BE Civil Engineering (IOE, TU) Design of RCC Structures (IOE, CE 702) Question Paper 2080 Nepal

Given: $b=300\ \text{mm}$, $d=500\ \text{mm}$, $M_u=320\ \text{kN·m}$, M20 ($f_{ck}=20\ \text{MPa}$), Fe415 ($f_y=415\ \text{MPa}$), $d'=50\ \text{mm}$.

Step 1 — Limiting moment of resistance (singly reinforced). For Fe415, $x_{u,max}/d = 0.48$, so $x_{u,max}=0.48\times500=240\ \text{mm}$.

$$M_{u,lim}=0.36\,f_{ck}\,b\,x_{u,max}(d-0.42x_{u,max})$$ $$=0.36\times20\times300\times240\times(500-0.42\times240)$$ $$=0.36\times20\times300\times240\times(500-100.8)$$ $$=518400\times399.2=206.9\times10^6\ \text{N·mm}=206.9\ \text{kN·m}$$

Since $M_u=320\ \text{kN·m} > M_{u,lim}=206.9\ \text{kN·m}$, compression steel is required (doubly reinforced).

Step 2 — Tension steel to balance $M_{u,lim}$ ($A_{st1}$).

$$A_{st1}=\frac{M_{u,lim}}{0.87\,f_y\,(d-0.42x_{u,max})}=\frac{206.9\times10^6}{0.87\times415\times399.2}$$ $$=\frac{206.9\times10^6}{144146}=1435\ \text{mm}^2$$

Step 3 — Remaining moment carried by the steel couple.

$$M_{u2}=M_u-M_{u,lim}=320-206.9=113.1\ \text{kN·m}$$

Step 4 — Compression steel $A_{sc}$. Strain in compression steel: $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_{u,max}}\right)=0.0035\left(1-\dfrac{50}{240}\right)=0.0035\times0.7917=0.00277$.

For Fe415 the design stress at $\varepsilon_{sc}=0.00277$ (from the IS stress–strain table; yield strain $\approx0.00380$) is $f_{sc}\approx351\ \text{MPa}$.

Net compression stress $=f_{sc}-0.45f_{ck}=351-9=342\ \text{MPa}$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-0.45f_{ck})(d-d')}=\frac{113.1\times10^6}{342\times(500-50)}=\frac{113.1\times10^6}{153900}=735\ \text{mm}^2$$

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

$$A_{st2}=\frac{A_{sc}(f_{sc}-0.45f_{ck})}{0.87f_y}=\frac{735\times342}{0.87\times415}=\frac{251370}{361.05}=696\ \text{mm}^2$$

Step 6 — Total tension steel.

$$A_{st}=A_{st1}+A_{st2}=1435+696=2131\ \text{mm}^2$$

Provide: Tension steel $\approx 2131\ \text{mm}^2$ — e.g. 4–25 mm bars (1963) + 2–16 mm (402) = 2365 mm² (OK), or 7–20 mm = 2199 mm². Compression steel $\approx 735\ \text{mm}^2$ — e.g. 4–16 mm = 804 mm².

Final answer: $\boxed{A_{sc}=735\ \text{mm}^2,\quad A_{st}=2131\ \text{mm}^2}$ (doubly reinforced section).

Given: $l_x=4.0\ \text{m}$ (short), $l_y=5.0\ \text{m}$ (long), LL $=3.0$, FF $=1.0\ \text{kN/m}^2$, M20, Fe415, $\alpha_x=0.099$, $\alpha_y=0.051$.

Step 1 — Trial depth. For a simply supported slab, span/depth $\approx 28$ (with modification factor for tension steel taken as 1 here for a first trial). $d=\dfrac{4000}{28}\approx143\ \text{mm}$. Take $d=140\ \text{mm}$, overall $D=140+15+10/2=160\ \text{mm}$ (15 mm cover, 10 mm bars). Use $D=160\ \text{mm}$, $d=140\ \text{mm}$.

Step 2 — Effective span. $l_x = \min(4.0+0.14,\ 4.0+c) = 4.14\ \text{m}$ (clear span + d). Use $l_x=4.14\ \text{m}$.

