Design of RCC Structures: Model Question 1

Question 1(a): Characteristic Load and Design Load

Define the characteristic load and design load used in structural design. [2 marks]

Characteristic load, F: the load value with a 95% probability of not being exceeded during the life of the structure. (IS 456:2000, Clause 36.2) For design under IS 456, the specified dead, imposed, wind and snow loads and the seismic forces from the relevant loading standards are adopted as characteristic loads.

Design load, F_d: the characteristic load multiplied by the partial safety factor appropriate to the nature of the loading and the limit state being checked:

F_d = γ_fF. (IS 456:2000, Clause 36.3.2)

The factor is not always 1.5: use the applicable combination and limit state. (IS 456:2000, Clause 36.4.1 and Table 18) An action already described as ultimate or factored in this paper must not be multiplied by 1.5 again.

Question 1(b): Transverse Reinforcement for Shear and Torsion

A reinforced concrete (RC) rectangular beam has a cross-section of 300 mm × 500 mm. The beam is reinforced with 4 bars of 16 mm diameter on the tension side and 2 bars of 12 mm diameter on the compression side. If the section is subjected to an ultimate bending moment of 60 kN·m, an ultimate twisting moment of 30 kN·m, and an ultimate shear force of 80 kN, design the required transverse (shear) reinforcement. Assume an effective cover of 50 mm. Use M20 grade concrete and Fe 415 steel. [7 marks]

Method and detailing basis: use the equivalent-action and closed-hoop method in Bhavikatti, Chapter 6, §6.13 and Example 6.7 (printed pp. 128–131 / supplied PDF pp. 135–138), checked against IS 456 §41. Take the given 50 mm effective cover to the corner-bar centres at both faces and sides: b₁ = 200 mm and d₁ = 400 mm. These are longitudinal-bar centre distances, not the concrete section dimensions.

1. Effective Section and Equivalent Shear

b = 300 mm; D = 500 mm; d = 500 − 50 = 450 mm.
A_st = 4π416² = 804.25 mm²; A_sc = 2π412² = 226.19 mm².
p_t = 100A_stbd = 0.596%.
V_e = V_u + 1.6T_ub (IS 456:2000, Clause 41.3.1)
= 80 + 1.6(30)0.300 = 240 kN.
τ_ve = V_ebd = 240,000300 × 450 = 1.778 N/mm². (IS 456:2000, Clause 41.3.1)

τ_c,max = 2.8 N/mm² for M20. (IS 456:2000, Table 20) Hence 1.778 < 2.8: the concrete section is not prohibited by the maximum shear-stress limit. (IS 456:2000, Clause 41.3.1)

For M20: at p_t = 0.50%, τ_c = 0.48; at 0.75%, τ_c = 0.56 N/mm². (IS 456:2000, Table 19)
τ_c = 0.48 + 0.59574 − 0.500.25(0.56 − 0.48) = 0.5106 N/mm².

2. Check Both Transverse-Steel Expressions

A_sv denotes the sum of the two effective leg areas of one closed hoop. Both of the following expressions must be satisfied:

A_svs ≥ T_ub₁d₁ × 0.87f_y + V_u2.5d₁ × 0.87f_y (IS 456:2000, Clause 41.4.3)
= 30 × 10⁶200 × 400 × 361.05 + 80,0002.5 × 400 × 361.05
= 1.2602 mm²/mm.
Also, A_svs ≥ (τ_ve − τ_c)b0.87f_y (IS 456:2000, Clause 41.4.3)
= (1.77778 − 0.51064)300361.05 = 1.0529 mm²/mm.

For two-legged 8 mm links, A_sv = 2π48² = 100.53 mm².
s ≤ 100.531.2602 = 79.77 mm. Adopt s = 75 mm.
Provided A_svs = 100.5375 = 1.3404 > 1.2602 mm²/mm.

The ordinary minimum is 0.4b0.87f_y = 0.3324 mm²/mm, also satisfied. (IS 456:2000, Clause 26.5.1.6) The shear spacing limit is min(0.75d, 300) = 300 mm. (IS 456:2000, Clause 26.5.1.5) The torsion limit is min[x₁, x₁ + y₁4, 300], where x₁, y₁ are the hoop dimensions. (IS 456:2000, Clause 26.5.1.7(a)) Even using the smaller enclosed-bar dimensions 200 and 400 mm gives a conservative limit of 150 mm. Thus 75 mm satisfies both spacing limits.

3. Longitudinal and Side-Face Detailing

M_t = T_u(1 + Db)1.7 (IS 456:2000, Clause 41.4.2)
= 30(1 + 500300)1.7 = 47.06 kN·m.
M_e1 = M_u + M_t = 60 + 47.06 = 107.06 kN·m. (IS 456:2000, Clause 41.4.2)
M_t < M_u, so additional opposite-face steel for a positive M_e2 is not required. (IS 456:2000, Clause 41.4.2.1)

As a completeness check, analysis of the given doubly reinforced section by the strain-compatible method used in Question 2(b) gives x_u = 102.21 mm and M_R = 117.71 kN·m > 107.06 kN·m. The given longitudinal reinforcement is adequate for the equivalent moment; it must still be enclosed and anchored correctly.

Because D = 500 > 450 mm, additional side-face reinforcement is required. (IS 456:2000, Clause 26.5.1.7(b)) Required total side-face area = 0.001bD = 150 mm², shared equally between the two faces. (IS 456:2000, Clause 26.5.1.3) Add one 10 mm longitudinal bar at mid-depth on each side: total 157.08 mm². The 200 mm vertical intervals between corner and middle bars are below min(300, b) = 300 mm. (IS 456:2000, Clause 26.5.1.3) Enclose all these bars within the closed hoops; properly develop the side bars beyond the torsion region.

Provide two-legged 8 mm closed rectangular hoops at 75 mm c/c throughout the region carrying the stated actions, enclosing 4×16 mm bottom bars, 2×12 mm top bars and one additional 10 mm side-face bar on each side. No beam span was supplied, so this is a sectional spacing design, not an invented bar count.

This is an IS 456 torsion design, not a substitute for the capacity-design shear check of a seismic beam.

