Module 3: Solid Mechanics Applications - Topic 3: Pressure Vessels

• longitudinal/axial:

  • acts parallel to the vessel’s long axis (length), it pulls the ends of a cylinder apart (σ_L​=pr/2t​)

• circumferential/tangential/hoop:

  • acts along the perimeter/circumference, tangent to the curved surface

  • it attempts to split the vessel open lengthwise (

    σ_h​=pr/t)

  • this is the critical, maximum stress

• radial:

  • acts perpendicular to the vessel wall, pointing straight from the center outward (σ_r)

  • assumed to be approximately 0 in thin-walled theory

What do the following terms mean: longitudinal, axial, circumferential, tangential, hoop, radial (in the context of pressure vessels)?

1. radial stress is negligible: the radial stress is assumed to be zero throughout the wall thickness

2. uniform stress distribution: the circumferential (hoop) and longitudinal (axial) stresses are uniform (constant) across the thickness of the vessel wall

3. no stress concentrations: the vessel wall is smooth and free from any geoemtric discontinuities, holes, or sudden changes in thickness that could causes localised stress concentrations

What are the 3 assumptions of the theory of thin-walled pressure vessels? (slide 98)

• a pressure vessel is considered thin-walled if the ratio of its wall thickness (t) to its inner radius (r_i) is less than or equal to 1/10 (or a thickness-to-diameter ratio <= 0.1

What is the engineering rule-of-thumb regarding wall thickness used to determine whether the vessel is “thin”? (slide 100)

• under a uniform hydrostatic pressure (p), the stress state is isotropic (the same in all directions) and purely compressive

• it is represented mathematically by the stress tensor:
  σ_ij​=−pδ_ij​

• which expands to the following matrix components:

  normal stresses:

  • σx​=σy​=σz​=−p (equal compression on all faces)

  shear stresses:

  • τxy​=τyz​=τzx​=0 (zero shear stress)

What does the stress state for a uniform pressure look like? (slide 101)

1. radial stress (σ_r)

   • inner wall: σ_r​=−p (the wall experiences direct compressive guage pressure from the inside)

   • outer wall: σ_r​=0 (exposed to atmospheric pressure, so gauge pressure is zero)

   • thin-walled assumption: because the thickness is so small (t«r), the variation across the wall is ignored, and radial stress is assumed to be negligible throughout:
     σr​≈0

tangential/hoop stress (σt) derivation:

• to find the tensile stress acting within the sphere’s walls, we cut the sphere exactly in half to create a free-body diagram of one hemisphere and balance the static forces

  1. identify the internal pressure force (F_p)

     • the internal pressure acts across the projected circular cross-sectional area of the cut sphere:
       F_P = pressure x projected area = p⋅(πr²)

  2. identify the resisting material force (F_m)

     • the tangential stress (σt) acts uniformly across the narrow ring-shaped cross-section of the sphere’s cut wall

     • the area of this thin ring is its circumference multiplied by thickness:

       F_m = stress x wall area σt​⋅(2πr⋅t)

  3. apply static equilibrium (ΣF=0)

     • for the hemisphere to remain intact, the bursting force from the pressure must perfectly balance the tensile force in the wall material:

       F_p = F_m

     • p⋅(πr²)=σt​⋅(2πrt)

  4. solve for tangential stress (σt)

     • cancel out pi and one r from both sides

       pr=2σ_t​t

       σt​=pr/2t​

Derive the expression for the tangential stress (and radial stress) in a thin-walled sphere, in terms of the (gauge) pressure, the average radius and the wall thickness (slides 102-107)

1. absolute maximum shear stress (out of plane)

   • τmax(abs)​=|σ_t​−σ_r|​​/2= (pr/2t - 0)/2​ = pr/4t​

     • significance: this shear stress acts on a plane angled at 45 degrees through the wall thickness

2. in-plane maximum shear stress

   • if you only look at the 2D surface of the sphere, the two principal stresses are equal (σ₁​=σ₂​=σ_t​)

     τ_{max(in-plane)}​=2σ_t​−σ_t​​=0

     • significance: there is zero shear stress acting purely along the surface plane of the sphere

What is the maximum shear stress in the thin-walled sphere?

