CalcHandbook

Circular Plate, Simply Supported Edge — Central Point Load

Plates · Circular · Edge Simply Supported All Round · Concentrated Load P at the Centre — Kirchhoff Thin-Plate Theory, Exact Axisymmetric Solution
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 simply supported all round
m
mm
kN
mm
MPa
—
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 simply supported all round (w = 0, Mr = 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; here every moment is sagging.
2

Deflection

Axisymmetric closed form — maximum at the centre, wmax → (3 + ν)·P·R²/16π(1 + ν)D for a true point load
w(r) = P16πD [ (3 + ν)·(R² − r²)/(1 + ν) + 2r²·ln(r/R) ]
+ P·c²16πD [ ln(r/R) − (1 − ν)·(1 − r²/R²)/2(1 + ν) ]
wmaxcentre, r = 0 — = (P/16πD)·[(3 + ν)R²/(1 + ν) − c²·ln(R/c) − (7 + 3ν)c²/4(1 + ν)] = —mm
2R / wmax = —
For a true point load wmax = (3 + ν)·P·R²/16π(1 + ν)D (Timoshenko §19) — (3 + ν)/(1 + ν) times the clamped plate, about 2.5 for ν = 0.3; the bearing circle reduces it only slightly. The edge rotates freely: w′(R) = −(P/8π(1 + ν)D)·(2R − c²/R). Thin-plate theory holds while wmax stays below about t/2.
Deflection along the diameter — section A–A (−R ≤ r ≤ R)
3

Bending moments and stresses

Mr = −D(w″ + ν·w′/r), Mt = −D(w′/r + ν·w″) — per unit width; the moment under the load governs, Mr vanishes at the edge
Mr(r) = P4π (1 + ν)·ln(R/r) + P·c²16π (1 − ν)·[ 1/r² − 1/R² ]
Mt(r) = P4π [ (1 + ν)·ln(R/r) + 1 − ν ] − P·c²16π (1 − ν)·[ 1/r² + 1/R² ]
Mloadunder the load, Mr = Mt — the maximum, see the formula in the report = —kN·m/m
Mt,edger = R, tangential — sagging, = (P(1 − ν)/8π)·(2 − c²/R²) → (1 − ν)·P/4π = —kN·m/m
Mr,edger = R, radial — the simply supported edge carries no moment = —kN·m/m
σload= 6·Mload/t² — bottom face in tension under the load, the maximum stress = —MPa
σt,edge= 6·Mt,edge/t² — tangential stress at the edge = —MPa
Both moments are sagging everywhere: Mr falls logarithmically from the load to zero at the edge, Mt only to (1 − ν)P/4π ≈ 0.056·P (ν = 0.3). The moment under the load exceeds the clamped-plate value by P/4π = 0.08·P at the same bearing radius, and grows like (1 + ν)·P/4π·ln(R/c) as the bearing circle shrinks (halving c adds about 0.07·P). The formulas above hold for c ≤ r ≤ R; inside the loaded circle both moments are quadratic in r.
Mr (radial) along the diameter — section A–A
Mt (tangential) along the diameter — section A–A
4

Shear and edge reaction

Qr = P/2πr outside the loaded circle (P·r/2πc² inside); the whole load goes to the supported edge — no corner forces, no twisting moments
Vedgereaction per unit length of the edge — = P/2πR = —kN/m
ΣVequilibrium check: Vedge·2πR = P = —kN vs —kN
Axisymmetry means Mrt = 0 everywhere, so the Kirchhoff edge reaction equals the shear Qr(R) exactly and there are no concentrated corner forces (unlike the simply supported rectangular plate). The edge carries no moment. The shear is largest next to the load, where it equals P/2πc.
5

At a given radius

Read w, Mr, Mt, Qr at any r — marked in amber on the diagrams
m
0R/2R
w= —mm Mr= —kN·m/m
Mt= —kN·m/m Qr= —kN/m