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A solid right cylinder of rock core is surrounded by four rods made of mild steel (all-thread rods). The rods are placed equidistantly around the core in a square formation. The tops and bottoms of the four rods are fixed to the top and bottom rigid plates, but the rock is just fitted snugly between the top and bottom plates. The rock is wrapped in an impermeable jacket. The whole system is submerged into a pressure vessel holding water and the water pressure is increased. When the water in the pressure vessel is increased to a certain amount, the fluid in the pore space of the rock is increased (pore pressure) accordingly to give the rock pressure support resulting in an effective stress on the rock. After reaching final pressures the temperature of the pressure vessel and all its contents is then increased.

Determine the following quantities: (a) the resulting load $P_s$ in the rock cylinder and $P_r$ in the steel rods; (b) the corresponding axial stresses $\sigma_s$, $\sigma_r$; and (c) the axial deformation $\delta$ of the assembly and (d) the principle stresses on the core.

fixture

Below is my initial attempt at this problem. I assumed that lateral strain is insignificant but I don't know if that is true. If it is not true then I suppose this problem would have to be solved with iterations? I am hoping someone can tell me if the methodology is correct or how to correctly solve this problem. I've included some values to solve with, but I would ultimately like to have the equations/methodology to solve for any values of the system. Thanks.

When trying to solve this problem I assumed the deformation of the rods are uniform throughout their volume, the rods are prismatic, the compressive load acts at the center of the rods, rods are homogeneous, isotropic, and behave linearly elastically. I made the same assumptions about the rock aswell.

The bottom rigid plate is a circle with a diameter $d_p=15$ in., each rod has a diameter $d_r=0.4$ in., and the core has diameter $d_s=7$ in. The rods have a Young's modulus $E_r=30,000$ ksi and the rock has a modulus of elasticity of $E_s=1,450$ ksi. The starting length of both the rods and the core is $L= 20$ in.

Before temperature is introduced to the system the entire fixture is subjected to the overburden water pressure $\sigma_{ob}=25,000$ psi and internal pore pressure $\sigma_{Pp}=20,000$ psi. Therefore the resulting load on the core and rods imposed by the plate is

$$P = \sigma_{ob} \cdot A_{plate} + \sigma_{Pp} \cdot A_s=\sigma_{ob} \cdot \left(A_{plate}^{bottom}-A_{plate}^{top}\right) + \sigma_{Pp} \cdot A_s $$

$$=\sigma_{ob} \cdot \left(A_s+4A_r\right) + \sigma_{Pp} \cdot A_s $$

$$ = (-25\, \text{ksi}) \left(\frac{\pi}{4} \cdot (7 \, \text{in.})^2 + \pi \cdot (0.4 \, \text{in.})^2 \right) + (20\, \text{ksi}) \left(\frac{\pi}{4} \cdot (7 \, \text{in.})^2 \right) \approx -205 \, \text{k}$$

The plate is subjected to the force $P$ and the the unkown compressive forces $P_s$ and $P_r$; thus the equation of equilibrium is:

$$\Sigma F_{vert} = 0$$ and $$4P_r + P_s - P = 0$$

This equation, which is the only nontrivial equilibrium equation available, contains two unkowns. Therefore, we conclude that the structure is statically indeterminate.

Because the end plates are rigid, the rock cylinder and steel rods must shorten by the same amount. Denoting the shortenings of the rock and steel parts by $\delta_s$ and $\delta_r$, respectively, we obtain the following equation of compatibility:

$$\delta_s = \delta_r$$

The changes in lengths of the cylinder and rods can be obtained from the general equation $\delta = PL/EA$. Therefore, in this problem the force-displacement relations are:

$$\delta_s = \frac{P_sL}{E_sA_s}$$ and $$\delta_r= \frac{P_rL}{E_rA_r}$$

We can now solve simultaneously the three stes of equations to obtain the axial forces in the rock cylinder and steel rods:

$$\frac{P_sL}{E_sA_s}=\frac{P_rL}{E_rA_r} \rightarrow P_r=P_s \frac{E_rA_r}{E_sA_s} \rightarrow 4P_s \frac{E_rA_r}{E_sA_s}+P_s-P=0$$

Solving for $P_s$ we have $$P_s=P \left(\frac{E_sA_s}{4E_rA_r+E_sA_s}\right)$$

Likewise, solving for $P_r$ we have $$P_r=P \left(\frac{E_rA_r}{4E_rA_r+E_sA_s}\right)$$

