Explanation Given: The cantilever beam is of length L . The beam is embedded at its right end and horizontal compressive force of P pounds is applied to its free left end. The deflection y ( x ) of the beam is satisfied by the differential equation as shown below: E I y ″ = − P y − w ( x ) x 2 Here, w ( x ) = w 0 x . Initial conditions are defined as follows: y ( 0 ) = 0 y ′ ( L ) = 0 Calculation: From the given information, the deflection of beam in vertical axis is satisfied by the differential equation as shown below: E I y ″ = − P y − w ( x ) x 2 Rewrite the differential equation as follows: E I y ″ = − P y − w ( x ) x 2 E I y ″ = − P y − ( w 0 x ) x 2 ( w ( x ) = w 0 x ) E I y ″ + P y = − w 0 x 2 2 y ″ + P E I y = − w 0 x 2 2 E I ( 1 ) The auxiliary equation for differential equation (1) is m 2 + P E I = 0 . Solve the auxiliary equation as follows: m 2 + P E I = 0 m 2 = − P E I m = ± P E I i Therefore, the complementary solution of the above differential equation is as shown below: y c = c 1 cos P E I x + c 2 sin P E I x Now, obtain the particular solution y p by the use of undetermined coefficients. Assume the particular solution as follows: y p = A x 2 + B x + C Differentiate the above equation on both sides with respect to x . y ′ p = 2 A x + B Differentiate the above equation on both sides with respect to x . y ″ p = 2 A Substitute the values of y p , y ′ p and y ″ p in the equation (1). ( 2 A ) + P E I ( A x 2 + B x + C ) = − w 0 x 2 2 E I 2 A + P A x 2 E I + P B x E I + P C E I = − w 0 x 2 2 E I ( 2 A + P C E I ) + P A E I x 2 + P B E I x = − w 0 2 E I x 2 Compare the corresponding coefficients of the above equation as follows: 2 A + P C E I = 0 ( a ) P A E I = − w 0 2 E I ( b ) P B E I = 0 ( c ) From equation (c), it is obtained that B = 0 . Solve the equation (b) as follows: P A E I = − w 0 2 E I A = − w 0 2 P From equation (b), it is obtained that A = − w 0 2 P . Substitute, the value of A in the equation (a) and obtain the value of C as follows: 2 A + P C E I = 0 P C E I = − 2 A P C E I = − 2 ( − w 0 2 P ) ( A = − w 0 2 P ) C = w 0 E I P 2 From equation (a), it is obtained that C = w 0 E I P 2

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 5.2, Problem 8E

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The solution of the differential equation EIy″=−Py−w(x)x2 if w(x)=w0x,0<x<L and initial condition are y(0)=0 and y′(L)=0.

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Chapter 5.2, Problem 7E

Chapter 5.2, Problem 9E

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Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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The solution of the differential equation E I y ″ = − P y − w ( x ) x 2 if w ( x ) = w 0 x , 0 < x < L and initial condition are y ( 0 ) = 0 and y ′ ( L ) = 0 .

Solution Summary: The author describes the differential equation EIy′′=-Py-w(x)x. The deflection of the beam is underset_

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The solution of the differential equation EIy″=−Py−w(x)x2 if w(x)=w0x,0<x<L and initial condition are y(0)=0 and y′(L)=0.

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Chapter 5 Solutions

The solution of the differential equation E I y ″ = − P y − w ( x ) x 2 if w ( x ) = w 0 x , 0 &lt; x &lt; L and initial condition are y ( 0 ) = 0 and y ′ ( L ) = 0 .

Solution Summary: The author describes the differential equation EIy′′=-Py-w(x)x. The deflection of the beam is underset_

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The solution of the differential equation EIy″=−Py−w(x)x2 if w(x)=w0x,0<x<L and initial condition are y(0)=0 and y′(L)=0.

Want to see the full answer?

Chapter 5 Solutions

The solution of the differential equation E I y ″ = − P y − w ( x ) x 2 if w ( x ) = w 0 x , 0 < x < L and initial condition are y ( 0 ) = 0 and y ′ ( L ) = 0 .