BE Civil Engineering (IOE, TU) Design of RCC Structures (IOE, CE 702) Question Paper 2077 Nepal

Step 1 — Limiting (balanced) moment of resistance.

For Fe415 steel, $x_{u,max}/d = 0.48$.

$$M_{u,lim} = 0.36\,f_{ck}\,b\,x_{u,max}(d - 0.42\,x_{u,max})$$

Using the standard coefficient $M_{u,lim}=0.138\,f_{ck}\,b\,d^2$ for Fe415:

$$M_{u,lim} = 0.138 \times 20 \times 300 \times 450^2 = 0.138 \times 20 \times 300 \times 202500$$ $$= 167.67 \times 10^6\ \text{N·mm} = 167.67\ \text{kN·m}$$

Since $M_u = 210\ \text{kN·m} > M_{u,lim} = 167.67\ \text{kN·m}$, compression steel IS required (doubly reinforced section).

Step 2 — Tension steel for the balanced part ($A_{st1}$).

$x_{u,max} = 0.48 d = 0.48 \times 450 = 216\ \text{mm}$.

$$A_{st1} = \frac{M_{u,lim}}{0.87\,f_y\,(d-0.42\,x_{u,max})}$$

Lever arm $= d - 0.42 x_{u,max} = 450 - 0.42 \times 216 = 450 - 90.72 = 359.28\ \text{mm}$.

$$A_{st1} = \frac{167.67\times10^6}{0.87 \times 415 \times 359.28} = \frac{167.67\times10^6}{129722} = 1292.5\ \text{mm}^2$$

Step 3 — Additional moment to be carried by the steel couple.

$$M_{u2} = M_u - M_{u,lim} = 210 - 167.67 = 42.33\ \text{kN·m}$$

Step 4 — Compression steel $A_{sc}$.

$d_c/d = 50/450 = 0.111$. For Fe415, the stress in compression steel $f_{sc}$ from IS 456 strain compatibility (interpolating the standard table, $d_c/d=0.10 \Rightarrow f_{sc}=353$, $d_c/d=0.15 \Rightarrow f_{sc}=342$):

$$f_{sc} \approx 353 - \frac{(0.111-0.10)}{0.05}(353-342) = 353 - 0.22\times11 \approx 350.6\ \text{N/mm}^2$$

Taking $f_{sc} = 351\ \text{N/mm}^2$:

$$A_{sc} = \frac{M_{u2}}{(f_{sc})(d-d_c)} = \frac{42.33\times10^6}{351 \times (450-50)} = \frac{42.33\times10^6}{351\times400} = \frac{42.33\times10^6}{140400}$$ $$A_{sc} = 301.5\ \text{mm}^2$$

Step 5 — Additional tension steel $A_{st2}$ to balance the compression steel force.

$$A_{st2} = \frac{A_{sc}\,f_{sc}}{0.87\,f_y} = \frac{301.5 \times 351}{0.87\times415} = \frac{105827}{361.05} = 293.1\ \text{mm}^2$$

Step 6 — Total tension steel.

$$A_{st} = A_{st1} + A_{st2} = 1292.5 + 293.1 = 1585.6\ \text{mm}^2$$

Step 7 — Bar selection.

Tension: provide 4–20 mm bars + 1–16 mm bar $= 4\times314 + 201 = 1457$ ... insufficient; use 5–20 mm bars $= 5\times314 = 1570\ \text{mm}^2$ plus 1–12 mm (113) $\Rightarrow 1683\ \text{mm}^2 > 1585.6\ \text{mm}^2$. Provide 4–20 mm + 2–16 mm = $4\times314+2\times201 = 1658\ \text{mm}^2 \ge 1585.6\ \text{mm}^2$.

Compression: provide 2–16 mm bars $= 2\times201 = 402\ \text{mm}^2 \ge 301.5\ \text{mm}^2$. OK.

Final answer (bold): Compression steel is required. $A_{st} = 1586\ \text{mm}^2$ (provide 4–20 + 2–16 mm), $A_{sc} = 302\ \text{mm}^2$ (provide 2–16 mm).

