( x ) = 0 (70) d42 w EI d p dx dx (84) EI42=− p() x dx kGA dx or, For the case of simply supported Timoshenko beam subject to a uniformly distributed transverse load of intensity p , the d dw 0 GAk() x − = p (71) differential equations of equilibrium simplify to the following dx dx system of two equations in ()x and w(x):
d d2 w 3 (72) dx() GAk −=2 p EI= p (85) dx dx dx3 0
Decoupling the differential equations of equilibrium 4 2 d w EI dp0 EI= p00 − = p (86) dx42kGA dx From,
Summary dM = Qx() (73) dx The equations of Timoshenko beam theory are:
dQ() x dQ() x =−px() (74) ==Q()() x w x (87) dx dx
Thus, dM() x ==M()() x Q x (88) dx 2 d dd −EI = Q() x = − EI 2 (75) dx dx dx d() x M() x = ()x = (89) dx EI Differentiating again,
du() x Q() x dd2 =u()() x = x − (90) (76) dx kGA −EI2 = − p() x dx dx
d 3 3. RESULTS (77) −EI3 = − p() x dx Illustrative Problem
d3 EI= p() x (78) The work considered as a specific application, a moderately dx3 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.
d2 d d2 w d 2 p kGA 2 −=2 2 (79) dxdx dx dx
d3 d 4 w d 2 p
kGA3−= 4 2 (80) dx dx dx Figure 1. Moderately thick rectangular beam under uniformly distributed transverse load p() x d42 w d p (81) kGA−=42 EI dx dx The boundary conditions are:
Multiplying by EI, Mx(== 0) 0 (91)
38 d 2 or, (x == 0) 0 (92) dl l 1 p0 For ()xc= = +2 = 0 (107) dx dx 2 EI 22
M( x== l ) 0 (93) 푝 푙2 0 + 푐 = 0 (108) 8 2 d or, (xl== ) 0 (94) 푝 푙2 dx 푐 = − 0 (109) 2 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 푝 푥3 푝 푙2 훼(푥) = ( 0 − 0 푥) (110) 퐸퐼 6 8
Mx==l 0 (95) ()2 Then
푑 푑푤 d 퐺퐴푘 (훼(푥) − ) = 푝(푥) = 푝 (111) or, ()x ==l 0 (96) 푑푥 푑푥 0 dx 2 Integration yields Mx= −l = 0 (97) ()2 dw GAk() x − = p0 x (112) dx d or, x = −l = 0 (98) dx ()2 dw px0 ()x − = (113) dx kGA Integration of Equation (85) yields:
dw px d 2 = ()x − 0 (114) (99) EI2 =+ p01 x c dx kGA dx
32 Integration of Equation (90) yields: dw 1 p0 x p 0 l x p0 x = − − (115) dx EI 68kGA 2 d px0 EI = +c12 x + c (100) dx 2 Integration yields
dw Integration of Equation (100) yields: w() x= dx dx xx22 32 1 p0 x p 0 l x p0 x EI() x = p0 + c 1 + c 2 x + c 3 (101) = − − dx (116) 62 EI 68kGA
where c1, c2, and c3 are constants of integration. 1 p x4 p l 2 x 2 p x2 The symmetrical nature of the problem demands (requires) w() x =0 − 0 −0 + c (117) 4 that ()x should be an odd function. Thus EI 24 16 2kGA
()() −xx = − (102) Applying the boundary conditions,
For 훼(푥) to be an odd function, wx==l 0 (118) ()2 c1 = 0 (103)
wx= −l = 0 (119) c3 = 0 (104) ()2 and We have,
px3 422 0 1 p00ll p l EI() x = + c2 x (105) wx=l =0 = − 6 ()2 EI 24 2 16 2
2 d 1 px2 p0 l =+0 c (106) −+ c4 (120) 2 22kGA dx EI 2
39 1 p l4 p l 4 p l 2 1+ 0− 0 − 0 +c = 0 (121) k = (131) 4 EI 384 64 8kGA 1.305+ 1.273
For a hollow cylinder of outer radius, r0 and inner radius, ri, p l241 −5 p l c =−00 (122) 4 8kGA EI 384 6(1+ )(1 + m )2 k = (132) (7+ 6 )(1 +mm )22 + (20 + 12 ) 24 p00 l5 p l c4 =+ (123) 8kGA 384 EI where m = b (133) a Hence, where ar= 0 (134) p x4 l 2 x 2 p x2 p l 25 p l 4 wx()=0 − −0 + 0 + 0 (124) EI 24 16 2kGA 8 kGA 384 EI and r0 is the outer radius, and
