Method of elastic line

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Berniece Lewis
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1 Method of elastic line To study the method of elastic line we shall first consider a beam of infinite length with a constant cross section (fig. 5.34). This beam rests on elastic soil and the deflection (y) of the footing at any point (x) from the origin is proportional to the soil reaction (q) at that point (q = k * y). Figure 5.34 Infinitely Long Beam on Elastic Soil The fundamental relationship is: And the general solution of the above equation can be given as: (5.1) Where: y : deflection of footing L e : Elastic length B : width of beam E : Modulus of elasticity of footing material I : Moment of inertia of footing K : Coefficient of sub grade reaction C 1, C 2, C 3 and C 4 : constants of integration. These constants are determined form boundary conditions for any particular case.

2 If we consider an infinitely long beam loaded by a single concentrated load (Q) as shown in fig. 5.34, and if we take the origin of coordinates at the point of application of the force, the case is symmetrical beam about origin the following boundary condition can be written: X = Y = 0 Deflection is zero X = 0 dy/dx = 0 Slope of deflection curve is zero X = 0 V = - Q/2 Shear force for the right part of the beam is equal to half of the applied concentrated force. From the above boundary condition we may determine integration constants as : C 1 = C 2 = 0 and C 3 = C 4 Therefore we obtain the following results : Deflection (5.2a) Slope (5.2b) Moment (5.2c) Shear (5.2d) A λx, B λx, C λx, and D λx values are given in table 5.1. The maximum moment and maximum deflection will occur at the origin. Their values are given below : X = 0 A λ0 = C λ0 = 1 max & max

3 Table 5.1 : coefficient for the solution of an infinite beam On an elastic foundation (Bowles, foundation analysis and design, page 239) λx A λx B λx C λx D λx λx A λx B λx C λx D λx

4 Deflection, moment, slope and shear equations for an infinitely long beam loaded with a uniformly distributed load, w, are given below Figure 5.35 Uniform Load on an Infinitely Long Beam For points A 1 For points A2 (5.3a) (5.3b) (5.3c) (5.3d) Note : upper sign when A z at left and lower sign when A z at right. If an infinitely long footing is subjected to a clockwise moment M 0 ( fig ) the equations are :

5 Figure 5.36 Infinitely long beam subjected to a moment (5.4a) (5.4b) (5.4c) (5.4d) Concentrated force ( Downward load is positive ) Equations for the part of the footing to the right of the load. Q(+) Sign convention is important. When the load acts downward, deflection is nominally positive both to right and left of load, but the real sign is determined by whether (A λx ) is positive or negative. Bending moment is shearing force is nominally positive to the

6 left but is influenced by the sign of (C λx ) shearing force is nominally positive to the left and negative to the right but is influenced by the sign of (D λx ). Moment (clockwise moment is positive ) Equation for the part of the footing to the right of the point of application of the moment. Mo(+) Note : the signs are reveres if the load is upward or the moment is anti-clockwise

7 In practice foundation-beams have finite length. A beam of finite length can also be investigated by the use of general equations of an infinitely long beam with the method of superposition as shown in fig Q 1 Q 2 w Beam of finite length A B (a) Q 1 Q 2 w Beam of finite length A B (b) M OA M OB Beam of finite length Q OA Q OB A L B (c) Figure 5.37 solution of a beam of finite length by superposition The footing of finite length is given in fig. 5.37a. This footing can be solved by superposing the solutions for the two kinds of loading of an infinitely long beam shown in fig. 5.37b and 5.37c. First, consider infinitely long beam (5.37b), and calculate bending moment (M A and M B ) and shear forces ( V A and V B ) at points A and B for the loads given in fig. 5.37a. For this purpose use equations 5.2c,5.2d and 5.2c, 5.2d. Create and conditions by introducing conditioning moments ( M OA and M OB ) and forces (Q OA and Q OB ) which reduce the calculated bending moments and shear forces at points A and B to zero. Therefore, the end conditioning moments and forces must produce bending moments (-M A and -M B ) and shear force (-V A and -V B ) at the ends. To create and conditions the following equation can be written :

8 + + = - M A + + = - M B (5.6) + + = - V A + - = - V B From the above equation MoA,MoB,QoA,QoB can be calculated. The deflection, bending moment and shear at any cross section of the beam of finite length shown in fig.5.37a, can be obtained by superposing fig. 5.37b and fig. 5.37c loading for beam infinite length. Method of successive Approximation This method is also known as the method of superposition. Baker (Raft foundation The soil line method of design, concrete publications Ltd.) has applied the principle of superposition to the combined footings. In this method, the column loads and bearing pressure and divided into three systems as shown in fig Each system must balance within itself so that the footing should be assumed to be the superposition of h=these three balance system. 1 st System : The first balance system of forces consist of the upward soil reaction and reaction calculated for a continuous beam, as shown a fig. 5.38b. It is first assumed that the footing is infinitely rigid, therefor soil pressure distribution is planar (Soil pressure distribution is uniform in case shown fig because column loads are symmetrical). The footing is treated as An invert continuous beam subjected to upward soil reactions. (Uniform soil pressure q in our case) and the reaction (n) at the column location are calculated. It is assumed that the deflection of the footing support is zero and midspan deflection are negligible small. It will be found (except for some very special loadings) that the magnitudes of the (R) reactions calculated are different from the

