be com ited (x² - 4Lx³ + 6L²x²) W y = 24EI where x = distance (m), E = the modulus of elasticity= 2 x 1011 Pa, I = moment of inertia = 3.3 x 10-4 m4, w = 12,000 N/m, and L = length = 4 m. This equation can be differentiated to yield the slope of the downward deflection as a function of x:
be com ited (x² - 4Lx³ + 6L²x²) W y = 24EI where x = distance (m), E = the modulus of elasticity= 2 x 1011 Pa, I = moment of inertia = 3.3 x 10-4 m4, w = 12,000 N/m, and L = length = 4 m. This equation can be differentiated to yield the slope of the downward deflection as a function of x:
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![**1.23** As depicted in Fig. P1.23, the downward deflection \( y \) (m) of a cantilever beam with a uniform load \( w \) (kg/m) can be computed as
\[ y = \frac{w}{24EI}(x^{4} - 4Lx^{3} + 6L^{2}x^{2}) \]
where \( x = \) distance (m), \( E = \) the modulus of elasticity \( = 2 \times 10^{11} \, \text{Pa}, \) \( I = \) moment of inertia \( = 3.3 \times 10^{-4} \, \text{m}^{4}, \) \( w = 12{,}000 \, \text{N/m}, \) and \( L = \) length \( = 4 \, \text{m}. \) This equation can be differentiated to yield the slope of the downward deflection as a function of \( x \):
\[ \frac{dy}{dx} = \frac{w}{24EI} (4x^{3} - 12Lx^{2} + 12L^{2}x) \]
If \( y = 0 \) at \( x = 0 \), use this equation with Euler’s method \((\Delta x = 0.125 \, \text{m})\) to compute the deflection from \( x = 0 \) to \( L \). Develop a plot of your results along with the analytical solution computed with the first equation.
**Figure P1.23 Description:**
The diagram illustrates a cantilever beam. The beam is fixed on one end and free on the other. It shows a uniform load \( w \) applied along the length of the beam, from \( x = 0 \) to \( x = L \). The beam experiences a downward deflection, depicted by a curved line, starting at zero deflection at the fixed end and increasing towards the free end, at \( x = L \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd20dfe5a-a4c1-4793-9f05-a80ad59a67d4%2F2f86494c-6382-4312-b831-0ed2273b16e6%2Farjhofn_processed.png&w=3840&q=75)
Transcribed Image Text:**1.23** As depicted in Fig. P1.23, the downward deflection \( y \) (m) of a cantilever beam with a uniform load \( w \) (kg/m) can be computed as
\[ y = \frac{w}{24EI}(x^{4} - 4Lx^{3} + 6L^{2}x^{2}) \]
where \( x = \) distance (m), \( E = \) the modulus of elasticity \( = 2 \times 10^{11} \, \text{Pa}, \) \( I = \) moment of inertia \( = 3.3 \times 10^{-4} \, \text{m}^{4}, \) \( w = 12{,}000 \, \text{N/m}, \) and \( L = \) length \( = 4 \, \text{m}. \) This equation can be differentiated to yield the slope of the downward deflection as a function of \( x \):
\[ \frac{dy}{dx} = \frac{w}{24EI} (4x^{3} - 12Lx^{2} + 12L^{2}x) \]
If \( y = 0 \) at \( x = 0 \), use this equation with Euler’s method \((\Delta x = 0.125 \, \text{m})\) to compute the deflection from \( x = 0 \) to \( L \). Develop a plot of your results along with the analytical solution computed with the first equation.
**Figure P1.23 Description:**
The diagram illustrates a cantilever beam. The beam is fixed on one end and free on the other. It shows a uniform load \( w \) applied along the length of the beam, from \( x = 0 \) to \( x = L \). The beam experiences a downward deflection, depicted by a curved line, starting at zero deflection at the fixed end and increasing towards the free end, at \( x = L \).
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