b = ri (135) 4 2 2 4 2 pp x l x5 l l 2 w() x =00 − + + − x (125) EI 24 16 384 2kGA 4 ri is the inner radius. For a thin-walled tube, 2 l − x2 ppx4 l 2 x 25 l 4 ()4 2(1+ ) wx()=00 − + + (126) k = (136) EI 24 16 384 2kbt E 4 + 2(1+ ) Maximum deflection 4 42 pl0 xx 35 wx()= − + The maximum deflection occurs at the center, and is: 24EI l 2 l 16 2 2 2 x 1 pl4 5 t 2(1+ ) 1 pl0 − 0 ()l 4 w(0) =+ (137) − (127) 24EI 16 l k 4 2k E bt 2(1+ ) 4 2 pl0 51t + w(0) =+ (138) 4 42 24EI 16 l 2 k pl0 xx 35 wx()= − + 24EI l2 l 16 For =0.25, 22 tx 2(1+ ) 1 −− l k l 4 4 2 pl0 5 t 1.25 w()()() x=+ w x w x (128) w(0) =+ (139) fs 24EI 16 l 2 5.08475 ()6 5 For a rectangular cross-section, k = 6 for =0. 4 2 pl0 5 t w(0) =+0.7375 (140) 10(1+ ) 24EI 16 l k = (129) 12+ 11 442 5p00 l −2 p l t k = 0.847457627 w(0)= + 3.0729166 10 (141) 5.08475 384EI EI l k for =0.25 6 T 5.098 w(0)=+ wfs (0) w (0) (142) k = for =0.30 6 5pl4 w (0) = 0 (143) For a circular cross-section, f 384EI
6(1+ ) 2 4 k = (130) −2 t pl0 76+ ws (0)= 3.0721966 10 (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 휇 = 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 pl4 pl4 pl4 end x = 0 and is fixed at the end x = l, as shown in Figure 2. l 0 0 0 EI EI EI T ws(0) ws(0) w (0) 0.005 1.302083 10−2 7.68229 10−7 1.30216 10−2 0.01 −2 −6 −2 1.302083 10 3.0729166 10 1.30239 10 0.02 1.302083 10−2 1.2291626 10−5 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 −2 −2 −2 1.302083 10 0.01966 10 1.32175 10 0.10 1.302083 10−2 3.0729166 10−4 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 0.20 −2 −2 −2 equilibrium considerations as 1.302083 10 0.12291666 10 1.425 10 0.40 1.302083 10−2 0.491666 10−2 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 휇 = 0.30 d −EI = M() x = − Px (150) 4 2 dx pl0 5t 1.3 w(0) =+ (145) 24EI 16 l 2 5.098 6 d EI= Px (151) dx 4 2 pl0 5 t w(0) =+0.765 (146) 24EI 16 l Integrating, d 442 EI dx= EI () x = Px dx (152) p00 l −2 t p l dx w(0)= + 3.1875 10 384EI l EI w(0)=+ w ( f ) w ( s ) (147) Px2 00 EI() x = + a (153) 2 1 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, 휇 = 0.30 (xl = ) = 0 (154) t pl4 pl4 pl4 l 0 0 0 EI EI EI We have T w0(f) w0(s) w (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 2 −2 −5 −2 Pl 0.05 1.302083 10 7.96875 10 1.3100517 10 a = − (156) 1 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 Px2 Pl 2 P() x 2− l 2 0.15 −2 −4 −2 EI() x = − = (157) 1.302083 10 7.171875 10 1.3738017 10 2 2 2 0.20 1.302083 10−2 0.1275 10−2 1.429583 10−2 0.40 1.302083 10−2 0.51 10−2 1.812083 10−2 P() x22− l =()x (158) 0.60 1.302083 10−2 1.1475 10−2 2.449583 10−2 2EI 0.80 1.302083 10−2 2.04 10−2 3.342083 10−2 1.0 −2 −2 −2 This yields 1.302083 10 3.1875 10 4.489583 10
41 Pxz Pl3 Pl =xx (159) w(0)= w ( x = 0) = + (172) I 3EI kGA
This result for 휎 is the same result obtained using the 3 푥푥 Pl3 E I (173) Euler-Bernoulli beam theory. w(0)=+ 1 2 From the second differential equation of equilibrium, 3EI kG Al
2 d dw p() x Pl3 31 E t − (x ) + = 0 (160) w(0)=+ 1 (174) dx dx kGA 3EI kG 12 l
Integration yields 3 2 Pl E t w(0)=+ 1 (175) dx p() x dx 34EI Gk l − ()x + = 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) 2 dx kGA Pl3 2(1+ ) t w(0)=+ 1 (177) 34EI k l dw Q() x P P() x22− l = − + ()x = − + (164) dx kGA kGA2 EI 2 Pl3 1+t w(0)=+ 1 (178) Integrating, 32EI k l
