If L_y < L_x, the sides are swapped automatically to keep β = L_y / L_x ≥ 1. Isotropic yield moment (same m_p in both orthogonal directions) is assumed.
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Load w rises 0 → w_u. The slab deflects and red yield lines (plastic hinges) form along the central and diagonal lines until the collapse mechanism develops. Cyan rim = four-edge simple supports / yellow arrows = applied load
For a four-edge simply supported rectangular slab under uniformly distributed load q, assuming the typical pattern with yield lines diverging from the centre to the four edge midpoints (X plus I shape), the upper bound method (Johansen) gives the ultimate load.
With aspect ratio β = L_y / L_x (≥1) and isotropic yield moment m_p (per unit width), the ultimate distributed load q_u is expressed with the coefficient α(β):
$$q_u = \alpha(\beta)\,\frac{m_p}{L_x^2}$$Textbook values (linearly interpolated): β=1 → α=24 (square), β=1.5 → α=18, β=2 → α=16, β=3 → α=12, β=∞ → α=8 (one-way slab).
Design distributed load q_d and equivalent point load P_u (A=L_x·L_y):
$$q_d = \frac{q_u}{\gamma}, \qquad P_u = q_u \cdot A$$For a square slab (β=1), q_u = 24 m_p / L_x². The larger β becomes, the more bending in the short direction dominates, and α decreases toward the one-way slab value of 8.
The animation and live values use Johansen's closed-form upper-bound solution derived from equating external work to internal work (μ = L_x/L_y):
$$w_u = \frac{24\,m}{L_x^2\left(\sqrt{3+\mu^2}-\mu\right)^2}$$Known-solution check: a square (μ=1) gives denominator (√4−1)²=1, so w_u = 24 m/L_x² (the textbook simply supported square slab). A one-way slab (μ→0) gives √3²=3, so w_u = 8 m/L_x².