BE Civil Engineering (IOE, TU) Design of RCC Structures (IOE, CE 702) Question Paper 2079 Nepal

Given data

• $b = 250\ \text{mm}$, $D = 450\ \text{mm}$, effective cover $= 50\ \text{mm}$ → $d = 450 - 50 = 400\ \text{mm}$
• $d' = 50\ \text{mm}$, $f_{ck} = 20\ \text{N/mm}^2$, $f_y = 415\ \text{N/mm}^2$
• $M_u = 180\ \text{kN·m}$

Step 1 — Limiting moment of resistance (singly reinforced)

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

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

$x_{u,max} = 0.48 \times 400 = 192\ \text{mm}$

$$M_{u,lim} = 0.36(20)(250)(192)\big(400 - 0.42\times192\big)$$ $$= 0.36(20)(250)(192)(400 - 80.64) = 345600 \times 319.36 = 110.37\times10^6\ \text{N·mm}$$ $$\boxed{M_{u,lim} = 110.37\ \text{kN·m}}$$

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

Step 2 — Tension steel for the limiting (balanced) part

$$A_{st1} = \frac{0.36\,f_{ck}\,b\,x_{u,max}}{0.87\,f_y} = \frac{0.36(20)(250)(192)}{0.87(415)} = \frac{345600}{361.05} = 957.3\ \text{mm}^2$$

Step 3 — Remaining moment

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

Step 4 — Compression steel

Depth ratio $d'/d = 50/400 = 0.125$. For Fe415 with $d'/d \approx 0.10\text{–}0.15$, take $f_{sc} \approx 353\ \text{N/mm}^2$ (from IS 456 stress-strain, value at $d'/d=0.125$).

$$A_{sc} = \frac{M_{u2}}{(f_{sc} - 0.45 f_{ck})(d - d')} = \frac{69.63\times10^6}{(353 - 0.45\times20)(400-50)}$$ $$= \frac{69.63\times10^6}{(353 - 9)(350)} = \frac{69.63\times10^6}{344\times350} = \frac{69.63\times10^6}{120400} = 578.3\ \text{mm}^2$$ $$\boxed{A_{sc} = 578.3\ \text{mm}^2}$$

Provide 3–16 mm bars $= 3\times201 = 603\ \text{mm}^2$ (> 578.3). OK.

Step 5 — Additional tension steel to balance compression steel

$$A_{st2} = \frac{A_{sc}(f_{sc} - 0.45 f_{ck})}{0.87 f_y} = \frac{578.3(344)}{0.87(415)} = \frac{198935}{361.05} = 551.0\ \text{mm}^2$$

Step 6 — Total tension steel

$$A_{st} = A_{st1} + A_{st2} = 957.3 + 551.0 = 1508.3\ \text{mm}^2$$ $$\boxed{A_{st} = 1508.3\ \text{mm}^2}$$

Provide 5–20 mm bars $= 5\times314 = 1570\ \text{mm}^2$ (> 1508.3). OK.

Detailing sketch (section)

   |<---- 250 ---->|
    ___________      d'=50
   | o   o   o |  <- 3-16 (compression)
   |           |
   |  D=450    |  d=400
   |           |
   | o o o o o |  <- 5-20 (tension)
    -----------

Final design: Tension steel 5–20 mm ($A_{st}=1570\ \text{mm}^2$); compression steel 3–16 mm ($A_{sc}=603\ \text{mm}^2$).

Step 1 — Trial depth (deflection control)

For a simply supported slab, basic span/depth $= 20$. Assume modification factor $\approx 1.4$ for tension steel.

$$d \approx \frac{\text{span}}{20\times1.4} = \frac{3600}{28} \approx 129\ \text{mm}$$

Adopt $d = 130\ \text{mm}$; with 15 mm cover and 10 mm bars, overall depth $D = 130 + 15 + 5 = 150\ \text{mm}$.

