Problem 2.15. Boussinesq formula

Derive solution, and check results.

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Step 1. Write Boussinesq solution in polar coordinate system. Give stresses, σ_φ, σ_r, τ_rφ.

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Only the radial stress is nonzero. 

$$\begin{gathered} {\sigma }_r=–\frac{2P/h}{\pi r}\cos \phi \\ {\sigma }_{\phi }={\tau }_{r\phi }=0 \end{gathered}$$

 

See Figure 2.34.

Step 2. Write stress transformation from r-φ coordinate system to x-y coordinate system. (Rotate r axis into x axis.)

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Angle of rotation is β = π/2-φ. Transformation is given below:

Eq.(2-9)

$\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}={\mathbf{T}}_{\mathbf{\sigma }}\left\{\begin{array}{c}{\sigma }_r\\ {\sigma }_{\phi }\\ {\tau }_{r\phi }\end{array}\right\}=\left[\begin{array}{ccc}{\cos }^2\beta & {\sin }^2\beta & 2\cos \beta \sin \beta \\ {\sin }^2\beta & {\cos }^2\beta & –2\cos \beta \sin \beta \\ –\cos \beta \sin \beta & \cos \beta \sin \beta & {\cos }^2\beta –{\sin }^2\beta \end{array}\right]\left\{\begin{array}{c}–\frac{2P/h}{\pi r}\cos \phi \\ 0\\ 0\end{array}\right\}$

Step 3. Express r and the trigonometrical functions of the above transformation in the function of x and y coordinates.

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Step 4. Express stresses in x –y coordinate system.

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Boussinesq solution in polar coordinate system is

$$\begin{gathered} {\sigma }_r=–\frac{2P/h}{\pi r}\cos \phi \\ {\sigma }_{\phi }={\tau }_{r\phi }=0 \end{gathered}$$

See Figure 2.34.

Stress, σ_r is transformed from r-φ coordinate system to x-y coordinate system. Angle of rotation is β = π/2-φ. The transformation  has the following form

Eq.(2-9)

Now we write r and the trigonometrical functions of the above equation in the function of x and y coordinates.

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ \cos \beta =\frac{x}{\sqrt{x^2+y^2}} \\ \sin \beta =\cos \phi =\frac{y}{\sqrt{x^2+y^2}} \end{gathered}$$

Substitution the above expressions into the stress transformation results in 

$$\begin{gathered} \left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{x^2}{x^2+y^2}& \frac{y^2}{x^2+y^2}& –2\frac{xy}{x^2+y^2}\\ \frac{y^2}{x^2+y^2}& \frac{x^2}{x^2+y^2}& 2\frac{xy}{x^2+y^2}\\ \frac{xy}{x^2+y^2}& –\frac{xy}{x^2+y^2}& \frac{x^2}{x^2+y^2}–\frac{y^2}{x^2+y^2}\end{array}\right]\left\{\begin{array}{c}–\frac{2P/h}{\pi \sqrt{x^2+y^2}}\frac{y}{\sqrt{x^2+y^2}}\\ 0\\ 0\end{array}\right\} \\ {\mathit{\sigma }}_{\mathbf{x}}=–\frac{2P/h}{\pi \sqrt{x^2+y^2}}\frac{y}{\sqrt{x^2+y^2}}\frac{x^2}{x^2+y^2}\mathbf{=}\mathbf{–}\frac{\mathbf{2}\mathit{P}\mathbf{/}\mathit{h}}{\mathit{\pi }{\left({\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\right)}^{\mathbf{2}}}{\mathit{x}}^{\mathbf{2}}\mathit{y} \\ {\mathit{\sigma }}_{\mathbf{y}}=–\frac{2P/h}{\pi \sqrt{x^2+y^2}}\frac{y}{\sqrt{x^2+y^2}}\frac{y^2}{x^2+y^2}\mathbf{=}\mathbf{–}\frac{\mathbf{2}\mathit{P}\mathbf{/}\mathit{h}}{\mathit{\pi }{\left({\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\right)}^{\mathbf{2}}}{\mathit{y}}^{\mathbf{3}} \\ {\mathit{\tau }}_{\mathit{x}\mathit{y}}=–\frac{2P/h}{\pi \sqrt{x^2+y^2}}\frac{y}{\sqrt{x^2+y^2}}\frac{xy}{x^2+y^2}\mathbf{=}\mathbf{–}\frac{\mathbf{2}\mathit{P}\mathbf{/}\mathit{h}}{\mathit{\pi }{\left({\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\right)}^{\mathbf{2}}}\mathit{x}{\mathit{y}}^{\mathbf{2}} \end{gathered}$$