Method of elastic line fig. 5

Advanced Foundation Engineering Prof. Ismat Ordamir

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:

(5.1)

And the general solution of the above equation can be given as:

Where:

y : deflection of footing

√

Le : 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

C1, C2, C3 and C4 : constants of integration. These constants are determined form

boundary conditions for any particular case.


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 :

C1 = C2 = 0 and C3 = C4

Therefore we obtain the following results :

Deflection

(5.2a)

Slope

(5.2b)

Moment

(5.2c)

Shear

(5.2d)

Aλx , Bλx , Cλx , and Dλx 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 → Aλ0 = Cλ0 = 1 → max

& max


Table 5.1 : coefficient for the solution of an infinite beam

On an elastic foundation

(Bowles , foundation analysis and design , page 239)

λx Aλx Bλx Cλx Dλx λx Aλx Bλx Cλx Dλx

0.0 1.0000 0.0000 1.0000 1.0000 4.10 -0.0231 -0.0136 0.0040 -0.0095

0.1 0.9907 0.0903 0.8100 0.9003 4.20 -0.0204 -0.0131 0.0057 -0.0074

0.2 0.9651 0.1627 0.6398 0.8024 4.30 -0.0179 -0.0124 0.0070 -0.0054

0.3 0.9267 0.2189 0.4888 0.7077 4.40 -0.0155 -0.0117 0.0079 -0.0038

0.4 0.8784 0.2610 0.3564 0.6174 4.50 -0.0132 -0.0109 0.0085 -0.0023

0.5 0.8231 0.2908 0.2415 0.5323 4.60 -0.0111 -0.0100 0.0089 -0.0011

0.6 0.7628 0.3099 0.1431 0.4560 4.70 -0.0092 -0.0091 0.0090 -0.0001

0.7 0.6997 0.3199 0.0590 0.3798 4.80 -0.0075 -0.0082 0.0089 0.0007

0.8 0.6354 0.3223 -0.0093 0.3131 4.90 -0.0059 -0.0073 0.0087 0.0014

0.9 0.5712 0.3185 -0.0657 0.2527 5.00 -0.0045 -0.0065 0.0084 0.0019

1.0 0.5083 0.3096 -0.1108 0.1088 5.10 -0.0033 -0.0056 0.0079 0.0023

1.1 0.4476 0.2967 -0.1457 0.1510 5.20 -0.0023 -0.0049 0.0075 0.0026

1.2 0.3899 0.2867 -0.1716 0.1001 5.30 -0.0011 -0.0042 0.0069 0.0028

1.3 0.3355 0.2626 -0.1897 0.0729 5.40 -0.0006 -0.0035 0.0064 0.0029

1.4 0.2849 0.2430 -0.2011 0.0419 5.50 0.0000 -0.0029 0.0058 0.0029

1.5 0.2384 0.2226 -0.2068 0.0158 5.60 0.0005 -0.0023 0.0052 0.0029

1.6 0.1959 0.2018 -0.2077 -0.0059 5.70 0.0010 -0.0018 0.0046 0.0028

1.7 0.1576 0.1812 -0.2047 -0.0235 5.80 0.0013 -0.0014 0.0041 0.0027

1.8 0.1234 0.1610 -0.1985 -0.0376 5.90 0.0015 -0.0010 0.0036 0.0025

1.9 0.0932 0.1415 -0.1899 -0.0484 6.00 0.0017 -0.0007 0.0031 0.0024

2.0 0.0667 0.1231 -0.1794 -0.0463 6.10 0.0018 -0.0004 0.0026 0.0022

2.1 0.0439 0.1057 -0.1675 -0.0618 6.20 0.0019 -0.0002 0.0022 0.0020

2.2 0.0244 0.0896 -0.1548 -0.0652 6.30 0.0019 0.0000 0.0018 0.0018

2.3 0.0080 0.0748 -0.1416 -0.0668 6.40 0.0018 0.0002 0.0015 0.0017

2.4 -0.0056 0.0613 -0.1282 -0.0669 6.50 0.0018 0.0003 0.0011 0.0015

2.5 -0.0166 0.0491 -0.1149 -0.0658 6.60 0.0017 0.0004 0.0009 0.0013

2.6 -0.0254 0.0383 -0.1019 -0.0636 6.70 0.0016 0.0005 0.0006 0.0011

2.7 -0.0320 0.0287 -0.0895 -0.0608 6.80 0.0015 0.0006 0.0004 0.0010

2.8 -0.0369 0.0204 -0.0777 -0.0573 6.90 0.0014 0.0006 0.0002 0.0008

2.9 -0.0403 0.0132 -0.0666 -0.0534 7.00 0.0013 0.0006 0.0001 0.0007

3.0 -0.0423 0.0070 -0.0563 -0.0493 7.10 0.0012 0.0006 0.0000 0.0006

3.1 -0.0431 0.0019 -0.0469 -0.0450 7.20 0.0010 0.0006 -0.0001 0.0005

3.2 -0.0431 -0.0024 -0.0383 -0.0407 7.30 0.0009 0.0006 -0.0002 0.0004

3.3 -0.0422 -0.0058 -0.0306 -0.0364 7.40 0.0008 0.0005 -0.0003 0.0003

