2.4 Pressure Vessels, Column Buckling & Spring Design

Key Takeaways

  • Thin-walled pressure vessels (ri/t≥10r_i / t \ge 10) exhibit uniform wall stresses: cylindrical vessels experience hoop stress σh=prt\sigma_h = \frac{p r}{t} (twice the longitudinal stress σL=pr2t\sigma_L = \frac{p r}{2t}), making longitudinal weld seams twice as critical as circumferential seams; spherical vessels experience uniform biaxial stress σ=pr2t\sigma = \frac{p r}{2t}.

  • Thick-walled cylinders (ri/t<10r_i / t < 10) require Lamé's equations to account for steep radial gradients; maximum tensile hoop stress and absolute maximum shear stress always occur at the innermost surface (r=rir = r_i).

  • Compound shrink-fit / press-fit assemblies create interface contact pressure pp that pre-compresses the inner cylinder, significantly increasing allowable internal operating pressure.

  • Column stability is governed by Euler's equation Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(KL)^2} for slender columns (KL/r≥CcKL/r \ge C_c), whereas intermediate columns (KL/r<Cc=2π2ESyKL/r < C_c = \sqrt{\frac{2\pi^2 E}{S_y}}) buckle inelastically and must be designed using Johnson's parabolic formulation.

  • Helical spring stress is amplified at the inner coil radius by the Wahl factor Kw=4C−14C−4+0.615CK_w = \frac{4C-1}{4C-4} + \frac{0.615}{C}; spring rate k=d4G8D3Nak = \frac{d^4 G}{8 D^3 N_a} and fundamental surge frequency govern dynamic performance.

Last updated: August 2026

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.


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 (σh=2σL\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 ri/t<10r_i / t < 10, stresses vary non-linearly across the cylinder wall. For a cylinder with internal radius rir_i, external radius ror_o, internal pressure pip_i, and external pressure pop_o:

σr(r)=piri2−poro2ro2−ri2−(pi−po)ri2ro2r2(ro2−ri2)\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)} σθ(r)=piri2−poro2ro2−ri2+(pi−po)ri2ro2r2(ro2−ri2)\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 (po=0p_o = 0):

σr(r)=piri2ro2−ri2(1−ro2r2)[Compressive: −pi at ri→0 at ro]\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] σθ(r)=piri2ro2−ri2(1+ro2r2)[Tensile: max at ri]\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] σθ,max=σθ(ri)=pi(ro2+ri2ro2−ri2)\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) τmax=σθ(ri)−σr(ri)2=pi(ro2ro2−ri2)\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 RR onto an inner shaft with diametral interference δ\delta generates contact interface pressure pp:

p=δR[1Eo(ro2+R2ro2−R2+νo)+1Ei(R2+ri2R2−ri2−νi)]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]}

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 (KK)

End ConditionsTheoretical KKRecommended Design KK (AISC)Effective Length (LeL_e)
Pinned - Pinned1.01.01.01.0LL
Fixed - Free (Flagpole)2.02.02.12.12.1L2.1 L
Fixed - Pinned0.7070.7070.800.800.80L0.80 L
Fixed - Fixed0.500.500.650.650.65L0.65 L
   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

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 (NaN_a) vs. Total Coils (NtN_t)

End TypeTotal Coils (NtN_t)Solid Length (LsL_s)Free Length (LfL_f)
PlainNaN_a(Nt+1)d(N_t + 1) dpNa+dp N_a + d
Plain & GroundNa+0.5N_a + 0.5NtdN_t dpNtp N_t
Squared (Closed)Na+2N_a + 2(Nt+1)d(N_t + 1) dpNa+3dp N_a + 3d
Squared & GroundNa+2N_a + 2NtdN_t dpNa+2dp N_a + 2d

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:

fn=12kmspring=d2πD2NaG2ρf_n = \frac{1}{2}\sqrt{\frac{k}{m_{\text{spring}}}} = \frac{d}{2 \pi D^2 N_a}\sqrt{\frac{G}{2\rho}}

4. Step-by-Step Worked Engineering Problem

Problem Statement

A heavy-duty hydraulic actuator cylinder with inner radius ri=60 mmr_i = 60\text{ mm} and outer radius ro=90 mmr_o = 90\text{ mm} is made of alloy steel (Sy=620 MPaS_y = 620\text{ MPa}). It operates at an internal pressure pi=80 MPap_i = 80\text{ MPa} (po=0p_o = 0).

Determine: (a) thickness ratio ri/tr_i / t and verify classification, (b) hoop stress σθ\sigma_\theta and radial stress σr\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=ro−ri=90−60=30 mmt = r_o - r_i = 90 - 60 = 30\text{ mm} rit=6030=2.0<10  ⟹  Thick-Walled Cylinder (Lameˊ’s Equations Required)\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=ri=60 mmr = r_i = 60\text{ mm})

σr(ri)=−pi=−80.0 MPa\sigma_r(r_i) = -p_i = -80.0\text{ MPa} σθ(ri)=pi(ro2+ri2ro2−ri2)=80(902+602902−602)=80(8100+36008100−3600)=80(117004500)=208.0 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=ro=90 mmr = r_o = 90\text{ mm})

σr(ro)=0 MPa\sigma_r(r_o) = 0\text{ MPa} σθ(ro)=pi(2ri2ro2−ri2)=80(2(3600)4500)=128.0 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: σL=piri2ro2−ri2=80(3600)4500=64.0 MPa\sigma_L = \frac{p_i r_i^2}{r_o^2 - r_i^2} = \frac{80(3600)}{4500} = 64.0\text{ MPa}):

