Plates, membranes, and shells

Take one beam idea into two surface directions

A beam model follows one long coordinate. A thin surface has two long directions and a short thickness. Its middle surface records where the material lies, while a thickness coordinate records the layers on either side. A plate starts with a flat middle surface; a shell starts with a curved one. A membrane model retains in-surface forces and neglects bending moments. These descriptions concern geometry and retained mechanics, rather than three unrelated kinds of material.

Start with two separate flat circular surfaces of radius a = 0.10 m, held around their rims. Apply the same static uniform normal load q = 10 Pa = 10 N/m². For each surface, let w be displacement along the load normal, positive downward in the drawing. The flat reference surface is w = 0, and r is distance from its centre. The disk area is πa², so each total applied normal force is qπa² = 0.1π N ≈ 0.314159 N.

The first surface is a homogeneous isotropic elastic plate, with thickness h = 0.50 mm, Young’s modulus E = 70 GPa and Poisson’s ratio ν = 0.30. It carries this load primarily by bending. The second is an ideal membrane with supplied uniform isotropic tensile force per unit length T = 100 N/m and negligible bending rigidity. Its tension is already present before q is applied, and changes in T are neglected. No membrane modulus or thickness is supplied; we will not infer the tension from its deflection.

For preparation, use Bending, torsion, and beam mechanics for bending and cut-body balance, and Stress, strain, and Hooke’s law for E. The force-per-length distinction and needed formulas are introduced here. Stress and strain tensors is an optional return path for the component notation; Derivatives, Taylor series, and local approximation helps with the derivative checks. On a first reading, follow the profiles, edge conditions and spherical cut before checking the fourth derivatives.

Keep force, stress and moment resultants distinct

Cut a small strip of a thin surface. Stress still means force per area and has unit Pa. Integrating an in-surface stress through the thickness instead gives force per unit cut length. With ζ measured through the thickness, N_xx = ∫[-h/2,h/2] σ_xx dζ. Its unit is Pa m = N/m. For a uniform tensile stress, N_xx = σ_xx h. A straight cut of length ℓ carrying constant N_xx transmits force N_xxℓ.

The stress can also have a first moment through the thickness: M_xx = ∫[-h/2,h/2] ζσ_xx dζ. We use this definition to fix its sign. It is a bending-moment resultant, with unit Pa m² = N, equivalently N m per metre of cut length. It is not a whole-beam torque measured in N m. Multiplying a constant M_xx by a cut length ℓ produces a bending moment of magnitude |M_xx|ℓ.

A constant tensile stress through the thickness gives nonzero N_xx but zero M_xx, because positive and negative ζ cancel in the first moment. An antisymmetric linear stress, tensile on one side and compressive on the other, can give zero N_xx but nonzero M_xx. A general thin structure can have both. The same integrals apply in the other surface direction and to in-surface shear components.

Out-of-surface shear also integrates to a force per cut length, commonly called Q. It can transmit the normal load in a bending plate. Neglecting its contribution to the leading deflection field does not set the equilibrating shear force to zero, just as in the beam model.

Give the flat plate two bending directions

For the plate, use the small-deflection Kirchhoff–Love approximation: material lines normal to the middle surface remain straight and normal, while their thickness change and transverse-shear contribution to deflection are neglected. Mid-surface stretching is also neglected in this bending calculation. The plate is thin compared with a, slopes are small, and the resulting w is small compared with h. This is a reduced elastic model, not a complete three-dimensional displacement solution.

Let x and y lie in the flat reference surface, and ζ point along the same normal as positive w. A rotation-compatible leading displacement is u_x = −ζw_x and u_y = −ζw_y, where a subscript x or y means a partial derivative. Thus ε_xx = −ζw_xx and ε_yy = −ζw_yy. Curvature changes strain with distance from the middle surface in each direction. A varying slope between the two directions can also produce a twisting response.

Poisson’s ratio ν is defined in a uniaxial test with free lateral surfaces by ν = −ε_transverse/ε_axial. It describes the coupling between normal strains in different directions. Under the leading isotropic plane-stress relation, σ_xx = E(ε_xx + νε_yy)/(1−ν²). Substitution and thickness integration give M_xx = −D(w_xx + νw_yy), with D = Eh³/[12(1−ν²)]. Similarly M_yy = −D(w_yy + νw_xx). The coupling by ν matters: the bending response in one surface direction depends on curvature in the other. Our stress-first-moment definition determines these signs.

