If L_y < L_x, the sides are swapped automatically to keep β = L_y / L_x ≥ 1. An isotropic yield moment (the same m_p in both orthogonal directions) is assumed.
While paused, moving a slider updates the results immediately.
The load w increases from 0 to w_u. The slab deflects, and red yield lines (plastic hinges) form along the centre and diagonals until a collapse mechanism develops. Cyan frame = four-edge simple support / yellow arrows = loading
For a rectangular slab simply supported on all four edges under a uniformly distributed load q, assuming yield lines that run from the centre region towards the corners and edges (an X plus I pattern), the upper-bound method (Johansen) gives the ultimate load.
With the aspect ratio β = L_y / L_x (≥1) and the isotropic yield moment m_p (per unit width), the ultimate distributed load q_u is expressed with the coefficient α(β):
$$q_u = \alpha(\beta)\,\frac{m_p}{L_x^2}$$Values from the closed form below: β=1 → α=24 (square), β=1.5 → α≈16.97, β=2 → α≈14.14, β=3 → α≈11.73, 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. As β grows, bending in the short direction dominates and α decreases toward the one-way slab value of 8.
The animation and values above use Johansen's closed-form upper-bound solution from external work = internal work (μ = L_x/L_y):
$$w_u = \frac{24\,m}{L_x^2\left(\sqrt{3+\mu^2}-\mu\right)^2}$$Check against known solutions: for a square, μ=1 gives a denominator (√4−1)²=1, so w_u = 24 m/L_x² (the textbook simply supported square slab). For a one-way slab, μ→0 gives √3²=3 and w_u = 8 m/L_x².