BE Civil Engineering (IOE, TU) Design of RCC Structures (IOE, CE 702) Question Paper 2078 Nepal

Given: $b=250\ \text{mm}$, $D=450\ \text{mm}$, effective cover $=50\ \text{mm}$ so effective depth $d = 450-50 = 400\ \text{mm}$; $d'=45\ \text{mm}$; $M_u=180\ \text{kN·m}$; M20 ($f_{ck}=20\ \text{N/mm}^2$), Fe415 ($f_y=415\ \text{N/mm}^2$).

Step 1 — Limiting moment of resistance. For Fe415, $x_{u,max}/d = 0.48$, so $x_{u,max}=0.48\times400=192\ \text{mm}$.

$$M_{u,lim}=0.36\,f_{ck}\,b\,x_{u,max}(d-0.42\,x_{u,max})$$ $$=0.36\times20\times250\times192\times(400-0.42\times192)$$

$0.42\times192=80.64\ \text{mm}$, so $(400-80.64)=319.36\ \text{mm}$.

$$M_{u,lim}=0.36\times20\times250\times192\times319.36 = 110.37\times10^6\ \text{N·mm}=110.37\ \text{kN·m}$$

Since $M_u=180>M_{u,lim}=110.37\ \text{kN·m}$ → doubly reinforced section required.

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

$$A_{st1}=\frac{0.36\,f_{ck}\,b\,x_{u,max}}{0.87\,f_y}=\frac{0.36\times20\times250\times192}{0.87\times415}=\frac{345600}{361.05}=957.2\ \text{mm}^2$$

Step 3 — Remaining moment and compression steel.

$$M_{u2}=M_u-M_{u,lim}=180-110.37=69.63\ \text{kN·m}$$

Stress in compression steel $f_{sc}$: $d'/d=45/400=0.1125$. For Fe415, interpolating IS 456 Table F values ($f_{sc}=353$ at $d'/d=0.10$, $329$ at $0.15$):

$$f_{sc}=353-\frac{(353-329)}{0.05}\times(0.1125-0.10)=353-24\times\tfrac{0.0125}{0.05}\times...$$

More directly: slope $=(353-329)/0.05=480$ per unit $d'/d$. $f_{sc}=353-480\times0.0125=353-6=347\ \text{N/mm}^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-0.45f_{ck})(d-d')}=\frac{69.63\times10^6}{(347-0.45\times20)(400-45)}$$

$f_{sc}-0.45f_{ck}=347-9=338\ \text{N/mm}^2$; $(d-d')=355\ \text{mm}$.

$$A_{sc}=\frac{69.63\times10^6}{338\times355}=\frac{69.63\times10^6}{119990}=580.3\ \text{mm}^2$$

Step 4 — Additional tension steel $A_{st2}$ to balance $A_{sc}$.

$$A_{st2}=\frac{A_{sc}(f_{sc}-0.45f_{ck})}{0.87f_y}=\frac{580.3\times338}{0.87\times415}=\frac{196141}{361.05}=543.2\ \text{mm}^2$$

Step 5 — Total tension steel.

$$A_{st}=A_{st1}+A_{st2}=957.2+543.2=1500.4\ \text{mm}^2$$

Step 6 — Provide bars. Tension: 3–25 mm = $3\times490.9=1472.6\ \text{mm}^2$ (slightly less); use 3–25 mm + 1–16 mm = $1472.6+201.1=1673.7\ \text{mm}^2 > 1500.4$ OK. (Alternatively 2–25 + 2–20 = 982+628 = 1610 mm².) Compression: 2–20 mm = $2\times314.2=628.4\ \text{mm}^2 > 580.3$ OK.

Check minimum/maximum steel: $A_{st,min}=0.85bd/f_y=0.85\times250\times400/415=204.8\ \text{mm}^2$ (OK); $A_{st,max}=0.04bD=0.04\times250\times450=4500\ \text{mm}^2$ (OK).

Answer: Provide 3–25 mm + 1–16 mm tension bars ($A_{st}=1673.7\ \text{mm}^2$) and 2–20 mm compression bars ($A_{sc}=628.4\ \text{mm}^2$).