Question 1(c): Critical Neutral-Axis Factor in WSM

Derive from first principles the expression for the critical neutral axis depth factor (k) for a rectangular singly reinforced section using the Working Stress Method (WSM). [3 marks]

At the critical or balanced working-stress section, the extreme concrete compressive stress and the tensile steel stress reach their respective permissible values, σ_cbc and σ_st, simultaneously. This is a permissible-stress condition, not a physical ultimate-failure state.

Let the effective depth be d, the neutral-axis depth x = kd, and m = E_sE_c be the modular ratio used by WSM. Plane sections remain plane, bond is perfect and the materials are treated as linear elastic at working loads. (IS 456:2000, Clause B-1.3) Hence the similar strain triangles give:

ε_cε_s = xd − x = k1 − k.
ε_c = σ_cbcE_c; ε_s = σ_stE_s.
Therefore, mσ_cbcσ_st = k1 − k.
mσ_cbc(1 − k) = kσ_st.

k = mσ_cbcσ_st + mσ_cbc = 11 + σ_stmσ_cbc. (Derived from IS 456:2000, Clause B-1.3)

For IS 456 WSM, use the code's effective modular ratio m = 2803σ_cbc, with stresses in N/mm². (IS 456:2000, Clause B-1.3(d)) It incorporates long-term effects and should not be replaced by E_s5000√f_ck in this derivation. Use the permissible concrete and steel stresses from the code tables. (IS 456:2000, Tables 21 and 22)

Force equilibrium: C = 12σ_cbcbx = A_stσ_st.
Lever arm: z = d − x3 = d(1 − k3). (Elastic triangular-stress-block derivation; Manual, PDF p. 7)

Question 2(a): Anchorage Bond, Flexural Bond and End Anchorage

Define Anchorage bond and Flexure bond and derive the equation L_d ≤ 1.3M₁V + L_o, where the symbols have their usual meaning. [4 marks]

Anchorage bond transfers the force in a reinforcing bar to the surrounding concrete over an embedded length, preventing withdrawal or end slip. Flexural bond transfers the change in longitudinal steel force caused by the variation of bending moment along a member. They describe different reasons for the same steel–concrete interface shear, not two independent strengths to add.

1. Development Length from Force Equilibrium

Force in one bar = πφ²4σ_s.
Bond resistance over L_d = πφL_dτ_bd.
Equating them: L_d = φσ_s4τ_bd. (IS 456:2000, Clause 26.2.1)

2. Flexural-Bond Relation

For a short length dx with approximately constant internal lever arm z, M = Tz. Thus dT = dMz = V dxz. If u is the sum of the bar perimeters, bond equilibrium gives uτ_bdx = dT:

τ_b = Vuz. (Derived from flexural equilibrium; Manual, PDF p. 77)
M₁ = A_sσ_sz; L_d = A_sσ_suτ_bd.
Therefore, L_dM₁V = Vuzτ_bd = τ_bτ_bd.

Consequently, the basic bond condition τ_b ≤ τ_bd corresponds to L_d ≤ M₁V. At a simple support or point of inflection, the available continuation/anchorage L_o is credited by the code:

L_d ≤ M₁V + L_o. (IS 456:2000, Clause 26.2.3.3(c))

L_d ≤ 1.3M₁V + L_o, only when the reinforcement ends are confined by a compressive reaction. (IS 456:2000, Clause 26.2.3.3(c), 30% allowance)

The 1.3 coefficient is a code allowance, not a number derived from elastic equilibrium alone. Without that confinement, retain the coefficient 1.0. This check also limits the bar diameter, because L_d increases with φ.

• L_d: development length corresponding to the design bar stress; σ_s = 0.87f_y at full design strength in LSM. (IS 456:2000, Clause 26.2.3.3(c))
• M₁: moment resistance with all reinforcement at the section assumed stressed to its design stress; it is not the zero applied moment at the support. (IS 456:2000, Clause 26.2.3.3(c))
• V: design shear force at the section.
• L_o: at a simple support, the anchorage beyond the support centre plus equivalent anchorage of hooks or mechanical anchorage. At an inflection point, the credited L_o is limited to max(d, 12φ). (IS 456:2000, Clause 26.2.3.3(c))

Question 2(b): Ultimate Moment Resistance of a Doubly Reinforced Beam

Calculate the ultimate moment of resistance (M_u) of a rectangular RC beam measuring 250 mm × 450 mm (overall depth). The beam is reinforced with 3–20 mm diameter bars on the tension side and 2–16 mm diameter bars on the compression side. Assume an effective cover of 50 mm on both sides. Use M20 concrete and Fe 415 steel. [8 marks]

Reference method: Subramanian, §5.6.2, printed pp. 165–166 / PDF pp. 179–180, and Example 5.15, printed pp. 195–196 / PDF pp. 209–210; Manual, doubly reinforced analysis, printed pp. 38–39 / PDF p. 21. Enforce strain compatibility and subtract the concrete displaced by compression steel. Do not use a limiting-section compression-steel table as though x_u were already known.

1. Section Properties

b = 250 mm; D = 450 mm; d = 400 mm; d′ = 50 mm.
A_st = 3π420² = 942.48 mm².
A_sc = 2π416² = 402.12 mm².
x_u,max = 0.48d = 192 mm. (IS 456:2000, Clause 38.1, limiting neutral-axis values)
f_st = 0.87(415) = 361.05 N/mm². (IS 456:2000, Clause 38.1 and Figure 23A)
T = f_stA_st = 340.282 kN.

2. Equilibrium with Compatible Compression-Steel Stress

ε_sc = 0.0035(1 − d′x_u). (Derived from IS 456:2000, Clause 38.1(a)–(b))
C = 0.36f_ckbx_u + (f_sc − f_cc)A_sc = T. (IS 456:2000, Clause 38.1; net-steel equilibrium: Subramanian §5.6.2)
Here f_sc comes from the design steel curve at ε_sc; f_cc is the concrete stress at the same level.

Using the rounded Fe415 stress–strain table, the relevant points are (ε, f) = (0.00192, 324.8) and (0.00241, 342.8), with f in N/mm². (Subramanian Table 5.2, based on IS 456:2000, Figure 23A) At convergence ε_sc > 0.002, so the manual's rounded concrete plateau f_cc = 0.446f_ck = 8.92 N/mm² applies. (IS 456:2000, Clause 38.1 and Figure 21; Manual rounding) Using 0.447 instead changes only the insignificant final rounding.

