Design of slabs and staircase

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.
• 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. 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 (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous short edge(s) (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 support: $A_{st} = 245$ mm²/m, $p_t = 0.169\%$; $\tau_c = 0.30$ MPa (IS 456 Table 19, M20). For $D = 170$ mm, $k = 1.26$ (cl. 40.2.1.1), so $k\tau_c = 0.382$ MPa. $\tau_v = 0.209 < k\tau_c = 0.382$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 205}{1000 \times 145} = 0.141\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 215.5$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 215$ N/mm²): $k_t = 1.10$ Permissible $l/d = 26 \times 1.10 = 28.6$; actual $l_x/d = 4145/145 = 28.6$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (103 mm²/m) run into the support: $M_1 = 5.29$ kN·m/m, $L_0 = \max(d, 12\phi) = 145$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5291593}{30310} + 145 = 372\ \text{mm}$$

Since 372 mm $<$ $L_d = 376$ mm, the bars must be anchored with a standard 90° bend/hook at the support (anchorage value 8φ = 64 mm); with a bend the available length becomes 436 mm. This is adequate, so the development length is satisfied with bent ends.

Torsion reinforcement at corners

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

• Corner with only one discontinuous edge (2 corners): half of the above = 0.375 × 204 = 76 mm²/m. 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 top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ 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 continuous edge 8 mm @ 205 c/c (short direction); 8 mm @ 245 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 5 kN/m², floor finish = 1.2 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 $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5000 + 150 = 5150\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5150}{4150} = 1.241 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.175 = 4.375 kN/m²
• Total working load $w$ = 4.375 + 1.2 + 5 = 10.575 kN/m²
• Factored load $w_u = 1.5 \times 10.575 = 15.862$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 15.862 \times 4.150^2 = 273.19$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 13.45$ kN·m/m. 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 $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 15.862 \times 4.150/2 = 32.91$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{32915}{1000 \times 150} = 0.219$ MPa Steel at support: $A_{st} = 258$ mm²/m, $p_t = 0.172\%$; $\tau_c = 0.31$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.382$ MPa. $\tau_v = 0.219 < k\tau_c = 0.382$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 217.2$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 217$ N/mm²): $k_t = 1.10$ Permissible $l/d = 26 \times 1.10 = 28.6$; actual $l_x/d = 4150/150 = 27.7$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d, 12\phi) = 150$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{32915} + 150 = 375\ \text{mm}$$

Since 375 mm $<$ $L_d = 376$ mm, the bars must be anchored with a standard 90° bend/hook at the support (anchorage value 8φ = 64 mm); with a bend the available length becomes 439 mm. This is adequate, so the development length is satisfied with bent ends.

Torsion reinforcement at corners

Corner size $l_x/5 = 4150/5 = 830$ mm (say 830 mm) in both directions.

• Corner with only one discontinuous edge (2 corners): half of the above = 0.375 × 210 = 79 mm²/m. 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@235 short span      |
  # bot: 8mm@235 long span       |
  #                             t|
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 235 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 $\ge$ 210 mm²/m is already satisfied by the bars provided.

Answer: $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 195 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4.5$ m (short), $l_y = 6.5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 4 kN/m², floor finish = 2.5 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 $d$.
• Beam width not given; 230 mm assumed.
• Panel type: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26 (case: interior panel (all four edges continuous)).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 195$ mm, so $d = 195 - 20 - 5 = 170$ mm for the short-span bars and $d_y = 170 - 10 = 160$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4500 + 170 = 4670\ \text{mm},\quad l_y = 6500 + 170 = 6670\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6670}{4670} = 1.428 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.195 = 4.875 kN/m²
• Total working load $w$ = 4.875 + 2.5 + 4 = 11.375 kN/m²
• Factored load $w_u = 1.5 \times 11.375 = 17.062$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 17.062 \times 4.670^2 = 372.11$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.43$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 170^2 \times 10^{-6} = 79.76$ kN·m/m $>$ $M_{max} = 19.19$ kN·m/m. 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 $bD$ = 234 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Check for shear