Step 3 — Loads (per m² ). Self weight $=25\times0.16=4.0\ \text{kN/m}^2$. Total service load $=4.0+1.0+3.0=8.0\ \text{kN/m}^2$. Factored load $w_u=1.5\times8.0=12.0\ \text{kN/m}^2$.

Step 4 — Design moments (per metre width).

$$M_x=\alpha_x\,w_u\,l_x^2=0.099\times12.0\times4.14^2=0.099\times12.0\times17.14=20.36\ \text{kN·m}$$ $$M_y=\alpha_y\,w_u\,l_x^2=0.051\times12.0\times17.14=10.49\ \text{kN·m}$$

Step 5 — Check depth for $M_x$. $M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times1000\times140^2=54.1\times10^6\ \text{N·mm}=54.1\ \text{kN·m}>20.36$. Depth OK.

Step 6 — Steel in short span (x).

$$M_x=0.87f_yA_{st}\,d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)$$

Using $A_{st}=\dfrac{0.5f_{ck}}{f_y}\left[1-\sqrt{1-\dfrac{4.6M_u}{f_{ck}bd^2}}\right]bd$:

$$\frac{4.6\times20.36\times10^6}{20\times1000\times140^2}=\frac{93.66\times10^6}{392\times10^6}=0.2389$$ $$A_{st,x}=\frac{0.5\times20}{415}\left[1-\sqrt{1-0.2389}\right]\times1000\times140$$ $$=0.02410\times(1-0.8724)\times140000=0.02410\times0.1276\times140000=430\ \text{mm}^2$$

Provide 10 mm @ 180 mm c/c → $A_{st}=\dfrac{78.5\times1000}{180}=436\ \text{mm}^2$. (short span)

Step 7 — Steel in long span (y). Effective depth $d_y=140-10=130\ \text{mm}$ (second layer).

$$\frac{4.6\times10.49\times10^6}{20\times1000\times130^2}=\frac{48.25\times10^6}{338\times10^6}=0.1428$$ $$A_{st,y}=\frac{0.5\times20}{415}\left[1-\sqrt{1-0.1428}\right]\times1000\times130$$ $$=0.02410\times(1-0.9259)\times130000=0.02410\times0.0741\times130000=232\ \text{mm}^2$$

Minimum steel $=0.12\%\times1000\times160=192\ \text{mm}^2$. So $A_{st,y}=232\ \text{mm}^2$ governs. Provide 8 mm @ 200 mm c/c → $\dfrac{50.3\times1000}{200}=251\ \text{mm}^2$. (long span)

Step 8 — Check spacing. Max spacing $=\min(3d,\ 300)=\min(420,300)=300\ \text{mm}$. All spacings OK.

Summary:

• Thickness $D=160\ \text{mm}$, $d=140\ \text{mm}$.
• Short span: 10 mm @ 180 mm c/c (436 mm²).
• Long span: 8 mm @ 200 mm c/c (251 mm²).
• Provide torsional/edge detailing; alternate bars bent up or separate top bars at supports as per practice.

Given: Column $400\times400\ \text{mm}$, service load $P=1200\ \text{kN}$, SBC $=200\ \text{kN/m}^2$, M20, Fe415.

Step 1 — Plan area. Add ~10% for self weight of footing: total $=1.1\times1200=1320\ \text{kN}$.

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

Provide a square footing $2.6\ \text{m}\times2.6\ \text{m}$ ($A=6.76\ \text{m}^2$).

Step 2 — Net upward (factored) soil pressure.

$$q_u=\frac{1.5\times1200}{2.6\times2.6}=\frac{1800}{6.76}=266.3\ \text{kN/m}^2.$$

Step 3 — Assume effective depth. Try $D=550\ \text{mm}$, cover $50$, 16 mm bars → $d\approx550-50-8=492$, take $d=490\ \text{mm}$.

