2.4 Pressure Vessels, Column Buckling & Spring Design

Pressure Vessels, Column Buckling & Spring Design

Mechanical systems rely heavily on pressure containment vessels, structural columns supporting compressive axial loads, and mechanical springs designed for energy absorption and motion control. Mastering the distinct governing equations for thin vs. thick pressure vessels, elastic vs. inelastic column instability, and Wahl-corrected spring mechanics is critical for the PE Mechanical exam.

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1. Pressure Vessels: Thin-Walled vs. Thick-Walled Analysis

+-----------------------------------------------------------------------------+
|                   PRESSURE VESSEL THICKNESS CRITERION                       |
|                                                                             |
|   [THIN-WALLED VESSEL]  ---> \frac{r_i}{t} \ge 10 \quad \left(\frac{t}{r_i} \le 0.10\right)  |
|                              - Stress assumed uniform across wall thickness |
|                              - Radial stress \sigma_r is neglected (\approx 0)      |
|                                                                             |
|   [THICK-WALLED VESSEL] ---> \frac{r_i}{t} < 10 \quad \left(\frac{t}{r_i} > 0.10\right)   |
|                              - Non-linear stress distribution (Lamé)        |
|                              - Radial stress \sigma_r must be included             |
+-----------------------------------------------------------------------------+

1. Thin-Walled Cylindrical & Spherical Vessels

+-----------------------------------------------------------------------------+
|                     THIN-WALLED VESSEL STRESS FORMULAS                      |
|                                                                             |
|   Cylindrical Vessel:                                                       |
|   - Hoop / Tangential Stress:    \sigma_h = \frac{p r}{t} = \frac{p d_i}{2t} |
|   - Longitudinal / Axial Stress: \sigma_L = \frac{p r}{2t} = \frac{p d_i}{4t} |
|   - Max In-Plane Shear Stress:   \tau_{\text{in-plane}} = \frac{\sigma_h - \sigma_L}{2} = \frac{p r}{4t} |
|   - Absolute Max Shear Stress:   \tau_{\text{max, abs}} = \frac{\sigma_h - (-p)}{2} \approx \frac{p r}{2t} |
|                                                                             |
|   Spherical Vessel:                                                         |
|   - Uniform Biaxial Stress:      \sigma_1 = \sigma_2 = \frac{p r}{2t}       |
|   - Absolute Max Shear Stress:   \tau_{\text{max, abs}} = \frac{p r}{4t}    |
+-----------------------------------------------------------------------------+

[!IMPORTANT] Welded Seam Integrity in Cylinders: Because hoop stress is exactly twice the longitudinal stress ($\sigma_h = 2 \sigma_L$), longitudinal weld seams endure double the tensile stress of circumferential (girth) seams. Consequently, longitudinal joints dictate vessel maximum allowable working pressure (MAWP).

2. Thick-Walled Cylinders (Lamé's Equations)

When $r_i / t < 10$, stresses vary non-linearly across the cylinder wall. For a cylinder with internal radius $r_i$, external radius $r_o$, internal pressure $p_i$, and external pressure $p_o$:

$\sigma_r(r) = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} - \frac{(p_i - p_o) r_i^2 r_o^2}{r^2 (r_o^2 - r_i^2)}$ $\sigma_\theta(r) = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} + \frac{(p_i - p_o) r_i^2 r_o^2}{r^2 (r_o^2 - r_i^2)}$

Special Case: Internal Pressure Only ($p_o = 0$): $\sigma_r(r) = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 - \frac{r_o^2}{r^2}\right) \quad [\text{Compressive: } -p_i \text{ at } r_i \to 0 \text{ at } r_o]$ $\sigma_\theta(r) = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 + \frac{r_o^2}{r^2}\right) \quad [\text{Tensile: max at } r_i]$

$\sigma_{\theta, \text{max}} = \sigma_\theta(r_i) = p_i \left(\frac{r_o^2 + r_i^2}{r_o^2 - r_i^2}\right)$ $\tau_{\text{max}} = \frac{\sigma_\theta(r_i) - \sigma_r(r_i)}{2} = p_i \left(\frac{r_o^2}{r_o^2 - r_i^2}\right)$