Step 2 — Effective span

Lesser of (clear span + d) and (clear span + wall width):

• $3.6 + 0.130 = 3.73\ \text{m}$
• $3.6 + 0.230 = 3.83\ \text{m}$

$$l_{eff} = 3.73\ \text{m}$$

Step 3 — Loads (per 1 m strip)

• Self weight $= 0.150 \times 25 = 3.75\ \text{kN/m}^2$
• Floor finish $= 1.0\ \text{kN/m}^2$
• Live load $= 3.0\ \text{kN/m}^2$
• Total service load $= 7.75\ \text{kN/m}^2$

$$w_u = 1.5 \times 7.75 = 11.625\ \text{kN/m}$$

Step 4 — Design moment

$$M_u = \frac{w_u\,l_{eff}^2}{8} = \frac{11.625\times3.73^2}{8} = \frac{11.625\times13.913}{8} = 20.22\ \text{kN·m}$$

Step 5 — Check depth

$$M_{u,lim} = 0.138\,f_{ck}\,b\,d^2 = 0.138(20)(1000)(130)^2 = 46.6\times10^6\ \text{N·mm} = 46.6\ \text{kN·m}$$

$M_u = 20.22 < 46.6$ kN·m → section is under-reinforced, depth OK.

Step 6 — Main steel

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

Solve: $20.22\times10^6 = 0.87(415)A_{st}(130)\left(1 - \frac{A_{st}(415)}{1000(130)(20)}\right)$

Let $A_{st}$ in mm². $0.87(415)(130) = 46936.5$.

$20.22\times10^6 = 46936.5\,A_{st}\left(1 - \frac{415 A_{st}}{2.6\times10^6}\right)$

First trial ignoring lever-arm term: $A_{st} \approx 20.22\times10^6/46936.5 = 430.8$. Refine: factor $= 1 - 415(431)/2.6\times10^6 = 1 - 0.0688 = 0.931$ → $A_{st} = 430.8/0.931 = 462.7\ \text{mm}^2$.

$$\boxed{A_{st} = 463\ \text{mm}^2/\text{m}}$$

Spacing of 10 mm bars ($a=78.5$ mm²): $s = 1000\times78.5/463 = 169.5\ \text{mm}$. Provide 10 mm @ 165 mm c/c ($A_{st,prov}=476\ \text{mm}^2$).

Max spacing $= \min(3d, 300) = \min(390,300)=300$ mm. OK.

Step 7 — Distribution steel

$$A_{st,dist} = 0.12\% \times b D = 0.0012\times1000\times150 = 180\ \text{mm}^2/\text{m}$$

8 mm bars ($a=50.3$ mm²): $s = 1000\times50.3/180 = 279\ \text{mm}$. Provide 8 mm @ 275 mm c/c.

Step 8 — Deflection check

$p_t = 100 A_{st,prov}/(bd) = 100(476)/(1000\times130) = 0.366\%$.

$f_s = 0.58 f_y (A_{st,req}/A_{st,prov}) = 0.58(415)(463/476) = 234\ \text{N/mm}^2$.

From IS 456 Fig.4, for $p_t=0.366\%$ and $f_s\approx234$, modification factor $k_t \approx 1.45$.

Permissible $(l/d) = 20\times1.45 = 29$. Actual $l/d = 3730/130 = 28.7 < 29$. Deflection OK.

Summary

• Slab thickness $D = 150\ \text{mm}$
• Main steel: 10 mm @ 165 mm c/c
• Distribution steel: 8 mm @ 275 mm c/c

Step 1 — Effective length and short-column check

Both ends held against rotation and position → $k = 0.65$ (IS 456 Table 28).

$$l_{ef} = 0.65 \times 3000 = 1950\ \text{mm}$$

Step 2 — Cross-section sizing

Use the short axially loaded column formula (with $A_{sc} = 0.015 A_g$):

$$P_u = 0.4 f_{ck} A_c + 0.67 f_y A_{sc} = 0.4 f_{ck}(A_g - A_{sc}) + 0.67 f_y A_{sc}$$

With $A_{sc} = 0.015 A_g$, $A_c = 0.985 A_g$:

$$P_u = 0.4(25)(0.985 A_g) + 0.67(415)(0.015 A_g) = (9.85 + 4.17) A_g = 14.02\,A_g$$ $$A_g = \frac{1600\times10^3}{14.02} = 114{,}123\ \text{mm}^2$$

Side $= \sqrt{114123} = 337.8\ \text{mm}$. Adopt $350\times350\ \text{mm}$ ($A_g = 122500\ \text{mm}^2$).