In the first trial, ε_sc < 0.002, so the parabolic concrete stress is used instead of its plateau. At the converged axis:

f_sc = 324.8 + 0.00201331 − 0.001920.00049(342.8 − 324.8) = 328.228 N/mm².
C_c,gross = 0.36(20)(250)(117.711) = 211.880 kN.
C_s,net = (328.228 − 8.92)(402.124)1000 = 128.401 kN.
C_c,gross + C_s,net = 340.282 kN = T.

3. Verify the Tensile Assumption

x_u = 117.711 < 192 mm.
ε_st = 0.0035(400 − 117.711)117.711 = 0.008394. (Strain compatibility, IS 456:2000, Clause 38.1)
Required minimum tensile strain = 0.002 + 4151.15 × 200,000 = 0.003804. (IS 456:2000, Clause 38.1(f))
0.008394 > 0.003804: the under-reinforced assumption is valid.

The compression steel stress is only about 328.2 N/mm², not 361.05 N/mm². Assuming it fully yields would fail strain compatibility and give the wrong neutral axis.

4. Moment Resistance

M_R = 0.36f_ckbx_u(d − 0.42x_u) + (f_sc − f_cc)A_sc(d − d′) (IS 456:2000, Annex G-1.2; net-steel form: Subramanian §5.6.2)
= 211.880(400 − 0.42 × 117.711)1000 + 128.401(400 − 50)1000
= 119.217 kN·m.

Ultimate design moment resistance M_u ≈ 119.2 kN·m. This is an analysis of the supplied reinforcement, not the limiting moment of a hypothetical balanced section.

Question 3(a): Control of Deflection and Cracking

Explain control of Deflection and control of Cracking in RCC structure design. [2 marks]

Deflection control: choose adequate member depth and stiffness so that service-load deformation does not damage partitions or finishes or impair use. Final deflection, including creep, shrinkage and temperature effects, is normally limited to span250. (IS 456:2000, Clause 23.2(a)) Deflection occurring after partitions or finishes are installed is normally limited to the smaller of span350 and 20 mm. (IS 456:2000, Clause 23.2(b)) Basic span-to-effective-depth ratios are 7 for cantilevers, 20 for simply supported members and 26 for continuous members, modified for reinforcement and other applicable conditions. (IS 456:2000, Clause 23.2.1 and Figures 4–6) Otherwise calculate deflection explicitly. (IS 456:2000, Clause 42.1 and Annex C)

Crack control: distribute sufficient reinforcement with suitable bar diameters, spacing and cover; detail anchorage, restraint and construction joints appropriately, and ensure proper placing and curing. Many well-distributed narrow cracks are preferable to a few wide cracks. The general surface-width limits are 0.3 mm where cracking is not harmful, 0.2 mm where exposure, moisture or contact with soil makes tensile cracking harmful, and 0.1 mm for particularly aggressive environments. (IS 456:2000, Clause 35.3.2) When a specific width must be demonstrated, calculate it rather than assuming that strength alone controls it. (IS 456:2000, Clause 43.1 and Annex F)

Question 3(b): Restrained Two-Way Slab with Two Adjacent Edges Discontinuous

Design a two-way reinforced concrete slab panel with two adjacent edges discontinuous and a clear span of 4.0 m × 6.0 m. The slab rests on 230 mm thick masonry walls and is subjected to a service live load of 3 kN/m² and a floor finish load of 1 kN/m². The corners of the slab are held down to prevent lifting. Use M20 grade concrete and Fe 415 steel to perform all necessary structural design checks and sketch the complete reinforcement detailing. [10 marks]

Reference-based design basis: follow Subramanian §10.3, especially printed pp. 371–374 / PDF pp. 385–388, and IS 456 Annex D. The two remaining edges are continuous into adjoining slab panels; assume approximately equal adjoining spans and loads, as required for the coefficient method. The masonry walls provide vertical bearing; the stated continuity and corner hold-down must come from the continuous slab and its supporting/tie system, not from unreinforced masonry tension. Assume an internal slab in mild exposure and nominal cover 20 mm. (IS 456:2000, Table 16) Take normal-weight RC at 25 kN/m³. Select a trial depth, then verify it below. No partition load beyond the stated loads is included.

1. Trial Depth, Effective Spans and Loads

Adopt D = 175 mm and 10 mm main bars.
Short-span bars form the outer bottom layer: d_x = 175 − 20 − 102 = 150 mm.
Long-span bars sit above them: d_y = 175 − 20 − 10 − 102 = 140 mm.
l_x = min(4.000 + 0.150, 4.000 + 0.230) = 4.150 m. (IS 456:2000, Clause 22.2(a)–(b))
l_y = min(6.000 + 0.140, 6.000 + 0.230) = 6.140 m. (IS 456:2000, Clause 22.2(a)–(b))
l_yl_x = 1.4795 < 2: two-way action. (IS 456:2000, Clause D-1.11)
Self-weight = 25(0.175) = 4.375 kN/m².
w_k = 4.375 + 1 + 3 = 8.375 kN/m².
w_u = 1.5(8.375) = 12.5625 kN/m². (IS 456:2000, Table 18, DL + LL)

2. Design Moments: Table 26, Case 4

Use the two adjacent edges discontinuous case. (IS 456:2000, Table 26, Case 4) Interpolate the short-span coefficients between ratios 1.4 and 1.5 with fraction 1.479518 − 1.40.1 = 0.79518.

M_x = α_xw_ul_x²; M_y = α_yw_ul_x². (IS 456:2000, Clause D-1.1)
Both directions use w_ul_x² = 216.3577 kN, even for the long-span moment.