Step 1 — Trial depth. For a simply supported two-way slab, span/depth $\approx 28$ (Fe415, modification factor included). Short span $l_x \approx 4.0\ \text{m}$. $d \approx 4000/28 = 143\ \text{mm}$; adopt $d=140\ \text{mm}$, with clear cover 20 mm and 10 mm bars → overall $D = 140+20+5 = 165\ \text{mm}$. Adopt $D=165\ \text{mm}$, $d_x=140\ \text{mm}$ (short span, outer layer), $d_y=140-10=130\ \text{mm}$ (long span, inner layer).

Step 2 — Effective spans. $l_x = 4.0+0.14 = 4.14\ \text{m}$; $l_y = 5.0+0.14 = 5.14\ \text{m}$. $l_y/l_x = 5.14/4.14 = 1.24 \approx 1.25$ → coefficients valid.

Step 3 — Loads (per m² , per metre strip). Self weight $= 0.165\times25 = 4.125\ \text{kN/m}^2$; finish $=1.0$; LL $=3.0$. Total $=8.125\ \text{kN/m}^2$. Factored $w_u = 1.5\times8.125 = 12.19\ \text{kN/m}^2$.

Step 4 — Factored moments (per metre width).

$$M_x = \alpha_x\,w_u\,l_x^2 = 0.099\times12.19\times4.14^2 = 0.099\times12.19\times17.14 = 20.69\ \text{kN·m}$$ $$M_y = \alpha_y\,w_u\,l_x^2 = 0.051\times12.19\times17.14 = 10.66\ \text{kN·m}$$

Step 5 — Check depth for $M_x$. $M_{u,lim}=0.138 f_{ck} b d^2 = 0.138\times20\times1000\times140^2 = 54.1\times10^6\ \text{N·mm}=54.1\ \text{kN·m} > 20.69$. Depth adequate (under-reinforced).

Step 6 — Steel in short span (x). Using $M_x=20.69\ \text{kN·m}$, $d=140$:

$$\frac{M_u}{bd^2}=\frac{20.69\times10^6}{1000\times140^2}=1.056\ \text{N/mm}^2$$ $$A_{st}=\frac{0.5f_{ck}}{f_y}\left[1-\sqrt{1-\frac{4.6M_u}{f_{ck}bd^2}}\right]bd$$

$\frac{4.6\times20.69\times10^6}{20\times1000\times140^2}=\frac{95.17\times10^6}{392\times10^6}=0.2428$; $\sqrt{1-0.2428}=\sqrt{0.7572}=0.8702$. $A_{st}=\frac{0.5\times20}{415}(1-0.8702)\times1000\times140=0.02410\times0.1298\times140000=438\ \text{mm}^2$. Provide 10 mm @ 175 mm c/c → $A_{st}=78.5\times1000/175=449\ \text{mm}^2 > 438$ OK.

Step 7 — Steel in long span (y). $M_y=10.66\ \text{kN·m}$, $d_y=130$: $\frac{4.6\times10.66\times10^6}{20\times1000\times130^2}=\frac{49.04\times10^6}{338\times10^6}=0.1451$; $\sqrt{1-0.1451}=0.9246$. $A_{st}=0.02410\times(1-0.9246)\times1000\times130=0.02410\times0.0754\times130000=236\ \text{mm}^2$. Min steel $=0.12\%\times bD=0.0012\times1000\times165=198\ \text{mm}^2$. Provide 8 mm @ 200 mm c/c → $50.3\times1000/200=251\ \text{mm}^2 > 236$ OK.

Step 8 — Checks. Max spacing $\le 3d=420\ \text{mm}$ or 300 mm → OK. Provide torsion reinforcement at corners (mesh of area $=\tfrac{3}{4}A_{st,x}$ over $l_x/5$) since corners are not held down.

Answer: $D=165\ \text{mm}$; short span 10 mm @ 175 mm c/c, long span 8 mm @ 200 mm c/c, plus corner torsion reinforcement.