9 magnitudes of the given column loads. Calculate bending moments (M1) for first balanced system. 2 nd System: In order to reduce the forces of the System (1) at the columns to the actual column loads, forces ( Q -R) are applied at the column. (Q -R) forces constitute the second balanced system and produce a deflection, y', that can be calculated. Bending moments (M 2 ) for the second system should be calculated (fig. 5.38c). Forces of system (2) cause a deflection y' which in turns creates no variation in the soil pressure distribution. 3 rd System: This variation in soil pressure distribution constitutes System (3) that is also a balanced system (upward and downward soil pressure areas are equal to each other). System (3). is shown in fig. 5.35d. A further deflection is imposed on the footing by System (3) and it is opposite in direction to that of a system (2). The true elastic line of the footing lies between the extremes defined by systems (2), (3). (y') was calculated from system(2). The overall variation in soil pressure imposed by the system (2) is (k B y'). If the deflection of system (3) is (y"1), then the deflection (y') is reduced by (y"1) and the overall variation in soil pressure is kb(y' y"1). The new deflection (y"2) for kb(y'-y"1) is imposed on the footing and is smaller than (y"1). This process is repeated in a series of successive approximation until a balance is achieved, when final deflection is yf= (y' y"n) (5.7)

10 Figure 5.38 Superposition Analysis of combined footing

11 If the elastic line assumed to be as as cubic parabola, then in order positive and negative pressure constituting System (3) Should be equal, there for variation in the soil pressure diagram is:. q 1 = Bky'at the center and(3q1) at the ends. The deflection (y"1) under this system of loading is: Y"1= (5.8) The deflection (y"2) under the soil pressure variation of kb (y'-y"1) is:.y"2= Y"1= (5.9) As the alternative to successive approximation can be given by direct substitution as follows:.y"n=y'-yf= (5.10) Therefor the final deflection is:.yf= (5.11) And the final variation in soil pressure is:.qn= (5.12) Therefore the method may be followed step by step as: a) Calculate (q) and calculate (R) reaction and (M1) moments, from system (1) b) Calculate (y') deflection and (M2) moments of the system (2). c) Calculate (yf) and (qn) from the above given equations,and calculate (M3) moments for cubic parabola loading. d) Determine final moment as M= M1+M2+M3 Soil pressure at the center is (q-qn) and at the end (q+3qn).

12 Soil pressure at any point can be determined from the variation of soil pressure along the length of cubic parabola. Instead of a cubic parabola variation in soil pressure a linear variation may be used without making big error (fig. 5.39). q q+q n Final soil pressure diagram Figure 5.39 linear soil pressure Distribution in system 3. The method giving may also be used when a footing has cantilever ends. For this purpose, the deflection due to system (2) loading is calculated by considering cantilever ends. The method can also be applied to a footing loader eccentrically. The (R) reactions at the columns will be calculated as reaction of a continuous inverted beam subjected an upward trapezoidal loading instead of a uniform soil pressure. Then calculated (R) reactions are corrected to the actual column loads. Combined footing may also be solved by the method of finite difference (Malter Numerical solution for beam on elastic foundation, Journal of soil mech. And Found. Div., ASCE,1958). In this method footing is treated as a flexural member consisting of section, usually of equal length. By the use of electronic computers, simultaneous equations can be solved easily. The main errors associated with elastic method is evaluating the coefficient of subgrade reaction that depends on the type soil, the size of the plate and the shape of the plate (Terzaghi evaluation of coefficient of subgrade reaction, Geotechmique, 1955). If the coefficient of subgrade reaction was not determined by plate loading tests and if the soil data was obtained using the standard penetration tests, there will be a difficulty, because there is not a direct conversion to (k) values. Elastic methods have not been widely used in the past, because the conventional rigid method is very simple and it permits fast and satisfactory result for practical

13 purposes. Although the elastic method may give an impression that the soil structure interaction is illustrated better than rigid method, the results of any method are not superior to each other as far true soil pressure distribution is concerned.

= 50 ksi. The maximum beam deflection Δ max is not = R B. = 30 kips. Notes for Strength of Materials, ET 200

= 50 ksi. The maximum beam deflection Δ max is not = R B. = 30 kips. Notes for Strength of Materials, ET 200 Notes for Strength of Materials, ET 00 Steel Six Easy Steps Steel beam design is about selecting the lightest steel beam that will support the load without exceeding the bending strength or shear strength