The coefficients apply to the middle three-quarters of each panel width. Each edge strip is one-eighth of that width. (IS 456:2000, Clauses D-1.2–D-1.3 and Figure 25) To simplify detailing, continue the selected bottom mesh across the full panel, and provide the stated top support bars across the corresponding support length. This also exceeds the edge-strip minimum. (IS 456:2000, Clause D-1.7)

3. Flexural Reinforcement

For a 1000 mm strip, M_u = 0.36f_ckbx_u(d − 0.42x_u). (IS 456:2000, Annex G-1.1)
A_st = 0.36f_ckbx_u0.87f_y. (Force equilibrium, IS 456:2000, Annex G-1.1)
Limiting moment for d = 150 mm: 0.138(20)(1000)(150²)10⁶ = 62.1 kN·m/m, far above the demand. (IS 456:2000, Annex G-1.1, Fe415 limit)
Minimum steel in each direction = 0.0012(1000)(175) = 210 mm²/m. (IS 456:2000, Clause 26.5.2.1)

The x top bars are nearest the top face; the y top bars are the inner layer where the top bars cross. All main bar spacings satisfy min(3d, 300 mm) = 300 mm. (IS 456:2000, Clause 26.3.3(b)(1)) Diameter 10 mm < D8 = 21.875 mm. (IS 456:2000, Clause 26.5.2.2) Thus the steel area, bar-size and crack-control spacing requirements are satisfied for the stated ordinary internal exposure.

4. One-Way Shear Checks in Both Directions

For a conservative strip check, assign the entire load to each direction in turn and use the fixed-propped continuous-end reaction 5w_ul8, without reducing to the critical section. This is a structural-analysis assumption, not a coefficient quoted from IS 456. In both directions it exceeds w_ul2 + M_negativel from the adopted Table 26 moments. It ignores beneficial two-way load sharing; it does not assert that both directions simultaneously carry the whole load.

V_x = 5w_ul_x8 = 32.584 kN/m; V_y = 5w_ul_y8 = 48.209 kN/m.
τ_vx = 32,5841000 × 150 = 0.217 N/mm². (IS 456:2000, Clause 40.1)
τ_vy = 48,2091000 × 140 = 0.344 N/mm². (IS 456:2000, Clause 40.1)
p_x = 0.2094%; p_y = 0.1870%.
Interpolation for M20 gives τ_cx = 0.3276 and τ_cy = 0.3096 N/mm². (IS 456:2000, Table 19)
For D = 175 mm, k = 1.25. (IS 456:2000, Clause 40.2.1.1)
0.217 < 1.25(0.3276) = 0.4095; 0.344 < 1.25(0.3096) = 0.3870 N/mm².

No shear reinforcement is required. Both stresses also lie below the solid-slab maximum, 12τ_c,max = 1.4 N/mm². (IS 456:2000, Clause 40.2.3 and Table 20) This is a wall-supported slab with no column punching perimeter or concentrated load in the question, so a flat-slab punching check does not arise.

5. Completed Deflection Check

Use the shorter span for a two-way slab. (IS 456:2000, Clause 24.1) At short-span mid-span:

f_s = 0.58f_yA_st,requiredA_st,provided (IS 456:2000, Figure 4)
= 0.58(415)228.60314.16 = 175.15 N/mm²; p_t = 0.2094%.

Use the more adverse rounded point p_t = 0.25% and f_s = 240 N/mm². Its ordinate is above 1.5; take the deliberately read-down modification factor 1.5. (IS 456:2000, Figure 4) Even using the conservative basic ratio 20 rather than claiming the continuous-member value 26:

Actual l_xd_x = 4150150 = 27.67.
Permitted ld = 20(1.5) = 30.0 > 27.67: deflection check passes. (IS 456:2000, Clause 23.2.1 and Figure 4)

No compression-steel benefit is claimed. The corresponding limits are 4150250 = 16.60 mm total and min(4150350, 20) = 11.86 mm after finishes. (IS 456:2000, Clause 23.2(a)–(b)) The modified-ratio method is the code's deemed-satisfactory check; these are limits, not calculated deflections. (IS 456:2000, Clause 42.1)

6. Corner Torsion and Edge Reinforcement

Name the corners A (two discontinuous edges), B and D (one discontinuous and one continuous edge), and C (two continuous edges). The maximum mid-span flexural steel requirement is 228.60 mm²/m.

• Corner A: each of four layers, top x, top y, bottom x and bottom y, requires 0.75(228.60) = 171.45 mm²/m. (IS 456:2000, Clause D-1.8) Provide 8 mm @ 250 mm each way, top and bottom, giving 201.06 mm²/m per layer, over an 850 × 850 mm corner zone. Required extent = l_x5 = 830 mm. (IS 456:2000, Clause D-1.8)
• Corners B and D: each layer requires half of 171.45 = 85.73 mm²/m. (IS 456:2000, Clause D-1.9) Provide 8 mm @ 300 mm each way, top and bottom, giving 167.55 mm²/m, over the same 850 mm extent.
• Corner C: no separate corner torsion mesh is required because both edges are continuous. (IS 456:2000, Clause D-1.10) Retain the normal top support and bottom reinforcement.
• Discontinuous x edge: top x bars must be at least half the provided mid-span steel, 314.162 = 157.08 mm²/m, extending at least 0.1l_x = 415 mm into the panel. (IS 456:2000, Clause D-1.6) Provide 8 mm @ 250 mm and adopt 450 mm extension.
• Discontinuous y edge: required top y steel = 261.802 = 130.90 mm²/m, extending at least 0.1l_y = 614 mm. (IS 456:2000, Clause D-1.6) Provide 8 mm @ 300 mm and adopt 650 mm extension.

In corner zones, provide the specified four-layer torsion mesh as additional steel, so there is no ambiguous double counting of the primary flexural bars. Place additional parallel 8 mm bars between the primary bars in the same two directional layers at each face, coordinating their positions without moving the main flexural layers inward. Keep all steel inside the cover envelope. Continue top steel over each continuous support into the neighbouring panel. For simple detailing, take all x top support bars at least 1300 mm into each panel, and all y top support bars at least 1850 mm into each panel, exceeding 0.3l_x and 0.3l_y, respectively. This also exceeds the 0.15l minimum for all bars. (IS 456:2000, Clause D-1.5)

7. Development Length and Wall Anchorage

For M20, τ_bd = 1.2 N/mm² for plain tension bars; increase by 60% for deformed bars, giving 1.92 N/mm². (IS 456:2000, Clause 26.2.1.1)

L_d(10 mm) = 10(361.05)4 × 1.92 = 470.12 mm. (IS 456:2000, Clause 26.2.1)
L_d(8 mm) = 376.09 mm.
Available straight embedment into a 230 mm support with 20 mm end cover = 210 mm > L_d(10)3 = 156.71 mm. (IS 456:2000, Clause 26.2.3.3(a))

Continue all bottom bars into the walls; this exceeds the required continuing fraction. (IS 456:2000, Clause 26.2.3.3(a)) Full L_d need not fit wholly within the wall if the support bond inequality is satisfied. With bars ending 20 mm from the outside wall face, L_o = 115 − 20 = 95 mm. The positive bearing reaction provides the confinement needed for the 1.3 factor.

L_d ≤ 1.3M₁V + L_o. (IS 456:2000, Clause 26.2.3.3(c), confined support ends)

Anchor top edge and corner bars with U-returns at the discontinuous edges. Keep the bars within the 175 mm RC slab over the 230 mm wall, not embedded in masonry. For each 8 mm bar, provide a 180 mm horizontal straight length from the inner wall face to the bend tangent, two standard 90° bends with 25 mm centreline radius, and a bottom return extending at least 400 mm back into the panel. Even for the inner directional layer, the vertical straight segment between bends is at least 57 mm. The equivalent anchorage from the support face is at least 180 + 57 + 2(8 × 8) + 180 + 400 = 945 mm > L_d(8) = 376.09 mm. (90°-bend credit: IS 456:2000, Clause 26.2.2.1(b)(1)) The outer vertical part lies 205 mm from the inner wall face, leaving 230 − 205 − 4 = 21 mm end cover.

The 450 mm and 650 mm extensions on the span side and the 850 mm corner meshes provide the required reinforcement extents; they are not a substitute for that end anchorage. Main top bars continue across the continuous supports and their extensions exceed the 470 mm development requirement. The simple-support bond table uses w_ul2 as a conservative reaction at the discontinuous edge, distinct from the larger continuous-edge shear used above. Jain §11.9, printed pp. 228–229 / PDF p. 92, illustrates the separate support-extension and bond-inequality checks.

Adopt a 175 mm slab, 20 mm nominal cover, bottom 10 mm @ 250 mm in x and 10 mm @ 300 mm in y. Provide top support, discontinuous-edge and corner meshes as scheduled above. Flexure, minimum steel, spacing, shear, deflection and support anchorage pass under the stated reference-method assumptions.

Question 4(a): Under-Reinforced Sections

Define an under-reinforced section. Why is it preferred in LSM design? [2 marks]

An under-reinforced flexural section has less tensile reinforcement than the limiting amount for its geometry, materials and compression reinforcement. In a singly reinforced rectangular beam this corresponds to x_u < x_u,max. The tension steel reaches substantial inelastic strain before the concrete compression zone reaches its crushing limit.

It is preferred because it gives ductile deformation and visible cracking/deflection before collapse, permits rotation and moment redistribution when properly detailed, and can dissipate energy. An over-reinforced section can fail by comparatively sudden concrete crushing before sufficient tensile-steel strain develops. Under-reinforcement alone does not prevent brittle shear, bond or joint failure; those modes must be designed and detailed separately.

ε_st at failure ≥ 0.002 + f_y1.15E_s. (IS 456:2000, Clause 38.1(f))
With ε_cu = 0.0035 and a linear strain diagram: (IS 456:2000, Clause 38.1(a)–(b))
x_u,maxd = 0.00350.0035 + 0.002 + f_y1.15E_s. (Derived from IS 456:2000, Clause 38.1)
Code values: Fe250, 0.53; Fe415, 0.48; Fe500, 0.46. (IS 456:2000, limiting neutral-axis values below Figure 23)

The explanation of failure mode follows Subramanian's flexural-analysis chapter; the code expresses the ductility requirement through strain rather than a narrative definition.

Question 4(b): Slender Column under Biaxial Bending

Design a reinforced cement concrete (RCC) column with a cross-sectional size of 400 mm × 600 mm and an unsupported length of 7.6 meters. The effective length factor (k) is given as 0.8. The column is subjected to a factored axial load of 1600 kN. In addition, it is subjected to a factored bending moment of 60 kN·m at the top and 40 kN·m at the bottom about the major axis, and 40 kN·m at the top and 30 kN·m at the bottom about the minor axis. The moments are acting in such a manner as to induce double curvature in the column. Use M30 grade concrete and Fe500 grade reinforcement. Design the column as per the relevant provisions of IS 456:2000. [10 marks]

Reference interpretation: adopt the no-sway/braced, no-transverse-load-between-ends model used by SP 16 §3.4 Example 9 (printed pp. 106–108 / PDF pp. 129–131) and the Manual §3.5 Example 3.4 (printed pp. 164–167 / PDF pp. 84–85). This is consistent with the supplied effective-length factor 0.8 but is still an explicit modelling assumption, not proof of the building's bracing. Take that factor about both axes. The given actions are already factored; no further 1.5 multiplier is applied. Do not reuse old IS 456 clause numbers printed in SP 16.

1. Slenderness and Minimum Eccentricity

b = 400 mm; D = 600 mm; l = 7600 mm.
l_ex = l_ey = 0.8(7600) = 6080 mm.
l_exD = 6080600 = 10.13 < 12: short about the major axis. (IS 456:2000, Clause 25.1.2)
l_eyb = 6080400 = 15.20 > 12: slender about the minor axis. (IS 456:2000, Clause 25.1.2)

Use the unsupported length, not the effective length, for minimum eccentricity:

e_min,x = max(7600500 + 60030, 20) = 35.20 mm. (IS 456:2000, Clause 25.4)
e_min,y = max(7600500 + 40030, 20) = 28.533 mm. (IS 456:2000, Clause 25.4)
P_ue_min,x = 56.320 kN·m; P_ue_min,y = 45.653 kN·m.