Step 1 — Plan size. Add ~10% for footing self-weight: total service load $=1.10\times1200=1320\ \text{kN}$. Required area $=1320/200=6.6\ \text{m}^2$. Side $=\sqrt{6.6}=2.57\ \text{m}$ → adopt $2.6\ \text{m}\times2.6\ \text{m}$ ($A=6.76\ \text{m}^2$).

Step 2 — Net upward factored soil pressure. 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 — Trial depth. Assume $d=500\ \text{mm}$, overall $D=560\ \text{mm}$ (60 mm cover to bar centroid).

Step 4 — Bending moment (at column face). Cantilever projection $=(2.6-0.4)/2=1.1\ \text{m}$. Per metre width:

$$M_u=q_u\times\frac{l^2}{2}=266.3\times\frac{1.1^2}{2}=266.3\times0.605=161.1\ \text{kN·m/m}$$

Total over 2.6 m width $=161.1\times2.6=418.9\ \text{kN·m}$ (steel computed per metre below).

Step 5 — Depth check for moment. $M_{u,lim}=0.138\times20\times1000\times500^2=690\times10^6=690\ \text{kN·m/m} \gg 161.1$ → depth ample for flexure; shear governs.

Step 6 — One-way shear check. Critical section at $d=0.5\ \text{m}$ from column face. Distance from footing edge to section $=1.1-0.5=0.6\ \text{m}$.

$$V_u=q_u\times0.6\times2.6=266.3\times0.6\times2.6=415.4\ \text{kN}$$ $$\tau_v=\frac{V_u}{b\,d}=\frac{415.4\times10^3}{2600\times500}=0.320\ \text{N/mm}^2$$

Assume $p_t\approx0.25\%$ → $\tau_c=0.36\ \text{N/mm}^2$ (M20, IS 456 Table 19) $>\tau_v$. One-way shear OK.

Step 7 — Two-way (punching) shear. Critical perimeter at $d/2=250\ \text{mm}$ from column faces: side $=400+500=900\ \text{mm}=0.9\ \text{m}$; perimeter $b_0=4\times900=3600\ \text{mm}$. Punching shear force $=q_u\times(A_{footing}-A_{crit})=266.3\times(6.76-0.9^2)=266.3\times(6.76-0.81)=266.3\times5.95=1584.5\ \text{kN}$.

$$\tau_v=\frac{1584.5\times10^3}{3600\times500}=0.880\ \text{N/mm}^2$$

Permissible $\tau_c'=k_s\times0.25\sqrt{f_{ck}}$; for square column $\beta_c=1$, $k_s=1$; $0.25\sqrt{20}=1.118\ \text{N/mm}^2 > 0.880$. Punching shear OK.

Step 8 — Flexural reinforcement (per metre). $M_u=161.1\ \text{kN·m}$, $d=500$: $\frac{4.6M_u}{f_{ck}bd^2}=\frac{4.6\times161.1\times10^6}{20\times1000\times500^2}=\frac{741.1\times10^6}{5000\times10^6}=0.1482$; $\sqrt{1-0.1482}=0.9229$.

$$A_{st}=\frac{0.5\times20}{415}(1-0.9229)\times1000\times500=0.02410\times0.0771\times500000=929\ \text{mm}^2/\text{m}$$

Min steel $=0.12\%\,bD=0.0012\times1000\times560=672\ \text{mm}^2/\text{m}<929$. Use $A_{st}=929\ \text{mm}^2/\text{m}$. Provide 16 mm @ 200 mm c/c → $201.1\times1000/200=1005\ \text{mm}^2/\text{m} > 929$ OK, both directions (square footing). Total bars per direction over 2.6 m: $\approx 14$ bars of 16 mm.

Step 9 — Development length. $L_d=\frac{0.87f_y\phi}{4\tau_{bd}}$; for M20 deformed bar $\tau_{bd}=1.2\times1.6=1.92$. $L_d=\frac{0.87\times415\times16}{4\times1.92}=\frac{5777}{7.68}=752\ \text{mm}$. Available from column face $=1100-(\text{end cover }50)=1050\ \text{mm}>752$ OK.

