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  • Deflection of Non-Uniform Beams Resting on a

    Non-Linear Elastic Foundation Using (GDQM)

    Ramzy M. Abumandour1, Islam M. Eldesoky

    1, Mohamed A. Safan

    2, R. M. Rizk-Allah

    1, and Fathi A.

    Abdelmgeed1

    1Basic Engineering Sciences Department, Faculty of Engineering, Menofia University, Egypt

    2Department of Civil Engineering, Faculty of Engineering, Menoufia University, Egypt

    Email: { ramzy_0000, eldesokyi, msafan2000, rizk_masoud, fathi_azeem}@yahoo.com

    AbstractIn this study, a method of a new technique of

    GDQM is presented for determining the deflection of a non-

    uniform beam resting on a non-linear elastic foundation,

    subjected to axial and transverse distributed force. The

    nonlinear subgrade model which describes the foundation

    includes the linear and nonlinear Winkler (normal)

    parameters and the linear Pasternak (shear) foundation

    parameter. The nonlinear 4th order differential equation of

    beam is solved using a new technique. In construction of

    numerical scheme a GDQM is used to transform the

    differential equation into a set of nonlinear algebraic

    equations. Then, these nonlinear algebraic equations solved

    by Newtons method. Comparison the present results with

    the previous solutions proves the accuracy of this

    combination.

    Index Termsnonlinear elastic foundation, deflection,

    GDQM and newtons method

    I. INTRODUCTION

    The analysis of beams resting on elastic foundation

    subjected to axial loading are very common in structural

    systems under actual operating conditions [1] and

    therefore an accurate and reliable method of analysis is

    required especially when the properties of their cross-

    section are variable. The applications for beams resting

    on elastic foundation are one of the important topics in

    many engineering fields due to its wide applications in

    engineering, such as mainly: In civil engineering e.g. in

    the design of structural components in buildings, aircrafts,

    ships, buried pipes and concrete pavement slabs and

    bridges and network of beams in the construction of floor

    systems for ships, buildings, and bridges, submerged

    floating tunnels, buried pipelines etc. In railway

    engineering, e.g. the design of railroad tracks. In

    mechanical engineering e.g. disc brake pad, shafts

    supported on ball, roller, or journal bearings, vibrating

    machines on elastic foundations.

    The work has been done on this subject is limited only

    to the linear response of uniform beams on linear elastic

    foundation [2], uniform beams on nonlinear elastic

    foundation [3]-[7] and non-uniform beams on nonlinear

    elastic foundation [8]. Numerical solutions for beams

    resting on elastic foundations using DQEM [9]. Some

    Manuscript received April 7, 2016; revised August 10, 2016.

    problems in structural analysis resting on fluid layer

    using GDQM [10]. Free vibration of uniform and non-

    uniform beams resting on fluid layer under axial force

    using the GDQM [11].

    In this paper, the numerical solution using a

    combination of a GDQM and Newtons method for

    solution the fourth order differential equation of beam

    under appropriate boundary conditions is nonlinear

    ordinary differential equations. Also, the effect of the

    linear and nonlinear Winkler (normal) parameters and the

    linear Pasternak (shear) foundation parameter is

    presented.

    II. FORMATION OF THE PROBLEM

    A. Formulation of the Problem

    For a general elastically end restrained non-uniform

    Euler Bernoulli beam of finite length L, resting on a non-

    linear-elastic foundation subjected to axial load and

    transverse forces, the dynamic flexural deflection v (x, t)

    satisfies the ordinary differential equation:

    2 2 2 23

    1 22 2 2 2A EI p k k

    t x x x

    2

    3 2( ek F x t x L

    x

    (1)

    Figure 1. Geometry of the beam resting on non-linear elastic foundation

    where, v (x,t) is the flexural deflection, is the density of

    the beam material, A is the area of the beam section, E(x)

    is the Young's modulus, I(x) is the moment of inertia,

    P(x,t) is the axial force, F(x,t) is the transverse force, k1 is

    the linear Winkler (normal) foundation parameter, k2 is

    the nonlinear Winkler (normal) foundation parameter and

    k3 is the linear Pasternak (shear) foundation parameter.

    Normalizing (1) then the non-dimensional governing

    equation of a Bernoulli-Euler beam as:

    International Journal of Structural and Civil Engineering Research Vol. 6, No. 1, February 2017

    2017 Int. J. Struct. Civ. Eng. Res.doi: 10.18178/ijscer.6.1.52-56

    52

    , ) 0, 0

  • 4 3 2 2

    4 3 2 2

    ( ) ( ) ( ) ( ) ( )( ) 2

    d W X dS X d W X d S X d W XS X

    dX dX dX dX dX

    23

    1 2 3 2

    ( )( ) ( ) ( ) 0, 0 1.

    d W XK W X K W X K F X X

    dX

    (2)

    where the non-dimensional coefficients are;

    42

    11

    0 0

    , , , ,k LV x pL

    W X P KL L EI EI

    26 3

    322 3

    0 0

    ( ), F(X) .

