The deflection (y) at adistance x alongabeamof Length Lis given by3.y =(5x'- 5x²+ 2x -1), whereE Iw is weight,E is modulus of elasticity andis the second momentNewton-areaThes e are allnon -Zero Constants .UsingRaphson method, find the distance (x). where the deflection1S zero, Consider X, = 1.5and showyour answer with%3Dfour decimal placesa ccuracy

Answered: The dea of Length y = E is modu area…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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The deflection (y) at adistance x along
abeam
of Length L
is given by
3.
y =
(5x'- 5x²+ 2x -1), where
E I
w is weight,
E is modulus of elasticity and
is the second moment
Newton-
area
Thes e are all
non -Zero Constants .Using
Raphson method, find the distance (x). where the deflection
1S zero, Consider X, = 1.5
and show
your answer with
%3D
four decimal places
a ccuracy

Expert Solution

Step 1

Given function:

$y = \frac{w L}{E I} (5 x^{3} - 5 x^{2} + 2 x - 1)$

Taking $x_{1} = 1.5$ .

Newton Raphson method is given by:

$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f' (x_{n})}$

$x_{n + 1} = x_{n} - \frac{(\frac{w L}{E I}) 5 {x_{n}}^{3} - 5 {x_{n}}^{2} + 2 x_{n} - 1}{(\frac{w L}{E I}) 15 {x_{n}}^{2} - 10 x_{n} + 2}$

Step 2

$x_{n + 1} = x_{n} - \frac{5 {x_{n}}^{3} - 5 {x_{n}}^{2} + 2 x_{n} - 1}{15 {x_{n}}^{2} - 10 x_{n} + 2}$

$x_{2} = x_{1} - \frac{5 {x_{1}}^{3} - 5 {x_{1}}^{2} + 2 x_{1} - 1}{15 {x_{1}}^{2} - 10 x_{1} + 2}$

$x_{2} = (1.5) - \frac{5 {(1.5)}^{3} - 5 {(1.5)}^{2} + 2 (1.5) - 1}{15 {(1.5)}^{2} - 10 (1.5) + 2}$

$= 1.5 - \frac{7.625}{20.75} = 1.13253012$

$x_{3} = 1.13253012 - \frac{2.114992116}{9.914065891} = 0.919197654$

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The dea of Length y = E is modu area Thes Raphson met iS zero, Co four decim