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konstantin123 [22]
3 years ago
5

For the data points (1,1),(2,1/2),(3,1/3),(4,1/4), finde the natural cubic spline.

Engineering
1 answer:
kow [346]3 years ago
7 0

Answer:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

Explanation:

A cubic has the form:

y = ax³ + bx² + cx + d

Given four points, we can write a system of equations:

1 = a + b + c + d

1/2 = 8a + 4b + 2c + d

1/3 = 27a + 9b + 3c + d

1/4 = 64a + 16b + 4c + d

Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]

First, we find the inverse of the coefficient matrix.  This is messy to do by hand, so let's use a calculator:

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]

Now we multiply by the solution matrix (again using a calculator):

-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

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An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
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Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

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The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

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L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

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L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

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∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

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V_{max} = 812.8 × 26/2 = 10566.4 lbs

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v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

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