2. End-Moment Rule and Slenderness Moment

M_u2 is the larger end moment and M_u1 is the smaller, taken negative for double curvature. The weighted initial moment must be at least 0.4M_u2; the final moment must also be at least M_u2. (IS 456:2000, Clause 39.7.1, Note 2)

Major-axis weighted moment = 0.6(60) − 0.4(40) = 20; lower bound = 0.4(60) = 24 kN·m. (IS 456:2000, Clause 39.7.1, Note 2)
The major axis is short: retain max(60, 56.320) = M_ux = 60 kN·m.
Minor-axis weighted moment = 0.6(40) − 0.4(30) = 12; lower bound = 0.4(40) = 16 kN·m. (IS 456:2000, Clause 39.7.1, Note 2)
Adopt initial M_iy = max(12, 16, 45.653) = 45.653 kN·m.
M_ay = P_ub2000(l_eyb)² (IS 456:2000, Clause 39.7.1)
= 1600(0.400)2000(6.0800.400)² = 73.9328 kN·m.
M_uy = max(45.653 + 73.933, 40) = 119.586 kN·m. (IS 456:2000, Clause 39.7.1, Note 2)

No additional major-axis slenderness moment is required. Taking the minor-axis eccentricity minimum as an initial-moment floor is conservative. The optional reduction is k_r = P_uz − P_uP_uz − P_b ≤ 1. (IS 456:2000, Clause 39.7.1.1) Choose k_r = 1.0, claiming no reduction. This k_r is not the given effective-length factor 0.8.

3. Trial Reinforcement and Actual Cover

Provide 8×20 mm longitudinal bars: four corner bars and one at the middle of each face. Nominal cover to the outside of the 10 mm ties = 40 mm. (IS 456:2000, Clause 26.4.2.1) Thus the actual longitudinal bar-centre cover is d′ = 40 + 10 + 202 = 60 mm.

A_sc = 8π420² = 2513.27 mm².
p = 100(2513.27)400 × 600 = 1.0472%.
pf_ck = 1.047230 = 0.03491.
P_uf_ckbD = 1,600,00030 × 400 × 600 = 0.22222.
d′D = 60600 = 0.10; d′b = 60400 = 0.15.

4. Actual SP 16 Chart Readings

Use charts for Fe500 with reinforcement distributed equally on four sides: d′D = 0.10 (SP 16, Chart 48) and d′D = 0.15 (SP 16, Chart 49). Use the lower pf_ck curve 0.02, even though 0.03491 is provided. At normalized axial load 0.222, read each intersection and round its moment ordinate down as follows:

These are deliberately conservative adopted chart values, not high-precision interpolated ordinates for pf_ck = 0.03491. The dimensions must be swapped for minor-axis bending. A chart capacity at a different axial load is not valid in the biaxial equation.

5. Biaxial Interaction Check

A_c = 240,000 − 2513.27 = 237,486.73 mm².
P_uz = 0.45f_ckA_c + 0.75f_yA_sc (IS 456:2000, Clause 39.6)
= 0.45(30)(237,486.73)1000 + 0.75(500)(2513.27)1000 = 4148.55 kN.
P_uP_uz = 0.38568.
α_n = 1 + 0.38568 − 0.20.8 − 0.2 = 1.30946. (Linear interpolation, IS 456:2000, Clause 39.6)
(M_uxM_ux1)^α_n + (M_uyM_uy1)^α_n ≤ 1.0 (IS 456:2000, Clause 39.6)
Calculated sum = (60302.4)^1.30946 + (119.586172.8)^1.30946
= 0.738 < 1.0: passes.

P_uz is the code quantity for the biaxial exponent, not the permissible axial load to use instead of this interaction check.

6. Longitudinal and Transverse Detailing

Minimum longitudinal steel = 0.008(240,000) = 1920 mm² < 2513.27 mm². (IS 456:2000, Clause 26.5.3.1(a))
Maximum = 0.06(240,000) = 14,400 mm²; the provided 1.047% is also comfortably below the practical 4% lap-zone recommendation. (IS 456:2000, Clause 26.5.3.1(a) and its Note)
Bar-centre spacing along the 600 mm faces = 600 − 2 × 602 = 240 mm.
Along the 400 mm faces = 400 − 2 × 602 = 140 mm.
Both < 300 mm. (IS 456:2000, Clause 26.5.3.1(g)) Minimum clear spacing = 140 − 20 = 120 mm.

Tie diameter ≥ max(6, 204) = 6 mm; adopt 10 mm. (IS 456:2000, Clause 26.5.3.2(c)(2))
Tie pitch ≤ min(400, 16 × 20, 300) = 300 mm; adopt 250 mm c/c. (IS 456:2000, Clause 26.5.3.2(c)(1))

Use an outer rectangular tie for the four corner bars and an inner diamond tie engaging the four face-middle bars at each tie level, with properly anchored ends. This gives every compression-face bar lateral restraint. (IS 456:2000, Clause 26.5.3.2(a)–(b)) The first-order shears from double-curvature end-moment equilibrium are only 60 + 407.6 = 13.16 kN and 40 + 307.6 = 9.21 kN; the nominal stresses are about 0.061 and 0.045 N/mm², respectively, so shear does not require a tighter tie pitch for the supplied actions.

400 × 600 mm column: 8×20 mm Fe500 longitudinal bars, 40 mm nominal cover, and 10 mm outer rectangular plus inner diamond ties at 250 mm c/c. Design actions are P_u = 1600 kN, M_ux = 60 kN·m and M_uy = 119.586 kN·m; conservative biaxial utilization = 0.738.

This answers the requested IS 456 column design. If it is part of a seismic moment-resisting frame, the building-level demands and IS 13920 confinement, joint and capacity-design checks must also be applied; Question 5(a) describes those separate requirements.

Question 5(a): Importance of Ductile Detailing

Describe the importance of ductile detailing in RC structures with neat sketches showing the ductile detailing requirements for beams and columns. [4 marks]

Ductile detailing enables an RC member and its connections to sustain repeated inelastic deformation under earthquake loading while retaining useful strength and gravity-load capacity. It confines the concrete core, restrains buckling of longitudinal bars, maintains anchorage, and prevents brittle shear, splice and joint failures from preceding the intended flexural yielding. Closely spaced hoops are therefore not merely nominal cage-holding steel.

