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Laplace Transform And

Its Application

Prepare by:Mayur prajapati

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Outline

Introduction Laplace transform Properties of Laplace transform Transform of derivatives Deflection of Beam Reference

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Introduction

The Laplace transform method is powerful method for solving linear ODEs and corresponding initial value problems, as well as systems of ODEs arising in engineering.

This method has two main advantages over usual methods of ODEs :

1) Problems are solved more directly, IVP without first determining general solution, and non-homogeneous ODEs without first solving the corresponding homogeneous ODE.

2) The use of Unit step function (Heaviside function) and Dirac’s delta make the method particularly powerful for problems with inputs (driving force) that have discontinuities or represent short impulse or complicated periodic function.

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Laplace transform

Definition : If f(t) is a function defined for all the Laplace

transform of f(t), denoted by is defined by,

provided that the integral exists. s is a parameter which may be a real or complex number.

0t f tL

0

stf t F s e f t

L

1f t F sL

The given function f(t) is called the Inverse transform of F(s) and is written by,

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Properties of Laplace transform

1) Linearity property : If a, b be any constants and f, g any functions of t, then

2) First shifting property : If then

af t bg t a f t b g t L L L

f t F sL

ate f t F s a L

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3) Multiplication by If

4) Change of scale property : If then,

nt

1n

nnn

dt f t F sds

L

f t F sL

f t F sL

1 sf at Fa a

L

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5) Division of f(t) by t : If then,

provided the integral exists.

f t F sL

1

s

f t F s dst

L

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Transform of derivatives

Laplace transform of the derivative of any order :

Let f, f ’, ..., be continuous for all and be piecewise continuous on every finite interval on the semi-axis

then transform of given by,

nf

1nf 0t nf

0t nf

1 2 10 ' 0 ... 0n n n n nf s f s f s f f L L

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Deflection of Beam

Consider a beam of length L and rectangular cross section and homogeneous elastic material (e.g., steel) shown in fig.

If we apply a load to beam in vertical plane through the axis of symmetry (the x-axis ) beam is bent.

It axis is curved into the so called elastic curve or deflection curve.

Consider a cross-section of the beam cutting the elastic curve in P and the neutral surface in the line AA’ .

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The bending moment M about AA’ is given by the Bernoulli-Euler law,

...(1)

where, E = modulus of elasticity of the beam I = moment of inertia of the cross-section AA’ R = radius of curvature of the elastic curve at P(x , y) If the deformation of the beam is small, the slop of the elastic

curve is also small so that we may neglect in the formula,

EIMR

32 2

2

2

1 dydx

Rd ydx

2( / )dy dx

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For small deflection, Hence,(1) bending moment

Shear force

Intensity of loading

The sum of moments about any section due to external forces on the left of the section, if anti-clock is taken as positive and if clockwise is taken as negative.

2

2

d yM EIdx

3

3

dM d yEIdx dx

2 4

2 4

d M d yEI q xdx dx

2 21/ ( / )R d y dx

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The most important supports corresponding boundary conditions are :

1) Simply supported :

There being no deflection and bending moment. We have, , ,

0 0y " 0 0y

" 0y L 0y L

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2) Clamped at x=0, free at x=L :

At x=0, the deflection and slop of the beam being both zero. At x=L, there are no bending moment and shear force. We have,

, 0 ' 0 0y y " "' 0y L y L

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3) Clamped at both ends :The deflection and the slop of the beam being both zero. we have

,

,

0 0y

0y L

' 0 0y

' 0y L

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Deflection of Beam Carrying Uniform Distributed Load A uniformly loaded beam of length L is supported at both ends.

The deflection y(x) is a function of horizontal position x & is given by the differential equation,

...(1)

4

4

1 ( )d y q xdx EI

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Where q(x) is the load per unit length at point x. We assume in this problem that q(x) = q (a constant).

The boundary conditions are (i) no deflection at x = 0 and x = L (ii) no bending moment of the beam at x = 0 and x = L.

no deflection at x= 0 and L

no bending moment at x= 0 and L

" 0 0y

0y L

" 0y L

0 0y

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Using Laplace transform, (1) becomes,

Applying Laplace transform

...(2) The boundary conditions give and

so (2) becomes,

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4 0d y qxdx EI

L L

4 3 2 10 ' 0 " 0 "' 0 0qs y x s y s y sy yEI s

L

0 0y " 0 0y

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...(3)

Here and are unknown constants, but they can be determined by using the remaining two boundary conditions and

Solving for ,(3) leads to

...(4)

4 2 1' 0 "' 0 0qs y x s y yEI s

L

' 0y "' 0y

0y L " 0y L

2 4 5

1 1 1' 0 "' 0 qy x y ys s EI s

L

y xL

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Apply inverse Laplace transform to equation (4) gives,

Hence ,

By simplifying,

...(5)

1 1 1 12 4 5

1 1 1' 0 "' 0 qy x y ys s s EI

L L L L L

1 1 12 4 5

1 1 1 1 1' 0 3! "' 0 4!3! 4!

qy ys s

y xEI s

L L L

3 4

' 0 "' 06 24x q xy x xy y

EI

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To use the boundary condition take the second derivative of (5),

The boundary condition implies,

...(6)

Using the last boundary condition with (7) in (6),

...(7)

" 0y L

2" "' 02

qy x xy xEI

" 0y L

"' 02

qy LEI

0y L

3

024qLy

EI

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Substituting (6) and (7) in (5) gives,

...(8)

The above equation gives deflection of the beam at distance x. To find maximum deflection, put x = L/2 in above equation,

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4 3

24 24q qL qLy x x x xEI EI EI

4 3 3

2 24 2 2 24 2L q L qL L qL Ly

EI EI EI

452 384L qLy

EI

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Reference

Advanced Engineering Mathematics 9th e, Erwin Kreyzig, 2014 Higher Engineering Mathematics, B. S. Grewal, 1998

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THANK YOU