$V_u = w_u l_x/2 = 17.062 \times 4.670/2 = 39.84$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{39841}{1000 \times 170} = 0.234$ MPa Steel at support: $A_{st} = 335$ mm²/m, $p_t = 0.197\%$; $\tau_c = 0.32$ MPa (IS 456 Table 19, M20). For $D = 195$ mm, $k = 1.21$ (cl. 40.2.1.1), so $k\tau_c = 0.392$ MPa. $\tau_v = 0.234 < k\tau_c = 0.392$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 251}{1000 \times 170} = 0.148\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 237.0$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 237$ N/mm²): $k_t = 1.07$ Permissible $l/d = 26 \times 1.07 = 27.8$; actual $l_x/d = 4670/170 = 27.5$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (126 mm²/m) run into the support: $M_1 = 7.59$ kN·m/m, $L_0 = \max(d, 12\phi) = 170$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{7594744}{39841} + 170 = 418\ \text{mm}$$

Since 418 mm $>$ $L_d = 376$ mm, development length is satisfied.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@200 short span      #
  # bot: 8mm@210 long span       #
  #                              #
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 195 mm ------|
   bottom 8mm @ 200 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 $\ge$ 234 mm²/m is already satisfied by the bars provided.

Answer: $D = 195$ mm; short-span bottom 8 mm @ 200 c/c; long-span bottom 8 mm @ 210 c/c; top steel over continuous edge 8 mm @ 150 c/c (short direction); 8 mm @ 210 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5.5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 5 kN/m², floor finish = 1.0 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 $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5500 + 150 = 5650\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5650}{4150} = 1.361 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.175 = 4.375 kN/m²
• Total working load $w$ = 4.375 + 1.0 + 5 = 10.375 kN/m²
• Factored load $w_u = 1.5 \times 10.375 = 15.562$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 15.562 \times 4.150^2 = 268.03$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.36$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 14.00$ kN·m/m. 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 $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 15.562 \times 4.150/2 = 32.29$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{32292}{1000 \times 150} = 0.215$ MPa Steel at support: $A_{st} = 272$ mm²/m, $p_t = 0.181\%$; $\tau_c = 0.31$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.391$ MPa. $\tau_v = 0.215 < k\tau_c = 0.391$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 230.7$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 231$ N/mm²): $k_t = 1.07$ Permissible $l/d = 26 \times 1.07 = 27.8$; actual $l_x/d = 4150/150 = 27.7$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d, 12\phi) = 150$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{32292} + 150 = 380\ \text{mm}$$

Since 380 mm $>$ $L_d = 376$ mm, development length is satisfied.

Torsion reinforcement at corners

Corner size $l_x/5 = 4150/5 = 830$ mm (say 830 mm) in both directions.

• Corner with only one discontinuous edge (2 corners): half of the above = 0.375 × 210 = 79 mm²/m. 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@235 short span      |
  # bot: 8mm@235 long span       |
  #                             t|
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 235 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 $\ge$ 210 mm²/m is already satisfied by the bars provided.

Answer: $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 185 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 2.5 kN/m², floor finish = 2.75 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 $d$.
• Beam width not given; 230 mm assumed.
• Panel type: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26 (case: interior panel (all four edges continuous)).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 6000 + 150 = 6150\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6150}{4150} = 1.482 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.175 = 4.375 kN/m²
• Total working load $w$ = 4.375 + 2.75 + 2.5 = 9.625 kN/m²
• Factored load $w_u = 1.5 \times 9.625 = 14.438$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.438 \times 4.150^2 = 248.65$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.48$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 13.09$ kN·m/m. 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 $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Check for shear

$V_u = w_u l_x/2 = 14.438 \times 4.150/2 = 29.96$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{29958}{1000 \times 150} = 0.200$ MPa Steel at support: $A_{st} = 251$ mm²/m, $p_t = 0.168\%$; $\tau_c = 0.30$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.377$ MPa. $\tau_v = 0.200 < k\tau_c = 0.377$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 215.8$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 216$ N/mm²): $k_t = 1.10$ Permissible $l/d = 26 \times 1.10 = 28.6$; actual $l_x/d = 4150/150 = 27.7$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d, 12\phi) = 150$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{29958} + 150 = 398\ \text{mm}$$

Since 398 mm $>$ $L_d = 376$ mm, development length is satisfied.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@235 short span      #
  # bot: 8mm@235 long span       #
  #                              #
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 235 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 $\ge$ 210 mm²/m is already satisfied by the bars provided.

