Problem 10.3. Navier solution

A rectangular plate of dimensions L_x × L_y is hinged at all four edges. The plate is subjected to a distributed load, which is uniform in one direction and sinusoidal load in the other direction: $p = p_{0} \sin (\frac{πy}{L_{y}})$ , where $p_{0} = 12 kN / m^{2}$ . The stiffness of the plate is D = 10 × 10⁶ Nm, the Poisson ratio is ν = 0.3. Determine the deflection and moment functions of the plate with the aid of the Navier solution. Apply three (nonzero) terms of the Fourier series expansion of the load.

Solve Problem

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Midspan deflection, w(L_x/2,L_y/2) [mm]=

Midspan bending moment, M_x(L_x/2,L_y/2) [kNm/m]=

Midspan bending moment, M_y(L_x/2,L_y/2) [kNm/m]=

Torsional moment at the corner, M_xy(0,0) [kNm/m]=

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Steps

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See steps of Example 10.2.

Step 1. Give Fourier series expansion of the load. Determine the coefficients of the first three terms.

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The load function is sinusoidal along y axis, the Fourier series expansion in x direction results in:

See Fourier series expansion in Figure 3.12.

p (x, y) = \underset{}{\sin \frac{π y}{L_{y}} \underset{p_{0}}{\underset{⏟}{\sum_{i = 1, 3, 5} p_{i} \sin \frac{i π x}{L_{x}}}}} where p_{i} = \frac{4 p_{0}}{i π} p_{1} = \frac{4 p_{0}}{π} = \frac{4 \times 12}{π} = 15.28 \times 10^{3} N / m^{2} p_{3} = \frac{4 p_{0}}{3 π} = \frac{4 \times 12}{3 π} = 5.09 \times 10^{3} N / m^{2} p_{5} = \frac{4 p_{0}}{5 π} = \frac{4 \times 12}{5 π} = 3.06 \times 10^{3} N / m^{2}

Step 2. Look for the solution in sinusoidal form. Write the Fourier series expansion of the deflection function.

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$w (x, y) = \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} w_{i} \sin \frac{i π x}{L_{x}}$

Step 3. Write the differential equation of the plate. Substitute the approximate load and deflection functions.

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Eq.(10-28)

D \frac{\partial^{4} w}{\partial x^{4}} + D \frac{\partial^{4} w}{\partial y^{4}} + 2 D \frac{\partial^{4} w}{\partial x^{2} \partial y^{2}} = p D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} [(\frac{i^{4} π^{4}}{L_{x}^{4}} + \frac{π^{4}}{L_{y}^{4}} + 2 \frac{i^{2} π^{4}}{L_{x}^{2} L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}}] = \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} p_{i} \sin \frac{i π x}{L_{x}}

Step 4. Compare the terms in the two sides of the equation. Express the unknown deflection function.

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Step 5. Give the moment functions.

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Table 10.1

M_{x} = - D \frac{\partial^{2} w}{\partial x^{2}} - ν D \frac{\partial^{2} w}{\partial y^{2}} = - D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} (\frac{i^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}} M_{y} = - ν D \frac{\partial^{2} w}{\partial x^{2}} - D \frac{\partial^{2} w}{\partial y^{2}} = - D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} (ν \frac{i^{2} π^{2}}{L_{x}^{2}} + \frac{π^{2}}{L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}} M_{x y} = - D (1 - ν) \frac{\partial^{2} w}{\partial x \partial y} = - D (1 - ν) \cos \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} \frac{i π^{2}}{L_{x}^{} L_{y}^{}} w_{i} \cos \frac{i π x}{L_{x}}

Step 6. Calculate the deflection and bending moment values at the midpoint and the torsional moment at the corner of the plate.

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$w (\frac{L_{x}}{2}, \frac{L_{y}}{2}) = \sin \frac{π}{2} (5.08 \times \sin \frac{π}{2} + 0.0667 \times \sin \frac{3 π}{2} + 0.00614 \times \sin \frac{5 π}{2}) = = 5.08 - 0.0667 + 0.00614 = 5.02 mm M_{x} (\frac{L_{x}}{2}, \frac{L_{y}}{2}) = - D ((\frac{π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 5.08 - (\frac{3^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 0.0667 + (\frac{5^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 0.00614) = = - 10 \times 10^{6} ((\frac{π^{2}}{6 . 0^{2}} + 0.3 \frac{π^{2}}{6 . 0^{2}}) 5.08 - (\frac{3^{2} π^{2}}{6 . 0^{2}} + 03 . \frac{π^{2}}{6 . 0^{2}}) 0.0667 + (\frac{5^{2} π^{2}}{6 . 0^{2}} + 0.3 \frac{π^{2}}{6 . 0^{2}}) 0.00614) \times 10^{- 6} = = 16.801 {kNm}^{2} / m$