Step 4 — Bending moment at column face. Cantilever projection $=\dfrac{2.6-0.4}{2}=1.1\ \text{m}$.

$$M_u=q_u\times\frac{l^2}{2}\times B=266.3\times\frac{1.1^2}{2}\times2.6=266.3\times0.605\times2.6=419.0\ \text{kN·m (full width)}$$

Per metre width: $M_u=q_u\dfrac{l^2}{2}=266.3\times0.605=161.1\ \text{kN·m/m}$.

Step 5 — Flexural steel (per m).

$$\frac{4.6M_u}{f_{ck}bd^2}=\frac{4.6\times161.1\times10^6}{20\times1000\times490^2}=\frac{741.1\times10^6}{4802\times10^6}=0.1543$$ $$A_{st}=\frac{0.5\times20}{415}\left[1-\sqrt{1-0.1543}\right]\times1000\times490$$ $$=0.02410\times(1-0.9196)\times490000=0.02410\times0.0804\times490000=950\ \text{mm}^2/\text{m}$$

Minimum $=0.12\%\times1000\times550=660\ \text{mm}^2$. So $A_{st}=950\ \text{mm}^2/\text{m}$. Provide 16 mm @ 200 mm c/c → $\dfrac{201\times1000}{200}=1005\ \text{mm}^2/\text{m}$ each way.

Step 6 — One-way (beam) shear. Critical section at $d$ from column face: distance from edge $=1.1-0.49=0.61\ \text{m}$.

$$V_u=q_u\times0.61\times2.6=266.3\times0.61\times2.6=422.4\ \text{kN}$$

Nominal shear stress $\tau_v=\dfrac{V_u}{bd}=\dfrac{422.4\times10^3}{2600\times490}=0.332\ \text{MPa}$. $p_t=\dfrac{100\times1005}{1000\times490}=0.205\%$. From IS 456 Table 19 (M20), $\tau_c\approx0.33\ \text{MPa}$. Since $\tau_v\le\tau_c$, one-way shear OK.

Step 7 — Two-way (punching) shear. Critical perimeter at $d/2$ from column face; side $=400+490=890\ \text{mm}=0.89\ \text{m}$. Perimeter $b_0=4\times890=3560\ \text{mm}$. Punching force $V_u=q_u\,[\,2.6^2-0.89^2\,]=266.3\,[6.76-0.792]=266.3\times5.968=1589\ \text{kN}$.

$$\tau_v=\frac{V_u}{b_0 d}=\frac{1589\times10^3}{3560\times490}=0.911\ \text{MPa}$$

Permissible $\tau_c=k_s\,0.25\sqrt{f_{ck}}$, $k_s=1$ (square), $=0.25\sqrt{20}=1.118\ \text{MPa}$. Since $\tau_v=0.911<1.118$, punching shear OK.

Summary:

• Footing $2.6\ \text{m}\times2.6\ \text{m}\times0.55\ \text{m}$.
• Reinforcement 16 mm @ 200 mm c/c both ways (1005 mm²/m).
• One-way and two-way shear checks satisfied.

Given: $P_u=2500\ \text{kN}$, M25 ($f_{ck}=25\ \text{MPa}$), Fe415 ($f_y=415\ \text{MPa}$), target $p=1.5\%$.

Step 1 — Design equation (short axially loaded column, IS 456 Cl 39.3, with min eccentricity within limits):

$$P_u=0.4f_{ck}A_c+0.67f_yA_{sc}$$

where $A_c=A_g-A_{sc}$ and $A_{sc}=0.015A_g$, so $A_c=0.985A_g$.

$$2500\times10^3=0.4\times25\times0.985A_g+0.67\times415\times0.015A_g$$ $$=9.85A_g+4.171A_g=14.02A_g$$ $$A_g=\frac{2500\times10^3}{14.02}=178{,}300\ \text{mm}^2$$

Side $=\sqrt{178300}=422\ \text{mm}$. Provide a $450\times450\ \text{mm}$ column ($A_g=202500\ \text{mm}^2$).

