Explanation Given: The expression for the load is w ( x ) = { w 0 , 0 ≤ x < L / 2 0 , L / 2 ≤ x < L . Concept: (1) The Laplace transformation of a linear differential equation with constant coefficient is express as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) The differential equation for the load distribution on cantilever is given as follows: E I d 4 y d x 4 = w ( x ) (3) The linearity property of the Laplace transformation is as below: If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } . (4) If F ( s ) = L { f ( t ) } and a > 0 , then following relations holds. L − 1 { e − a s F ( s ) } = f ( t − a ) U ( t − a ) L − 1 { e − a s s } = U ( t − a ) Definition used: The given expression of the load is as follows: w ( x ) = { w 0 , 0 < x < L / 2 0 , L / 2 < x < L The unit step function for the above step function is written as follows: w ( x ) = w 0 − w 0 U ( x − L / 2 ) = w 0 { 1 − U ( x − L / 2 ) } Substitute the value of w ( x ) in equation of load distribution on cantilever E I d 4 y d x 4 = w ( x ) . E I d 4 y d x 4 = w ( x ) E I d 4 y d x 4 = w 0 { 1 − U ( x − L / 2 ) } Use the definition (1) and take the Laplace transform on the both side of the above expression. s 4 L { y } − s 3 y ( 0 ) − s 2 y ′ ( 0 ) − s y ″ ( 0 ) − y ‴ ( 0 ) = w 0 E I ⋅ 1 s { 1 − e − s L 2 } Substitute the y ( 0 ) = 0 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = c 1 and y ‴ ( 0 ) = c 2 in above expression and solve. s 4 L { y } − s c 1 − c 2 = w 0 E I ⋅ 1 s { 1 − e − s L 2 } s 4 L { y } = s c 1 + c 2 + w 0 E I ⋅ 1 s { 1 − e − s L 2 } L { y } = c 1 s 3 + c 2 s 4 + w 0 E I ⋅ 1 s 5 { 1 − e − s L 2 } Use the definition (3) and (4) to obtain the inverse Laplace transform on both sides of above expression

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 80E

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The equation of the deflection y(x) for the expression of load w(x)={w0, 0≤x<L/20, L/2≤x<L on cantilever.

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Chapter 4.3, Problem 79E

Chapter 4.3, Problem 81E

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The equation of the deflection y ( x ) for the expression of load w ( x ) = { w 0 , 0 ≤ x < L / 2 0 , L / 2 ≤ x < L on cantilever.

Solution Summary: The author explains the Laplace transformation of a linear differential equation with constant coefficient.

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The equation of the deflection y(x) for the expression of load w(x)={w0, 0≤x<L/20, L/2≤x<L on cantilever.

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Chapter 4 Solutions

The equation of the deflection y ( x ) for the expression of load w ( x ) = { w 0 , 0 ≤ x &lt; L / 2 0 , L / 2 ≤ x &lt; L on cantilever.

Solution Summary: The author explains the Laplace transformation of a linear differential equation with constant coefficient.

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The equation of the deflection y(x) for the expression of load w(x)={w0, 0≤x<L/20, L/2≤x<L on cantilever.

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Chapter 4 Solutions

The equation of the deflection y ( x ) for the expression of load w ( x ) = { w 0 , 0 ≤ x < L / 2 0 , L / 2 ≤ x < L on cantilever.