Pressure Sharing Design by Bedrock for Steel Penstock

 


Model of Penstock for Pressure Sharing Design by Bedrock

Model of penstock for Pressure Sharing Design by Bedrock is shown in below figure. The formula for Design is derived using the assumption shown in below table, and it is a formura by Elastic Theory.

fig
Model of Penstock for Pressure Sharing Design by Bedrock

Item Description
Penstock (steel shell) Elastic body which can transmit only curcumferential stress
Filling Concrete Elastic body which can transmit only compressive stress in the radial direction because of cracks in the concrete due to the internal water pressure. Plastick deformation coefficient is considered to evaluate the boid between the steel shell and concrete.
Bedrock Elastic body which can transmit both stresses in the circumferential and radial directions, alsoo compressive and tensile stresses, under the condition of sufficient initial stress in the underground.
Void between steel shell and concrete Void between steel shell and concrete includes the Plastic displacements of concrete and bedrock, temperature change of steel shell and shrinkage of the concrete. This void is considered under the condition of low internal water pressure.
Basic equation Basic equation is derived using the assumption that the displacement of the steel shell in the radial direction due to the internal water pressure is equal to the sum of the void between the steel shell and concrete, complessive displacements of concrete and bedrock.


Notation

$t$ Thickness of the steel shell    $\delta_s$ Elastic displacement of steel shell in the radius direction
$r_s$ Internal diameter of the steel shell    $\delta_c$ Elastic displacement of the concrete in the radius direction
$r_c$ Excavated diameter of the tunnel    $\delta_g$ Elastic displacement of the bedrock in the radius direction
$P$ Design internal water pressure    $\delta_0$ Void between the concrete and the steel shell
$P_s$ Internal pressure shared by the steel shell    $\Delta_c$ Plastic displacement of the concrete
$\lambda$ Pressure sharing ratio by the bedrock    $\Delta_g$ Plastic displacement of the bedrock
$\Delta T$ Temperature change of the steel shell    $\Delta_s$ Void due to the temperature change of the steel shell
$\alpha_s$ Coefficient of thermal expantion of the steel    $\Delta_{sd}$ Shrinkage of the concrete
$E_s$ Elastic modulus of the steel    $\sigma_s$ Stress of the steel shell in the Circumferential direction
$E_c$ Elastic modulus of the concrete    $\sigma_c$ Stress of the concrete in the radius direction
$E_g$ Elastic modulus of the bedrock    $\sigma_r$ Stress of the bedrock in the radius direction
$m_g$ Poisson's number of the bedrock($=1/\nu_g$)   $\sigma_{\theta}$ Stress of the bedrock in the circumferential direction
$\nu_g$ Poisson's ratio of the bedrock    $\varepsilon_s$ Strain of the steel shell in the circumferential direction
$\beta_c$ Plastic deformation modulus of the concrete    $\varepsilon_c$ Strain of the cincrete in the radius direction
$\beta_g$ Plastic deformation modulus of the bedrock    $\varepsilon_g$ Strain of the bedrock in the radius direction
$D$ Internal diameter of the steel shell    $\sigma_{sa}$ Allowable stress of the steel shell
$D_R$ Excavated diameter of the tunnel   


Basic equation

Define the displacements of the steel shell, concrete and bedrock in the radial direction as follow.

\begin{align} &\text{Elastic displacement of the steel shell} &~& \delta_s=\delta_c+\delta_g+\delta_0 \tag{1} \\ &\text{Void between the concrete and the steel shell} &~& \delta_0=\Delta_c+\Delta_g+\Delta_s+\Delta_{sd} \tag{2} \\ &\text{Internal water pressure shared by the steel shell} &~& P_s=P(1-\lambda) \tag{3} \end{align}


Elastic displacements of the components in the radial direction

Elastic displacement of the steel shell

When it seems that the steel shell is a thin wall which can transmit only circumferential stress, following equation can be obtained taking into consideration of equiliblium of small element.

\begin{equation} 2\sigma_s\sin\left(\frac{\Delta\theta}{2}\right)\cdot t=P_s\cdot r_s\cdot\Delta\theta \end{equation}

Above can be approximated as follow because of small angle $\Delta\theta$,

\begin{equation} 2\sigma_s\left(\frac{\Delta\theta}{2}\right)\cdot t=P_s\cdot r_s\cdot\Delta\theta \end{equation}

Then, the circumferential stress of the steel shell is expressed below,

\begin{equation} \sigma_s=\frac{P_s\cdot r_s}{t} \tag{4} \end{equation}

and the circumferential strain of the steel shell is expressed below,

\begin{equation} \varepsilon_s=\frac{\Delta_s}{r_s} \end{equation}

as a result, the displacement of the steel shell in the radial direction is obtained as follow.

