Problem 10.4. Strip method – Solution Manual

Apply the strip method to approximate the bending moment diagrams of the rectangular plates given in the Figures below in both directions. All plates are subjected to uniform load: p = 6 kN/m².
a) Three edges are built-in, the fourth is hinged.
b) All four edges are built-in.
c) Three edges are hinged, the fourth is built-in.

The deflections of beams are given in Table 3.5.

When one end is built-in and the other is simply supported, the midspan deflection is:

$w_{0} = \frac{2}{384} \frac{p L^{4}}{E I}$

Solve Problem

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Problem a)

Load in x direction, p_x[kN/m/m]=

Problem b)

Load in x direction, p_x[kN/m/m]=

Problem c)

Load in x direction, p_x[kN/m/m]=

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Steps

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Strip method is described in Section 10.3.

Load is assumed to be carried by orthogonal strips in x and y directions. The deflections of the strips are identical at the middle of the plate.

Problem a)

Step 1. Draw the strips, show their supports, write their deflection at the midspan.

Show stripsHide strips

$w_{x} = \frac{2}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I}$

Step 2. Distribute the load between x and y directions. Determine the load parts from the equality of the strips’ deflections.

Show load partsHide load parts

$\begin{array}{r} p = p_{x} + p_{y} \\ w_{x} = w_{y} \to \frac{2}{384} \frac{p_{x} L_{x}^{4}}{E I} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I} \end{array}\} \to \begin{matrix} p_{x} = 4.765 kN / m / m \\ p_{y} = 1.235 kN / m / m \end{matrix}$

Step 3. Draw approximate moment diagrams.

Show moment diagramsHide moment diagrams

$M_{x, \max}^{-} = \frac{p_{x} L_{x}^{2}}{8} = - \frac{4.765 \times 3 . 0^{2}}{8} = - 5.36 kNm / m M_{x, \max}^{+} = \frac{9 p_{x} L_{x}^{2}}{128} = - \frac{9 \times 4.765 \times 3 . 0^{2}}{128 \times 8} = 3.02 kNm / m M_{y, \max}^{-} = \frac{p_{y} L_{y}^{2}}{12} = - \frac{1.235 \times 5 . 0^{2}}{12} = - 2.57 kNm / m M_{y, \max}^{-} = \frac{p_{y} L_{y}^{2}}{24} = \frac{1.235 \times 5 . 0^{2}}{24} = 1.29 kNm / m$

Problem b)

Step 1. Draw the strips, show their supports, write their deflection at the midspan.

Show stripsHide strips

$w_{x} = \frac{1}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I}$

Step 2. Distribute the load between the x and y directions. Determine the load parts from the equality of the strips’ deflections.

Show load partsHide load parts

$\begin{array}{r} p = p_{x} + p_{y} \\ w_{x} = w_{y} \to \frac{1}{384} \frac{p_{x} L_{x}^{4}}{E I} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I} \end{array}\} \to \begin{matrix} p_{x} = 0.794 kN / m / m \\ p_{y} = 5.206 kN / m / m \end{matrix}$

Step 3. Draw approximate moment diagrams.

Show moment diagramsHide moment diagrams

$M_{x, \max}^{-} = \frac{p_{x} L_{x}^{2}}{12} = - \frac{0.794 \times 8 . 0^{2}}{12} = - 4.24 kNm / m M_{x, \max}^{+} = \frac{p_{x} L_{x}^{2}}{24} = - \frac{0.794 \times 8 . 0^{2}}{24} = 2.12 kNm / m M_{y, \max}^{-} = \frac{p_{y} L_{y}^{2}}{12} = - \frac{5.206 \times 5 . 0^{2}}{12} = - 10.84 kNm / m M_{y, \max}^{-} = \frac{p_{y} L_{y}^{2}}{24} = \frac{5.206 \times 5 . 0^{2}}{24} = 5.42 kNm / m$

Problem c)

Step 1. Draw the strips, show their supports, write their deflection at the midspan.

Show stripsHide strips

$w_{x} = \frac{5}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{2}{384} \frac{p_{y} L_{y}^{4}}{E I}$

Step 2. Distribute the load between x and y directions. Determine the load parts from the equality of the strips’ deflections.

Show load partsHide load parts

$\begin{array}{r} p = p_{x} + p_{y} \\ w_{x} = w_{y} \to \frac{5}{384} \frac{p_{x} L_{x}^{4}}{E I} = \frac{2}{384} \frac{p_{y} L_{y}^{4}}{E I} \end{array}\} \to \begin{matrix} p_{x} = 0.566 kN / m / m \\ p_{y} = 5.434 kN / m / m \end{matrix}$

Step 3. Draw approximate moment diagrams.

Show moment diagramsHide moment diagrams

$M_{x, \max}^{+} = \frac{p_{x} L_{x}^{2}}{8} = - \frac{0.566 \times 7 . 0^{2}}{8} = 3.46 kNm / m M_{y, \max}^{+} = \frac{p_{y} L_{y}^{2}}{8} = - \frac{5.434 \times 5 . 0^{2}}{8} = - 16.98 kNm / m M_{y, \max}^{-} = \frac{9 p_{y} L_{y}^{2}}{128} = - \frac{9 \times 5.434 \times 5 . 0^{2}}{128} = 9.57 kNm / m$

Results

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Strip method is described in Section 10.3.

Load is assumed to be carried by orthogonal strips in x and y directions. The deflections of the strips are identical at the middle of the plate.

Problem a)

Strips, their supports and their deflections at the midspan are shown in the Figure.

$w_{x} = \frac{2}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I}$

The load is distributed between the x and y directions assuming the equality of the strips’ deflections.

The approximate moment diagrams are:

$M_{x, \max}^{-} = \frac{p_{x} L_{x}^{2}}{8} = - \frac{4.765 \times 3 . 0^{2}}{8} = - 5.36 kNm / m M_{x, \max}^{+} = \frac{9 p_{x} L_{x}^{2}}{128} = - \frac{9 \times 4.765 \times 3 . 0^{2}}{128 \times 8} = 3.02 kNm / m M_{y, m a x}^{-} = \frac{p_{y} L_{y}^{2}}{12} = - \frac{1.235 \times 5 . 0^{2}}{12} = - 2.57 kNm / m M_{y, \max}^{-} = \frac{p_{y} L_{y}^{2}}{24} = \frac{1.235 \times 5 . 0^{2}}{24} = 1.29 kNm / m$

Problem b)

Strips, their supports and their deflections at the midspan are shown in the Figure.

$w_{x} = \frac{1}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{1}{384} \frac{p_{y} L_{y}^{4}}{E I}$

The load is distributed between x and y directions assuming the equality of the strips’ deflections.

The approximate moment diagrams are:

Problem c)

Strips, their supports and their deflections at the midspan are shown in the Figure.

$w_{x} = \frac{5}{384} \frac{p_{x} L_{x}^{4}}{E I} w_{y} = \frac{2}{384} \frac{p_{y} L_{y}^{4}}{E I}$

The load is distributed between x and y directions assuming the equality of the strips’ deflections.

The approximate moment diagrams are: