Circular Plate, Clamped Edge — Central Point Load

1

Geometry, material and load

Radius R, thickness t, E, ν, the load P and the radius c of the small circle it bears on — edge built in all round

Radius R

m

Thickness t

mm

Central load P

kN

Loaded circle radius c

mm

Modulus E

MPa

Poisson's ratio ν

—

Dflexural rigidity per unit width = E·t³12(1 − ν²) = —kN·m

c / Rbearing radius over plate radius — sets the peak moment = —

Model: Kirchhoff thin plate (t ≪ R, deflection ≪ t), edge built in all round (w = 0, dw/dr = 0 at r = R), load P at the centre spread uniformly over a small circle of radius c (a true point load gives a finite deflection but a logarithmically infinite moment under it — the bearing radius, typically the contact area or about half the plate thickness, sets the peak moment). The axisymmetric problem has the exact closed-form solution of Timoshenko & Woinowsky-Krieger §19: a polynomial inside the loaded circle joined to the r²·ln r solution outside. Sign convention: w positive downward, M positive when the bottom face is in tension; the built-in edge carries a negative (hogging) radial moment.

2

Deflection

Axisymmetric closed form — maximum at the centre, w_max → P·R²/16πD for a true point load

w(r) = P16πD [ R² − r² + 2r²·ln(r/R) ]

+ P·c²16πD [ ln(r/R) + ½ − r²/2R² ]

w_maxcentre, r = 0 — = (P/16πD)·[R² − ¾c² − c²·ln(R/c)] = —mm

2R / w_max = —

For a true point load w_max = P·R²/16πD (Timoshenko §19); the bearing circle reduces it only slightly (by c²·ln(R/c) against R²). The simply supported plate would deflect (3 + ν)/(1 + ν) times more. Thin-plate theory holds while w_max stays below about t/2.

Deflection along the diameter — section A–A (−R ≤ r ≤ R)

3

Bending moments and stresses

M_r = −D(w″ + ν·w′/r), M_t = −D(w′/r + ν·w″) — per unit width; the moment under the load and the built-in edge govern

M_r(r) = P4π [ (1 + ν)·ln(R/r) − 1 ] + P·c²16π [ (1 − ν)/r² + (1 + ν)/R² ]

M_t(r) = P4π [ (1 + ν)·ln(R/r) − ν ] − P·c²16π [ (1 − ν)/r² − (1 + ν)/R² ]

M_r,edgebuilt-in edge, r = R — hogging, = −(P/8π)·(2 − c²/R²) → −P/4π = —kN·m/m

M_t,edgebuilt-in edge, tangential — = ν·M_r,edge = —kN·m/m

M_centreunder the load, M_r = M_t — sagging, = (P(1 + ν)/4π)·[ln(R/c) + c²/4R²] = —kN·m/m

σ_edge= 6·|M_r,edge|/t² — top face in tension at the built-in edge = —MPa

σ_centre= 6·M_centre/t² — bottom face in tension under the load = —MPa

The edge moment tends to −P/4π = −0.0796·P whatever the bearing radius; the moment under the load grows like (1 + ν)·P/4π·ln(R/c) as the bearing circle shrinks (halving c adds about 0.07·P for ν = 0.3). M_r changes sign at r = R·e^−1/(1+ν) ≈ 0.46R for ν = 0.3; M_t at r = R·e^−ν/(1+ν) ≈ 0.79R. The formulas above hold for c ≤ r ≤ R; inside the loaded circle both moments are quadratic in r.

M_r (radial) along the diameter — section A–A

M_t (tangential) along the diameter — section A–A

4

Shear and edge reaction

Q_r = P/2πr outside the loaded circle (P·r/2πc² inside); the whole load goes to the built-in edge — no corner forces, no twisting moments

V_edgereaction per unit length of the edge — = P/2πR = —kN/m

ΣVequilibrium check: V_edge·2πR = P = —kN vs —kN

Axisymmetry means M_rt = 0 everywhere, so the Kirchhoff edge reaction equals the shear Q_r(R) exactly and there are no concentrated corner forces. The edge also carries the hogging moment M_r,edge per unit length. Unlike the uniform-load case the shear is largest next to the load, where it equals P/2πc.

5

At a given radius

Read w, M_r, M_t, Q_r at any r — marked in amber on the diagrams

r (from the centre)

m

0R/2R

w= —mm M_r= —kN·m/m

M_t= —kN·m/m Q_r= —kN/m