Step 2 — Longitudinal steel for the provided section. Recompute with $A_g=202500$:

$$2500\times10^3=0.4\times25(202500-A_{sc})+0.67\times415\,A_{sc}$$ $$2500\times10^3=10\times202500-10A_{sc}+278.05A_{sc}$$ $$2500000=2025000+268.05A_{sc}$$ $$A_{sc}=\frac{475000}{268.05}=1772\ \text{mm}^2$$

Check minimum $0.8\%$ of $A_g=1620\ \text{mm}^2$ (OK). Provide 8–20 mm bars $=8\times314=2513\ \text{mm}^2$, or 4–25 mm = 1963 mm² (closer). Use 4–25 mm = 1963 mm² ($p=0.97\%$, > min, satisfies $A_{sc}$ needed 1772).

Step 3 — Lateral ties (IS 456 Cl 26.5.3.2). Tie diameter $\ge\max\left(\dfrac{\phi_{long}}{4},\ 6\ \text{mm}\right)=\max(25/4=6.25,\ 6)=6.25\Rightarrow$ use 8 mm ties. Spacing $\le\min$ of:

• least lateral dimension $=450\ \text{mm}$,
• $16\times\phi_{long}=16\times25=400\ \text{mm}$,
• $300\ \text{mm}$.

Governing $=300\ \text{mm}$. Provide 8 mm ties @ 300 mm c/c (close ties near lap/ends as required).

Summary:

• Column $450\times450\ \text{mm}$.
• Longitudinal: 4–25 mm bars (1963 mm²).
• Ties: 8 mm @ 300 mm c/c.

Given: $b_f=1000\ \text{mm}$, $D_f=120\ \text{mm}$, $b_w=300\ \text{mm}$, $d=600\ \text{mm}$, $A_{st}=2500\ \text{mm}^2$, M20, Fe415.

Step 1 — Assume NA in flange; find $x_u$. Equate tension and compression (treat flange like a rectangular section of width $b_f$):

$$0.36f_{ck}b_f x_u=0.87f_yA_{st}$$ $$0.36\times20\times1000\times x_u=0.87\times415\times2500$$ $$7200\,x_u=902{,}625\Rightarrow x_u=125.4\ \text{mm}$$

Since $x_u=125.4\ \text{mm} > D_f=120\ \text{mm}$, the neutral axis lies in the web — analyse as a T-section.

Step 2 — Check $D_f$ vs $x_u$ to choose flange stress block. For IS 456, when $D_f > 0.43x_u$ we use an equivalent flange depth $y_f=0.15x_u+0.65D_f$ (but only if $D_f/d \le 0.2$, here $120/600=0.2$, OK). We must first locate $x_u$ for the T-section.

Step 3 — Force balance for the T-section. Total tension $T=0.87f_yA_{st}=0.87\times415\times2500=902{,}625\ \text{N}$. Compression $=$ web block $+$ flange block:

$$C=0.36f_{ck}b_w x_u+0.45f_{ck}(b_f-b_w)y_f,\quad y_f=0.15x_u+0.65D_f$$

Set $C=T$:

$$0.36\times20\times300\,x_u+0.45\times20\times(1000-300)(0.15x_u+0.65\times120)=902625$$ $$2160x_u+9\times700\times(0.15x_u+78)=902625$$ $$2160x_u+6300(0.15x_u+78)=902625$$ $$2160x_u+945x_u+491400=902625$$ $$3105x_u=411225\Rightarrow x_u=132.4\ \text{mm}$$

Step 4 — Check ductility. $x_{u,max}=0.48d=288\ \text{mm}>132.4$, so the section is under-reinforced (steel yields). Good. Recompute $y_f=0.15\times132.4+0.65\times120=19.86+78=97.9\ \text{mm}$ (and $y_f<D_f=120$, consistent).

Step 5 — Ultimate moment of resistance (take moments of the two compressive blocks about tension steel):

$$M_u=0.36f_{ck}b_w x_u\left(d-0.42x_u\right)+0.45f_{ck}(b_f-b_w)y_f\left(d-\tfrac{y_f}{2}\right)$$

Web part: $0.36\times20\times300\times132.4\times(600-0.42\times132.4)$ $=2160\times132.4\times(600-55.6)=285{,}984\times544.4=155.7\times10^6\ \text{N·mm}$. Flange part: $0.45\times20\times700\times97.9\times(600-48.95)$ $=6300\times97.9\times551.05=616{,}770\times551.05=339.9\times10^6\ \text{N·mm}$.

$$M_u=155.7+339.9=495.6\ \text{kN·m}$$

Final answer: $\boxed{M_u\approx 495.6\ \text{kN·m}}$ (neutral axis in the web, under-reinforced section).