   Stress Distribution Across Thick Cylinder Wall (p_o = 0):

       Stress
         ^
         |  +-- \sigma_\theta (Hoop Stress, Maximum Tensile at r_i)
         |  |   \
         |  |    \---\
         |  |         \--- \sigma_\theta(r_o)
       0-+--+---------------------------------------------> Radius r
         |  r_i                                r_o
         |  |         /--- \sigma_r(r_o) = 0
         |  |    /---/
         |  +-- \sigma_r(r_i) = -p_i (Radial Stress, Compressive)

3. Interference / Shrink-Fit Pressures

Pressing or heat-shrinking an outer collar with nominal interface radius $R$ onto an inner shaft with diametral interference $\delta$ generates contact interface pressure $p$:

$p = \frac{\delta}{R \left[ \frac{1}{E_o}\left(\frac{r_o^2 + R^2}{r_o^2 - R^2} + \nu_o\right) + \frac{1}{E_i}\left(\frac{R^2 + r_i^2}{R^2 - r_i^2} - \nu_i\right) \right]}$

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2. Column Buckling & Stability: Euler vs. Johnson Criteria

Structural columns under axial compression can fail by material yielding (short columns) or geometric bifurcation buckling (slender columns).

+-----------------------------------------------------------------------------+
|                        COLUMN BUCKLING REGIMES                              |
|                                                                             |
|   Slenderness Ratio:              SR = \frac{K L}{r_g}                      |
|                                                                             |
|   Radius of Gyration:             r_g = \sqrt{\frac{I}{A}}                  |
|                                                                             |
|   Critical Transition Ratio:      C_c = \sqrt{\frac{2 \pi^2 E}{S_y}}        |
|                                                                             |
|   1. Slender Columns (SR \ge C_c)  ---> Euler Elastic Buckling              |
|      P_{cr} = \frac{\pi^2 E I}{(K L)^2} = \frac{\pi^2 E A}{(K L / r_g)^2}     |
|                                                                             |
|   2. Intermediate Columns (SR < C_c) ---> Johnson Inelastic Parabola        |
|      P_{cr} = A \left[ S_y - \frac{S_y^2}{4 \pi^2 E}\left(\frac{K L}{r_g}\right)^2 \right] |
+-----------------------------------------------------------------------------+

Column End-Fixity Effective Length Factors ($K$)

   Critical Buckling Stress \sigma_{cr}
     ^
     |  S_y ----------------- Material Yield Strength
     |     \ 
     |      \   Johnson Parabolic Equation (SR < C_c)
     |       \ 
     |        * Transition Point (C_c, S_y/2)
     |         \ 
     |          \---\   Euler Hyperbola (SR >= C_c)
     |               \-------
     +---------+-------------------+-----------------------------> Slenderness Ratio KL/r
              0                   C_c

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3. Helical Compression & Extension Spring Mechanics

+-----------------------------------------------------------------------------+
|                        HELICAL SPRING PARAMETERS                            |
|                                                                             |
|   Spring Index:            C = \frac{D}{d} \quad (\text{Recommended: } 4 \le C \le 12) |
|                                                                             |
|   Wahl Correction Factor:  K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}    |
|                                                                             |
|   Peak Torsional Shear:    \tau_{\text{max}} = K_w \frac{8 F D}{\pi d^3}    |
|                                                                             |
|   Spring Deflection:       y = \frac{8 F D^3 N_a}{d^4 G}                    |
|                                                                             |
|   Spring Rate:             k = \frac{F}{y} = \frac{d^4 G}{8 D^3 N_a}        |
|                                                                             |
|   Energy Stored:           U = \frac{1}{2} k y^2 = \frac{F^2}{2 k}          |
+-----------------------------------------------------------------------------+

Active Coils ($N_a$) vs. Total Coils ($N_t$)

Spring Surge & Fundamental Critical Frequency

To prevent internal resonant surging (valve bounce and catastrophic fatigue in high-speed machinery), the spring's fundamental natural frequency must be at least 13 to 20 times the operating cycle frequency:

$f_n = \frac{1}{2}\sqrt{\frac{k}{m_{\text{spring}}}} = \frac{d}{2 \pi D^2 N_a}\sqrt{\frac{G}{2\rho}}$

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4. Step-by-Step Worked Engineering Problem

Problem Statement

A heavy-duty hydraulic actuator cylinder with inner radius $r_i = 60\text{ mm}$ and outer radius $r_o = 90\text{ mm}$ is made of alloy steel ($S_y = 620\text{ MPa}$). It operates at an internal pressure $p_i = 80\text{ MPa}$ ($p_o = 0$).