Answer: $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 200 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 500$ MPa.
• Live load = 4 kN/m², floor finish = 1.3 kN/m², partition = 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 $d$.
• Concrete grade not stated; M20 assumed.
• Imposed load 4 kN/m² is taken as the live load; floor finish 1.3 and partition 1.5 kN/m² are added as dead loads.
• Panel type: one long edge discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: one long edge discontinuous).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 180$ mm, so $d = 180 - 20 - 5 = 155$ mm for the short-span bars and $d_y = 155 - 10 = 145$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 155 = 4155\ \text{mm},\quad l_y = 5000 + 155 = 5155\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5155}{4155} = 1.241 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.180 = 4.500 kN/m²
• Total working load $w$ = 4.500 + 1.3 + 4 + 1.5 = 11.300 kN/m²
• Factored load $w_u = 1.5 \times 11.300 = 16.950$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 16.950 \times 4.155^2 = 292.63$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.133\, f_{ck} b d^2 = 0.133 \times 20 \times 1000 \times 155^2 \times 10^{-6} = 63.91$ kN·m/m $>$ $M_{max} = 15.81$ kN·m/m. 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 $bD$ = 216 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous long edge(s) (short-span direction): 50% of mid-span steel = 109 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 16.950 \times 4.155/2 = 35.21$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{35214}{1000 \times 155} = 0.227$ MPa Steel at support: $A_{st} = 245$ mm²/m, $p_t = 0.158\%$; $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 180$ mm, $k = 1.24$ (cl. 40.2.1.1), so $k\tau_c = 0.365$ MPa. $\tau_v = 0.227 < k\tau_c = 0.365$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 219}{1000 \times 155} = 0.141\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 243.6$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 244$ N/mm²): $k_t = 1.05$ Permissible $l/d = 26 \times 1.05 = 27.3$; actual $l_x/d = 4155/155 = 26.8$. Since actual < permissible, deflection is satisfied.

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): $L_d = \dfrac{0.87 \times 500 \times 8}{4 \times 1.92} = 453$ mm Alternate bottom bars (109 mm²/m) run into the support: $M_1 = 7.24$ kN·m/m, $L_0 = \max(d, 12\phi) = 155$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{7237864}{35214} + 155 = 422\ \text{mm}$$

Since 422 mm $<$ $L_d = 453$ mm, the bars must be anchored with a standard 90° bend/hook at the support (anchorage value 8φ = 64 mm); with a bend the available length becomes 486 mm. This is adequate, so the development length is satisfied with bent ends.

Torsion reinforcement at corners

Corner size $l_x/5 = 4155/5 = 831$ mm (say 840 mm) in both directions.

• Corner with only one discontinuous edge (2 corners): half of the above = 0.375 × 216 = 81 mm²/m. 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)
  +==============================+
  #                              #
  # bot: 8mm@230 short span      #
  # bot: 8mm@230 long span       #
  #t                            t#
  +------------------------------+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 180 mm ------|
   bottom 8mm @ 230 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 $\ge$ 216 mm²/m is already satisfied by the bars provided.

Answer: $D = 180$ mm; short-span bottom 8 mm @ 230 c/c; long-span bottom 8 mm @ 230 c/c; top steel over continuous edge 8 mm @ 205 c/c (short direction); 8 mm @ 230 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 3 kN/m², floor finish = 1 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 $d$.
• Support width not given; 230 mm assumed.
• Panel type: two adjacent edges discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: two adjacent edges discontinuous).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5000 + 150 = 5150\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5150}{4150} = 1.241 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.175 = 4.375 kN/m²
• Total working load $w$ = 4.375 + 1 + 3 = 8.375 kN/m²
• Factored load $w_u = 1.5 \times 8.375 = 12.562$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 12.562 \times 4.150^2 = 216.36$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 13.42$ kN·m/m. 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 $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous long edge(s) (short-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 12.562 \times 4.150/2 = 26.07$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{26067}{1000 \times 150} = 0.174$ MPa Steel at support: $A_{st} = 258$ mm²/m, $p_t = 0.172\%$; $\tau_c = 0.31$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.382$ MPa. $\tau_v = 0.174 < k\tau_c = 0.382$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 215.5$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 215$ N/mm²): $k_t = 1.10$ Permissible $l/d = 26 \times 1.10 = 28.6$; actual $l_x/d = 4150/150 = 27.7$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d, 12\phi) = 150$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{26067} + 150 = 435\ \text{mm}$$