Answer: Footing 2.6 m × 2.6 m, overall depth 560 mm ($d=500$ mm), reinforcement 16 mm @ 200 mm c/c both ways.

Step 1 — Design equation (short column, minimum eccentricity satisfied).

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

where $A_c = A_g - A_{sc}$. Let $A_{sc}=0.015A_g$ (1.5%), so $A_c=0.985A_g$.

$$2000\times10^3 = 0.4\times25\times0.985A_g + 0.67\times415\times0.015A_g$$ $$=9.85A_g + 4.171A_g = 14.02A_g$$ $$A_g = \frac{2000\times10^3}{14.02} = 142654\ \text{mm}^2$$

Side $=\sqrt{142654}=377.7\ \text{mm}$ → adopt $400\ \text{mm}\times400\ \text{mm}$ ($A_g=160000\ \text{mm}^2$).

Step 2 — Required steel for adopted section. With $A_g=160000$: $0.4\times25\times(160000-A_{sc})+0.67\times415\times A_{sc}=2000\times10^3$. $10(160000-A_{sc})+278.05A_{sc}=2\,000\,000$ $1\,600\,000-10A_{sc}+278.05A_{sc}=2\,000\,000$ $268.05A_{sc}=400\,000$ → $A_{sc}=1492\ \text{mm}^2$. This is $1492/160000=0.93\%$ (> min 0.8%, < max 6%). Provide 4–25 mm = $4\times490.9=1963.5\ \text{mm}^2$ (one bar each corner; 1.23%), comfortably above requirement and meeting minimum.

Step 3 — Check minimum eccentricity. $e_{min}=\frac{L}{500}+\frac{D}{30}$. For $L\approx3000\ \text{mm}$: $e_{min}=3000/500+400/30=6+13.3=19.3\ \text{mm}$. Limit $0.05D=0.05\times400=20\ \text{mm}>19.3$ → axial formula valid.

Step 4 — Lateral ties. Tie diameter $\ge$ greater of (i) $\phi_{long}/4=25/4=6.25\ \text{mm}$, (ii) 6 mm → use 8 mm ties. Pitch $\le$ least of: (a) least lateral dimension $=400\ \text{mm}$; (b) $16\times\phi_{long}=16\times25=400\ \text{mm}$; (c) 300 mm. Governing $=300\ \text{mm}$ → provide 8 mm ties @ 300 mm c/c (reduce to 200 mm near splice/end zones).

Answer: Column 400 mm × 400 mm, M25/Fe415, longitudinal 4–25 mm bars ($A_{sc}=1963.5\ \text{mm}^2$), lateral 8 mm ties @ 300 mm c/c.

Step 1 — Nominal shear stress.

$$\tau_v=\frac{V_u}{b\,d}=\frac{260\times10^3}{300\times500}=1.733\ \text{N/mm}^2$$

Step 2 — Check against maximum. For M20, $\tau_{c,max}=2.8\ \text{N/mm}^2 > 1.733$ → section size adequate.

Step 3 — Design shear strength of concrete $\tau_c$.

$$p_t=\frac{100A_{st}}{bd}=\frac{100\times1257}{300\times500}=0.838\%$$

From IS 456 Table 19 (M20): at $p_t=0.75\%$, $\tau_c=0.56$; at $1.00\%$, $\tau_c=0.62$. Interpolate at 0.838%:

$$\tau_c=0.56+\frac{0.62-0.56}{1.00-0.75}(0.838-0.75)=0.56+0.24\times0.088=0.56+0.0211=0.581\ \text{N/mm}^2$$

Shear carried by concrete $V_c=\tau_c b d=0.581\times300\times500=87150\ \text{N}=87.15\ \text{kN}$.

Step 4 — Shear to be carried by stirrups.

$$V_{us}=V_u-V_c=260-87.15=172.85\ \text{kN}$$

Since $\tau_v>\tau_c$, shear reinforcement is required.