1. radial stress (σr)

   • derivation/assumption: like the sphere, σ_r varies from -p at the inner wall to 0 at the outer wall

   • because t«r, this variation is negligible compared to the massive tensile stresses

   • result:
     σ_r​≈0

2. tangential/hoop stress (σ_t​ or σ_h)​

   • to isolate the hoop stress, making a cutting plane lengthwise along a section of the cylinder of length L

     1. bursting force from fluid pressure (F_P)

        • the internal pressure acts on the rectangular projected area of the cut fluid core:
          Fp​=Pressure×Projected Area=p⋅(2r⋅L)

     2. resisting force from wall material (F_m)

        • the tensile hoop stress acts uniformly across the two cut narrow rectangular wall strips:

          F_m​=Stress×Material Area=σ_t​⋅(2⋅t⋅L)

     3. static equilibrium (ΣFy​=0)

        Fp​=Fm​⟹p⋅(2rL)=σt​⋅(2tL)

        • result: cancel 2 and L from both sides:
          σ_t​=pr​/t

3. longitudinal/axial stress (σ_L​ or σ_a​)

   • to isolate longitudinal stress, make a transverse cut perpendicular to the cylinder’s long axis

     1. axial force from fluid pressure (F_p)

        • the pressure pushes against the circular end-cap of the cylinder:
          Fp​=Pressure×Projected Circular Area=p⋅(πr2)

     2. resisting force from wall (material F_m)

        • the longitudinal stress acts along the cross-section of the cut circular thin wall ring:
          F_m​=Stress×Ring Area=σ_L​⋅(2πr⋅t)

     3. static equilibrium (ΣFx​=0)

        • F_p​=F_m​⟹p⋅(πr²)=σ_L​⋅(2πrt)

        • result: cancel pi and one r from both sides:
          σ_L​=pr/2t​

        • key relationship: in a thin-walled cylinder, hoop stress is exactly twice the magnitude of longitudinal stress

          (σ_t​=2σ_L​)

Derive expressions for the tangential and longitudinal stresses (and radial stress)

in a thin-walled cylinder, in terms of the (gauge) pressure, the average radius and

the wall thickness (slides 108-111)

1. Absolute Maximum Shear Stress (Out-of-Plane)

This is the true maximum shear stress experienced by the cylinder wall material, calculated by looking at the plane cutting through the wall thickness between the largest stress (σt​) and smallest stress (σr​)

$$\tau_{\max(\text{abs})} = \frac{\sigma_{\max} - \sigma_{\min}}{2} = \frac{\sigma_t - \sigma_r}{2} = \frac{\frac{pr}{t} - 0}{2} = \frac{pr}{2t}$$

• Orientation: Acts on a plane angled at 45 degrees through the thickness of the wall.

2. In-Plane Maximum Shear Stress

This is the maximum shear stress acting purely along the flat outer surface plane of the cylinder, calculated using the two surface principal stresses (σ_t​ and σ_L).

$$\tau_{\max(\text{in-plane})} = \frac{\sigma_t - \sigma_L}{2} = \frac{\frac{pr}{t} - \frac{pr}{2t}}{2} = \frac{pr}{4t}$$

• Orientation: Acts on a plane angled at 45 degrees relative to the longitudinal axis on the vessel's surface.

What is the maximum shear stress in the thin-walled cylinder? (slides 112-113)

Step 1: Define Initial and Final Circumference

Let the initial radius of the vessel be r

When pressurized, the radius undergoes a small outward radial displacement u, making the new radius (r + u)

• Initial Circumference (C₀):

  $$C_0 = 2\pi r$$

• Final Circumference (C_f):

  $$C_f = 2\pi(r + u) = 2\pi r + 2\pi u$$

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Step 2: Calculate the Change in Circumference (ΔC)

The total elongation around the circumference is the final length minus the initial length:

$$\Delta C = C_f - C_0 = (2\pi r + 2\pi u) - 2\pi r$$

$$\Delta C = 2\pi u$$

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Step 3: Apply the Definition of Normal Strain

Tangential strain (εt) is defined as the change in circumference divided by the original circumference:

$$\varepsilon_t = \frac{\Delta C}{C_0}$$

Substitute the expressions from Steps 1 and 2:

$$\varepsilon_t = \frac{2\pi u}{2\pi r}$$

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Final Geometric Relationship:

Canceling out 2pi yields the clean, fundamental equation:

$$\varepsilon_t = \frac{u}{r}$$

Derive a relationship between the tangential/circumferential strain and the radial

displacement in a thin-walled pressure vessel (slide 114)

1. Finding Strains using Hooke's Law (Stress-Strain Matrices)

By applying Generalized Hooke’s Law for plane stress (σ_r​≈0), we calculate the normal strains from our known thin-walled stresses:

• For a Cylinder:

  • Hoop Strain (ε_t​):

    ε_t​=1/E​(σ_t​−νσ_L​)=1/E​(pr/t​−νpr/2t​)=pr/2Et​(2−ν)

  • Longitudinal Strain (εL​):

    ε_L​=1/E​(σ_L​−νσt​)=1/E​(pr/2t​−νpr​/t)=pr/2Et​(1−2ν)

• For a Sphere: Because σ1​=σ2​=pr/2t​, the tangential strain in all directions is:

  ε_t​=1/E​(σt​−νσt​)=pr/2Et​(1−ν)

2. Finding Displacements (Strain-Displacement Relations)

Once you have calculated the numeric values for the strains, use geometry to determine how much the physical dimensions change:

• Change in Radius (Δr or Radial Displacement u): Utilize the relationship derived on slide 114 (ε_t​=ru​):

  Δr=u=r⋅ε_t​

• Change in Circumference (ΔC): The total change in perimeter length scales directly with the initial circumference:

  ΔC=C₀​⋅ε_t​=2πr⋅ε_t​

• Change in Cylinder Length (ΔL): The total elongation along the axis depends on the longitudinal strain:

  ΔL=L₀​⋅ε_L​

Variables Reminder: E=Young’s Modulus, ν=Poisson’s Ratio, r=Average Radius, t=Wall Thickness.

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2. Finding Displacements (Strain-Displacement Relations)

Once you have calculated the numeric values for the strains, use geometry to determine how much the physical dimensions change:

• Change in Radius (Δr or Radial Displacement u): Utilize the relationship derived on slide 114 (εt​=ru​):

  Δr=u=r⋅εt​

• Change in Circumference (ΔC): The total change in perimeter length scales directly with the initial circumference:

  ΔC=C0​⋅εt​=2πr⋅εt​

• Change in Cylinder Length (ΔL): The total elongation along the axis depends on the longitudinal strain:

  ΔL=L0​⋅εL​

Variables Reminder: E=Young’s Modulus, ν=Poisson’s Ratio, r=Average Radius, t=Wall Thickness.

Given the stresses, be able to use the stress-strain matrices to evaluate the strains

for thin-walled vessels; be able to use the strain-displacement relations to

evaluate the displacements in these vessels (change in radius & circumference)

(slides 115-118)

1. the signs of the strains

   • tangential/hoop strain (ε_t): positive (+)

   • longitudinal/axian strain (ε_L): positive (+)

   • radial strain (ε_r): negative (-)

2. physical sense

   • ε_t​ & ε_L​ (+): internal pressure pushes outward, physically expanding both the vessel’s circumference and its length (tensile stretching)

   • ε_r​ (−): the wall material is directly compressed (squeezed) by the internal pressure

     • simultaneously, stretching lengthwise and circumferentially forces the wall to thin out to conserve volume (Poisson’s ratio effect)

For an internally pressurised pressure vessel (thin or thick walled), which of the

strains will be positive and which negative? Does this make physical sense?

1. the critical stress

   • tangential (hoop) stress: this remains the largest tensile stress acting on the material, making it the most likely to cause failure or damage

2. location of maximum damage

   • at the inner wall (bore): damage occurs at the innermost surface (r = r_i)

3. physical reason

   • unlike thin-walled vessels where stress is uniform, thick-walled cylinders experience a steep gradient across their walls:

     • hoop stress (σ_t): is at its absolute maximum at the inner surface drops significantly toward the outer surface

     • radial stress (σ_r): is also at its peak compression (equal to -p) at the inner surface and drops to 0 at the outside​

   • because the highest tensile hoop stress and the highest compressive radial stress peak simultaneously at the inner wall, the material experiences its maximum shear stress (τ_max) right at the inner surface, making it the most vulnerable zone​

For an internally pressurised thick-walled cylinder, which of the stresses is likely

to cause the most damage? Where in the vessel will this occur?