σ1=σθ=208.0 MPa,σ2=σL=64.0 MPa,σ3=σr=−80.0 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} σ′=12(σ1−σ2)2+(σ2−σ3)2+(σ3−σ1)2\sigma' = \frac{1}{\sqrt{2}}\sqrt{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2} σ′=12(208.0−64.0)2+(64.0−(−80.0))2+(−80.0−208.0)2\sigma' = \frac{1}{\sqrt{2}}\sqrt{(208.0 - 64.0)^2 + (64.0 - (-80.0))^2 + (-80.0 - 208.0)^2} σ′=12(144.0)2+(144.0)2+(−288.0)2=1220736+20736+82944=12124416=249.4 MPa\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=Syσ′=620249.4=2.49n = \frac{S_y}{\sigma'} = \frac{620}{249.4} = 2.49

5. Common Exam Traps & PE Pro-Tips

  • Trap 1 — Using Thin-Wall Formula on Thick Vessels: Using σh=pr/t\sigma_h = p r / t on a cylinder with ri/t=2r_i/t = 2 gives σh=80(60)/30=160 MPa\sigma_h = 80(60)/30 = 160\text{ MPa}—an underestimation of 23% compared to the true peak stress of 208 MPa208\text{ MPa}!
  • Trap 2 — Column Boundary Factor Confusion: Double-check whether the problem requests theoretical KK or AISC recommended design KK. For fixed-pinned, theoretical is 0.7070.707 while design is 0.800.80.
  • Trap 3 — Inactive Coils in Springs: Squared and ground ends have 2 inactive coils (Na=Nt−2N_a = N_t - 2). Failing to subtract inactive coils causes a direct error in spring stiffness kk.
Test Your Knowledge

A cylindrical steel pressure vessel with inner diameter di=1200 mmd_i = 1200\text{ mm} and wall thickness t=12 mmt = 12\text{ mm} operates at an internal gauge pressure of p=1.8 MPap = 1.8\text{ MPa}. What are the nominal hoop stress σh\sigma_h and longitudinal stress σL\sigma_L in the vessel wall?

A

σh=45.0 MPa,σL=22.5 MPa\sigma_h = 45.0\text{ MPa}, \sigma_L = 22.5\text{ MPa}

B

σh=180.0 MPa,σL=90.0 MPa\sigma_h = 180.0\text{ MPa}, \sigma_L = 90.0\text{ MPa}

C

σh=45.0 MPa,σL=90.0 MPa\sigma_h = 45.0\text{ MPa}, \sigma_L = 90.0\text{ MPa}

D

σh=90.0 MPa,σL=45.0 MPa\sigma_h = 90.0\text{ MPa}, \sigma_L = 45.0\text{ MPa}

Test Your Knowledge

An ASTM A36 structural steel column (E=200 GPa,Sy=250 MPaE = 200\text{ GPa}, S_y = 250\text{ MPa}) with cross-sectional area A=2400 mm2A = 2400\text{ mm}^2 and radius of gyration rg=40 mmr_g = 40\text{ mm} has an unsupported length L=3.2 mL = 3.2\text{ m} with fixed-pinned ends (K=0.70K = 0.70). Which buckling model governs, and what is the critical compressive buckling load PcrP_{cr}?

A

Euler slender column formula governs with Pcr=1480 kNP_{cr} = 1480\text{ kN}

B

Johnson parabolic intermediate column formula governs with Pcr=540 kNP_{cr} = 540\text{ kN}

C

Euler slender column formula governs with Pcr=320 kNP_{cr} = 320\text{ kN}

D

Direct compressive yielding governs with Pcr=600 kNP_{cr} = 600\text{ kN}

Test Your Knowledge

A helical compression spring is wound from wire of diameter d=4 mmd = 4\text{ mm} (G=79.3 GPaG = 79.3\text{ GPa}) with mean coil diameter D=32 mmD = 32\text{ mm} and 12 active coils. If an axial load F=400 NF = 400\text{ N} is applied, what is the Wahl stress concentration factor KwK_w and the peak shear stress τmax\tau_{\text{max}}?

A

Kw=1.052,τmax=428.5 MPaK_w = 1.052, \tau_{\text{max}} = 428.5\text{ MPa}

B

Kw=1.320,τmax=537.4 MPaK_w = 1.320, \tau_{\text{max}} = 537.4\text{ MPa}

C

Kw=1.184,τmax=603.6 MPaK_w = 1.184, \tau_{\text{max}} = 603.6\text{ MPa}

D

Kw=1.250,τmax=636.6 MPaK_w = 1.250, \tau_{\text{max}} = 636.6\text{ MPa}

Test Your Knowledge

A thick-walled steel cylinder with inner radius ri=50 mmr_i = 50\text{ mm} and outer radius ro=100 mmr_o = 100\text{ mm} is subjected to internal pressure pi=120 MPap_i = 120\text{ MPa} (po=0p_o = 0). Where does the maximum tensile hoop stress occur, and what is its magnitude?

A

At the inner radius (r=50 mmr = 50\text{ mm}) with σθ=200.0 MPa\sigma_\theta = 200.0\text{ MPa}

B

At the outer radius (r=100 mmr = 100\text{ mm}) with σθ=120.0 MPa\sigma_\theta = 120.0\text{ MPa}

C

At the mid-wall radius (r=75 mmr = 75\text{ mm}) with σθ=160.0 MPa\sigma_\theta = 160.0\text{ MPa}

D

At the outer radius (r=100 mmr = 100\text{ mm}) with σθ=80.0 MPa\sigma_\theta = 80.0\text{ MPa}

Sections you finish are checked off in the contents.