Given: $b=250$, $d=450$, $V_u=200\ \text{kN}$, $\tau_c=0.56\ \text{MPa}$, M20, Fe415, 2-legged 8 mm stirrups ($A_{sv}=2\times50.3=100.6\ \text{mm}^2$).

Step 1 — Nominal shear stress.

$$\tau_v=\frac{V_u}{bd}=\frac{200\times10^3}{250\times450}=1.778\ \text{MPa}$$

Step 2 — Check against maximum. For M20, $\tau_{c,max}=2.8\ \text{MPa}$. Since $\tau_v=1.778<2.8$, section size is adequate.

Step 3 — Shear carried by concrete.

$$V_{uc}=\tau_c\,bd=0.56\times250\times450=63{,}000\ \text{N}=63.0\ \text{kN}$$

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

Step 4 — Shear to be carried by stirrups.

$$V_{us}=V_u-V_{uc}=200-63.0=137.0\ \text{kN}$$

Step 5 — Stirrup spacing.

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.6\times450}{137{,}000}$$ $$=\frac{16{,}338{,}000}{137{,}000}=119.3\ \text{mm}$$

Step 6 — Maximum spacing checks. $\min(0.75d,\ 300)=\min(337.5,300)=300\ \text{mm}$. Minimum-steel spacing: $s_{v,max}=\dfrac{0.87f_yA_{sv}}{0.4b}=\dfrac{0.87\times415\times100.6}{0.4\times250}=\dfrac{36{,}322}{100}=363\ \text{mm}$. Governing design spacing $=119.3\ \text{mm}$.

Provide 8 mm 2-legged stirrups @ 110 mm c/c.

Final answer: $\boxed{8\ \text{mm 2-legged stirrups @ 110 mm c/c}}$.

Given: $\phi=20\ \text{mm}$, Fe415 ($f_y=415\ \text{MPa}$), $\tau_{bd}=1.2\ \text{MPa}$ (plain, M20).

Step 1 — Increase bond stress for deformed bars. For HYSD bars in tension, IS 456 allows a 60% increase:

$$\tau_{bd}=1.2\times1.6=1.92\ \text{MPa}$$

Step 2 — Development length formula.

$$L_d=\frac{\phi\,\sigma_s}{4\,\tau_{bd}}=\frac{\phi\times0.87f_y}{4\tau_{bd}}$$

Step 3 — Substitute.

$$L_d=\frac{20\times0.87\times415}{4\times1.92}=\frac{20\times361.05}{7.68}=\frac{7221}{7.68}=940\ \text{mm}$$

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

Final answer: $\boxed{L_d\approx 940\ \text{mm}\ (\approx 47\phi)}$.

Physical significance. Development length is the minimum embedded length of a bar required so that the bar can develop its full design stress ($0.87f_y$) through bond (adhesion, friction and mechanical interlock of ribs) between steel and concrete, before it can slip or be pulled out. It ensures the force in the bar is safely transferred to the surrounding concrete, preventing bond failure and guaranteeing the assumed flexural capacity is reached. It governs bar curtailment, lap splices and anchorage at supports.

Limit state method (LSM). A limit state is a state beyond which a structure or a structural element becomes unfit for its intended use. LSM is a semi-probabilistic design philosophy that aims to ensure, with an acceptable probability, that the structure will not reach any relevant limit state during its design life. It uses characteristic values of loads and material strengths together with partial safety factors to account for variability and uncertainty. It combines the safety of ultimate load (plastic) methods with the in-service control of working stress methods.

Limit state of collapse (ultimate). Concerns the safety of the structure against total or partial collapse. Sub-categories: flexure, shear, torsion, compression, bond and overall stability/buckling. Exceeding it endangers life.