Beam Requirements

• Geometry and longitudinal bars: width at least 200 mm, width-to-depth ratio preferably greater than 0.3, and depth not greater than one-quarter of clear span, subject to the code's stated exception. (IS 13920:2016, Clauses 6.1.1–6.1.3) Provide at least two 12 mm bars each at top and bottom. (IS 13920:2016, Clause 6.2.1(a)) Minimum longitudinal ratio ρ_min = 0.24√f_ckf_y. (IS 13920:2016, Clause 6.2.1(b)) Maximum on a face ρ_max = 0.025. (IS 13920:2016, Clause 6.2.2)
• Moment reversal: bottom longitudinal steel at a joint face is at least half the top steel there. (IS 13920:2016, Clause 6.2.3) Steel on either face at any section is at least one-quarter of the maximum support top steel. (IS 13920:2016, Clause 6.2.4) Fully anchor beam bars in joints; at an exterior joint the prescribed anchorage is L_d,tension + 10d_b minus the 90°-bend allowance. (IS 13920:2016, Clause 6.2.5 and Figure 2)
• Shear links: vertical closed links, diameter at least 8 mm. (IS 13920:2016, Clauses 6.3.1–6.3.2) Within 2d of each joint face, spacing s ≤ min(d4, 8d_b,min, 100 mm). (IS 13920:2016, Clause 6.3.5) Place the first link within 50 mm of the joint face. (IS 13920:2016, Clause 6.3.5.1) Elsewhere s ≤ d2. Provide similar close links on either side of any other likely yielding section. (IS 13920:2016, Clause 6.3.5.2)
• Capacity design: design for the larger of analysis shear and the equilibrium shear associated with end plastic-hinge capacities plus gravity load. (IS 13920:2016, Clause 6.3.3) Exclude concrete, inclined links and bent-up bars from the seismic beam shear-capacity calculation. (IS 13920:2016, Clause 6.3.4)
• Splices and hooks: no lap in the joint, within 2d of its face, or within the specified potential-yielding regions; no more than 50% of a face's bars spliced at one section; links over laps at no more than 150 mm. (IS 13920:2016, Clause 6.2.6.1) Use a 135° hook and tail of at least max(6d_link, 65 mm), embedded in the confined core. (IS 13920:2016, Clause 6.3.1)

Column and Joint Requirements

• Geometry and hierarchy: minimum column dimension is the larger of 300 mm and 20 times the largest beam-bar diameter passing through or anchoring in the joint; rectangular aspect ratio at least 0.45. (IS 13920:2016, Clauses 7.1.1–7.1.2) Apply the strong-column/weak-beam check: the sum of column nominal moment strengths at a joint is at least 1.4 times that of the beams, with the specific code exceptions. (IS 13920:2016, Clause 7.2.1) Under the relevant seismic load combinations the factored axial stress limit is 0.40f_ck. (IS 13920:2016, Clause 7.1)
• Hoops and cross-ties: closed hoops with 135° hooked ends and tails at least max(6d_link, 65 mm). (IS 13920:2016, Clause 7.4.1) Minimum link diameter is 8 mm for longitudinal bars up to 32 mm, and 10 mm for larger bars. (IS 13920:2016, Clause 7.4.2(a)) Parallel link legs are at most 300 mm apart; add cross-ties or overlapping hoops where required, engaging peripheral longitudinal bars. (IS 13920:2016, Clause 7.4.2(b)–(c))
• Confinement length: at either end, l_o ≥ max(larger section dimension, clear height6, 450 mm). (IS 13920:2016, Clause 8.1(a)) Within this length, s_v ≤ min(one-quarter of the least section dimension, 6d_b,min, 100 mm), and calculate the required hoop area. (IS 13920:2016, Clause 8.1(b)–(c)) Ordinary column-zone hoop spacing may not exceed half the least lateral dimension. (IS 13920:2016, Clause 7.4.2(d))
• Splices: lap only in the central half of clear column height, outside joints and outside the prohibited 2d beam-face zones. Splice at most 50% at one section and use links at no more than 100 mm over the lap. Bars larger than 32 mm require mechanical splicing under the supplied code. (IS 13920:2016, Clause 7.3.2.1)
• Special cases: extend confinement at least 300 mm into a footing. (IS 13920:2016, Clause 8.2) Provide full-height confinement for a contraflexure point outside the middle half or significant stiffness variation. (IS 13920:2016, Clauses 8.3–8.4) Joint shear and joint confinement also require separate checks.

For rectangular links, satisfy both lower bounds, equivalent to taking their maximum:

A_sh ≥ 0.18s_vhf_ckf_y(A_gA_k − 1), (IS 13920:2016, Clause 8.1(c)(2))
and A_sh ≥ 0.05s_vhf_ckf_y. (IS 13920:2016, Clause 8.1(c)(2))

Here h is the longer outside dimension of a rectangular link, not exceeding 300 mm; it can be reduced by introducing the prescribed cross-ties. A_k is the confined-core area measured to the outer link dimensions and remains the overall core area when cross-ties are used. A_sh is the area of the bar forming the link, not automatically the two-leg A_sv used in ordinary beam shear design. (IS 13920:2016, Clause 8.1(c)(2), symbol definitions) The 0.05 term is not interchangeable with an older code's confinement expression.

Question 5(b): Isolated Footing under Axial Load and Major-Axis Moment

Design an isolated footing to support a rectangular reinforced concrete column of size 400 mm × 600 mm, subjected to a factored axial load of 2000 kN and a factored bending moment of 150 kN·m along the major axis. The safe bearing capacity of soil is 200 kN/m². The column is reinforced with 10 bars of 20 mm diameter. Use M25 grade concrete and Fe415 grade steel for both the column and the footing. Also check the load bearing capacity of the column footing joint. [8 marks]

Reference method and missing data: follow Subramanian Example 15.5, printed pp. 630–632 / PDF pp. 644–646: centre the footing on the column, use service loads for SBC, then the nonuniform net factored reaction for structural design. Assume the stated factored actions came from 1.5(DL + LL), as that example does. Treat SBC = 200 kN/m² as gross allowable pressure at the founding level. Adopt founding depth 1.5 m and soil unit weight 20 kN/m³ from the Manual's footing example, printed pp. 270–273 / PDF pp. 134–135; assume no groundwater uplift. Adopt 75 mm bottom and side cover and a levelling course, following the footing details in Subramanian. This exceeds the 50 mm code minimum. (IS 456:2000, Clause 26.4.2.2) These are examination assumptions, not a site investigation.

The column is centred geometrically; its 600 mm dimension is parallel to the direction in which the pressure varies. The moment is not removed by silently moving the column or footing. A square pad is selected for simple uniform reinforcement distribution. Take the supplied column actions at its junction with the top of the footing.