$M_{y} (\frac{L_{x}}{2}, \frac{L_{y}}{2}) = - D ((ν \frac{π^{2}}{L_{x}^{2}} + \frac{π^{2}}{L_{y}^{2}}) 5.08 - (ν \frac{3^{2} π^{2}}{L_{x}^{2}} + \frac{π^{2}}{L_{y}^{2}}) 0.0667 + (ν \frac{5^{2} π^{2}}{L_{x}^{2}} + \frac{π^{2}}{L_{y}^{2}}) 0.00601) = = 17.57 kNm / m M_{x y} (0, 0) = - D (1 - ν) (\frac{π^{2}}{L_{x} L_{y}} 5.08 + \frac{3 π^{2}}{L_{x} L_{y}} 0.0667 + \frac{5 π^{2}}{L_{x} L_{y}} 0.00601) = 10.2 kNm / m$

Results

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See steps of Example 10.2.

The load function is sinusoidal along y axis, the Fourier series expansion in x direction is:

See Fourier series expansion in Figure 3.12.

p (x, y) = \underset{}{\sin \frac{π y}{L_{y}} \underset{p_{0}}{\underset{⏟}{\sum_{i = 1, 3, 5} p_{i} \sin \frac{i π x}{L_{x}}}}} where p_{i} = \frac{4 p_{0}}{i π} p_{1} = \frac{4 p_{0}}{π} = \frac{4 \times 12}{π} = 15.28 \times 10^{3} N / m^{2} p_{3} = \frac{4 p_{0}}{3 π} = \frac{4 \times 12}{3 π} = 5.09 \times 10^{3} N / m^{2} p_{5} = \frac{4 p_{0}}{5 π} = \frac{4 \times 12}{5 π} = 3.06 \times 10^{3} N / m^{2}

We are looking for the solution in sinusoidal form. The Fourier series expansion of the deflection function is:

$w (x, y) = \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} w_{i} \sin \frac{i π x}{L_{x}}$

The approximate load and deflection functions are substituted into the differential equation of the plate.

Eq.(10-28)

D \frac{\partial^{4} w}{\partial x^{4}} + D \frac{\partial^{4} w}{\partial y^{4}} + 2 D \frac{\partial^{4} w}{\partial x^{2} \partial y^{2}} = p D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} [(\frac{i^{4} π^{4}}{L_{x}^{4}} + \frac{π^{4}}{L_{y}^{4}} + 2 \frac{i^{2} π^{4}}{L_{x}^{2} L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}}] = \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} p_{i} \sin \frac{i π x}{L_{x}}

Comparing the terms in the two sides of the equation the unknown deflection function can be determined:

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The moment functions are

Table 10.1

M_{x} = - D \frac{\partial^{2} w}{\partial x^{2}} - ν D \frac{\partial^{2} w}{\partial y^{2}} = - D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} (\frac{i^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}} M_{y} = - ν D \frac{\partial^{2} w}{\partial x^{2}} - D \frac{\partial^{2} w}{\partial y^{2}} = - D \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} (ν \frac{i^{2} π^{2}}{L_{x}^{2}} + \frac{π^{2}}{L_{y}^{2}}) w_{i} \sin \frac{i π x}{L_{x}} M_{x y} = - D (1 - ν) \frac{\partial^{2} w}{\partial x \partial y} = - D (1 - ν) \sin \frac{π y}{L_{y}} \sum_{i = 1, 3, 5} \frac{i π^{2}}{L_{x}^{} L_{y}^{}} w_{i} \sin \frac{i π x}{L_{x}}

The deflection and the bending moment values at the midpoint and the torsional moment at the corner of the plate are calculated.

$M_{x} (\frac{L_{x}}{2}, \frac{L_{y}}{2}) = - D ((\frac{π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 5.08 - (\frac{3^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 0.0667 + (\frac{5^{2} π^{2}}{L_{x}^{2}} + ν \frac{π^{2}}{L_{y}^{2}}) 0.00614) == - 10 \times 10^{6} ((\frac{π^{2}}{6 . 0^{2}} + 0.3 \frac{π^{2}}{6 . 0^{2}}) 5.08 - (\frac{3^{2} π^{2}}{6 . 0^{2}} + 03 . \frac{π^{2}}{6 . 0^{2}}) 0.0667 + (\frac{5^{2} π^{2}}{6 . 0^{2}} + 0.3 \frac{π^{2}}{6 . 0^{2}}) 0.00614) \times 10^{- 6} = = 16.801 {kNm}^{2} / m$