Since 435 mm $>$ $L_d = 376$ mm, development length is satisfied.

Torsion reinforcement at corners

Corner size $l_x/5 = 4150/5 = 830$ mm (say 830 mm) in both directions.

• Corner with both edges discontinuous (1 corner): top and bottom meshes, each layer 3/4 of $A_{st}$ of mid-span = 0.75 × 210 = 158 mm²/m (not less than the minimum steel). Provide 8 mm @ 235 c/c in both directions, top and bottom, over 830 mm × 830 mm.
• Corner with only one discontinuous edge (2 corners): half of the above = 0.375 × 210 = 79 mm²/m. 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@235 short span      |
  # bot: 8mm@235 long span       |
  #t                            T|
  +------------------------------+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 235 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 $\ge$ 210 mm²/m is already satisfied by the bars provided.

Answer: $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 195 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied.

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 3 kN/m², floor finish = 2 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 $d$.
• Support (beam) width not given; 230 mm assumed.
• Panel type: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26 (case: interior panel (all four edges continuous)).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 165$ mm, so $d = 165 - 20 - 5 = 140$ mm for the short-span bars and $d_y = 140 - 10 = 130$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 140 = 4140\ \text{mm},\quad l_y = 5000 + 140 = 5140\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5140}{4140} = 1.242 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.165 = 4.125 kN/m²
• Total working load $w$ = 4.125 + 2 + 3 = 9.125 kN/m²
• Factored load $w_u = 1.5 \times 9.125 = 13.688$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.688 \times 4.140^2 = 234.60$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 140^2 \times 10^{-6} = 54.10$ kN·m/m $>$ $M_{max} = 10.48$ kN·m/m. 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 $bD$ = 198 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Check for shear

$V_u = w_u l_x/2 = 13.688 \times 4.140/2 = 28.33$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{28333}{1000 \times 140} = 0.202$ MPa Steel at support: $A_{st} = 219$ mm²/m, $p_t = 0.156\%$; $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 165$ mm, $k = 1.27$ (cl. 40.2.1.1), so $k\tau_c = 0.372$ MPa. $\tau_v = 0.202 < k\tau_c = 0.372$ MPa, hence the slab is safe in shear, no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 201}{1000 \times 140} = 0.144\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 191.7$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 192$ N/mm²): $k_t = 1.14$ Permissible $l/d = 26 \times 1.14 = 29.6$; actual $l_x/d = 4140/140 = 29.6$. Since actual < permissible, deflection is satisfied.

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 Alternate bottom bars (101 mm²/m) run into the support: $M_1 = 5.01$ kN·m/m, $L_0 = \max(d, 12\phi) = 140$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5005823}{28333} + 140 = 370\ \text{mm}$$

Since 370 mm $<$ $L_d = 376$ mm, the bars must be anchored with a standard 90° bend/hook at the support (anchorage value 8φ = 64 mm); with a bend the available length becomes 434 mm. This is adequate, so the development length is satisfied with bent ends.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@250 short span      #
  # bot: 8mm@250 long span       #
  #                              #
  +==============================+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 165 mm ------|
   bottom 8mm @ 250 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 $\ge$ 198 mm²/m is already satisfied by the bars provided.

Answer: $D = 165$ mm; short-span bottom 8 mm @ 250 c/c; long-span bottom 8 mm @ 250 c/c; top steel over continuous edge 8 mm @ 230 c/c (short direction); 8 mm @ 250 c/c (long direction). Shear, deflection and development length are satisfied.