Problem 10.1. Deflection of a one-way plate

Constant load, p  [kN/m²]=

Midspan deflection, w(L/2)=

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Moment, M_x(0)=

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Moment, M_y(0)=

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Problem a)

Boundary conditions are given in Table 10.2.

Step 1.  Check boundary conditions at built-in edge, x = 0.

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Step 2.  Check boundary conditions at built-in edge, x = L.

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Problem b)

Step 1.  Give the load function deriving the fourth derivative of the deflection.

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Higher order derivatives of the deflection function are:

$$\begin{gathered} \frac{{\partial }^{2}w\left(x\right)}{\partial x^2}=24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D} \\ \frac{{\partial }^{3}w\left(x\right)}{\partial x^3}=48\frac{x}{24D}–12\frac{L}{12D} \\ \frac{{\partial }^{4}w\left(x\right)}{\partial x^4}=\frac{48}{24D} \end{gathered}$$

The load function obtained from by the DE of the one-way slab is constant:

Eq.(10-29)

${\mathit{p}}_{\mathbf{z}}=D\frac{{\partial }^{4}w}{\partial x^4}=\frac{48}{24}\mathbf{=}\mathbf{2}$

When the unit of the length is [m] and the stiffness of the slab is given in [kNm²/m], then the load is in [kN/m²] while the deflection is in [m].

Problem c)

Step 1.  Calculate midspan deflection at x = L/2.

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Problem d)

Step 1. Give the moments from the second derivatives of the deflection function.

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See Table 10.1.

$$\begin{gathered} M_x=D{\kappa }_x+\nu D{\kappa }_y=–D\frac{{\partial }^{2}w}{\partial x^2}+\nu D\frac{{\partial }^{2}w}{\partial y^2}=–D\frac{{\partial }^{2}w}{\partial x^2}=–D\left(24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D}\right) \\ \to {\mathit{M}}_{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}=–D\left(24\frac{0^2}{24D}–12\frac{L\times 0}{12D}+4\frac{L^2}{24D}\right)=–\frac{L^2}{6}\mathbf{=}\mathbf{–}\frac{\mathit{p}{\mathit{L}}^{\mathbf{2}}}{\mathbf{12}} \\ {M}_y=\nu D{\kappa }_x+D{\kappa }_y=–D\nu \frac{{\partial }^{2}w}{\partial x^2}+\nu D\frac{{\partial }^{2}w}{\partial y^2}=–D\nu \frac{{\partial }^{2}w}{\partial x^2}=–\nu D\left(24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D}\right) \\ \to {\mathit{M}}_{\mathit{y}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{–}\mathit{\nu }\frac{\mathit{p}{\mathit{L}}^{\mathbf{2}}}{\mathbf{12}} \end{gathered}$$

Note that the derivatives of the deflection function with respect to to y is zero.

Problem a)

Boundary conditions are given in Table 10.2.

The deflection function fulfills the boundary conditions at x = 0:

$$\begin{gathered} w\left(x\right)=2\frac{x^4}{24D}–2\frac{L{x}^3}{12D}+2\frac{L^2{x}^2}{24D} \to w\left(0\right)=0 \\ \frac{\partial w\left(x\right)}{\partial x}=8\frac{x^3}{24D}–6\frac{L{x}^2}{12D}+4\frac{L^{2}x}{24D} \to {\overline{)\frac{\partial w}{\partial x}}}_{x=0}=0 \end{gathered}$$

The deflection function fulfills the boundary conditions at x = L:

$$\begin{gathered} w\left(x\right)=2\frac{x^4}{24D}–2\frac{L{x}^3}{12D}+2\frac{L^2{x}^2}{24D} \to w\left(L\right)=\frac{L^4}{24D}(2–4+2)=0 \\ \frac{\partial w\left(x\right)}{\partial x}=8\frac{x^3}{24D}–6\frac{L{x}^2}{12D}+4\frac{L^{2}x}{24D} \to {\overline{)\frac{\partial w}{\partial x}}}_{x=L}==\frac{L^3}{24D}(8–12+4)=0 \end{gathered}$$

Problem b)

Higher order derivatives of the deflection function are:

$$\begin{gathered} \frac{{\partial }^{2}w\left(x\right)}{\partial x^2}=24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D} \\ \frac{{\partial }^{3}w\left(x\right)}{\partial x^3}=48\frac{x}{24D}–12\frac{L}{12D} \\ \frac{{\partial }^{4}w\left(x\right)}{\partial x^4}=\frac{48}{24D} \end{gathered}$$

The load function obtained from by the DE of the one-way slab is constant:

Eq.(10-29)

${\mathit{p}}_{\mathbf{z}}=D\frac{{\partial }^{4}w}{\partial x^4}=\frac{48}{24}\mathbf{=}\mathbf{2}$

When the unit of the length is [m] and the stiffness of the slab is given in [kNm²/m], then the load is in [kN/m²] while the deflection is in [m].

Problem c)

Midspan deflection at x = L/2 is

$$\begin{gathered} \mathit{w}\left(\frac{\mathit{L}}{\mathbf{2}}\right)=2\frac{{\left({\displaystyle \frac{L}{2}}\right)}^4}{24D}–2\frac{L{\left(\frac{L}{2}\right)}^3}{12D}+2\frac{L^2{\left(\frac{L}{2}\right)}^2}{24D}=2\frac{L^4}{2^4\times 24D}–2\frac{L^4}{2^3\times 12D}+2\frac{L^4}{4\times 24D}= \\ =\frac{2}{384}\frac{L^4}{D}\mathbf{=}\mathbf{ }\frac{\mathbf{1}}{\mathbf{384}}\frac{\mathit{p}{\mathit{L}}^{\mathbf{4}}}{\mathit{D}} \end{gathered}$$

Problem d)

The moments are given by the second derivatives of the deflection function.

See Table 10.1.

$$\begin{gathered} M_x=D{\kappa }_x+\nu D{\kappa }_y=–D\frac{{\partial }^{2}w}{\partial x^2}+\nu D\frac{{\partial }^{2}w}{\partial y^2}=–D\frac{{\partial }^{2}w}{\partial x^2}=–D\left(24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D}\right) \\ \to {\mathit{M}}_{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}=–D\left(24\frac{0^2}{24D}–12\frac{L\times 0}{12D}+4\frac{L^2}{24D}\right)=–\frac{L^2}{6}\mathbf{=}\mathbf{–}\frac{\mathit{p}{\mathit{L}}^{\mathbf{2}}}{\mathbf{12}} \\ {M}_y=\nu D{\kappa }_x+D{\kappa }_y=–D\nu \frac{{\partial }^{2}w}{\partial x^2}+\nu D\frac{{\partial }^{2}w}{\partial y^2}=–D\nu \frac{{\partial }^{2}w}{\partial x^2}=–\nu D\left(24\frac{x^2}{24D}–12\frac{Lx}{12D}+4\frac{L^2}{24D}\right) \\ \to {\mathit{M}}_{\mathit{y}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{–}\mathit{\nu }\frac{\mathit{p}{\mathit{L}}^{\mathbf{2}}}{\mathbf{12}} \end{gathered}$$

Note that the derivatives of the deflection function with respect to y is zero.