Limit state of serviceability. Concerns performance under service (working) loads so the structure remains usable and durable. Sub-categories: deflection (limits e.g. span/250 for total deflection), cracking (crack width typically ≤ 0.3 mm for normal exposure), and vibration/durability.

Partial safety factors for loads ($\gamma_f$), IS 456 Table 18 (collapse / serviceability):

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

• Concrete: $\gamma_{mc}=1.5$, so design strength $=f_{ck}/1.5=0.667f_{ck}$.
• Steel: $\gamma_{ms}=1.15$, so design strength $=f_y/1.15=0.87f_y$.

These higher factors on concrete reflect its greater variability and the importance of brittle compression failure compared with the more ductile, better-controlled steel.

Given: $l=3.2\ \text{m}$, LL $=4.0$, FF $=1.0\ \text{kN/m}^2$, M20, Fe415.

Step 1 — Trial depth (deflection control). Simply supported basic span/depth $=20$; with modification factor $\approx1.4$ for under-reinforced slabs, $\dfrac{l}{d}\approx28$. $d=\dfrac{3200}{28}=114\ \text{mm}$. Take $d=115\ \text{mm}$; with 15 mm cover and 10 mm bars, $D=115+15+5=135\ \text{mm}$. Use $D=135\ \text{mm}$, $d=115\ \text{mm}$.

Step 2 — Loads (per m² , 1 m strip). Self weight $=25\times0.135=3.375\ \text{kN/m}^2$. Total service $=3.375+1.0+4.0=8.375\ \text{kN/m}^2$. Factored $w_u=1.5\times8.375=12.56\ \text{kN/m}$ (per m width).

Step 3 — Design moment.

$$M_u=\frac{w_u l^2}{8}=\frac{12.56\times3.2^2}{8}=\frac{12.56\times10.24}{8}=16.08\ \text{kN·m/m}$$

Step 4 — Check depth. $M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times1000\times115^2=36.5\times10^6=36.5\ \text{kN·m}>16.08$. OK.

Step 5 — Main steel.

$$\frac{4.6M_u}{f_{ck}bd^2}=\frac{4.6\times16.08\times10^6}{20\times1000\times115^2}=\frac{73.97\times10^6}{264.5\times10^6}=0.2797$$ $$A_{st}=\frac{0.5\times20}{415}\left[1-\sqrt{1-0.2797}\right]\times1000\times115$$ $$=0.02410\times(1-0.8487)\times115000=0.02410\times0.1513\times115000=419\ \text{mm}^2/\text{m}$$

Minimum $=0.12\%\times1000\times135=162\ \text{mm}^2$ (OK, 419 governs). Provide 10 mm @ 180 mm c/c → $\dfrac{78.5\times1000}{180}=436\ \text{mm}^2/\text{m}$.

Step 6 — Distribution steel. $=0.12\%\times1000\times135=162\ \text{mm}^2/\text{m}$. Provide 8 mm @ 250 mm c/c ($201\ \text{mm}^2$).

Summary: $D=135\ \text{mm}$; $M_u=16.08\ \text{kN·m/m}$; main steel 10 mm @ 180 c/c; distribution 8 mm @ 250 c/c.

Given: $G=250\ \text{mm}$, $R=160\ \text{mm}$, span $=3.0\ \text{m}$, waist $t=150\ \text{mm}$, LL $=3.0$, finishes $=1.0\ \text{kN/m}^2$, M20, Fe415.

Step 1 — Self weight of waist slab on slope, projected to plan. Inclined length per going $=\sqrt{G^2+R^2}=\sqrt{250^2+160^2}=\sqrt{62500+25600}=\sqrt{88100}=296.8\ \text{mm}$. Waist self weight on plan $=25\times t\times\dfrac{\sqrt{G^2+R^2}}{G}=25\times0.15\times\dfrac{296.8}{250}=25\times0.15\times1.187=4.45\ \text{kN/m}^2$.

Step 2 — Self weight of steps (triangular). $=25\times\dfrac{R}{2}=25\times\dfrac{0.16}{2}=2.0\ \text{kN/m}^2$.