Step 3 — Slenderness check

$$\frac{l_{ef}}{B} = \frac{1950}{350} = 5.57 < 12 \Rightarrow \textbf{short column. OK}$$

Step 4 — Longitudinal steel (re-evaluate for adopted size)

$$P_u = 0.4(25)(A_g - A_{sc}) + 0.67(415)A_{sc}$$ $$1600\times10^3 = 10(122500 - A_{sc}) + 278.05 A_{sc}$$ $$1600000 = 1225000 - 10 A_{sc} + 278.05 A_{sc} = 1225000 + 268.05 A_{sc}$$ $$A_{sc} = \frac{1600000 - 1225000}{268.05} = \frac{375000}{268.05} = 1399\ \text{mm}^2$$

Minimum steel $= 0.8\% A_g = 980\ \text{mm}^2$; $1399 > 980$, OK.

$$\boxed{A_{sc} = 1399\ \text{mm}^2}$$

Provide 8–16 mm bars $= 8\times201 = 1608\ \text{mm}^2$ ($1.31\%$ of $A_g$, between 0.8% and 4%). OK.

Step 5 — Lateral ties

Tie diameter $= \max(\tfrac{1}{4}\times16,\ 6) = \max(4, 6) = 6\ \text{mm}$. Use 8 mm ties.

Pitch $= \min$ of:

• least lateral dimension $= 350\ \text{mm}$
• $16 \times \phi_{long} = 16\times16 = 256\ \text{mm}$
• $300\ \text{mm}$

Governing $= 256$ mm → provide 8 mm ties @ 250 mm c/c.

Detailing sketch

  350 x 350 column
   _______________
  | o   o   o   o |   8-16 mm bars
  |               |   one bar at each corner +
  | o           o |   intermediate bars tied
  |               |
  | o   o   o   o |   8 mm ties @ 250 c/c
   ---------------

Summary

• Section: $350\times350\ \text{mm}$
• Longitudinal: 8–16 mm ($A_{sc}=1608\ \text{mm}^2$)
• Ties: 8 mm @ 250 mm c/c

Step 1 — Plan area of footing

Add ~10% for self weight of footing. Total service load $= 1.10\times1200 = 1320\ \text{kN}$.

$$A_{req} = \frac{1320}{200} = 6.6\ \text{m}^2$$

Side $= \sqrt{6.6} = 2.57\ \text{m}$. Provide $2.6\ \text{m} \times 2.6\ \text{m}$ ($A = 6.76\ \text{m}^2$).

Step 2 — Net upward soil pressure (factored)

For structural design, use the factored column load (exclude footing self weight from soil-pressure used for shear/moment, applied over plan area):

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

Step 3 — One-way (flexural) shear check

Critical section is at distance $d = 0.5\ \text{m}$ from the column face.

Distance from footing edge to critical section:

$$a = \frac{B - b_{col}}{2} - d = \frac{2.6 - 0.4}{2} - 0.5 = 1.1 - 0.5 = 0.6\ \text{m}$$

Shear force (per full width $2.6$ m):

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

Nominal shear stress:

$$\tau_v = \frac{V_u}{B\,d} = \frac{415.4\times10^3}{2600\times500} = 0.319\ \text{N/mm}^2$$

For M20 with assumed $p_t \approx 0.25\%$, $\tau_c \approx 0.36\ \text{N/mm}^2$ (IS 456 Table 19).

$\tau_v = 0.319 < \tau_c = 0.36$. One-way shear safe.

Step 4 — Two-way (punching) shear check

Critical perimeter at $d/2 = 250\ \text{mm}$ from column face. Side of critical square:

$$b_0 = b_{col} + d = 400 + 500 = 900\ \text{mm} = 0.9\ \text{m}$$

Perimeter $= 4\times0.9 = 3.6\ \text{m} = 3600\ \text{mm}$.