\begin{equation} \delta_s=r_s\cdot\varepsilon_s=r_s\frac{\sigma_s}{E_s}=\frac{P_s\cdot r_s^2}{E_s\cdot t} \tag{5} \end{equation}

Elastic displacement of the concrete

In the concrete body, tensile stress in the circumferential direction and compressive stress in the radius direction are act due to the displacement of the steel shell by internal water pressyre. However, it seems that the concrete can transmit only the compressive stress in the radial direction because the tensile strength of the cincrete is small and cracks are occured by the displacement of the steel shell.

\begin{equation} 2\pi r\sigma_c=2\pi r_s(P-P_s) \end{equation}
\begin{equation} \sigma_c=\frac{(P-P_s)r_s}{r} \qquad \text{(compressibe stress)} \tag{6} \end{equation}

So, the strain of the concrete is expressed as follow,

\begin{equation} \varepsilon_c=\frac{\sigma_c}{E_c}=\frac{(P-P_s)r_s}{E_c\cdot r} \end{equation}

where, positive is the compressive stress.

The displacement can be obtained as a result of integration of the strain of the concrete, and it is expressed below.

\begin{equation} \delta_c=\int_{r_s}^{r_c}\varepsilon_c dr=\int_{r_s}^{r_c}\frac{(P-P_s)r_s}{E_c\cdot r}dr=\frac{(P-P_s)r_s}{E_c}\ln\left(\frac{r_c}{r_s}\right) \tag{7} \end{equation}

Elastic displacement of the bedrock

fig

Since the initial stress acts to the bedrock in the underground, it seems that the bedrock can be transmit both of the compressive stress in the radius direction and the tensile stress in the circumferential direction. When we consider the general theorem of thick cylinder shown in the left figure, the stress in the radius direction $\sigma_r$ and the stress in the circumferential $\sigma_{\theta}$ are expressed below.

\begin{align} &\sigma_r=\cfrac{a^2}{b^2-a^2}\left(1-\cfrac{b^2}{r^2}\right)\cdot P_a-\cfrac{b^2}{b^2-a^2}\left(1-\cfrac{a^2}{r^2}\right)\cdot P_b \tag{8}\\ &\sigma_{\theta}=\cfrac{a^2}{b^2-a^2}\left(1+\cfrac{b^2}{r^2}\right)\cdot P_a-\cfrac{b^2}{b^2-a^2}\left(1+\cfrac{a^2}{r^2}\right)\cdot P_b \tag{9} \end{align}

where, positive value means tensile stress.

In this case, it seems that the penstock is installed in the deep area of the underground and the outer boundary of thr ground seems infinity. Then the boundary conditions are

\begin{align} & a=r_c & \qquad & b\rightarrow\infty \notag \\ & P_a=\frac{(P-P_s)r_s}{r_c} & \qquad & P_b=0 \notag \end{align}

From above, the stresses of bedrock in the radial and curcumferential directions are expressed as follow.

\begin{align} &\sigma_r=-\frac{(P-P_s)\cdot r_s\cdot r_c}{r^2} & \qquad\text{(compressive stress)} \tag{10}\\ &\sigma_{\theta}=\frac{(P-P_s)\cdot r_s\cdot r_c}{r^2} & \qquad\text{(tensile stress)} \tag{11} \end{align}

And the strain of bedrock in the radial direction is

\begin{equation} \varepsilon_g=\frac{(1+\nu_g)\sigma_r}{E_g}=(1+\nu_g)\frac{(P-P_s)r_s r_c}{E_g}\frac{1}{r^2} \end{equation}

where compressive stress is positive value.


(Note)

Above equation can be used in the both conditions of plane stress and plane strain, because of $\sigma_r=-\sigma_{\theta}$ .

\begin{align} &\text{Plane stress} & \varepsilon_g&=\frac{\sigma_r-\nu_g\sigma_{\theta}}{E_g} &=\frac{(1+\nu_g)\sigma_r}{E_g} \notag \\ &\text{plane strain} & \varepsilon_g&=\frac{(1-\nu_g^2)\sigma_r-\nu_g(1+\nu_g)\sigma_{\theta}}{E_g}&=\frac{(1+\nu_g)\sigma_r}{E_g} \notag \end{align}

The displacement of bedrock can be obtained as an integrated value of the strain of bedrock.

\begin{equation} \delta_g=\int_{r_c}^{\infty}\varepsilon_g dr=(1+\nu_g)\frac{(P-P_s)r_s}{E_g} \tag{12} \end{equation}


