CalcHandbook

Rectangular Plate, Simply Supported — Central Point Load

Plates · Rectangular · All Four Edges Simply Supported · Concentrated Load P at the Centre — Kirchhoff Thin-Plate Theory, Navier Solution
1

Geometry, material and load

a = short side, b = long side (b ≥ a); thickness, E, ν, the load P and the side u of the small square it bears on
m
m
mm
kN
mm
MPa
—
Dflexural rigidity per unit width = E·t³12(1 − ν²) = —kN·m
b/aaspect ratio = —
Model: Kirchhoff thin plate (t ≪ a, deflection ≪ t), all four edges simply supported (w = 0, Mn = 0), load P at the centre spread uniformly over a small square u × u (a true point load gives a finite deflection but a logarithmically infinite moment under it — the bearing area, typically the contact patch or about the plate thickness, is what sets the peak moment). Solved exactly with the Navier double sine series (odd terms up to m, n = 299) — no coefficient tables, any aspect ratio and any ν. Sign convention: w positive downward, M positive when the bottom face is in tension.
2

Deflection

Navier series — maximum at the centre, wmax = α·P·a²/D (Timoshenko Table 23)
w(x, y) = 16Pπ⁶D·u² ΣmΣn sin(mπ/2)·sin(nπ/2)·sin(mπu/2a)·sin(nπu/2b)·sin(mπx/a)·sin(nπy/b)m·n·(m²/a² + n²/b²)² m, n = 1, 3, 5, … (patch u × u at the centre)
wmaxcentre (a/2, b/2) — = α·P·a²/D, α = — = —mm a / wmax = —
α is the Timoshenko Table 23 coefficient for a concentrated load (0.01160 for a square plate, 0.01651 for b/a = 2); the small bearing patch changes the deflection only in the 4th decimal. Thin-plate theory holds while wmax stays below about t/2.
Deflection along the short span — section A–A (y = b/2, 0 ≤ x ≤ a)
Deflection along the long span — section B–B (x = a/2, 0 ≤ y ≤ b)
3

Bending moments and stresses

Mx = −D(w,xx + ν·w,yy), My = −D(w,yy + ν·w,xx) — per unit width
Mx,maxunder the load, spanning the short side — = β·P, β = — = —kN·m/m
My,maxunder the load, spanning the long side — = β₁·P, β₁ = — = —kN·m/m
σx,max= 6·Mx/t² — centre, top/bottom face = —MPa
σy,max= 6·My/t² = —MPa
The peak moment grows like (1+ν)·P/4π·ln(a/u) as the bearing square shrinks — halving u adds about 0.07·P for ν = 0.3. Square plate, ν = 0.3, u = 0.1a: β = β₁ = 0.284; u = 0.05a: 0.356. Away from the load the moments are insensitive to u.
Mx along the short span — section A–A (y = b/2, 0 ≤ x ≤ a)
My along the long span — section B–B (x = a/2, 0 ≤ y ≤ b)
4

Edge reactions and corner forces

Kirchhoff edge reaction V = Q + ∂Mxy/∂s; corners must be held down
Vx,maxmiddle of the long edges (x = 0, a) — = δ·P/a, δ = — = —kN/m
Vy,maxmiddle of the short edges (y = 0, b) — = δ₁·P/a, δ₁ = — = —kN/m
Rcorner= 2·Mxy at each corner, downward (anchorage) — = n·P, n = — = —kN
ΣVequilibrium check: edge reactions − 4·Rcorner = P = —kN vs —kN
Twisting moments Mxy along the edges become extra distributed reactions and concentrated corner forces pulling the plate up; without hold-downs the corners lift and the moments rise. For a square plate at ν = 0.3 the corner force is n = 0.122·P (Timoshenko §36). The equilibrium sum closes to within ~0.5 %.
5

At a given point

Read w, Mx, My, Mxy anywhere — marked in amber on the diagrams
m
0a/2a
m
0b/2b
w= —mm Mx= —kN·m/m
My= —kN·m/m Mxy= —kN·m/m