Check minimum steel: $A_{st,min}=0.85 bd/f_y = 0.85\times300\times450/415 = 276\ \text{mm}^2 <$ provided. Max steel $0.04bD = 0.04\times300\times500 = 6000\ \text{mm}^2 >$ provided. OK.

Step 1 — Trial depth from span/depth ratio.

For simply supported slab, basic $L/d = 20$. Take modification factor $\approx 1.4$ (lightly stressed) $\Rightarrow L/d \approx 28$.

Effective span (lesser of clear span + d, or c/c of supports). Assume $d \approx 110\ \text{mm}$: effective span $= 3000 + 110 = 3110\ \text{mm}$ or c/c $= 3000+230 = 3230\ \text{mm}$. Use $L = 3110\ \text{mm}$ (clear + d governs).

Required $d = 3110/28 = 111\ \text{mm}$. Adopt $D = 140\ \text{mm}$, clear cover 20 mm, 10 mm bars $\Rightarrow d = 140 - 20 - 5 = 115\ \text{mm}$.

Re-check effective span $= 3000 + 115 = 3115\ \text{mm} \approx 3.115\ \text{m}$.

Step 2 — Loads (per 1 m width).

• Self weight $= 0.14 \times 1 \times 25 = 3.5\ \text{kN/m}$
• Floor finish $= 1.0\ \text{kN/m}$
• Live load $= 3.0\ \text{kN/m}$
• Total service $= 7.5\ \text{kN/m}$
• Factored $w_u = 1.5 \times 7.5 = 11.25\ \text{kN/m}$

Step 3 — Design moment.

$$M_u = \frac{w_u L^2}{8} = \frac{11.25 \times 3.115^2}{8} = \frac{11.25 \times 9.703}{8} = \frac{109.16}{8} = 13.64\ \text{kN·m}$$

Step 4 — Check depth for moment.

$$d_{req} = \sqrt{\frac{M_u}{0.138\,f_{ck}\,b}} = \sqrt{\frac{13.64\times10^6}{0.138\times20\times1000}} = \sqrt{\frac{13.64\times10^6}{2760}} = \sqrt{4942} = 70.3\ \text{mm}$$

$70.3\ \text{mm} < 115\ \text{mm}$ provided — section is under-reinforced, OK.

Step 5 — Area of tension steel.

$$M_u = 0.87 f_y A_{st} d\left(1 - \frac{A_{st} f_y}{b d f_{ck}}\right)$$

Solve: $13.64\times10^6 = 0.87\times415\times A_{st}\times115\left(1-\frac{A_{st}\times415}{1000\times115\times20}\right)$

Let first approximate with lever arm $0.9d$: $A_{st}\approx \dfrac{M_u}{0.87 f_y (0.9 d)} = \dfrac{13.64\times10^6}{0.87\times415\times103.5} = \dfrac{13.64\times10^6}{37372} = 365\ \text{mm}^2$.

Refine using quadratic: $A_{st} f_y/(b d f_{ck}) = 365\times415/(1000\times115\times20)=0.0658$; lever-arm factor $=1-0.0658=0.934$; $A_{st}=13.64\times10^6/(0.87\times415\times115\times0.934)=13.64\times10^6/38783=351.7\ \text{mm}^2$.

Converged: $A_{st} \approx 352\ \text{mm}^2$ per metre.

Step 6 — Main bar spacing (10 mm bars, area = 78.5 mm²).

$$s = \frac{1000 \times 78.5}{352} = 223\ \text{mm}$$

Max spacing $= \min(3d, 300) = \min(345,300) = 300\ \text{mm}$. Provide 10 mm bars @ 200 mm c/c ($A_{st,prov}=1000\times78.5/200=392.5\ \text{mm}^2$).

Step 7 — Distribution steel.

$$A_{dist} = 0.12\% \times b D = 0.0012 \times 1000 \times 140 = 168\ \text{mm}^2$$

Using 8 mm bars (50.3 mm²): $s = 1000\times50.3/168 = 299\ \text{mm}$; max $=\min(5d,450)=\min(575,450)=450$. Provide 8 mm @ 250 mm c/c.

Step 8 — Deflection check.

$p_t = 100\times392.5/(1000\times115)=0.341\%$. $f_s = 0.58 f_y (A_{st,req}/A_{st,prov}) = 0.58\times415\times352/392.5 = 215.8\ \text{N/mm}^2$. Modification factor $\approx 1.5$. Allowable $L/d = 20\times1.5 = 30 >$ actual $3115/115 = 27.1$. OK.

Final answer (bold): Slab $D = 140\ \text{mm}$, $d=115\ \text{mm}$. Main steel: 10 mm @ 200 mm c/c; Distribution: 8 mm @ 250 mm c/c.

Step 1 — Design equation (short axially loaded column, IS 456 Cl. 39.3).

$$P_u = 0.4\,f_{ck}\,A_c + 0.67\,f_y\,A_{sc}$$

where $A_c = A_g - A_{sc}$. With $A_{sc} = 0.02 A_g$, $A_c = 0.98 A_g$.

$$1800\times10^3 = 0.4\times25\times0.98 A_g + 0.67\times415\times0.02 A_g$$ $$1800\times10^3 = 9.8 A_g + 5.561 A_g = 15.361 A_g$$ $$A_g = \frac{1800\times10^3}{15.361} = 117180\ \text{mm}^2$$

Side $= \sqrt{117180} = 342.3\ \text{mm}$. Adopt $350 \times 350\ \text{mm}$ ($A_g = 122500\ \text{mm}^2$).

Step 2 — Slenderness check.

Effective length (braced, assume both ends partially restrained, $k\approx0.85$): $l_e = 0.85\times3200 = 2720\ \text{mm}$.

$$\frac{l_e}{D} = \frac{2720}{350} = 7.77 < 12 \Rightarrow \textbf{short column. OK.}$$

Step 3 — Recompute required steel for adopted section.

$$P_u = 0.4 f_{ck}(A_g - A_{sc}) + 0.67 f_y A_{sc}$$ $$1800\times10^3 = 0.4\times25\times(122500 - A_{sc}) + 0.67\times415\times A_{sc}$$ $$1800\times10^3 = 10\times(122500 - A_{sc}) + 278.05 A_{sc}$$ $$1800\times10^3 = 1225000 - 10 A_{sc} + 278.05 A_{sc}$$ $$1800000 - 1225000 = 268.05 A_{sc}$$ $$575000 = 268.05 A_{sc} \Rightarrow A_{sc} = 2145\ \text{mm}^2$$

Step 4 — Longitudinal bars.

Provide 8–20 mm bars $= 8\times314 = 2513\ \text{mm}^2 > 2145\ \text{mm}^2$.

$p = 100\times2513/122500 = 2.05\%$ — within $0.8\%$ to $6\%$. OK. Provide 8–20 mm bars.

Step 5 — Lateral ties (IS 456 Cl. 26.5.3.2).

Tie diameter $\ge \max(\frac{1}{4}\phi_{long}, 6\ \text{mm}) = \max(5, 6) = 6\ \text{mm}$. Use 8 mm ties.

Pitch $\le \min$ of:

• least lateral dimension $= 350\ \text{mm}$
• $16 \times \phi_{long} = 16\times20 = 320\ \text{mm}$
• $300\ \text{mm}$

Governing $= 300\ \text{mm}$. Provide 8 mm ties @ 300 mm c/c. With 8 bars, provide internal cross-ties / open ties so that no bar is more than 135° from a tied bar.

Final answer (bold): Column $350\times350\ \text{mm}$, M25, 8–20 mm longitudinal bars (2.05%), 8 mm ties @ 300 mm c/c. Short column ($l_e/D = 7.77$).

Step 1 — Plan area of footing.

Self weight of footing taken as 10% of column load $= 120\ \text{kN}$. Total service load $= 1200 + 120 = 1320\ \text{kN}$.

$$A_{req} = \frac{1320}{200} = 6.6\ \text{m}^2 \Rightarrow \text{side} = \sqrt{6.6} = 2.57\ \text{m}$$

Adopt $2.6\ \text{m} \times 2.6\ \text{m}$ footing ($A = 6.76\ \text{m}^2$).

Step 2 — Net upward soil pressure (factored, for structural design).

Factored column load $= 1.5\times1200 = 1800\ \text{kN}$.

$$q_u = \frac{1800}{2.6\times2.6} = \frac{1800}{6.76} = 266.3\ \text{kN/m}^2$$

Step 3 — Bending moment (at column face).

Cantilever projection $= (2.6 - 0.4)/2 = 1.1\ \text{m}$.

$$M_u = q_u \times \frac{l^2}{2} \times B = 266.3 \times \frac{1.1^2}{2}\times 2.6 = 266.3\times0.605\times2.6 = 418.9\ \text{kN·m}$$

(per full 2.6 m width.)

Step 4 — Depth from moment.

$$d = \sqrt{\frac{M_u}{0.138 f_{ck} B}} = \sqrt{\frac{418.9\times10^6}{0.138\times20\times2600}} = \sqrt{\frac{418.9\times10^6}{7176}} = \sqrt{58375} = 241.6\ \text{mm}$$

Punching shear usually governs depth; adopt $d = 500\ \text{mm}$, $D = 550\ \text{mm}$ (cover 50 mm).

Step 5 — Two-way (punching) shear check.

Critical perimeter at $d/2 = 250\ \text{mm}$ from column face: side $= 400 + d = 400 + 500 = 900\ \text{mm}$.

Perimeter $b_0 = 4\times900 = 3600\ \text{mm}$.

Punching force $V_p = q_u(A_{footing} - A_{critical}) = 266.3\times(2.6^2 - 0.9^2) = 266.3\times(6.76-0.81) = 266.3\times5.95 = 1584.5\ \text{kN}$.

Nominal punching stress $\tau_v = \dfrac{V_p}{b_0 d} = \dfrac{1584.5\times10^3}{3600\times500} = \dfrac{1584.5\times10^3}{1.8\times10^6} = 0.880\ \text{N/mm}^2$.

Permissible $\tau_c = k_s\,0.25\sqrt{f_{ck}}$; for square column $\beta_c=1$, $k_s = 1$. $\tau_c = 0.25\sqrt{20} = 0.25\times4.472 = 1.118\ \text{N/mm}^2$.

$\tau_v = 0.880 < \tau_c = 1.118$ — punching shear OK.

Step 6 — One-way (flexural) shear check.

Critical section at distance $d = 500\ \text{mm}$ from column face. Distance from footing edge $= 1.1 - 0.5 = 0.6\ \text{m}$.

$V_u = q_u \times 0.6 \times 2.6 = 266.3\times0.6\times2.6 = 415.4\ \text{kN}$.

$\tau_v = \dfrac{V_u}{B d} = \dfrac{415.4\times10^3}{2600\times500} = 0.320\ \text{N/mm}^2$.

Provide $p_t \approx 0.30\%$ (see Step 7). For M20, $p_t=0.30\% \Rightarrow \tau_c \approx 0.387\ \text{N/mm}^2 > 0.320$ — one-way shear OK.

Step 7 — Flexural reinforcement.

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

$\dfrac{4.6 M_u}{f_{ck} B d^2} = \dfrac{4.6\times418.9\times10^6}{20\times2600\times500^2} = \dfrac{1.927\times10^9}{1.3\times10^{10}} = 0.1482$.

$\sqrt{1-0.1482} = \sqrt{0.8518} = 0.9229$.

$A_{st} = \dfrac{0.5\times20}{415}(1-0.9229)\times2600\times500 = \dfrac{10}{415}\times0.0771\times1.3\times10^6 = 0.02410\times0.0771\times1.3\times10^6$

$= 0.02410\times100230 = 2416\ \text{mm}^2$ (over 2.6 m width).

Check minimum: $A_{st,min}=0.0012 B D = 0.0012\times2600\times550 = 1716\ \text{mm}^2 < 2416$. Governs the larger value, use $2416\ \text{mm}^2$.

