**1 Answer**

written 5.6 years ago by |

When the ratio of inner diameter of the cylinder to the thickness is less than 15, the cylinder is called as a **'thick-walled'** cylinder or simply **'thick'** cylinder.

**Examples**-Hydraulic Cylinders, high pressure pipes, gun barrel

**Formulae list:-**

(1) $\sigma_{t}$ = Tangential stress(Tensile)

$\sigma_{t}$ = +$\frac{P_{i}(D_{o}^2+D_{i}^2)}{D_{o}^2-D_{i}^2}$

(2) Radial stress (Compressive), $\sigma_{r}$ $\;\;\Bigg\}$Principal stresses at inner surface of cylinder

$\sigma_{r} = -P_{i}$

(3) Axial stress (Tensile), $\sigma_{1}$

$\sigma_{1}$ = $\frac{P_{i}\,D_{i}^2} {D_{o}^2 - D_{i}^2}$

**Lames Eqn:**

(1) When the material of the cylinder is brittle, such as cast iron or cast steel, lames eqn. is used to determine wall thickness.

(2) It is based on maximum principle stress theory of failure.

(3) Experimental investigations suggest that maximum principle stress theory gives good predictions for brittle materials.

The three principle stresses at the inner surface of the cylinder are as follows-

(i) $\sigma_{t}$ = +$\frac{P_{i}(D_{o}^2+D_{i}^2)}{D_{o}^2-D_{i}^2}$-----(a)

(ii) $\sigma_{r} = -P_{i}$ -----(b)

(iii) $\sigma_{1}$ = +$\frac{P_{i}\,D_{i}^2} {D_{o}^2 - D_{i}^2}$-----(c)

Therefore, $\sigma_{t}$ > $\sigma_{1}$ > $\sigma_{r}$

Since, $\sigma_{t}$ is maximum principle stress , it is the criteria for design.

From eqn (a),

$\frac{\sigma_{t}}{P_{i}} = \frac{D_{o}^2+D_{i}^2}{D_{o}^2-D_{i}^2}$

By componendo dividendo,

$\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}} = \frac{(D_{o}^2 + D_{i}^2) + (D_{o}^2 - D_{i}^2)}{(D_{o}^2+D_{i}^2) - (D_{o}^2 - D_{i}^2)}$

$\therefore \frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}} = \frac{D_{o}^2}{D_{i}^2}$

$\therefore$ $\frac{D_{o}}{D_{i}}$ = $\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}}$

Substitute, $D_{o} = D_{i} + 2t$ in above eqn,

$\therefore$ $\frac{D_{i} + 2t}{D_{i}}$ = $\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}}$

$\therefore$ 1 + 2$\;[\frac{t}{D_{i}}]$ = $\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}}$

$\therefore$ t = $\frac{D_{i}}{2}\;[\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}}$ - 1] ... Lames Eqn.

where, $\sigma_{t}$ = $\frac{S_{ut}}{FOS}$

**Q.1 A hydraulic cylinder, made of gray cast iron( S$_{ut}$ = 300N/m)is subjected to an internal pressure of 15MPa. If the inner and outside diameters of the cylinders are 200mm and 240mm respectively, determine the factor of safety,using lames eqn.**

**If the cylinder pressure is further increased by 50%, what will be the factor of safety?**

Soln: Given $P_{i} = 15MPa = 15N/mm^2$

$D_{i} = 200\ mm$

$D_{o} = 240\ mm$

$S_{ut} = 300\ N/mm^2$

Soln:

**CASE I**

$t = \frac{D_{o} - D_{i}}{2} = \frac{240 - 200}{2} = 20\ mm$

$t = \frac{D_{i}}{2}\;[\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}} - 1]$ ... Lames Eqn.

$\therefore 20 = \frac{200}{2}\;[\sqrt{\frac{\sigma_{t} + 15}{\sigma_{t} - 15}} - 1]$

$\therefore \sigma_{t} = 83.1818\ N/mm^2$

$\sigma_{t}$ = $\frac{S_{ut}}{FOS}$

$\therefore 83.1818 = \frac{300}{FOS}$

$\therefore \text{FOS} = 3.6$

**CASE II**

$P_{i} = 15 + [\frac{50}{100}\;(15)]$

$\therefore P_{i} = 22.5\ MPa = 22.5\ N/mm^2$

By Lames Eqn,

$20 = \frac{200}{2}\;[\sqrt{\frac{\sigma_{t} + 22.5}{\sigma_{t} - 22.5}} - 1]$

$\therefore \sigma_{t} = 124.7727\ N/mm^2$

$\sigma_{t}$ = $\frac{S_{ut}}{FOS}$

$\therefore 124.7727 = \frac{300}{FOS}$

$\therefore \text{FOS} = 2.4$

**Q.2 A piston rod of hydraulic cylinder exerts an operating force of 10KN. The friction due to piston packing and stuffing box is equivalent to 10% of operating force. The pressure in the cylinder A is 10MPa. The cylinder is made of cast iron FG200 [S$_{ut}$ = 200N/mm$^2$] and the factor of safety is 5. Determine the diameter and thickness of wall.**

**Given:** $P_{i} = 10\ MPa$

$S_{ut} = 200\ N/mm^2$

**Solution:**

Piston force = 10kN

Friction effect = 10% of piston force

$= \frac{10}{100} \times 10$

$= 1\ kN$

$\therefore$ Total force on piston = Piston force + Friction effect

$= 10 +1$

$= 11\ kN$

$P_{i} = \frac{Piston\;force}{A_{us}}$

$\therefore 10 = \frac{11 \times 10^3}{\frac{\pi}{4}\; (D_{i})^2}$

$\therefore D_{i} = 37.4 mm \approx 40\ mm$

$\sigma_{t} = \frac{S_{ut}}{FOS} = \frac{200}{5} = 40\ N/mm^2$

By lames eqn, $t = \frac{D_{i}}{2}\;[\sqrt{\frac{\sigma_{t} + P_{i}}{\sigma_{t} - P_{i}}} - 1]$

$\therefore t = \frac{40}{2}\;[\sqrt{\frac{40 + 10}{40 - 10}} - 1]$

$\therefore t = 5.8198 \approx 6\ mm.$

$t = \frac{D_{o} - D_{i}}{2}$

$\therefore 6 = \frac{D_{o} - 40}{2}$

$\therefore D_{o} = 52\ mm$