Void between the steel shell and the concrete

In case that the plastic displacements of th econcrete and the bedrock is proportional to the elastic displacements of each material, and we use the constants of proportionality $\beta_c$ and $\beta_g$ , the plastic displacements of the concrete and the bedrock can be espressed as follow.

\begin{align} \Delta_c&=\beta_c\frac{(P-P_s)r_s}{E_c}\ln\left(\frac{r_c}{r_s}\right) \notag \\ \Delta_g&=\beta_g(1+\nu_g)\frac{(P-P_s)r_s}{E_g} \end{align}

The void due to the temperature change of steel shell is

\begin{equation} \Delta_s=\alpha_s\cdot\Delta T\cdot r_s \end{equation}

From above, the void between the steel shell and the concrete can be obtained as follow, by adding the shrinkage of the concrete $\Delta_{sd}$ .

\begin{align} \delta_0&=\Delta_c+\Delta_g+\Delta_s+\Delta_{sd} \notag \\ &=\beta_c\frac{(P-P_s)r_s}{E_c}\ln\left(\frac{r_c}{r_s}\right) +\beta_g(1+\nu_g)\frac{(P-P_s)r_s}{E_g}+\alpha_s\cdot\Delta T\cdot r_s+\Delta_{sd} \tag{13} \end{align}


Calculation of the pressure sharing ratio by the bedrock

Substituting the equations (5)(7)(12)(13) in (1), equation (14) can be obtained.

\begin{equation} \frac{P_s\cdot r_s^2}{E_s\cdot t} =(1+\beta_c)\frac{(P-P_s)r_s}{E_c}\ln\left(\frac{r_c}{r_s}\right) +(1+\beta_g)(1+\nu_g)\frac{(P-P_s)r_s}{E_g} +\alpha_s\cdot\Delta T\cdot r_s+\Delta_{sd} \tag{14} \end{equation}

The pressure sharing ratio by the bedrock $\lambda$ can be obtained by eliminating $P_s$ in the equations (3) and (14), and it is shown below as an euqation (15).

\begin{equation} \lambda=\cfrac{1-\cfrac{E_s}{P}\cdot\alpha_s\cdot\left(\Delta T+\cfrac{\Delta_{sd}}{\alpha_s r_s}\right)\cdot\cfrac{t}{r_s}} {1+(1+\beta_c)\cdot\cfrac{E_s}{E_c}\cdot\cfrac{t}{r_s}\cdot\ln\left(\cfrac{r_c}{r_s}\right) +(1+\beta_g)\cdot\cfrac{E_s}{E_g}\cdot(1+\nu_g)\cdot\cfrac{t}{r_s}} \tag{15} \end{equation}

In addition, the stresses of the steel shell, the concrete and the bedrock can be calculated as follows. These equations are obtained using the the pressure sharing ratio by the bedrock $\lambda$ instead of $P_s$ in the equations (4)(6)(10)(11).

\begin{align} &\sigma_s=\frac{P\cdot r_s}{t}\cdot(1-\lambda) &\text{stress of the steel shell} \tag{16}\\ &\sigma_c=-\lambda\cdot P\cdot\frac{r_s}{r} &\text{stress of the concrete} \tag{17}\\ &\sigma_r=-\lambda\cdot P\cdot\frac{r_s\cdot r_c}{r^2} &\text{stress of the bedrock in the radial direction} \tag{18}\\ &\sigma_{\theta}=\lambda\cdot P\cdot\frac{r_s\cdot r_c}{r^2} &\text{stress of the bedrock in the circumferential direction} \tag{19} \end{align}

In above equations, positive values mean tensile stresses.

After we rewrite the equations (15)(16) using the tunnel excavated diameter $D_0$ and Poisson's number $m_g~(=1/\nu_g)$ , we can obtain the same equation as indicated in the 'Technical Standards for Gates and Penstocks' in japan. (In this web page, the shrinkage of the concrete $\Delta_{sd}$ is added in the equation.)

\begin{align} &\sigma_s=\frac{PD}{2t}(1-\lambda) \tag{20}\\ &\lambda=\cfrac{1-\cfrac{E_s}{P}\cdot\alpha_s\cdot\left(\Delta T+\cfrac{2\Delta_{sd}}{\alpha_s D}\right)\cdot\cfrac{2t}{D}}{1+(1+\beta_c)\cdot\cfrac{E_s}{E_c}\cdot\cfrac{2t}{D}\cdot\ln\left(\cfrac{D_R}{D}\right)+(1+\beta_g)\cdot\cfrac{E_s}{E_g}\cdot\cfrac{1+m_g}{m_g}\cdot\cfrac{2t}{D}} \tag{21} \end{align}