$p_t = 100\times2416/(2600\times500)=0.186\%$. (Slightly below the 0.30% assumed in shear; recheck $\tau_c$ at $p_t=0.19\%\approx0.32\ \text{N/mm}^2$, marginal — adopt $p_t=0.25\%$, i.e. $A_{st}=0.0025\times2600\times500=3250\ \text{mm}^2$ to satisfy one-way shear comfortably.)

Using 16 mm bars (201 mm²): number $= 3250/201 = 16.2 \Rightarrow$ 17 bars; spacing $= 2600/16 = 162\ \text{mm}$. Provide 16 mm @ 160 mm c/c both ways.

Final answer (bold): Footing $2.6\times2.6\ \text{m}$, $D=550\ \text{mm}$ ($d=500\ \text{mm}$); 16 mm @ 160 mm c/c both ways. Punching ($\tau_v=0.88<1.12$) and one-way shear satisfied.

Step 1 — Geometry.

Riser $R = 150\ \text{mm}$, tread $T = 300\ \text{mm}$.

Inclined length of one step $= \sqrt{R^2 + T^2} = \sqrt{150^2 + 300^2} = \sqrt{22500+90000} = \sqrt{112500} = 335.4\ \text{mm}$.

Step 2 — Trial waist thickness.

Take $L/d \approx 25$ (simply supported, modification factor for steel). $d \approx 3000/25 = 120\ \text{mm}$. Adopt waist slab $D = 175\ \text{mm}$ (measured perpendicular to slope), cover 20 mm, 12 mm bars $\Rightarrow d = 175-20-6 = 149\ \text{mm}$.

Step 3 — Loads on going (per 1 m width, expressed on horizontal plan).

Waist slab self weight (on slope, projected to plan):

$$w_{waist} = 25 \times \frac{D}{1000}\times\frac{\sqrt{R^2+T^2}}{T} = 25\times0.175\times\frac{335.4}{300} = 25\times0.175\times1.118 = 4.892\ \text{kN/m}^2$$

Steps (triangular, average thickness $R/2$):

$$w_{steps} = 25\times\frac{R}{2}/1000 = 25\times0.075 = 1.875\ \text{kN/m}^2$$

Finish $= 1.0\ \text{kN/m}^2$; Live $= 3.0\ \text{kN/m}^2$.

Total service $= 4.892 + 1.875 + 1.0 + 3.0 = 10.767\ \text{kN/m}^2$.

Factored $w_u = 1.5\times10.767 = 16.15\ \text{kN/m}$ (per 1 m width).

Step 4 — Design moment.

Treating the going as simply supported over $L = 3.0\ \text{m}$:

$$M_u = \frac{w_u L^2}{8} = \frac{16.15\times3.0^2}{8} = \frac{16.15\times9}{8} = \frac{145.35}{8} = 18.17\ \text{kN·m}$$

Step 5 — Check depth.

$$d_{req} = \sqrt{\frac{M_u}{0.138 f_{ck} b}} = \sqrt{\frac{18.17\times10^6}{0.138\times20\times1000}} = \sqrt{\frac{18.17\times10^6}{2760}} = \sqrt{6584} = 81.1\ \text{mm}$$

$81.1 < 149$ provided — OK.

Step 6 — Main reinforcement.

Approximate lever arm $0.9d = 134.1\ \text{mm}$:

$$A_{st} \approx \frac{M_u}{0.87 f_y (0.9d)} = \frac{18.17\times10^6}{0.87\times415\times134.1} = \frac{18.17\times10^6}{48417} = 375.3\ \text{mm}^2$$

Refine: $A_{st} f_y/(b d f_{ck}) = 375.3\times415/(1000\times149\times20)=0.0523$; lever-arm factor $0.9477$; $A_{st}=18.17\times10^6/(0.87\times415\times149\times0.9477)=18.17\times10^6/50985=356.4\ \text{mm}^2$.

$A_{st} \approx 357\ \text{mm}^2$ per metre.