Determine: (a) thickness ratio $r_i / t$ and verify classification, (b) hoop stress $\sigma_\theta$ and radial stress $\sigma_r$ at the inner and outer surfaces, and (c) the factor of safety against static yielding at the inner wall using Distortion Energy (von Mises).

Step 1: Vessel Classification

$t = r_o - r_i = 90 - 60 = 30\text{ mm}$ $\frac{r_i}{t} = \frac{60}{30} = 2.0 < 10 \implies \text{Thick-Walled Cylinder (Lamé's Equations Required)}$

Step 2: Stress Components at Inner Radius ($r = r_i = 60\text{ mm}$)

$\sigma_r(r_i) = -p_i = -80.0\text{ MPa}$ $\sigma_\theta(r_i) = p_i \left(\frac{r_o^2 + r_i^2}{r_o^2 - r_i^2}\right) = 80 \left(\frac{90^2 + 60^2}{90^2 - 60^2}\right) = 80 \left(\frac{8100 + 3600}{8100 - 3600}\right) = 80 \left(\frac{11700}{4500}\right) = 208.0\text{ MPa}$

Step 3: Stress Components at Outer Radius ($r = r_o = 90\text{ mm}$)

$\sigma_r(r_o) = 0\text{ MPa}$ $\sigma_\theta(r_o) = p_i \left(\frac{2 r_i^2}{r_o^2 - r_i^2}\right) = 80 \left(\frac{2(3600)}{4500}\right) = 128.0\text{ MPa}$

Step 4: von Mises Effective Stress at Inner Wall

Principal stresses (assuming closed ends: $\sigma_L = \frac{p_i r_i^2}{r_o^2 - r_i^2} = \frac{80(3600)}{4500} = 64.0\text{ MPa}$): $\sigma_1 = \sigma_\theta = 208.0\text{ MPa}, \quad \sigma_2 = \sigma_L = 64.0\text{ MPa}, \quad \sigma_3 = \sigma_r = -80.0\text{ MPa}$

$\sigma' = \frac{1}{\sqrt{2}}\sqrt{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}$ $\sigma' = \frac{1}{\sqrt{2}}\sqrt{(208.0 - 64.0)^2 + (64.0 - (-80.0))^2 + (-80.0 - 208.0)^2}$ $\sigma' = \frac{1}{\sqrt{2}}\sqrt{(144.0)^2 + (144.0)^2 + (-288.0)^2} = \frac{1}{\sqrt{2}}\sqrt{20736 + 20736 + 82944} = \frac{1}{\sqrt{2}}\sqrt{124416} = 249.4\text{ MPa}$

Step 5: Factor of Safety

$n = \frac{S_y}{\sigma'} = \frac{620}{249.4} = 2.49$

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5. Common Exam Traps & PE Pro-Tips

• Trap 1 — Using Thin-Wall Formula on Thick Vessels: Using $\sigma_h = p r / t$ on a cylinder with $r_i/t = 2$ gives $\sigma_h = 80(60)/30 = 160\text{ MPa}$—an underestimation of 23% compared to the true peak stress of $208\text{ MPa}$!
• Trap 2 — Column Boundary Factor Confusion: Double-check whether the problem requests theoretical $K$ or AISC recommended design $K$. For fixed-pinned, theoretical is $0.707$ while design is $0.80$.
• Trap 3 — Inactive Coils in Springs: Squared and ground ends have 2 inactive coils ($N_a = N_t - 2$). Failing to subtract inactive coils causes a direct error in spring stiffness $k$.