Step 5 — Spacing of vertical stirrups. Use 8 mm 2-legged stirrups, $A_{sv}=2\times50.3=100.5\ \text{mm}^2$.

$$s_v=\frac{0.87\,f_y\,A_{sv}\,d}{V_{us}}=\frac{0.87\times415\times100.5\times500}{172.85\times10^3}=\frac{18\,143\,000}{172850}=104.9\ \text{mm}$$

Adopt 8 mm 2-legged stirrups @ 100 mm c/c near support.

Step 6 — Maximum spacing checks. $0.75d=0.75\times500=375\ \text{mm}$; absolute max 300 mm. Provided 100 mm < 300 mm OK. Minimum shear reinforcement spacing: $s_v\le\frac{0.87f_yA_{sv}}{0.4b}=\frac{0.87\times415\times100.5}{0.4\times300}=\frac{36287}{120}=302\ \text{mm}$.

Step 7 — Spacing in low-shear zone. Where $\tau_v\le\tau_c$, provide nominal stirrups at min(0.75d, 300, 302)=300 mm.

Answer: Provide 8 mm two-legged vertical stirrups @ 100 mm c/c in the high-shear zone near supports, increasing to 300 mm c/c in the central low-shear region.

Limit state method (LSM). A structure is designed so that it does not reach any of the limit states — conditions beyond which it ceases to fulfil its intended function. Two principal categories:

• Limit state of collapse (strength): flexure, shear, torsion, compression, bond — guards against structural failure.
• Limit state of serviceability: deflection, cracking, vibration — guards against loss of usability.

LSM uses characteristic values and partial safety factors (a semi-probabilistic approach), giving a uniform, rational margin of safety.

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

Partial safety factors for materials ($\gamma_m$): concrete $\gamma_c = 1.5$; steel $\gamma_s = 1.15$. Hence design strengths: concrete $0.446 f_{ck}\approx0.45f_{ck}$ and steel $f_y/1.15=0.87f_y$.

Contrast with working stress method (WSM): WSM keeps stresses within permissible (elastic) limits using a single factor of safety on material strength and assumes linear stress distribution; it is conservative and does not reflect the true ultimate strength or non-linear behaviour. LSM accounts separately for load and material uncertainties and checks both strength and serviceability, giving more economical and reliable designs.

Development length ($L_d$). The minimum length of a bar required to transfer the bar force to the surrounding concrete through bond, so that the bar can develop its design stress without slipping (or before bond failure occurs). It ensures the calculated tension/compression at any section can be safely anchored.

Derivation. Equate the tensile force in the bar to the bond resistance over length $L_d$:

$$\frac{\pi}{4}\phi^2\,\sigma_s = \pi\,\phi\,L_d\,\tau_{bd}$$

where $\sigma_s=0.87f_y$ (design stress) and $\tau_{bd}$ is the design bond stress. Solving:

$$\boxed{L_d=\frac{\phi\,\sigma_s}{4\,\tau_{bd}}=\frac{0.87\,f_y\,\phi}{4\,\tau_{bd}}}$$

Numeric computation. For M25, plain-bar design bond stress $\tau_{bd}=1.4\ \text{N/mm}^2$ (IS 456 Cl. 26.2.1.1). For deformed bars in tension, increase by 60%: $\tau_{bd}=1.4\times1.6=2.24\ \text{N/mm}^2$.

$$L_d=\frac{0.87\times415\times20}{4\times2.24}=\frac{7221}{8.96}=805.9\ \text{mm}$$

Answer: $L_d \approx \mathbf{806\ mm}$ (about $40\phi$).

Step 1 — Assume neutral axis within the flange. Equate compression to tension:

$$0.36\,f_{ck}\,b_f\,x_u = 0.87\,f_y\,A_{st}$$ $$x_u=\frac{0.87\times415\times2200}{0.36\times20\times1000}=\frac{794\,310}{7200}=110.3\ \text{mm}$$

Since $x_u=110.3\ \text{mm} < D_f=120\ \text{mm}$, the neutral axis lies within the flange → the beam behaves as a rectangular section of width $b_f$.