Using 12 mm bars (113 mm²): spacing $s = 1000\times113/357 = 316\ \text{mm}$; max $=\min(3d,300)=300$. Provide 12 mm @ 300 mm c/c (or 10 mm @ 200 c/c giving 392 mm²). Adopt 12 mm @ 250 mm c/c ($A_{st,prov}=452\ \text{mm}^2$).

Step 7 — Distribution steel (along the flight width).

$$A_{dist} = 0.12\% bD = 0.0012\times1000\times175 = 210\ \text{mm}^2$$

Using 8 mm bars (50.3 mm²): $s = 1000\times50.3/210 = 240\ \text{mm}$. Provide 8 mm @ 200 mm c/c.

Final answer (bold): Waist slab $D = 175\ \text{mm}$ ($d=149\ \text{mm}$); main bars 12 mm @ 250 mm c/c; distribution 8 mm @ 200 mm c/c. Provide adequate anchorage into landing beams and continue alternate bars at supports.

Philosophy of limit state method.

The limit state method is a semi-probabilistic design philosophy that aims to ensure that a structure remains fit for use throughout its design life with an acceptably low probability that it reaches any of its limit states — conditions beyond which it no longer satisfies the design performance requirements. It combines the safety of the ultimate load (plastic) approach with the serviceability assurance of the working stress approach by applying separate partial safety factors to loads (which may be larger than expected) and to material strengths (which may be smaller than expected).

Partial safety factors for loads ($\gamma_f$) — IS 456:2000 (Table 18):

Partial safety factors for materials ($\gamma_m$):

• Concrete: $\gamma_m = 1.5$ (design strength $= f_{ck}/1.5$).
• Steel: $\gamma_m = 1.15$ (design strength $= f_y/1.15 = 0.87 f_y$).

Limit state of collapse vs serviceability:

• Limit state of collapse (ultimate): concerns the safety of the structure against failure — flexure, shear, torsion, compression, bond. The structure must not collapse under factored loads. Stresses approach material strengths; large safety factors used.
• Limit state of serviceability: concerns the satisfactory performance under service (working) loads — limiting deflection (span/250 or span/350 + 20 mm), crack width (typically $\le 0.3\ \text{mm}$), and vibration. Load factors of 1.0 are used.

Key distinction (bold): Collapse limit state guards against structural failure (uses factored loads and reduced material strengths); serviceability limit state ensures comfort and durability under working loads (deflection and cracking).

Step 1 — Nominal shear stress.

$$\tau_v = \frac{V_u}{b\,d} = \frac{200\times10^3}{250\times450} = \frac{200000}{112500} = 1.778\ \text{N/mm}^2$$

Step 2 — Check against maximum permissible shear stress.

For M20, $\tau_{c,max} = 2.8\ \text{N/mm}^2$. Since $\tau_v = 1.778 < 2.8$, section size is adequate (no need to revise dimensions).

Step 3 — Compare with concrete shear strength.

$\tau_c = 0.56\ \text{N/mm}^2$ (given for $p_t=0.75\%$). Since $\tau_v > \tau_c$, shear reinforcement is required.

Step 4 — Shear to be carried by stirrups.

$$V_{us} = V_u - \tau_c\,b\,d = 200\times10^3 - 0.56\times250\times450$$ $$= 200000 - 63000 = 137000\ \text{N} = 137\ \text{kN}$$

Step 5 — Spacing of vertical stirrups.

Use 8 mm 2-legged stirrups: $A_{sv} = 2\times50.3 = 100.6\ \text{mm}^2$.

$$s_v = \frac{0.87\,f_y\,A_{sv}\,d}{V_{us}} = \frac{0.87\times415\times100.6\times450}{137000}$$

Numerator $= 0.87\times415 = 361.05$; $\times100.6 = 36321.6$; $\times450 = 16344738$.

$$s_v = \frac{16344738}{137000} = 119.3\ \text{mm}$$

Step 6 — Check maximum spacing limits (IS 456 Cl. 26.5.1.5).

• $0.75 d = 0.75\times450 = 337.5\ \text{mm}$
• $300\ \text{mm}$

Governing $= 300\ \text{mm} > 119.3\ \text{mm}$ — calculated spacing governs.