Step 3 — Total load on going (per m² of plan).

$$w=4.45+2.0+1.0+3.0=10.45\ \text{kN/m}^2$$

Per metre width $w=10.45\ \text{kN/m}$. Factored $w_u=1.5\times10.45=15.68\ \text{kN/m}$.

Step 4 — Design moment (simply supported flight).

$$M_u=\frac{w_u l^2}{8}=\frac{15.68\times3.0^2}{8}=\frac{15.68\times9}{8}=17.64\ \text{kN·m/m}$$

Step 5 — Main steel. $d=150-15-5=130\ \text{mm}$ (15 cover, 10 mm bars).

$$\frac{4.6M_u}{f_{ck}bd^2}=\frac{4.6\times17.64\times10^6}{20\times1000\times130^2}=\frac{81.14\times10^6}{338\times10^6}=0.2401$$ $$A_{st}=\frac{0.5\times20}{415}\left[1-\sqrt{1-0.2401}\right]\times1000\times130$$ $$=0.02410\times(1-0.8718)\times130000=0.02410\times0.1282\times130000=402\ \text{mm}^2/\text{m}$$

Provide 10 mm @ 190 mm c/c → $\dfrac{78.5\times1000}{190}=413\ \text{mm}^2/\text{m}$. Distribution steel $=0.12\%\times1000\times150=180\ \text{mm}^2$ → 8 mm @ 250 c/c (201 mm²).

Final answer: $\boxed{M_u=17.64\ \text{kN·m/m},\ \text{main steel }10\ \text{mm @ 190 c/c}}$.

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

• Minimum: $\dfrac{A_s}{bd}=\dfrac{0.85}{f_y}$, i.e. for Fe415, $A_{s,min}=\dfrac{0.85}{415}bd=0.205\%\,bd$. This guards against sudden brittle failure once concrete cracks.
• Maximum: $A_{s,max}=0.04\,bD$ (4% of gross area) to avoid congestion and ensure proper compaction/anchorage.

(b) Nominal cover for durability (IS 456 Cl 26.4 / Table 16). Cover protects steel from corrosion and gives fire resistance. Typical nominal covers: mild exposure 20 mm, moderate 30 mm, severe 45 mm, very severe 50 mm, extreme 75 mm. For beams a minimum of 25–30 mm is common; for footings cast against earth, 50–75 mm. Cover must be at least the bar diameter.

(c) Spacing of main bars (IS 456 Cl 26.3).

• Minimum horizontal spacing $\ge$ the greatest of: bar diameter, or (max aggregate size + 5 mm) — typically $\ge$ aggregate size $+5\ \text{mm}$, ensuring concrete flows around bars.
• Maximum spacing in slabs: main steel $\le \min(3d,\ 300\ \text{mm})$; distribution steel $\le \min(5d,\ 450\ \text{mm})$. This controls cracking and ensures load distribution.

(d) Curtailment / anchorage at simple supports (IS 456 Cl 26.2.3).

• At a simple support, the positive (bottom) reinforcement should satisfy $\dfrac{M_1}{V}+L_o \ge L_d$, where $M_1$ is the moment of resistance of the bars continued, $V$ the shear at the support, $L_o$ the anchorage beyond the centre of support (the standard relaxation allows $1.3M_1/V + L_o \ge L_d$ for confined supports).
• At least one-third of the positive steel must extend into a simple support; bars must be anchored a length $\ge L_d/3$ (or with hooks). Curtailed bars must extend a distance $\ge \max(d,\ 12\phi)$ beyond the theoretical cut-off point.

Why detailing matters. Analysis assumes forces are transferred between concrete and steel exactly as designed; only correct detailing — adequate anchorage, cover, spacing, bar continuity, ductile reinforcement layout — makes that assumption true in the built structure. Poor detailing leads to bond/anchorage failure, corrosion, wide cracks, brittle behaviour and unsafe load paths, even when member sizes and steel areas are adequate. Detailing is what converts a correct calculation into a safe, durable structure, and is especially critical for ductility and energy dissipation in seismic zones such as Nepal.