A uniform horizontal beam of length, L. units, having a simple support at its left end and a fixed support at its right end will be distorted, due to its own weight, into a curve: y=f(x) as shown in the figure. 0 L y=f(x) This curve is called the deflection curve of the beam and it satisfies the differential equation: E-Idy w along with the four boundary conditions: y(0) = 0, y(L) = 0, y" (0) = 0 y' (L) = 0 Problem: Substitute each boundary condition into the appropriate equation: (2), (3) or (4) from page 17.7 of the article: Topic 17: The Deflection Equation of a Uniform Beam and then solve for the coefficients, C. i=1,2,3,4, to verify that the beam's equation is: y= [2x*-3Lx³ + L³x] 4RFI
A uniform horizontal beam of length, L. units, having a simple support at its left end and a fixed support at its right end will be distorted, due to its own weight, into a curve: y=f(x) as shown in the figure. 0 L y=f(x) This curve is called the deflection curve of the beam and it satisfies the differential equation: E-Idy w along with the four boundary conditions: y(0) = 0, y(L) = 0, y" (0) = 0 y' (L) = 0 Problem: Substitute each boundary condition into the appropriate equation: (2), (3) or (4) from page 17.7 of the article: Topic 17: The Deflection Equation of a Uniform Beam and then solve for the coefficients, C. i=1,2,3,4, to verify that the beam's equation is: y= [2x*-3Lx³ + L³x] 4RFI
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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
Transcribed Image Text:Integrating a second time, we find that
SO
>
SE.I.
E.I.
Integrate a third time to obtain:
E.I.
E.I
E.I.
Integrate a fourth time to get:
E I
E.I.
E·I· y =
d³y
dx³
dy
dx
dy dx = w
dx
dx =
W
dx = "
dx
6
W
24
d³ y
dx.³
=
d'y
dx²
- dx = √ w-x dx + [ c, dx
W
2
W
6
= w.x + C₁
x² +
C₁₂
6
+
fx
[x dx + G₂₁₂ • ¹ a
1 dx
W
·•ƒ x² dx + G₂ • Sx
x² dx +
·x² + G·x + C₂
2
x dx
+ .f x² dx
+²³+
+ƒ
•x² + C₂ ·x + C₂
C₂ dx
c₂.f
-x² + C₁-x + C₂
(3)
x dx
+ C₂ dx
(4)
(1)
(2)
The values of the coefficients: C₁, C₂, C3, C4 are dependent upon the types of supports that are
provided to the beam at its ends.
17.3
17.4
![A uniform horizontal beam of length, L units, having a simple support at its left end
and a fixed support at its right end will be distorted, due to its own weight, into a curve:
y=f(x) as shown in the figure.
L
y = f(x)
This curve is called the deflection curve of the beam and it satisfies the differential equation:
E- I dy = w along with the four boundary conditions:
dx
y(0) = 0,
y(L) = 0,
y" (0) = 0
y'(L) = 0
W
4RFI
X
Problem: Substitute each boundary condition into the appropriate equation: (2), (3) or (4) from
page 17.7 of the article: Topic 17: The Deflection Equation of a Uniform Beam and then
solve for the coefficients, C₁, i = 1,2,3,4, to verify that the beam's equation is:
y= [2x-3Lx³ + L³x]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f989c57-8a7c-4041-9fe7-2f824b6b0127%2Fcca40d90-33c6-44ca-8ee8-673da806dfaf%2Fzwz6s9d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A uniform horizontal beam of length, L units, having a simple support at its left end
and a fixed support at its right end will be distorted, due to its own weight, into a curve:
y=f(x) as shown in the figure.
L
y = f(x)
This curve is called the deflection curve of the beam and it satisfies the differential equation:
E- I dy = w along with the four boundary conditions:
dx
y(0) = 0,
y(L) = 0,
y" (0) = 0
y'(L) = 0
W
4RFI
X
Problem: Substitute each boundary condition into the appropriate equation: (2), (3) or (4) from
page 17.7 of the article: Topic 17: The Deflection Equation of a Uniform Beam and then
solve for the coefficients, C₁, i = 1,2,3,4, to verify that the beam's equation is:
y= [2x-3Lx³ + L³x]
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