Check minimum shear reinforcement spacing: $s_{v,min} = \dfrac{0.87 f_y A_{sv}}{0.4 b} = \dfrac{361.05\times100.6}{0.4\times250} = \dfrac{36322}{100} = 363\ \text{mm}$. Provided spacing is far below — OK.

Final answer (bold): Provide 8 mm 2-legged vertical stirrups @ 110 mm c/c (round down from 119 mm) in the high-shear zone.

Definition. Development length $L_d$ is the minimum length of a reinforcing bar that must be embedded in concrete beyond a critical section so that the bar can develop its full design stress ($0.87 f_y$) through bond, without bond/slip failure.

Derivation. Equating the tensile force in the bar to the bond resistance over the embedded length:

$$\text{Bar force} = \frac{\pi}{4}\phi^2 (0.87 f_y)$$ $$\text{Bond resistance} = (\pi \phi L_d)\,\tau_{bd}$$

Setting them equal:

$$\frac{\pi}{4}\phi^2 (0.87 f_y) = \pi \phi L_d \tau_{bd} \;\Rightarrow\; \boxed{L_d = \frac{\phi\,(0.87 f_y)}{4\,\tau_{bd}}} = \frac{0.87 f_y \phi}{4\tau_{bd}}$$

Bond stress values (M20).

• Plain bars: $\tau_{bd} = 1.2\ \text{N/mm}^2$.
• Deformed bars in tension: $\tau_{bd} = 1.2\times1.6 = 1.92\ \text{N/mm}^2$.
• Deformed bars in compression: $\tau_{bd} = 1.92\times1.25 = 2.40\ \text{N/mm}^2$.

Development length in tension ($\phi = 20\ \text{mm}$, Fe415):

$$L_d = \frac{0.87\times415\times20}{4\times1.92} = \frac{7221}{7.68} = 940.2\ \text{mm}$$

$\Rightarrow$ $L_d \approx 940\ \text{mm} \approx 47\phi$.

Development length in compression:

$$L_{d,c} = \frac{0.87\times415\times20}{4\times2.40} = \frac{7221}{9.6} = 752.2\ \text{mm}$$

$\Rightarrow$ $L_{d,c} \approx 752\ \text{mm} \approx 38\phi$.

Final answer (bold): $L_d$ (tension) $= 940\ \text{mm}$ ($\approx 47\phi$); $L_d$ (compression) $= 752\ \text{mm}$ ($\approx 38\phi$).

Step 1 — Assume neutral axis within the flange and find $x_u$.

Equate tensile and compressive forces (treat as a rectangular section of width $b_f$):

$$0.36 f_{ck} b_f x_u = 0.87 f_y A_{st}$$ $$x_u = \frac{0.87 f_y A_{st}}{0.36 f_{ck} b_f} = \frac{0.87\times415\times1800}{0.36\times20\times1000} = \frac{649890}{7200} = 90.3\ \text{mm}$$

Step 2 — Check NA position.

$x_u = 90.3\ \text{mm} < D_f = 100\ \text{mm}$ — the neutral axis lies within the flange, so the section behaves as a rectangular beam of width $b_f = 1000\ \text{mm}$.

Check under-reinforced: $x_{u,max} = 0.48 d = 0.48\times500 = 240\ \text{mm} > 90.3\ \text{mm}$ — under-reinforced, steel yields. OK.

Step 3 — Ultimate moment of resistance.

Lever arm $= d - 0.42 x_u = 500 - 0.42\times90.3 = 500 - 37.9 = 462.1\ \text{mm}$.

$$M_u = 0.87 f_y A_{st}(d - 0.42 x_u) = 0.87\times415\times1800\times462.1$$

$0.87\times415 = 361.05$; $\times1800 = 649890$; $\times462.1 = 300324...$

$$M_u = 649890\times462.1 = 300325000\ \text{N·mm} = 300.3\ \text{kN·m}$$

Verification via concrete couple: $C = 0.36\times20\times1000\times90.3 = 650160\ \text{N} \approx T = 649890\ \text{N}$ (consistent).