Substituting our values we can solve for the resulting loads:

$$P_r=(-205 \, \text{k}) \left(\frac{(30000 \, \text{ksi})(\frac{\pi}{4}(0.4 \, \text{in.})^2)}{4(30000 \, \text{ksi})(\frac{\pi}{4}(0.4 \, \text{in.})^2)+(1450 \, \text{ksi})(\frac{\pi}{4}(7 \, \text{in.})^2)}\right) \approx -11 \, \text{k}$$

Similarily, plugging values in for $P_s$ we get: $$P_s \approx -161 \, \text{k}$$

The change in length of the core and rods due to the pressure increase will be: $$\delta_r=\delta_s=\frac{PL}{4E_rA_r+E_sA_s}=\frac{(-205 \, \text{k})(20 \, \text{in.})}{4(30000 \, \text{ksi})(\frac{\pi}{4}(0.4 \, \text{in.})^2)+(1450 \, \text{ksi})(\frac{\pi}{4}(7 \, \text{in.})^2)} \approx -0.057 \, \text{in.}$$

At this point the temperature is increased to $400\,^{\circ}\mathrm{F}$. The thermal strain relation $\epsilon_T$ induced on a body is $$\epsilon_T = \alpha (\Delta T)$$ where $\alpha$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature. We assumed $\alpha_r = 6.5 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1}$ and $\alpha_s = 3 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1}$.

The equivalent stress induced due to thermal expansion is: $$\sigma_{eq} = E\alpha(\Delta T)$$

And the equivalent load can be calculated as: $$P_{T} = \sigma_{eq} \cdot A$$

Substituting values we can solve for the equivalent loads due to temperature increase in the rods and rock: $$P_T^s = (1450 \, \text{ksi})(3 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1})(400\,^{\circ}\mathrm{F})(\frac{\pi}{4}(7 \, \text{in.})^2) \approx 67 \, \text{k}$$

$$P_T^r = (30000 \, \text{ksi})(6.5 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1})(400\,^{\circ}\mathrm{F})(\frac{\pi}{4}(0.4 \, \text{in.})^2) \approx 9.8 \, \text{k}$$

We then add these loads to the loads due to compression to get the final loads:

$$P_r + P_T^r = -11 \, \text{k} + 9.8 \, \text{k} \approx -1.1 \, \text{k (compression)}$$ $$P_s + P_T^s = -161 \, \text{k} + 67 \, \text{k} \approx -94.4 \, \text{k (compression)}$$

With these values we can calculate the axial stresses: $$\sigma_s = \frac{P_s}{A_s} = \frac{-94.4 \, \text{k}}{\frac{\pi}{4}(7 \, \text{in.})^2} \approx -2.45 \, \text{ksi}$$

$$\sigma_r = \frac{P_r}{A_r} = \frac{-1.1 \, \text{k}}{\frac{\pi}{4}(0.4 \, \text{in.})^2} \approx -8.76 \, \text{ksi}$$

We now solve for the resulting length change in the rods and core do to the temperature increase:

$$\delta_r=\delta_{r}^i+\alpha_r(\Delta T)L'= -0.057 \, \text{in.} + (6.5 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1})(400\,^{\circ}\mathrm{F})(20 \, \text{in.} - 0.057 \, \text{in.}) \approx -0.006 \, \text{in.}$$

$$\delta_s=\delta_{s}^i+\alpha_s(\Delta T)L'= -0.057 \, \text{in.} + (3 \times 10^{-6} \,^{\circ}\mathrm{F}^{-1})(400\,^{\circ}\mathrm{F})(20 \, \text{in.} - 0.057 \, \text{in.}) \approx -0.034 \, \text{in.}$$

From this result we can conclude that the expansion in length of the rods due to the temperature increase causes a seperation between the bottom core face and the bottom fixture plate. This result may have came about due to incorrect assumptions on the values for the young's modulus and/or coefficient of thermal expansion. This separation will cause a fluid connection between overburden pressure and pore pressure which is not acceptable. If the resulting axial stress on the rock is to be a 1000 psi less than the resulting rock lateral stress, what would be the required material properties for the rock and rods?

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  • $\begingroup$ If the rods have a diameter, how can they be prismatic? $\endgroup$ – user80551 Aug 1 '13 at 18:06
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    $\begingroup$ @user80551 A prismatic bar is a structural member having a straight longitudinal axis and constant cross section throughout its length. $\endgroup$ – Armadillo Aug 1 '13 at 18:45

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