    k Lk L f x LK K

    EI EI Eand

    I

    Equation (1) is a 4th

    order ordinary differential

    equation with inertia ratio S(X) = (1+1X)2

    . In the case

    of non-uniform beam will study one case of inertia ratio

    S(X); 1 = 0.5 and 2 = 1.0. It requires 4 boundary

    conditions, two at X = 0, and two at X = 1. In the present

    work, the following three types of boundary conditions

    are considered:

    For ClampedClamped (CC),

    (0)(0) 0dW

    WdX

    and

    ( )( ) 0

    dW LW L

    dX (3)

    For SimplySimply (SS);

    2

    2

    (0)(0 ) 0

    d WW

    dXand

    2

    2

    ( )( ) 0

    dW LW L

    dX (4)

    For ClampedSimply (CS);

    (0)(0) 0

    dWW

    dXand

    2

    2

    ( )( ) 0

    dW LW L

    dX (5)

    B. Method of Solution

    The coordinates of the grid points are chosen

    according to Chebyshev-Gauss-Lobatto as:

    1 1( ) 1 cos , 1,2, , .

    2 1

    iX i i N

    N

    (6)

    Applying the GDQ method to (2) yields;

    (1) (2)

    , , ,

    1 1 1

    2( ) ( ) ( )N N N

    i i j j i i j j i i j j

    j j j

    S X D W S X C W S X B W

    3

    , 1 2 3 ,

    1 1

    N N

    i j j i i i j j

    j j

    P B W K W K W K B W

    ( ) 0, 1,2, , .iF X i N (7)

    where Wi, i=1, 2, 3,, N, is the functional value at the

    grid Xi, Bij, Cij and Dij is the weighting coefficient matrix

    of the second, third and fourth order derivatives. S(2)

    (Xi)

    and S(1)

    (Xi) are the second and first order derivatives of

    S(X) at Xi.

    Similarly, the derivatives on the boundary conditions

    can be discretized by the GDQ method, as a result, the

    numerical boundary conditions can be written as:

    1 0.W (8-1)

    ( 0)

    1,

    1

    0.N

    n

    k k

    k

    C W

    (8-2)

    0.NW (9-1)

    ( 1)

    ,

    1

    0.N

    n

    N k k

    k

    C W

    (9-2)

    where n0, n1 may be taken as either 1 or 2. By choosing

    the value of n0 and n1, (8) and (9) can give the following

    four sets of boundary conditions,

    n0 = 1, n1 = 1 (CC) supported,

    n0 = 1, n1 = 2 (CS) supported.

    n0 = 2, n1 = 1 (SC) supported,

    n0 = 2, n1 = 2 (SS) supported.

    Equations (8-1) and (9-1) can be easily substituted into

    the governing equation. This is not the case for (8-2) and

    (9-2) which, can couple these two equations together to

    give two solutions, W2 and WN-1, as: 2

    2

    3

    11. .

    N

    k

    k

    W AXK WAXN

    (10)

    2

    1

    3

    1. .

    N

    N k

    k

    W AXKN WAXN

    (11)

    where; ( 0) ( 1) ( 0) ( 1)

    1, , 1 1, 1 ,1 ,n n n n

    k N N N N kC C C CAXK ( 0) ( 1) ( 0) ( 1)

    1,2 , 1, ,2 ,n n n n

    N k k NC C C andCAXKN

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

    ,2 1, 1 1,2 , 1.n n n n

    N N N NC C C CAXN

    According to (10) and (11), W2 and WN-1 are expressed

    in terms of W3, W4,,WN-2. Substituting (8-1), (9-1), (10)

    and (11) into (7) gives:

    2 2 2(1) (2)

    , , ,

    3 3 3

    2( ) ( ) ( )N N N

    i i j j i i j j i i j j

    j j j

    S X D W S X C W S X B W

    2 23

    , 1 2 3 ,

    3 3

    1,N N

    i j j i i i j j

    j j

    P B W K W K W K B W

    3,4, , 2.i N (12)

    TABLE I. DEFLECTION OF NON-UNIFORM SIMPLYSIMPLY BEAM RESTING ON A NONLINEAR ELASTIC FOUNDATION UNDER UNIFORMLY

    DISTRIBUTED FORCE (K1 = 1.0, K2 = 1.0, K3 = 1.0, P=1.0, F=1.0).

    X W(GDQM) W (exact) W (C-C)

    Uniform (CC) Uniform (CC) 1 = 0.5, 2 = 1.0

    0 0 0 0

    0.012536 -6.38E-06 -6.38E-06 -5.84E-06

    0.049516 -9.23E-05 -9.23E-05 -8.35E-05

    0.109084 -0.000393538 -0.000393538 -0.00035

    0.188255 -0.000973019 -0.000973019 -0.00085

    0.283058 -0.001715963 -0.001715963 -0.00146

    0.38874 -0.002352658 -0.002352658 -0.00195

    0.5 -0.002604167 -0.002604167 -0.0021

    0.61126 -0.002352658 -0.002352658 -0.00184