Punching shear force = total upward force outside the critical area:

$$V_p = q_u\big(B^2 - b_0^2\big) = 266.3\,(2.6^2 - 0.9^2) = 266.3(6.76 - 0.81) = 266.3\times5.95 = 1584.5\ \text{kN}$$

Punching shear stress:

$$\tau_{v,p} = \frac{V_p}{b_0^{(perim)}\,d} = \frac{1584.5\times10^3}{3600\times500} = 0.880\ \text{N/mm}^2$$

Permissible punching stress:

$$\tau_{c,p} = k_s\,(0.25\sqrt{f_{ck}}),\quad k_s = 1\ (\beta_c = 1\ \text{for square column})$$ $$\tau_{c,p} = 0.25\sqrt{20} = 0.25\times4.472 = 1.118\ \text{N/mm}^2$$

$\tau_{v,p} = 0.880 < \tau_{c,p} = 1.118$. Punching shear safe.

Summary

Provide footing $2.6\ \text{m} \times 2.6\ \text{m}$, effective depth $500\ \text{mm}$ — adequate in shear.

Step 1 — Aspect ratio

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

Step 2 — Moment coefficients (IS 456 Table 27, simply supported, corners not held)

For $l_y/l_x = 1.25$:

• $\alpha_x = 0.084$ (short span)
• $\alpha_y = 0.059$ (long span)

Step 3 — Design moments

$$M_x = \alpha_x\,w_u\,l_x^2 = 0.084\times12\times4.0^2 = 0.084\times12\times16 = 16.13\ \text{kN·m}$$ $$M_y = \alpha_y\,w_u\,l_x^2 = 0.059\times12\times16 = 11.33\ \text{kN·m}$$

(Note: both use $l_x^2$ per IS 456.)

$$\boxed{M_x = 16.13\ \text{kN·m/m}, \quad M_y = 11.33\ \text{kN·m/m}}$$

Step 4 — Check depth adequacy (short span)

$$M_{u,lim} = 0.138 f_{ck} b d_x^2 = 0.138(20)(1000)(150)^2 = 62.1\times10^6\ \text{N·mm} = 62.1\ \text{kN·m}$$

$M_x = 16.13 < 62.1$. Depth OK (under-reinforced).

Step 5 — Short-span steel

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

$0.87(415)(150) = 54157.5$.

First trial: $A_{st} \approx 16.13\times10^6 / 54157.5 = 297.8\ \text{mm}^2$.

Lever-arm factor: $1 - \dfrac{415(298)}{1000(150)(20)} = 1 - \dfrac{123670}{3\,000\,000} = 1 - 0.0412 = 0.959$.

$A_{st} = 297.8 / 0.959 = 310.5\ \text{mm}^2$.

$$\boxed{A_{st,x} = 311\ \text{mm}^2/\text{m}}$$

Minimum steel $= 0.12\%\,bD = 0.0012\times1000\times175 = 210\ \text{mm}^2$; $311 > 210$, OK.

Using 10 mm bars ($a=78.5$ mm²): $s = 1000\times78.5/311 = 252\ \text{mm}$. Provide 10 mm @ 250 mm c/c in the short span ($A_{st,prov}=314\ \text{mm}^2$).

Max spacing $= \min(3d, 300) = \min(450,300)=300$ mm. OK.

Summary

• $M_x = 16.13\ \text{kN·m/m}$, $M_y = 11.33\ \text{kN·m/m}$
• Short-span steel: 10 mm @ 250 mm c/c

Step 1 — Nominal shear stress

$$\tau_v = \frac{V_u}{b\,d} = \frac{250\times10^3}{300\times500} = 1.667\ \text{N/mm}^2$$

Step 2 — Check against maximum permissible

For M20, $\tau_{c,max} = 2.8\ \text{N/mm}^2$.

$\tau_v = 1.667 < 2.8$. Section size adequate (no crushing of web concrete).

Step 3 — Shear to be carried by stirrups

Since $\tau_v = 1.667 > \tau_c = 0.62$, shear reinforcement is required.

$$V_{us} = (\tau_v - \tau_c)\,b\,d = (1.667 - 0.62)(300)(500) = 1.047\times150000 = 157{,}050\ \text{N} = 157.05\ \text{kN}$$

Step 4 — Spacing of stirrups

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

$$s_v = \frac{0.87 f_y A_{sv} d}{V_{us}} = \frac{0.87(415)(100.6)(500)}{157050} = \frac{18{,}160{,}815}{157050} = 115.6\ \text{mm}$$

Step 5 — Maximum spacing limits

• $0.75 d = 0.75\times500 = 375\ \text{mm}$
• $300\ \text{mm}$
• Minimum-shear-reinforcement spacing: $s_{v} \le \dfrac{0.87 f_y A_{sv}}{0.4\,b} = \dfrac{0.87(415)(100.6)}{0.4(300)} = \dfrac{36321}{120} = 302.7\ \text{mm}$

Governing maximum $= 300\ \text{mm}$. Design spacing $115.6$ mm governs.