Final answer (bold): Neutral axis $x_u = 90.3\ \text{mm}$ (within flange); ultimate moment of resistance $M_u \approx 300.3\ \text{kN·m}$.

Step 1 — Spans and aspect ratio.

Short span $l_x = 4.0\ \text{m}$, long span $l_y = 5.0\ \text{m}$.

$$\frac{l_y}{l_x} = \frac{5.0}{4.0} = 1.25$$

For a simply supported slab with corners not held down (IS 456 Table 27), $\alpha_x = 0.084$, $\alpha_y = 0.059$ (given).

Step 2 — Mid-span moment in the short span direction.

$$M_x = \alpha_x\,w_u\,l_x^2 = 0.084\times12\times4.0^2 = 0.084\times12\times16$$ $$= 0.084\times192 = 16.13\ \text{kN·m per metre}$$

Step 3 — Mid-span moment in the long span direction.

$$M_y = \alpha_y\,w_u\,l_x^2 = 0.059\times12\times16 = 0.059\times192 = 11.33\ \text{kN·m per metre}$$

(Note: both moments use $l_x^2$ as per the IS 456 coefficient method.)

Step 4 — Remarks on reinforcement direction.

• $M_x = 16.13\ \text{kN·m/m}$ governs the shorter span reinforcement, placed in the bottom-most (outer) layer for maximum effective depth.
• $M_y = 11.33\ \text{kN·m/m}$ governs the longer span reinforcement, placed above the short-span bars.
• Since corners are not held down, no special corner (torsion) reinforcement is required, but at least 50% of mid-span steel should be carried to the supports.

Final answer (bold): $M_x = 16.13\ \text{kN·m/m}$ (short span), $M_y = 11.33\ \text{kN·m/m}$ (long span).

(a) Minimum and maximum tension reinforcement in beams (IS 456 Cl. 26.5.1).

• Minimum: $\dfrac{A_{st}}{b d} = \dfrac{0.85}{f_y}$, i.e. $A_{st,min} = \dfrac{0.85\,b\,d}{f_y}$. For Fe415 this is about $0.205\%$ of $bd$. This prevents sudden brittle failure immediately after cracking.
• Maximum: $A_{st,max} = 0.04\,b\,D$ (4% of gross area), to avoid congestion and ensure proper compaction. The same limit applies to compression steel.

(b) Curtailment of flexural reinforcement (IS 456 Cl. 26.2.3).

Bars may be curtailed where no longer required for moment, but a bar must extend beyond the theoretical cut-off point by the greater of the effective depth $d$ or $12\phi$. Curtailment in the tension zone is permitted only if one of the code conditions is met (e.g. shear at the cut-off $\le 2/3$ of permissible, or extra stirrups provided). At least one-third of positive-moment steel must continue into a simple support and be anchored for $L_d/3$.

(c) Lap splices (IS 456 Cl. 26.2.5).

• Lap length in tension $= L_d$ or $30\phi$, whichever is greater (for flexural tension); in compression $= L_d$ in compression or $24\phi$, whichever is greater.
• Laps should be staggered; not more than 50% of bars spliced at one section. Splices should be away from sections of maximum stress (e.g. mid-span for sagging, support for hogging). Bars larger than 36 mm are preferably welded or mechanically coupled rather than lapped.

(d) Nominal cover for durability and fire (IS 456 Cl. 26.4 / Table 16, 16A).

• Nominal cover protects steel from corrosion and provides fire resistance. Typical values: mild exposure 20 mm, moderate 30 mm, severe 45 mm, very severe 50 mm, extreme 75 mm.
• For beams a minimum cover of 25–30 mm is common; for footings 50 mm.
• Higher grade of concrete with lower w/c ratio is required for severe exposure.
• Fire resistance: cover increases with required fire rating (e.g. 1 h beam $\ge 20\ \text{mm}$, 4 h $\ge 70\ \text{mm}$).

Key point (bold): Proper detailing — adequate minimum steel, correct anchorage/curtailment, staggered laps, and durability-based cover — is essential to realize the strength and ductility assumed in design.