D is the plate’s flexural rigidity. Its unit is Pa m³ = N m, whereas the whole-beam EI has unit N m². For the supplied h = 0.00050 m, E = 70 × 10⁹ Pa and ν = 0.30, D = 125/156 N m ≈ 0.801282 N m. At fixed E and ν, doubling h multiplies D by eight; this is a geometric change, not a change in E.

Check one clamped circular plate profile

The ideal plate clamp holds both normal displacement and edge slope: w(a) = 0 and w_r(a) = 0. For a smooth axisymmetric profile, w_r(0) = 0 at the centre. The constant-D, bending-only normal-load equation is DΔ²w = q. Here Δw = w_xx + w_yy, and applying Δ twice gives Δ²w = w_xxxx + 2w_xxyy + w_yyyy. This fourth-order operator records bending balance in two surface directions.

Supply the candidate w_P(r) = C(a²−r²)², with r² = x²+y². It already has zero displacement and slope at r = a. Expanding its polynomial shows Δ²w_P = 64C: the three fourth-derivative terms contribute 24C, 16C and 24C. Therefore C = q/(64D), and w_P(r) = q(a²−r²)²/(64D). This check fixes the coefficient and sign without solving a general plate boundary-value problem.

The centre value is w_P(0) = qa⁴/(64D) = 0.0000195 m = 0.0195 mm. It is largest there, decreasing to zero at the rim. The predicted centre deflection divided by thickness is 0.039, while h/a = 0.005. The largest slope magnitude is about 0.0003002. These values are consistency checks for the chosen small-deflection model, not universal accuracy thresholds.

At the centre, w_xx = w_yy = −4Ca² = −0.0078 m⁻¹. Consequently M_xx = M_yy = 0.008125 N. The corresponding σ_xx on the ζ = +h/2 side is +6M_xx/h² = +0.195 MPa, with −0.195 MPa on the opposite side and zero at the middle surface. This is a bending stress distribution: its thickness-integrated N_xx is zero, while its first moment is not.

Let supplied tension carry the membrane load

For the membrane, keep T = 100 N/m positive and constant, and neglect bending moments. Two nearly opposite tension forces on a small patch acquire a normal resultant when their directions differ. For small slopes, the x-direction contribution per area is T w_xx and the y-direction contribution is T w_yy. Normal balance is TΔw + q = 0, or −TΔw = q. At a downward displacement maximum the curvature sum is negative, so tension supplies an upward restoring force.

For an axisymmetric smooth profile, Δw = w_rr + w_r/r away from r = 0, with the regular limit used at the centre. The equation becomes (r w_r)_r = −qr/T. Integrating once and excluding a singular centre term gives w_r = −qr/(2T). Integrating again and enforcing the fixed rim w(a) = 0 gives w_M(r) = q(a²−r²)/(4T).

Its centre displacement is w_M(0) = qa²/(4T) = 0.00025 m = 0.250 mm. Its rim slope is w_r(a) = −qa/(2T) = −0.005. A membrane whose bending moment has been neglected does not enforce the plate’s zero-slope clamp condition. Here “fixed rim” means no normal displacement; the second-order equation uses that edge condition and centre regularity.

Small slope supports the normal-force approximation, but does not by itself establish that T changes negligibly. Constant supplied pretension is a separate assumption of this example. An initially slack sheet, or one whose stretching substantially creates or changes its tension, needs a coupled deformation and in-surface constitutive calculation. Substituting an unknown deformation-generated tension into this linear formula would leave the problem unsolved.

Compare shape, displacement and rim balance

The same a and q give different actual displacements because the retained restoring mechanisms and edge conditions differ. Our assigned membrane centre moves 500/39 ≈ 12.8205 times as far as the plate centre. This ratio describes these two model parameter sets; it is not a universal statement that every membrane is more compliant than every plate.

A separate shape comparison divides each profile by its own centre value. With s = r/a, the plate has w_P/w_P(0) = (1−s²)² and the membrane has w_M/w_M(0) = 1−s². Both start at 1 and finish at 0, but the plate meets its rim horizontally while the membrane has a finite slope. Normalisation removes the displacement magnitude; a dimensionless shape plot alone cannot show the 12.8205 centre-displacement ratio.

Both rims still balance the same total applied normal force. For the membrane, the small-slope normal force per unit rim length is T w_r(a) = −qa/2 = −0.5 N/m. Multiplication by 2πa gives −qπa² = −0.1π N, opposite the applied load. The membrane’s in-plane tension remains T = 100 N/m; −0.5 N/m is only its small normal component at the rim.

