Ribbed slabSame path — four changes

The section — what each dimension is (30 seconds)

• b_f = B + b_w — the block width plus one rib width: the distance from one rib axis to the next rib axis.
• In bending, the concrete at the bottom hardly works — the tension is taken by the reinforcement. With ribs, most of this concrete is replaced by light blocks.
• t_eq — the height of the concrete alone in the section, without the blocks, as if all the concrete were spread into one layer. The +1 in the formula — one more cm for extra safety.
• Default when the section is not given: t_f = 5, b_w = 15, b_f = 65 cm. h is always entered. Did not pass? Increase h by 1 to 3 cm and calculate again — t_f stays constant.

Full exercise: 6.3 + 6.3 m, given section

3 supports, L = 6.3 + 6.3 m, q = 0.3, Δg = 0.2, B30 limestone. Section: h = 30, t_f = 5, b_w = 15, b_f = 65.

• L₀
  0.8 · 630 = 504 cm
• Equivalent t
  5 + 15·(30 − 5)/65 + 1 = 11.77 cm
• S.w
  (2.5·11.77 + 0.5·(30 − 11.77)) / 100 = 0.385 t/m²
• K₁₁
  h/t_f = 6, b_f/b_w = 4.33 → 4.3 → from the table K₁₁ = 0.747
• Thickness
  504 / (0.747 · 25.41 · 1) = 26.55 → 27 ≤ 30 passes ✓
• Load per rib
  1.299 · 0.65 = 0.844 t/m
• Moments
  Over B: −0.125 · 0.844 · 6.3² = −4.187 · In the span: 0.07 · 0.844 · 6.3² = 2.345 t·m — per rib
• Top reinforcement
  Rectangular section b_w = 15: ω = 0.359 → 4.345 cm² → 4Φ12 in the rib
• Bottom reinforcement
  T-section b_f = 65: ω → 0.1, x = 2.7 ≤ t_f = 5 → 2.102 cm² → 2Φ12 in the rib

That's it. 4Φ12 at the top over the support, 2Φ12 at the bottom — in every rib.

How to choose n·Φ: area per rib = number of bars × area of one bar: 2Φ12 = 2 · 1.131 = 2.26, 4Φ12 = 4.52 cm². Choose ≥ the governing A_s — with 2 or 4 bars only (an even number).

K₁₁ table — the rows around h/t_f = 6

K₁₁ in a ribbed slab (part of the table)

h/tf	4.2	4.3	4.4
5.8	0.752	0.748	0.744
6.0	0.751	0.747	0.743
6.2	0.750	0.746	0.742

How to read: round to the nearest row and column — no interpolation. 4.33 → column 4.3.

All the calculation lines

Thickness

• Slab thickness — givenh = 30 cm
• Equivalent thicknesst_eq = t_f + b_w · (h − t_f)b_f + 1 = 5 + 15 · (30 − 5)65 + 1 = 11.77 cm
• Self-weightS.w = 2.5 · t_eq + 0.5 · (h − t_eq)100 = 2.5 · 11.77 + 0.5 · (30 − 11.77)100 = 0.385 t/m²
• Service loadf_ser = S.w + Δg + q = 0.385 + 0.2 + 0.3 = 0.885 t/m²
• Load coefficientK₁₂ = 24.4∛(f_ser) = 24.4∛(0.885) = 25.41
• Section ratios for the K₁₁ tableht_f = 305 = 6
• b_fb_w = 6515 = 4.33
• Ribbed slab coefficient — from the tableK₁₁ = 0.747
• Aggregate and concrete coefficientK₁₃ = 1
• Required thicknessh_req = L₀(max)K₁₁ · K₁₂ · K₁₃ = 5040.747 · 25.41 · 1 = 26.55 cm
• Thickness check (rounding up)h_req = 26.55 → 27 ≤ h = 30 cm ✓

Loads

• Dead loadg = Δg + S.w = 0.2 + 0.385 = 0.585 t/m²
• Maximum design loadf_d,max = 1.4 · g + 1.6 · q = 1.4 · 0.585 + 1.6 · 0.3 = 1.299 t/m²
• Design load on a rib (width b_f in metres)f_d,max per rib = f_d,max · b_f = 1.299 · 0.65 = 0.844 t/m

Top reinforcement

• Negative moment · top reinforcementM_d = 4.187 t·m
• Effective depthd = h − 3 = 30 − 3 = 27 cm
• Minimum reinforcementA_s,min = ρ · b_w · d = 0.0013 · 15 · 27 = 0.527 cm²
• Compression width — b_w (negative moment)ω = 1 − √(1 − 2 · M_d · 10⁴b_w · d² · f_cd) = 1 − √(1 − 2 · 4.187 · 10⁴15 · 27² · 13) = 0.359
• Required reinforcementA_s,req = M_d · 10⁵d · (1 − 0.5ω) · f_sd = 4.187 · 10⁵27 · (1 − 0.5 · 0.359) · 4350 = 4.345 cm²
• Governing reinforcementA_s = max(A_s,min , A_s,req) = 4.345 cm²
• Bar selection for a rib4Φ12 → A_s = 4.52 ≥ 4.35 cm² ✓

Bottom reinforcement

• Positive moment · bottom reinforcementM_d = 2.345 t·m
• Effective depthd = h − 3 = 30 − 3 = 27 cm
• Minimum reinforcementA_s,min = ρ · b_w · d = 0.0013 · 15 · 27 = 0.527 cm²
• Compression width — b_f (positive moment)ω = 1 − √(1 − 2 · M_d · 10⁴b_f · d² · f_cd) = 1 − √(1 − 2 · 2.345 · 10⁴65 · 27² · 13) = 0.0388
• Depth of the compression zonex = ω · d = 0.1 · 27 = 2.7 cm
• The compression zone is within the flangex = 2.7 ≤ t_f = 5 cm ✓
• Required reinforcementA_s,req = M_d · 10⁵0.95 · d · f_sd = 2.345 · 10⁵0.95 · 27 · 4350 = 2.102 cm²
• Governing reinforcementA_s = max(A_s,min , A_s,req) = 2.102 cm²
• Bar selection for a rib2Φ12 → A_s = 2.26 ≥ 2.1 cm² ✓

Why sometimes b_f and sometimes b_w? (30 seconds)

• Positive moment (in the span): the compression is at the top — in the wide flange → the section is a T of width b_f. Check that x = ω·d stays within the flange.
• Negative moment (over a support): the compression is at the bottom — only in the narrow rib → rectangular section of width b_w, and ω no more than 0.4. Came out higher? A capacity problem — the solution: 'flipping a block' next to the support (the rib widens there), and the calculation continues with ω = 0.4.
• In positive moment, when the compression is within the flange, the formula gets shorter — the lever arm is approximately 0.95·d: A_s = M_d·10⁵ / (0.95·d·f_sd).

5 traps

• b_f in cm in the load per rib. 1.299 · 65? No — 1.299 · 0.65.
• K₁₁ = 1 as in the solid slab. In a ribbed slab — from the table.
• ω with b_f over the support. Negative moment → b_w.
• One bar in a rib. Minimum 2.
• Forgetting x ≤ t_f. Without the check the short formula is not allowed.