Problem 2.16. Infinite half space

Point 1

Radial stress, ${\sigma }_r \left[\mathrm{kN}/{\mathrm{m}}^2\right] =$

CHECK

Point 2

Radial stress, ${\sigma }_r \left[\mathrm{kN}/{\mathrm{m}}^2\right] =$

CHECK

Point 3

Radial stress, ${\sigma }_r \left[\mathrm{kN}/{\mathrm{m}}^2\right] =$

CHECK

Failure load from the relevant failure criterion (minimum failure load of Rankine, Tresca and von Mises criterion), F_t [kN] CHECK

Point 1

Step 1. Calculate radial stress from load F₁ using the Boussinesq formula. 

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Step 2. Calculate radial stress from load F₂ using the Boussinesq formula. 

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Step 3. Give total stresses of Point 1. Illustrate stress’ direction. Name the stress state of the point.

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Point 2

Step 1. Calculate radial stress from load F₁  and F₁ using the Boussinesq formula. 

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Point 3

Step 1. Calculate radial stress from load F₁ using the Boussinesq formula. 

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Step 2. Calculate radial stress from load F₂ using the Boussinesq formula. 

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Step 3. Summarize stresses of Point 3 arising from loads F₁ and F₂. Illustrate stress’ direction. Name the stress state of the point.

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Step 4. Check resistance with Rankine failure criterion.

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Step 5. Check resistance with Tresca failure criterion.

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Step 6. Check resistance with von Mises failure criterion.

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Point 1

The distance of Point 1 from the point of application of load F₁ is 2a. The angle between the axis of the load and the polar coordinate axis, r is 45°. Radial stress from load F₁ is given by the Boussinesq formula:

${\sigma }_{r,1}^1=–\frac{2F}{\pi 2a}\cos 45°=–\frac{2\times 250}{\pi \times 2\times 1.2}\frac{\sqrt{2}}{2}=–46.89\frac{\mathrm{kN}}{{\mathrm{m}}^2} \left(\mathrm{compression}\right)$ 

Radius of the stress arises from load F₂ is perpendicular to the load’s axis, thus no stress arise from F₂ (cos 90^o = 0, Boussinesq formula results in zero). The total stress is

${\mathit{\sigma }}_{\mathbf{r}}^{\mathbf{1}}={\sigma }_{r,1}^1+0\mathbf{=}\mathbf{–}\mathbf{46}\mathbf{.}\mathbf{89}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)}$ 

The point is subjected to uniaxial compression.

Point 2

For Point 2 the rotation angle, φ= 90° for both loads, thus no stress arises at point 2 (cos 90^o = 0, Boussinesq formula results in zero stress).

Point 3 

The distance of Point 3 from the point of application of load F₁ is a. The angle between the axis of the load and the polar coordinate axis, r is 135^°.

${\sigma }_{r,1}^3=–\frac{2F}{\pi a}\cos 135°=–\frac{2\times 250}{\pi \times 1.2}\left(–\frac{\sqrt{2}}{2}\right)=93.78\frac{\mathrm{kN}}{{\mathrm{m}}^2} \left(\mathrm{tension}\right)$ 

The distance of Point 3 from the point of application of load F₂ is a. The angle between the axis of the load and the polar coordinate axis, r is -135^°. Radial stress from load F₂ is:${\sigma }_{r,2}^3=–\frac{2F}{\pi a}\cos (–135°)=–\frac{2\times 250}{\pi \times 1.2}\left(–\frac{\sqrt{2}}{2}\right)=93.78\frac{\mathrm{kN}}{{\mathrm{m}}^2} \left(\mathrm{tension}\right)$ 

Summation of the above stresses is

${\mathit{\sigma }}_{\mathbf{r}}^{\mathbf{3}}={\sigma }_{r,1}^3+{\sigma }_{r,2}^3=\mathbf{187}\mathbf{.}\mathbf{56}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)}$ 

The point is subjected to uniaxial tension.

Check resistance with Rankine failure criterion.

Equation

$\left|{\sigma }_1\right|=\left|{\sigma }_r^3\right|=187.56 \frac{\mathrm{kN}}{{\mathrm{m}}^2}\le f=200\frac{\mathrm{kN}}{{\mathrm{m}}^2}$ 

$\left|{\sigma }_2\right|=0$

The material is safe. The failure load is

$\begin{array}{c}f=\left|{\sigma }_1^{\max }\right|=\left|–\frac{2F_t}{\pi a}\cos 135°–\frac{2F_t}{\pi a}\cos \left(–135°\right)\right|=2\frac{2F_t}{\pi a}\frac{\sqrt{2}}{2}\\ \to {\mathit{F}}_{\mathbf{t}}=\frac{f\pi a}{4\sqrt{2}/2}=\frac{200\pi \times 1.2}{4\sqrt{2}/2}\mathbf{=}\mathbf{266}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{kN}\end{array}$

Check resistance with Tresca failure criterion.

$$\begin{gathered} \left|{\sigma }_1\right|=\left|{\sigma }_r^1\right|=187.56 \frac{\mathrm{kN}}{{\mathrm{m}}^2}\le 200 \frac{\mathrm{kN}}{{\mathrm{m}}^2} \\ \left|{\sigma }_1\right|=0 \\ \left|{\tau }^{\max }\right|=\left|\frac{{\sigma }_1}{2}\right|=93.78 \frac{\mathrm{kN}}{{\mathrm{m}}^2}\le \frac{f}{2}=100\frac{\mathrm{kN}}{{\mathrm{m}}^2} \end{gathered}$$

The material is safe. The failure load is equivalent to that of Rankine criterion:

$\left.\begin{array}{r}f=\left|{\sigma }_1^{\max }\right|=\left|2\frac{2F_t}{\pi a}\cos 135°\right|\\ \frac{ f}{2}=\left|{\tau }_{}^{\max }\right|=\left|\frac{{\sigma }_1^{\max }}{2}\right|\end{array}\right\} \to {\mathit{F}}_{\mathbf{t}}=\frac{f\pi a}{4\sqrt{2}/2}=\frac{200\pi \times 1.2}{4\sqrt{2}/2}\mathbf{=}\mathbf{266}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{kN}$

Check resistance with von Mises failure criterion.

${\sigma }_1^2+{\sigma }_2^2–{\sigma }_1{\sigma }_2={\sigma }_r^2=35178.8 \frac{{\mathrm{kN}}^2}{{\mathrm{m}}^4}\le f^2=40000\frac{{\mathrm{kN}}^2}{{\mathrm{m}}^4}$

The material is safe. The failure load is

$f^2={\sigma }_1^2+{\sigma }_2^2–{\sigma }_1{\sigma }_2={\sigma }_r^2 \to {\mathit{F}}_{\mathbf{t}}=\frac{f\pi a}{4\sqrt{2}/2}=\frac{200\pi \times 1.2}{4\sqrt{2}/2}\mathbf{=}\mathbf{266}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{kN}$

The failure load results in the same values for all the failure criteria.