a cantilever beam of length l is embedded at its right end and horizontal tensile force of p pou

First, we need to solve the homogeneous part of the differential equation, which is EIy'' = Py.

A cantilever beam of length l is embedded at its right end...

A cantilever beam of length L is embedded at its right end, and a horizontal tensile force of P pounds is applied to its free left end. When the origin is taken at its free end, as shown in the figure below, the deflection y(x) of the beam can be shown to satisfy the differential equation EIy" = Py + W(x)z. Find the deflection of the cantilever beam if w(x) = WoX, 0 < x < L, and Y(0) = 0, Y'(L) = 0.

Transcript

00:01 Hello students.

00:02 Today we will discuss about this question.

00:04 In this question we need to consider the differential equation that is eiy -dabeldash that is equals to py minus omega x divided by t.

00:15 Our objective is to find the deflection of cantilever beam is omega of x that is equal to omega 0 x 0 is less than x and x is less than x and y of the is equal to 0 and y -dice of l is equal to 0 so here we need to find y of x so first of us the differential equation can be written as e i y -double dash that is equal to p y minus omega x multiplied by x divided by 2 that is equal to p y minus omega 0 x squared by 2 so y double dash minus p divide by e i y that is equal to minus omega 0 divide by 2 e i x square this will be supposed equation number one now the ordinary equation of the differential equation that will be n square minus p divide by e i that is equal to zero so that implies that n that is equal to pluser minus square root of p divided by e i the root are irrational numbers thus the complementary solution of the above differential equation that is y c that is equal to c1 for hyperbolic of square root of p divided by ei x plus p to sign hyperbolic of square root of p divided by ei.

01:42 Now use the method of undetermined coefficient to find the particular solution yp of the given differential equation.

01:54 So here we assume that yp that is equal to ax square plus bx plus c, then y dash of p that is equal to 2ax plus and then y double dash of p that is equals to 2a.

02:08 So substitute this value in the given differential equation that is y -double dash minus p divide by eiy that is equals to minus omega divided by 2ei x -square so therefore we can write 2a here sorry we can write 2a minus p divided by ei in bracket a x square plus bx plus c that is equal to minus omega 0 divided by 2e i x squared so therefore we will get 2a minus p c divided by e i minus ap divided by e x squared minus bp divided by e i x that is equal to minus omega 0 2 e i x square now equate the corresponding coefficient so here we will get minus ap p divided by ei that is equals to minus omega 0 divided by 2ei that implies that a is equal to omega 0 divide by 2p and minus bp divided by ei that is equal to 0 that implies that b that is equal to 0.

03:25 And here we have given that 2a minus pc divided by ei that is equal to 0.

03:31 So p c divided by ei that is equals to 2a...

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