Problem 2.12. Plate’s loads on the edge

Normal force acting on edge 1, N₁, [kN/m]=

Shear force acting on edge 1, V₁, [kN/m]=

Normal force acting on edge 2, N₂, [kN/m]=

Shear force acting on edge 2, V₂, [kN/m]=

Normal force acting on edge 3, N₃, [kN/m]=

Shear force acting on edge 2, V₃, [kN/m]=

Step 1. Write equilibrium equations and prove that the surface of the plate is unloaded.

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Equilibrium equations are given in Table 2.2.

$$\begin{gathered} \frac{\partial {\sigma }_x}{\partial x}+\frac{{\displaystyle \partial {\tau }_{xy}}}{{\displaystyle \partial y}}+\frac{{\displaystyle \partial {\tau }_{xz}}}{{\displaystyle \partial z}}+p_x=0 \\ \frac{{\displaystyle \partial {\sigma }_y}}{{\displaystyle \partial y}}+\frac{{\displaystyle \partial {\tau }_{xy}}}{{\displaystyle \partial x}}+\frac{{\displaystyle \partial {\tau }_{yz}}}{{\displaystyle \partial z}}+p_y=0 \\ \frac{{\displaystyle \partial {\sigma }_z}}{{\displaystyle \partial z}}+\frac{{\displaystyle \partial {\tau }_{xz}}}{{\displaystyle \partial x}}+\frac{{\displaystyle \partial {\tau }_{yz}}}{{\displaystyle \partial y}}+p_z=0 \end{gathered}$$

Substitute constant non zero stresses σ_x = 60 N/mm² and σ_y = -60 N/mm² into the above equations. Since the derivatives of the constant stresses and all the other stresses are also zero, substitution results in

$p_x=0, p_y=0, p_z=0$ 

Step 2. Give the edge loads in the global x-y coordinate system.

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Step 3. Give the edge loads in the local coordinate systems attached to the edges.

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Results of the stress transformation into the rotated coordinate systems is given below and they are illustrated in the Figure (degrees of the rotations are 45°, 0°,-45°)

Edge 1

$$\begin{gathered} \mathbf{\sigma }\mathbf{‘}=\left\{\begin{array}{c}\mathrm{\sigma }{‘}_x\\ \mathrm{\sigma }{‘}_y\\ \mathrm{\tau }{‘}_{xy}\end{array}\right\}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{45}}\mathbf{\sigma }=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ –60\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_1=0 \\ {v}_1=–\mathrm{\tau }{‘}_{xy}=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\mathit{N}}_{\mathbf{1}}\mathbf{=}\mathbf{0} \end{gathered}$$

Since the shear stress is negative, the direction of the load is in the opposite direction, its value is

${\mathit{V}}_{\mathbf{1}}=t{v}_1=20\times 60\mathbf{=}\mathbf{1200}\frac{\mathbf{N}}{\mathbf{mm}}$

Edge 2

$$\begin{gathered} \mathbf{\sigma }=\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_2=–{\sigma }_y=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {v}_2=0 \\ {\mathit{N}}_{\mathbf{2}}=t{n}_2=20\times 60=\mathbf{1200}\frac{\mathbf{N}}{\mathbf{m}\mathbf{m}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$

Edge 3

$$\begin{gathered} \mathbf{\sigma }\mathbf{‘}\mathbf{‘}=\left\{\begin{array}{c}\mathrm{\sigma }‘{‘}_x\\ \mathrm{\sigma }‘{‘}_y\\ \mathrm{\tau }‘{‘}_{xy}\end{array}\right\}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{–}\mathbf{45}}\mathbf{\sigma }=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 60\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_3=0 \\ {v}_3=\mathrm{\tau }‘{‘}_{xy}=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\mathit{V}}_{\mathbf{3}}=t{v}_3=20\times 60\mathbf{=}\mathbf{1200}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{N}}_{\mathbf{3}}\mathbf{=}\mathbf{0} \end{gathered}$$

We get the same edge loads if the resultants of f_x and f_y  are calculated at all edges. As it was shown before compression and tension results in pure shear in the 45° direction.

See Figure 2.4.

Step 4. In which special stress-state is the plate in?

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The equilibrium equations are the following

See Table 2.2.

We substitute the constant non zero stresses σ_x = 60 N/mm² and σ_y = -60 N/mm² into the above equations. Since the derivatives of the constant stresses and all the other stresses are also zero, substitution results in

$p_x=0, p_y=0, p_z=0$ 

these results prove that the surface is unloaded.

Loads must equilibrate the stresses of the element at the edge of the plate. The edge loads in the global x-y coordinate system are shown in the Figure below.

Resultant of the distributes loads are obtained by the multiplication of them with the thickness of the plate.

$$\begin{gathered} F_x=f_{x}t={\sigma }_{x}t=60\times 20=1200\frac{\mathrm{N}}{\mathrm{mm}} \\ {F}_y=f_{y}t=–{\sigma }_{y}t=60\times 20=1200\frac{\mathrm{N}}{\mathrm{mm}} \end{gathered}$$

The edge loads in the local coordinate systems attached to the edges can be obtained by performing stress transformation. Transformations into the rotated coordinate systems are given below and the results are illustrated in the Figure (degrees of the rotations are 45°, 0°,-45°)

Edge 1

$$\begin{gathered} \mathbf{\sigma }\mathbf{‘}=\left\{\begin{array}{c}\mathrm{\sigma }{‘}_x\\ \mathrm{\sigma }{‘}_y\\ \mathrm{\tau }{‘}_{xy}\end{array}\right\}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{45}}\mathbf{\sigma }=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ –60\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_1=0 \\ {v}_1=–\mathrm{\tau }{‘}_{xy}=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\mathit{N}}_{\mathbf{1}}\mathbf{=}\mathbf{0} \end{gathered}$$

Since the shear stress is negative, the direction of the load is in the opposite direction, its value is

${\mathit{V}}_{\mathbf{1}}=t{v}_1=20\times 60\mathbf{=}\mathbf{1200}\frac{\mathbf{N}}{\mathbf{mm}}$

Edge 2

$$\begin{gathered} \mathbf{\sigma }=\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_2=–{\sigma }_y=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {v}_2=0 \\ {\mathit{N}}_{\mathbf{2}}=t{n}_2=20\times (–60)=\mathbf{–}\mathbf{1200}\frac{\mathbf{N}}{\mathbf{m}\mathbf{m}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$

Edge 3

$$\begin{gathered} \mathbf{\sigma }\mathbf{‘}\mathbf{‘}=\left\{\begin{array}{c}\mathrm{\sigma }‘{‘}_x\\ \mathrm{\sigma }‘{‘}_y\\ \mathrm{\tau }‘{‘}_{xy}\end{array}\right\}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{–}\mathbf{45}}\mathbf{\sigma }=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}60\\ –60\\ 0\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 60\end{array}\right\}\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {n}_3=0 \\ {v}_3=\mathrm{\tau }‘{‘}_{xy}=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\mathit{V}}_{\mathbf{3}}=t{v}_3=20\times 60\mathbf{=}\mathbf{1200}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{N}}_{\mathbf{3}}\mathbf{=}\mathbf{0} \end{gathered}$$

We get the same edge loads if the resultants of f_x and f_y  are calculated at all edges. As it was shown before compression and tension results in pure shear in the 45° direction.

See Figure 2.4.

The plate is subjected to pure shear.