For the bending plate, the reconstructed radial normal-shear resultant is Q_r = −D d(Δw_P)/dr = −qr/2. It also reaches −0.5 N/m at the rim. Its clamped edge additionally transmits bending moments to hold the slope. Matching total reaction force does not determine the profile or show that the two surfaces have the same local stress field.

Change one parameter and recheck the model

Keep the plate radius, normal load, E and ν fixed, but double h from 0.50 to 1.00 mm. D rises from 125/156 to 250/39 N m, and the predicted centre deflection falls from 0.0195 to 0.0024375 mm. In the membrane model, keeping a and q fixed but doubling supplied T to 200 N/m halves the centre displacement to 0.125 mm. Bending rigidity and tensile force per length enter different equations and have different units.

The formulas also have different radius dependence at fixed other parameters: the plate centre value is proportional to a⁴, while the membrane value is proportional to a². This is a comparison within the stated boundary and loading models. Altering span can also alter the regime in which the assumptions hold.

For example, formally doubling the original plate radius gives sixteen times the centre displacement, 0.312 mm. With unchanged h = 0.50 mm, w_P(0)/h would become 0.624. The bending-only calculation no longer has its original small-displacement-relative-to-thickness consistency. The algebraic scaling is correct, but it does not establish an accurate prediction for that changed plate; mid-surface stretching can matter.

Use initial curvature to balance a spherical shell

Now take a separate complete spherical shell with middle-surface radius R = 0.20 m and thickness h_s = 0.001 m. Apply a uniform internal pressure difference p = 1000 Pa relative to the outside. At leading thin-wall order, use R for both the middle-surface radius and the pressure-surface radius; their small thickness difference is omitted in this model. The shell is thin, h_s/R = 0.005. Use the leading thin-wall membrane approximation, homogeneous isotropic linear elasticity and small deformation. There are no openings, concentrated supports or other loads in this ideal symmetric problem.

Cut the sphere into two hemispheres. Pressure acts outward normal to each point of the curved surface. Its leading resultant along the cut axis on one hemisphere is p times the projected circular area, using the model radius R: F_p = pπR². It is not p times the curved hemisphere area 2πR², because the differently directed pressure forces must be projected before addition.

By spherical symmetry the tensile in-surface force resultant is the same in two perpendicular tangent directions; call it N_s. Around the great-circle cut, its axial component sums to 2πR N_s and opposes the pressure resultant. Balance gives 2πR N_s = pπR², hence N_s = pR/2. Numerically N_s = 100 N/m; both opposing hemisphere resultants have magnitude 40π N ≈ 125.664 N.

The curved geometry lets tangent forces have a net normal component without first bending an initially flat surface into shape. For this sphere the membrane normal balance is p = N_s(1/R + 1/R) = 2N_s/R. The two terms represent its two equal surface curvatures. This local relation and the hemisphere balance agree, but they do not turn every curved structure into a pure membrane under every load.

Read the sphere’s tensile stress and small expansion

Divide the tensile force per unit length by thickness: σ_s = N_s/h_s = pR/(2h_s) = 100 000 Pa = 0.100 MPa. To leading thin-wall order the two tangent normal stresses are equal and tensile throughout the wall, while bending moments are neglected. A normal stress from pressure exists through a real wall; the thin-wall plane-stress approximation neglects it relative to the larger tangent stresses.

Supply E_s = 70 GPa and ν_s = 0.30 for this separate shell model. Equal biaxial tangent stress gives ε_s = (σ_s − ν_sσ_s)/E_s = (1−ν_s)σ_s/E_s = 10⁻⁶ in each tangent direction. Using σ_s/E_s alone would apply a uniaxial formula even though the other tangent stress is present.

For uniform spherical expansion, every small arc length scales with the radius. Thus ε_s = δR/R to first order, and δR = Rε_s = 2 × 10⁻⁷ m = 0.0002 mm = 0.2 μm. This numerical expansion is too small to depict as a visibly swollen sphere at the reference drawing scale; a value label or a separate graph preserves its magnitude.

The pressure load, shell dimensions and moduli are stipulated model inputs. The calculation finds the leading symmetric membrane response. It does not resolve a wall near a hole, a joint, a support, a local pressure patch or a different curvature, where stress variation and bending can be important.