Step 2 — Check $x_u$ against limiting value. $x_{u,max}=0.48d=0.48\times550=264\ \text{mm}$. Since $x_u=110.3<264$, section is under-reinforced (steel yields). Valid.

Step 3 — Ultimate moment of resistance. Taking the lever arm to the centroid of the rectangular stress block:

$$M_u=0.87\,f_y\,A_{st}(d-0.42\,x_u)$$

$0.42\times110.3=46.33\ \text{mm}$; $(d-0.42x_u)=550-46.33=503.67\ \text{mm}$.

$$M_u=0.87\times415\times2200\times503.67=794\,310\times503.67=400.1\times10^6\ \text{N·mm}$$ $$\boxed{M_u \approx 400.1\ \text{kN·m}}$$

Answer: Neutral axis depth $x_u = \mathbf{110.3\ mm}$ (inside flange); ultimate moment of resistance $M_u = \mathbf{400.1\ kN·m}$.

Step 1 — Self weight of waist slab on slope (per m² of plan). Inclined length per tread $=\sqrt{R^2+T^2}=\sqrt{150^2+300^2}=\sqrt{22500+90000}=\sqrt{112500}=335.4\ \text{mm}$. Waist self weight along slope $=0.200\times25=5.0\ \text{kN/m}^2$ (on sloping area). On plan: $5.0\times\frac{335.4}{300}=5.0\times1.118=5.59\ \text{kN/m}^2$.

Step 2 — Weight of steps (triangular). Equivalent thickness $=R/2=75\ \text{mm}$ → $0.075\times25=1.875\ \text{kN/m}^2$ on plan.

Step 3 — Total load on flight (per m² plan). Waist (on plan) $5.59$ + steps $1.875$ + finish $1.0$ + LL $3.0 = 11.47\ \text{kN/m}^2$. Factored $w_u=1.5\times11.47=17.20\ \text{kN/m}^2$ → 17.20 kN/m per metre width.

Step 4 — Load on landing (per m² plan). Landing slab self weight $=0.200\times25=5.0$ + finish $1.0$ + LL $3.0=9.0\ \text{kN/m}^2$; factored $=1.5\times9.0=13.5\ \text{kN/m}^2$.

Step 5 — Effective span and maximum moment. Effective span $=$ going $+$ landing $=2.7+1.2=3.9\ \text{m}$ (centre-to-centre of supports, simplified). For a simply supported member with the heavier flight load assumed over the full span (conservative design simplification):

$$M_{max}=\frac{w_u\,L^2}{8}=\frac{17.20\times3.9^2}{8}=\frac{17.20\times15.21}{8}=\frac{261.6}{8}=32.7\ \text{kN·m per metre width}$$

Answer: Factored flight load $=\mathbf{17.20\ kN/m^2}$; maximum design moment $M_{max}\approx\mathbf{32.7\ kN·m/m}$ (landing factored load $=13.5\ \text{kN/m}^2$).

Step 1 — Trial depth. Span/depth for simply supported (Fe415) basic $=20$; with a modification factor $\approx1.4$ for low $p_t$, take effective ratio $\approx26$. $d\approx3200/26=123\ \text{mm}$; adopt $d=125\ \text{mm}$, clear cover 20 mm, 10 mm bars → overall $D=125+20+5=150\ \text{mm}$.

Step 2 — Effective span. $l=3.2+0.125=3.325\ \text{m}$ (clear span + d), or clear span + bearing; use $l=3.325\ \text{m}$.

Step 3 — Loads (per m width). Self weight $=0.150\times25=3.75$; finish $1.0$; LL $4.0$ → total $8.75\ \text{kN/m}^2$. Factored $w_u=1.5\times8.75=13.125\ \text{kN/m}$.

Step 4 — Factored moment.

$$M_u=\frac{w_u l^2}{8}=\frac{13.125\times3.325^2}{8}=\frac{13.125\times11.06}{8}=\frac{145.1}{8}=18.14\ \text{kN·m/m}$$

Step 5 — Depth check. $M_{u,lim}=0.138\times20\times1000\times125^2=43.1\times10^6=43.1\ \text{kN·m}>18.14$ → OK.