Result

Provide 8 mm two-legged stirrups @ 110 mm c/c near supports (round down from 115.6 mm), increasing spacing toward mid-span where shear reduces (up to 300 mm max).

Development length formula

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

where $\sigma_s = 0.87 f_y$ (stress at yield, LSM), $\tau_{bd}$ is the design bond stress.

Step 1 — Design bond stress (tension)

For M20 plain bars, IS 456 Table gives $\tau_{bd} = 1.2\ \text{N/mm}^2$. For deformed bars in tension, increase by 60%:

$$\tau_{bd} = 1.2 \times 1.6 = 1.92\ \text{N/mm}^2$$

Step 2 — Development length in tension

$$\sigma_s = 0.87\times415 = 361.05\ \text{N/mm}^2$$ $$L_d = \frac{20\times361.05}{4\times1.92} = \frac{7221}{7.68} = 940.2\ \text{mm}$$ $$\boxed{L_d \approx 940\ \text{mm} \approx 47\,\phi}$$

Step 3 — Development length in compression

For bars in compression, the design bond stress is increased by a further 25%:

$$\tau_{bd,comp} = 1.92\times1.25 = 2.4\ \text{N/mm}^2$$ $$L_{d,comp} = \frac{20\times361.05}{4\times2.4} = \frac{7221}{9.6} = 752.2\ \text{mm} \approx 38\,\phi$$

Comment

The development length in compression ($\approx 752$ mm, $38\phi$) is about 20% shorter than in tension ($\approx 940$ mm, $47\phi$), because the higher permissible bond stress under compression reduces the embedment needed to transfer the bar force to the surrounding concrete.

Step 1 — Inclined length per going

$$\text{slope length} = \sqrt{G^2 + R^2} = \sqrt{250^2 + 160^2} = \sqrt{62500 + 25600} = \sqrt{88100} = 296.8\ \text{mm}$$

Step 2 — Self weight of waist slab on plan

Waist slab weight along the slope, projected on plan:

$$w_{waist} = \gamma\times t\times\frac{\sqrt{G^2+R^2}}{G} = 25\times0.150\times\frac{296.8}{250}$$ $$= 25\times0.150\times1.187 = 4.45\ \text{kN/m}^2\ \text{(of plan area)}$$

Step 3 — Weight of steps (triangular)

Weight of one step per metre width, smeared over the going:

$$w_{steps} = \gamma\times\frac{R}{2} = 25\times\frac{0.160}{2} = 25\times0.080 = 2.0\ \text{kN/m}^2$$

Step 4 — Total dead load (on plan)

$$w_{DL} = w_{waist} + w_{steps} + \text{finishes} = 4.45 + 2.0 + 0.6 = 7.05\ \text{kN/m}^2$$

Step 5 — Total service and factored load

$$w_{service} = w_{DL} + LL = 7.05 + 3.0 = 10.05\ \text{kN/m}^2$$ $$w_u = 1.5\times10.05 = 15.08\ \text{kN/m}^2$$ $$\boxed{w_u \approx 15.1\ \text{kN/m}^2\ \text{(on plan)}}$$

This factored load (kN/m² of horizontal plan area) is used with the horizontal span of the flight to compute the bending moment for design of the waist slab.