Finally, the required thickness of the steel shell ' $t$ ' can be calculated using the allowable atress of the steel plate $\sigma_{sa}$ instead of $\sigma_s$ and by eliminating the $\lambda$ in the equations (20)(21).

\begin{equation} t=\cfrac{\cfrac{PD}{\sigma_{sa}}\left\{(1+\beta_c)\cdot\cfrac{E_s}{E_c}\cdot\cfrac{1}{D}\cdot\ln\left(\cfrac{D_R}{D}\right)+(1+\beta_g)\cdot\cfrac{E_s}{E_g}\cdot\cfrac{1+m_g}{m_g}\cdot\cfrac{1}{D}+\cfrac{E_s}{P}\cdot\alpha_s\cdot\left(\Delta T+\cfrac{2\Delta_{sd}}{\alpha_s D}\right)\cdot\cfrac{1}{D}\right\}-1}{(1+\beta_c)\cdot\cfrac{E_s}{E_c}\cdot\cfrac{2}{D}\cdot\ln\left(\cfrac{D_R}{D}\right)+(1+\beta_g)\cdot\cfrac{E_s}{E_g}\cdot\cfrac{1+m_g}{m_g}\cdot\cfrac{2}{D}} \tag{22} \end{equation}


In the case of the limited boundary of the bedrock

The equations (10)(11) are derived under the condition of infinity bedrock area. In this section, we consider the condition that the bedrock has the limited boundary ( $r=r_g$ ) and the bedrock has the darial displacement. In this case, the boundary conditions are expressed below.

\begin{align} & a=r_c & \qquad & b=r_g \notag \\ & P_a=\frac{(P-P_s)r_s}{r_c} & \qquad & P_b=0 \notag \end{align}

The stresses of bedrock in the radial and circumferential direction can be expressed below.

\begin{align} &\sigma_r=\frac{(P-P_s)r_s r_c}{r_g^2-r_c^2}\left(1-\frac{r_g^2}{r^2}\right) \notag \\ &\sigma_{\theta}=\frac{(P-P_s)r_s r_c}{r_g^2-r_c^2}\left(1+\frac{r_g^2}{r^2}\right) \notag \end{align}

The strains of the bedrock in the radial direction are shown below, under the cases for plane stress and plane strain.

\begin{equation} \varepsilon_g=A\cdot\sigma_r-B\cdot\sigma_{\theta}=\frac{r_s r_c(P-P_s)}{r_g^2-r_c^2}\left\{(A-B)-(A+B)\frac{r_g^2}{r^2}\right\} \qquad \text{(tension: positive)} \end{equation}
\begin{equation} \left\{\begin{array}{lllll} \text{Plane stress} & \quad & A=\cfrac{1}{E_g} & \quad & B=\cfrac{\nu_g}{E_g} \\ \text{Plane strain} & \quad & A=\cfrac{1-\nu_g^2}{E_g} & \quad & B=\cfrac{\nu_g(1+\nu_g)}{E_g} \\ \end{array}\right. \notag \end{equation}

The displacement of bedrock becomes below, which can be obtained as an integrated value of the strain of the bedrock.

\begin{equation} \delta_g=\int_{r_c}^{r_g}\varepsilon_g dr=\frac{r_s r_c(P-P_s)}{r_g+r_c}\left\{(A+B)\frac{r_g}{r_c}-(A-B)\right\} \end{equation}

After this, the same process as obtaining the equation (14)(15) can be adopted, and we can get the pressure sharing ratio by the bedrock under the condition of limited boundary of the bedrock.

\begin{equation} \lambda=\cfrac{1-\cfrac{E_s}{P}\cdot\alpha_s\cdot\left(\Delta T+\cfrac{\Delta_{sd}}{\alpha_s r_s}\right)\cdot\cfrac{t}{r_s}} {1+(1+\beta_c)\cdot\cfrac{E_s}{E_c}\cdot\cfrac{t}{r_s}\cdot\ln\left(\cfrac{r_c}{r_s}\right) +(1+\beta_g)\cdot\cfrac{E_s\cdot r_c}{r_g+r_c}\cdot\cfrac{t}{r_s}\cdot\left\{(A+B)\cfrac{r_g}{r_c}-(A-B)\right\}} \tag{23} \end{equation}
\begin{equation} \left\{\begin{array}{lllll} \text{Plane stress} & \quad & A+B=\cfrac{1+\nu_g}{E_g} & \quad & A-B=\cfrac{1-\nu_g}{E_g} \\ \text{Plane strain} & \quad & A+B=\cfrac{1+\nu_g}{E_g} & \quad & A-B=\cfrac{(1+\nu_g)(1-2\nu_g)}{E_g} \\ \end{array}\right. \notag \end{equation}


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inserted by FC2 system