Step 6 — Main steel. $\frac{4.6M_u}{f_{ck}bd^2}=\frac{4.6\times18.14\times10^6}{20\times1000\times125^2}=\frac{83.44\times10^6}{312.5\times10^6}=0.2670$; $\sqrt{1-0.2670}=\sqrt{0.7330}=0.8562$.

$$A_{st}=\frac{0.5\times20}{415}(1-0.8562)\times1000\times125=0.02410\times0.1438\times125000=433\ \text{mm}^2/\text{m}$$

Provide 10 mm @ 175 mm c/c → $78.5\times1000/175=449\ \text{mm}^2/\text{m}>433$ OK.

Step 7 — Distribution steel. $=0.12\%\,bD=0.0012\times1000\times150=180\ \text{mm}^2/\text{m}$. Provide 8 mm @ 250 mm c/c → $50.3\times1000/250=201\ \text{mm}^2/\text{m}>180$ OK.

Step 8 — Deflection (span/depth) check. $p_t=\frac{100\times449}{1000\times125}=0.359\%$. Service stress $f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=0.58\times415\times\frac{433}{449}=232.1\ \text{N/mm}^2$. From IS 456 Fig. 4, modification factor $\approx1.5$ for $p_t=0.36\%$, $f_s\approx232$. Allowable $(l/d)=20\times1.5=30$. Actual $l/d=3325/125=26.6 < 30$ → deflection OK.

Answer: $D=150\ \text{mm}$ ($d=125$); main 10 mm @ 175 mm c/c, distribution 8 mm @ 250 mm c/c; deflection satisfied ($l/d=26.6<30$).

(a) Minimum and maximum reinforcement in beams (IS 456:2000).

• Minimum tension steel: $\dfrac{A_s}{bd}\ge\dfrac{0.85}{f_y}$, i.e. for Fe415, $A_{s,min}=\dfrac{0.85bd}{415}=0.205\%\,bd$. This prevents sudden brittle failure when concrete cracks.
• Maximum tension (and compression) steel: $A_{s,max}=0.04\,bD$ (4% of gross area) to avoid congestion and ensure proper concreting.
• Side-face reinforcement required when web depth $>750\ \text{mm}$: $\ge0.1\%$ of web area on each face, spacing $\le300\ \text{mm}$.

(b) Anchorage at simply supported ends (IS 456 Cl. 26.2.3.3). At a simple support, the positive-moment tension bars must satisfy:

$$L_d\le\frac{1.3M_1}{V}+L_o$$

where $M_1$ = moment of resistance of the section with the bars continuing into the support, $V$ = shear at support, and $L_o$ = sum of anchorage beyond the centre of support (effective end anchorage / hook). At least one-third of the mid-span positive steel must extend into a simple support for a length $\ge L_d/3$. Standard 90° or 135° hooks (bend + extension) provide the anchorage value $L_o$ — a U-bar or standard hook of $\ge8\phi$ straight after a 90° bend.

Sketch (text): tension bars enter the support, bent up with a standard hook; the embedded length plus hook supplies the required anchorage.

(c) Curtailment of tension reinforcement. Bars are curtailed where no longer required for moment, but with extra rules to cover the shifted tension envelope:

• Extend each bar beyond its theoretical cut-off point by the greater of $d$ (effective depth) or $12\phi$.
• A bar must not be curtailed in a tension zone unless one of these is met: (i) shear at the cut-off $\le\tfrac{2}{3}$ of permissible, or (ii) extra stirrups are added over a distance $0.75d$, or (iii) the continuing bars provide at least twice the area required for flexure at that point.
• Anchor the curtailed bar for a full $L_d$ from the point of maximum stress.

Sketch (text): the bending-moment diagram is shifted horizontally by $d$ (the 'tension shift'); curtailment points are read off the shifted diagram so that bond and diagonal-tension demands are safely covered.