3.4 -0.0408 -0.0085 -0.0237 -0.0323 7.50 0.0007 0.0005 -0.0003 0.0002

3.5 -0.0389 -0.0106 -0.0177 -0.0283 7.60 0.0006 0.0005 -0.0004 0.0001

3.6 -0.0366 -0.0121 -0.0124 -0.0245 7.70 0.0005 0.0004 -0.0004 0.0001

3.7 -0.0341 -0.0131 -0.0079 -0.0210 7.80 0.0004 0.0004 -0.0004 0.0000

3.8 -0.0314 -0.0137 -0.0040 -0.0177 7.90 0.0004 0.0004 -0.0004 0.0000

3.9 -0.0286 -0.0139 -0.0008 -0.0147 8.00 0.0003 0.0003 -0.0004 0.0000

4.0 -0.0258 -0.0139 0.0019 -0.0120 9.00 0.0000 0.0000 -0.0001 -0.0001


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 A1

For points A2

(5.3a)

(5.3b)

(5.3c)

(5.3d)

Note : upper sign when Az at left and lower sign when Az at right .

If an infinitely long footing is subjected to a clockwise moment M0 ( fig. 5.36 ) the

equations are :


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 (Aλx)

is positive or negative. Bending moment is shearing force is nominally positive to the


left but is influenced by the sign of (Cλx) shearing force is nominally positive to the

left and negative to the right but is influenced by the sign of (Dλx) .

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


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. 5.37

Q1 Q2

Beam of finite length w

A B (a)

Q1 Q2

Beam of finite length w

A B (b)

Beam of finite length MOB MOA

QOA QOB

A B (c)

L

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 (MA and

MB) and shear forces ( VA and VB ) 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 ( MOA and MOB ) and

forces (QOA and QOB ) 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 (-MA and -MB) and shear force (-VA and -VB) at the ends.

To create and conditions the following equation can be written :


+

+

= - MA

+

+

= - MB

+

+

= - VA

+

-

= - VB

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. 5.38. Each system

must balance within itself so that the footing should be assumed to be the

superposition of h=these three balance system.

1st 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. 5.38 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 mid-

span 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

(5.6)


magnitudes of the given column loads. Calculate bending moments (M1) for first

balanced system.

2nd

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 (M2) 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.

3rd

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)


Figure 5.38 Superposition Analysis of combined footing


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:

. q1 =

Bky'at the center and(3q1) at the ends.

The deflection (y"1) under this system of loading is:

Y"1= -0.00176

(5.8)

The deflection (y"2) under the soil pressure variation of kB (y'-y"1) is:

.y"2= Y"1= -0.00176

(5.9)

As the alternative to successive approximation can be given by direct substitution as

follows:

.y"n=y'-yf=-0.00176

(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).


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+qn

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


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.

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Method of elastic line fig. 5