Load transferFrom the slab to the beams — in one line

Why 0.6 to the inner beam? (30 seconds)

• The slab is continuous over the inner beam — there it 'hangs' more → a larger share of the load.
• On the moments page — the reactions of 3 supports: 0.375 at the end, 0.625 + 0.625 in the middle. 0.4 and 0.6 — the same order, rounded.
• Note: the c here is a transfer coefficient — not the c of the equivalent span. And L is the real span, not L₀.
• An inner beam receives from both sides — from every span that rests on it.

The slab from the track: 3 beams

• Beam A
  0.875 · (0.4 · 4) = 1.4 t/m
• Beam B
  0.875 · (0.6 · 4 + 0.6 · 4) = 4.2 t/m — from both spans
• Beam C
  0.875 · (0.4 · 4) = 1.4 t/m
• Check
  1.4 + 4.2 + 1.4 = 7 = 0.875 · 8 all the load arrived ✓

That's it. Every beam knows how many t per m it carries — and we move on to its dimensions.

And with 4 beams — spans 3 + 4 + 3 m, f_ser = 1.0: the middle span is an inner one → 0.5 to each side.

• A, D
  1.0 · 0.4 · 3 = 1.2 t/m
• B, C
  1.0 · (0.6 · 3 + 0.5 · 4) = 3.8 t/m
• Check
  1.2 + 3.8 + 3.8 + 1.2 = 10 = 1.0 · 10 ✓

Twist: a cantilever

An inner beam of a ribbed slab: on one side a cantilever of 1.8 m, on the other side an inner span of 6.3 m, f_ser = 0.847.

• Segment ⁦5–6⁩q_5–6 = f_ser · Σ(c · L) = 0.847 · (1 · 1.8 + 0.5 · 6.3) = 4.193 t/m

The cantilever: coefficient 1.0 — all the load on it pours onto one beam, because it has no support on the other side.

4 pitfalls

• A load that doesn't fit the step. For the beam dimensions — f_ser; for its reinforcement — F_d,max.
• 0.4 and 0.6 on the wrong side. 0.6 always goes to the inner (continuous) side.
• Forgetting one side. Inner beam = the sum from both spans.
• L in cm. Here L is in metres — q comes out in t per m.