dw −−P P() x22 l 5.08475 dx= dw = w() x = + dx (165) For =0.25, k = dx kGA2 EI 6 2 Pl3 1.25 t 3 w(0)=+ 1 (179) 5.08475 −Px P x 2 32EI l w() x = + −l x + a2 (166) 6 kGA23 EI 3 2 Pl 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)=+ wfs (0) w (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 휇 = 0.25) 3 −Pl P l 3 w( x= l ) = 0 = + −l + a2 (168) t Pl3 Pl3 Pl3 kGA23 EI l EI EI EI wf(0) ws(0) w(0) 3 Pl P2 l 0.005 0.333333 −6 0.33333 a = − − 6.145833 10 2 kGA23 EI 0.01 0.333333 2.45833 10−5 0.333358 33 Pl2 Pl Pl Pl 0.02 0.333333 9.8333 10−5 0.333432 a2 = + = + (169) kGA63 EI kGA EI 0.05 0.333333 6.14583 10−4 0.333948
0.08 0.333333 1.57333 10−3 0.33491 Thus, 0.10 0.333333 −3 0.335792 2.45833 10 0.15 0.333333 −3 0.338865 P x33 Px Pl Pl 5.53125 10 w() x = −l2 x − + + (170) 0.20 0.333333 9.8333 10−3 0.343167 2EI 3 kGA3 EI kGA 0.40 0.333333 0.039333 0.372666 0.60 0.333333 0.0885 0.421833 P x3 l 2 l 3 P 0.80 0.333333 0.157333 0.490667 w() x = −x + +()l − x 1.0 0.333333 0.245833 0.5791666 EI 6 2 3 kGA
w()()() x=+ wfs x w x (171)
42 5.098 and (84). Two specific illustrative cases of the closed form For =0.3, k = 6 analytical solution of the Timoshenko beam equations were considered and solved. They were: (i) moderately thick beam of rectangular cross-section 3 2 Pl 1.3 t l w(0)=+ 1 (181) simply supported at x = and subject to a uniformly 32EI 5.098 l 2 6 distributed load of intensity p0, and (ii) moderately thick rectangular cantilever beam of span l 2 Pl3 t and beam b and thickness t with a point load P, applied at w(0)=+ 1 0.765 (182) 3EI l x=0, and fixed (clamped) at x=l. Successive integration of the governing ODE was used to obtain the general solution for the unknown rotation 훼(푥) 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 훼(푥) be an odd function, imposed by the free end (for 휇 = 0.30) symmetry of the problem was used to obtain two integration
constants c1 and c3 as Equations (103) and (104). The third t Pl3 Pl3 Pl3 l unknown integration constant c2 was obtained using the EI EI EI boundary conditions at the simply supported ends as Equation wf(0) ws(0) w(0) (109). The solution for 훼(푥) 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 0.02 0.33333 −4 0.33344 Equation (115) with respect to x gave the general solution for 1.02 10 w(x) as Equation (117). Application of the boundary 0.05 0.33333 6.375 10−4 0.33397 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) as Equation (125). The solution for w(x) is observed to be 0.10 0.33333 2.55 10−3 0.33588 expressed in terms of a flexural component wf(x) and a shear 0.15 0.33333 5.7375 10−3 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 휇 = 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 휇 = 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 w is observed to depend on the ratio of the 4. DISCUSSION max beam thickness to the beam span l. Values of w (0), w (0) and f s 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 휇 = 0.25 and 휇 = 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 휇 = 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 휇 = 0.25. For the components were assumed as Equations (15)-(17), and the moderately thick cantilever beam, the unknown rotation 훼(푥) 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 휇 = 0.25, maximum deflection expression 훼(푥) and derivatives of the two unknown functions. The was found as Equation (180); while for 휇 = 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 휇 = 0.25, and 휇 = 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 훼(푥) 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 terms of the square of the ratio of the beam thickness t to Mechanics 33: 335-340. 