Step 1 — Effective flange width (isolated T-beam, IS 456 Cl. 23.1.2)

$$b_f = \frac{l_0}{\dfrac{l_0}{b} + 4} + b_w$$

where $l_0 = 6000\ \text{mm}$ (simply supported span), actual flange $b = 1500\ \text{mm}$.

$$b_f = \frac{6000}{\dfrac{6000}{1500} + 4} + 250 = \frac{6000}{4 + 4} + 250 = \frac{6000}{8} + 250 = 750 + 250 = 1000\ \text{mm}$$

But $b_f$ must not exceed the actual width $b = 1500\ \text{mm}$. So:

$$\boxed{b_f = 1000\ \text{mm}}$$

Step 2 — Depth of neutral axis (assume within flange)

Equate compression and tension:

$$0.36 f_{ck} b_f x_u = 0.87 f_y A_{st}$$ $$x_u = \frac{0.87 f_y A_{st}}{0.36 f_{ck} b_f} = \frac{0.87(415)(1600)}{0.36(20)(1000)} = \frac{577680}{7200} = 80.2\ \text{mm}$$

Step 3 — Verify assumption

$x_u = 80.2\ \text{mm} < D_f = 120\ \text{mm}$. NA lies within the flange → behaves as a rectangular section of width $b_f$. Assumption valid.

Also check $x_{u,max} = 0.48 d = 0.48\times500 = 240\ \text{mm}$; $x_u = 80.2 < 240$ → under-reinforced. OK.

Step 4 — Moment of resistance

$$M_u = 0.87 f_y A_{st}\,(d - 0.42 x_u) = 0.87(415)(1600)\big(500 - 0.42\times80.2\big)$$ $$= 577680\,(500 - 33.7) = 577680\times466.3 = 269.4\times10^6\ \text{N·mm}$$ $$\boxed{M_u = 269.4\ \text{kN·m}}$$

Limit state method (LSM)

The limit state method is a design philosophy in which a structure is designed so that it does not reach any of the relevant limit states — conditions beyond which the structure or a member ceases to fulfil the function for which it was designed. It is a semi-probabilistic approach using characteristic loads/strengths and partial safety factors, combining the safety of the ultimate load method with the serviceability of the working stress method.

Limit state of collapse (ultimate)

Concerned with the strength and stability of the structure under the worst factored loads. The structure must not collapse. Sub-categories: flexure, shear, torsion, compression, bond. The probability of reaching this state must be acceptably low.

Limit state of serviceability

Concerned with the performance under normal (service) loads — the structure must remain fit for use. Includes limits on:

• Deflection (≈ span/250 final deflection)
• Cracking (crack width typically ≤ 0.3 mm in normal exposure)
• Vibration, durability, etc.

Partial safety factors (IS 456:2000)

For loads ($\gamma_f$):

For materials ($\gamma_m$, collapse):

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

These factors account for variability in materials, loads, workmanship, and the consequences of failure.

1. Nominal cover

Clear cover protects steel from corrosion and provides fire resistance.

• For beams (moderate exposure): nominal cover $= 30\ \text{mm}$ (mild $=20$ mm, severe $=45$ mm).
• Cover shall not be less than the diameter of the bar.

2. Minimum and maximum tension steel

Minimum (IS 456 Cl. 26.5.1.1):

$$\frac{A_{st,min}}{b\,d} = \frac{0.85}{f_y} \Rightarrow A_{st,min} = \frac{0.85\,b\,d}{f_y}$$

For Fe415, this gives $p_{t,min} \approx 0.205\%$.

Maximum tension (and compression) steel: $A_{st,max} = 0.04\,b\,D$ (4% of gross area), to avoid congestion and ensure proper compaction.

3. Spacing of reinforcing bars

To allow concrete to flow and bond:

• Minimum horizontal spacing $= \max(\text{bar diameter},\ d_{agg} + 5\ \text{mm})$ where $d_{agg}$ is the maximum aggregate size.
• Minimum vertical spacing $= \max(15\ \text{mm},\ \tfrac{2}{3}d_{agg},\ \text{bar diameter})$.
• Maximum spacing limits control cracking (see Cl. 26.3.3).

4. Curtailment and anchorage

• Bars are curtailed where no longer required for moment, but extended beyond the theoretical cut-off point by $\max(d,\ 12\phi)$.
• At simple supports, positive-moment steel must satisfy:

$$\frac{M_1}{V} + L_0 \ge L_d$$

where $M_1$ is moment capacity of bars at the support, $V$ is shear, $L_0$ is anchorage beyond centre of support, and $L_d$ is development length.

• At least one-third of positive-moment steel must extend into a simple support.

These requirements ensure strength, ductility, durability, and crack control in the beam.