Choose the retained mechanics before the formula

A thin elastic shell can carry both in-surface force and bending moment. “Membrane response” names the part dominated by in-surface force; it does not mean that a curved shell has no bending rigidity. Initial curvature couples normal displacement and in-surface deformation, and boundary conditions or nonuniform loads can require moments and changes of curvature. Our complete uniformly pressurised sphere avoids those disturbances by assumption.

Likewise, a flat plate can acquire membrane stresses when it stretches during larger transverse deflection, or can have imposed in-plane tension before loading. A model retaining both constant pretension and bending has, under the corresponding linear assumptions, DΔ²w − TΔw = q. It reduces to our separate equations when one restoring contribution is neglected. With geometry-dependent tension or larger deformation, that fixed-coefficient equation is also insufficient.

Before using a thin-structure result, identify the initial surface geometry, thickness, material law, in-surface force state, normal load, edge conditions and deformation regime. Our plate includes bending and reconstructed normal shear; our flat membrane includes supplied tension but no moments; our sphere uses a symmetric curved membrane balance. None of the supplied elastic constants or equilibrium calculations specifies a failure load or permissible operating pressure.

Worked examples

Try it yourself

1. A straight 30 mm cut carries a uniform tensile resultant N_xx = 100 N/m. What tensile force does it transmit? Does the unit N/m make N_xx a stress?

   Show the answer and reasoning

   3 N; N_xx is a force per cut length, not stress in Pa.

   Use ℓ = 0.030 m: N_xxℓ = 100(0.030) N. Recovering a uniform through-thickness stress would additionally require a thickness, σ_xx = N_xx/h.

2. Both circular profiles have w(a) = 0. Which model also has w_r(a) = 0, and what rim slope does the supplied-T membrane actually give?

   Show the answer and reasoning

   The clamped bending plate has zero rim slope; the membrane gives w_r(a) = −0.005.

   The plate solution contains (a²−r²)², whose first derivative vanishes at the rim. The membrane contains a²−r² and uses only fixed normal displacement there, with centre regularity; neglected moments do not enforce zero edge rotation.

3. Keep the original loading and radii. Double the plate thickness at fixed E and ν, and independently double the membrane’s supplied T. What new centre displacements follow?

   Show the answer and reasoning

   Plate: 0.0024375 mm. Membrane: 0.125 mm.

   D is proportional to h³, so twice h multiplies D by eight and divides w_P by eight. w_M is inversely proportional to T, so twice T halves its centre displacement. The two changes do not alter the same model parameter.

4. For the supplied sphere, a proposed cut balance uses pressure force p(2πR²), treating the curved hemisphere area as if every pressure arrow pointed along the cut axis. What is the correct axial pressure resultant, and what N_s follows?

   Show the answer and reasoning

   The axial pressure resultant is pπR² = 40π N, and N_s = pR/2 = 100 N/m.

   Pressure is normal to the curved surface; its axial components integrate over the projected disk area πR². The opposing rim force is 2πR N_s. Using the curved area without projection would incorrectly double the result.

5. The formal bending-only formula gives sixteen times the original centre displacement if the plate radius doubles at fixed q, h, E and ν. Compute that value and w_P(0)/h. Does the scaling alone establish an accurate changed-plate prediction?

   Show the answer and reasoning

   The formal value is 0.312 mm, with w_P(0)/h = 0.624; the scaling alone does not establish its accuracy.

   a⁴ gives the factor 16, but the new displacement is no longer small relative to the 0.50 mm thickness in the original bending-only sense. Mid-surface stretching and the retained model must be reconsidered; a correct algebraic ratio does not verify the changed regime.

What to carry forward

• A plate starts flat, a shell starts curved, and a membrane approximation retains in-surface forces while neglecting bending moments.
• Thickness integration turns stress in Pa into N/m; its first moment gives bending moment per cut length in N m/m. Resultants are not whole-section forces or torques.
• A bending plate uses D = Eh³/[12(1−ν²)] and a fourth-order load equation. A supplied-tension membrane uses T in N/m and a second-order equation; their edge slope conditions differ.
• Our plate and membrane balance the same total normal load but have different actual displacements, normalised shapes and local stress mechanisms.
• For a complete uniformly pressurised thin sphere, N_s = pR/2 and σ_s = pR/(2h_s); symmetry enables membrane balance, while other loading and boundary conditions can require bending.
• Identify geometry, tension, bending rigidity, boundary conditions and deformation regime before applying or scaling a thin-structure formula.