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. contribution of the shear deformation to the resultant [13] Dahake AG, Ghugal YM. (2012). Flexure of thick simply (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 (vii) the contribution of shear deformation to the overall [14] Ghugal YM, Dahake AG. (2012). Flexure of thick beams 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 휇 = 0.25. 335. [15] Levinson M. (1981). A new rectangular beam theory. Journal of Sound and Vibration 74: 81-87. REFERENCES https://doi.org/10.1016/0022-460x(81)90493-4. [16] Krishna Murty AV. (1984). Toward a consistent beam [1] Ike CC, Ikwueze EU. (2018). Ritz method for the theory. AIAA Journal 22: 811-816. analysis of statically indeterminate Euler-Bernoulli [17] Baluch MH, Azad AK, Khidir MA. (1984). Technical beams. Saudi Journal of Engineering and Technology theory of beams with normal strain. Journal of the (SJEAT) 3(3): 133-140. Engineering Mechanics, Proceedings of the ASCE 110: https://doi.org/10.21276/sjeat20183.3.3 1233-1237. [2] Ike CC, Ikwueze EU. (2018). Fifth degree Hermittian [18] Bhimaraddi A, Chandreshekhara K. (1993). polynomial shape functions for the finite element Observations on higher-order beam theory. Journal of analysis of clamped simply supported Euler-Bernoulli Aerospace Engineering, Proceedings of ASCE, beam. American Journal of Engineering Research (AJER) Technical Note 6: 403-413. 7(4): 97-105. [19] Brickford WB. (1982). A consistent higher order beam [3] Ghugal YM, Dahake AG. (2013). Flexure of simply theory. Development in Theoretical Applied Mechanics supported thick beams using refined shear deformation SECTAM 11: 137-150. theory. World Academy of Science, Engineering and [20] Touratier M. (1991). An efficient standard plate theory. Technology, International Journal of Civil, Architectural, International Journal of Engineering Science 29(8): 901- Structural and Construction Engineering 7(1): 82-91. 916. https://doi.org/10.1999/1307-6892/9996869 [21] Vlasov VZ, Leontiev UN. (1996). 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44 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 wx() derivative of w(x) with respect to x l length of beam aa12 constants of integration b width of beam c1 c 2 c 3 c 4 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) integral ()x rotation due to shear deformation double integral xz transverse shear angle ()x rotation due to bending deformation triple integral or volume integral x longitudinal coordinate axis of beam d w transverse deflection ordinary derivative with respect to x dx xx,, yy zz normal strains
,, shear strains xy yz xz Subscripts xx,, yy zz normal stresses f flexural xy,, yz xz shear stresses 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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