Slab three edges continuous, one short edge discontinuous (5x4 m, 250 beam)

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Grades are not stated in the question; M20 concrete and Fe415 steel are assumed.
• Live load = 4 kN/m², floor finish = 1.5 kN/m². Unit weight of concrete 25 kN/m³. Clear cover 20 mm (mild exposure, IS 456 Table 16); 10 mm bars assumed for working out $d$.
• Panel type: one short edge discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: one short edge discontinuous).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 170$ mm, so $d = 170 - 20 - 5 = 145$ mm for the short-span bars and $d_y = 145 - 10 = 135$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 145 = 4145\ \text{mm},\quad l_y = 5000 + 145 = 5145\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5145}{4145} = 1.241 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.170 = 4.250 kN/m²
• Total working load $w$ = 4.250 + 1.5 + 4 = 9.750 kN/m²
• Factored load $w_u = 1.5 \times 9.750 = 14.625$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.625 \times 4.145^2 = 251.27$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 145^2 \times 10^{-6} = 58.03$ kN·m/m $>$ $M_{max} = 12.37$ kN·m/m. The section is under-reinforced; $d$ is adequate for flexure.

Reinforcement

$$A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_u}{f_{ck} b d^2}}\right] b d$$

Minimum steel = 0.12% of $bD$ = 204 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions.

Top steel at the discontinuous edge (for the negative moment due to partial fixity, IS 456 Annex D) extends $0.1 l$ into the span. Along the discontinuous short edge (long-span direction): 50% of mid-span steel = 103 mm²/m; provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 14.625 \times 4.145/2 = 30.31$ kN/m

$\tau_v = \dfrac{V_u}{bd} = \dfrac{30310}{1000 \times 145} = 0.209$ MPa

Steel at the support: $A_{st} = 245$ mm²/m, $p_t = 0.169\%$, so $\tau_c = 0.30$ MPa (IS 456 Table 19, M20). For $D = 170$ mm, $k = 1.26$ (cl. 40.2.1.1), so $k\tau_c = 0.38$ MPa.

$\tau_v = 0.209 < k\tau_c = 0.38$ MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 205}{1000 \times 145} = 0.141\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{184}{205} = 216$ N/mm².

Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 216$ N/mm²): at such a low steel percentage the curve reaches its upper limit, so $k_t = 2.0$.

Permissible $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 4145/145 = 28.6$. Since actual < permissible, the provided section satisfies the deflection criterion.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ MPa (M20, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 1.92} = 376$ mm

The check $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)) is made at the discontinuous short edge, where the long-span bottom bars end. Alternate bars are bent up, so 103 mm²/m reach the support at $d_y = 135$ mm: $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 4.92$ kN·m/m. Taking $V = 30.31$ kN/m (conservative), $1.3\dfrac{M_1}{V} = 1.3 \times \dfrac{4.92 \times 10^6}{30310} = 211$ mm.

$L_0$ = straight length beyond the centre of the support $= 250/2 - 25$ (end cover) $= 100$ mm, plus the anchorage value of any hook or bend:

• straight bars: $211 + 100 = 311$ mm $< 376$ mm;
• with a 90° bend ($8\phi = 64$ mm): $375$ mm, just below 376 mm;
• with a standard U-hook ($16\phi = 128$ mm): $211 + 100 + 128 = 439$ mm $> 376$ mm.

So the long-span bottom bars are given standard hooks at the discontinuous edge, and the development length is then satisfied.

Torsion reinforcement at corners

Corner size $l_x/5 = 4145/5 = 829$ mm (say 830 mm) in both directions.

• Corners with one edge discontinuous and the other continuous (2 corners, at the ends of the discontinuous short edge): 3/8 of the mid-span steel = 0.375 × 204 = 76 mm²/m in each layer. Provide 8 mm @ 300 c/c in each direction, top and bottom.
• No torsion steel is needed at corners where both edges are continuous (IS 456 D-1.10).

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                             t|
  # bot: 8mm@245 short span      |
  # bot: 8mm@245 long span       |
  #                             t|
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  t : torsion mesh 8mm@300 top+bottom, 830 x 830 mm

SECTION ALONG SHORT SPAN (supports and mid-span)
  top bars over      top bars over
  continuous edge    continuous edge
  8mm@205            8mm@205
   ~~~~~~~            ~~~~~~~
  |------ slab D = 170 mm ------|
   bottom 8mm @ 245 c/c, alt. bars bent up
  [support]            [support]

• Bottom bars: short-span bars in the lower layer, long-span bars above them; alternate bars may be bent up at $0.1 l$ from the support.
• Top bars over continuous supports extend $0.3 l_x$ from the face of the support; all bars have 20 mm cover.
• Distribution/minimum steel $\ge$ 204 mm²/m is already satisfied by the bars provided.

Answer: $D = 170$ mm; short-span bottom 8 mm @ 245 c/c; long-span bottom 8 mm @ 245 c/c; top steel over the continuous long edges 8 mm @ 205 c/c (short direction) and over the continuous short edge 8 mm @ 245 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edge; torsion mesh 8 mm @ 300 c/c at the two corners. Shear and deflection are satisfied; development length is satisfied with standard hooks at the discontinuous edge.