1. Service Pressure and Plan Size

P_s = 20001.5 = 1333.333 kN; M_s = 1501.5 = 100 kN·m. (Assumed DL + LL combination: IS 456:2000, Table 18)
Adopt a trial uniform thickness D = 800 mm; its development-length justification is checked below.
Uniform footing and backfill surcharge, conservatively covering the entire plan:
q_w = 25(0.8) + 20(1.5 − 0.8) = 34.0 kN/m².

A preliminary 10% self-weight allowance, as used for trial sizing in the reference, is not enough for this selected thickness and founding depth. With a 3.0 m square pad, the actual-weight check gives q_s,max = 1333.3339 + 6(100)3³ + 34 = 204.37 > 200 kN/m². Increase the size.

Adopt L = B = 3.2 m; A = 10.24 m².
q_s,max, q_s,min = P_sA + q_w ± 6M_sBL² (Elastic bearing distribution; Subramanian, Example 15.5)
= 130.208 + 34.000 ± 18.311 kN/m².
q_s,max = 182.519 < 200 kN/m². (SBC requirement: IS 456:2000, Clause 34.1)
q_s,min = 145.898 > 0: no loss of soil contact.

The direct column eccentricity M_uP_u = 0.075 m < L6 = 0.533 m; including the centred footing and backfill weight reduces it further. (Middle-third check; Subramanian, Example 15.5) The SBC check uses service pressure, not the larger factored pressure.

2. Net Factored Soil-Pressure Distribution

The uniform upward reaction caused by footing/backfill weight cancels its equal downward distributed load in the structural free body. Therefore, with coordinate r measured from the footing centre toward the high-pressure edge:

q_u(r) = P_uBL + 12M_uBL³r (Linear-pressure equilibrium; Subramanian, Example 15.5)
= 195.3125 + 17.16614r kN/m², with r in metres.
q_u,max = 222.778 kN/m²; q_u,min = 167.847 kN/m².
At the high-pressure column face r = 0.300 m: q_f = 200.462 kN/m².

The pressure integrates to P_u = 2000 kN and its moment about the centre is 150 kN·m. A uniform P_uA pressure alone would omit the given moment.

3. Bending Moments and Reinforcement

Projection in the pressure-varying direction: a_x = 3.2 − 0.62 = 1.3 m.
Transverse projection: a_y = 3.2 − 0.42 = 1.4 m.
d_x = 800 − 75 − 162 = 717 mm.
d_y = 800 − 75 − 16 − 162 = 701 mm.

Take the critical bending section at the column face and integrate the rectangular plus triangular pressure over the cantilever. (IS 456:2000, Clause 34.2.3.1–34.2.3.2) With g = 17.16614 kN/m³:

M_x,+ = B[q_fa_x²2 + ga_x³3] (Pressure integration; Subramanian, Example 15.5)
= 3.2[200.462(1.3²)2 + 17.16614(1.3³)3]
= 582.278 kN·m.
The low-pressure face gives M_x,− = 473.972 kN·m.
M_y = Lq_u,averagea_y²2 (Face-moment equilibrium, IS 456:2000, Clause 34.2.3)
= 3.2(195.3125)(1.4²)2 = 612.500 kN·m.

For transverse bending, the varying component integrates to zero over the full symmetric length. Using x and y here denotes bar directions; the column's major-axis moment creates pressure variation in x.

M_u = 0.36f_ckbx_u(d − 0.42x_u). (IS 456:2000, Annex G-1.1)
A_st = 0.36f_ckbx_u0.87f_y. (Force equilibrium, IS 456:2000, Annex G-1.1)
Required A_st,x = 2287.72 mm² across 3.2 m.
Required A_st,y = 2465.70 mm² across 3.2 m.
Minimum each way = 0.0012(3200)(800) = 3072 mm², governing both directions. (IS 456:2000, Clauses 34.5.1 and 26.5.2.1)

Provide 17 bars of 16 mm diameter in each direction, area 17π416² = 3418.05 mm² each way. With 75 mm side cover, the actual centre spacing is 3200 − 2 × 75 − 1617 − 1 = 189.625 mm, not more than 200 mm. Distribute both sets uniformly because the footing is square. (IS 456:2000, Clause 34.3.1(b)) Place x bars in the bottom layer and y bars immediately above. Spacing is below min(3d, 300) = 300 mm. (IS 456:2000, Clauses 26.3.3(b)(1) and 34.5.1) Minimum clear spacing is 173.625 mm.

4. One-Way Shear at d from the Column Faces

Check one-way shear at a distance d from the column face. (IS 456:2000, Clause 34.2.4.1(a); Subramanian, Example 15.5)

High-pressure residual cantilever: a_x − d_x = 1.3 − 0.717 = 0.583 m.
V_x = 3.2[222.778(0.583) − 17.16614(0.583²)2] = 406.280 kN.
τ_vx = 406,2803200 × 717 = 0.1771 N/mm². (IS 456:2000, Clause 40.1)
V_y = 3.2(195.3125)(1.4 − 0.701) = 436.875 kN.
τ_vy = 436,8753200 × 701 = 0.1948 N/mm². (IS 456:2000, Clause 40.1)

p_x = 100(3418.05)3200 × 717 = 0.1490%.
p_y = 100(3418.05)3200 × 701 = 0.1524%.
For M25, τ_cx = 0.290 and τ_cy = 0.2917 N/mm². (IS 456:2000, Table 19)
Both shear stresses are lower: one-way shear passes.

No thin-slab enhancement is used: D > 300 mm, so k = 1.0. (IS 456:2000, Clause 40.2.1.1) The opposite x-face has lower pressure and therefore a lower shear demand.

5. Two-Way Punching, Including Moment Transfer

Use the smaller effective depth d = 0.701 m for a conservative perimeter check. The closed perimeter at d2 from the column is L_p = 0.600 + 0.701 = 1.301 m by B_p = 0.400 + 0.701 = 1.101 m. (IS 456:2000, Clauses 34.2.4.1(b) and 31.6.1)

u_o = 2(1.301 + 1.101) = 4.804 m.
V_p = P_u − q_u,averageL_pB_p
= 2000 − 195.3125(1.301)(1.101) = 1720.234 kN.
τ_average = 1720.2344.804 × 0.701 × 1000 = 0.5108 N/mm². (IS 456:2000, Clause 31.6.2.1; units converted to N/mm²)

The pressure-gradient term integrates to zero over the centred perimeter area. Conservatively transfer the entire 150 kN·m through a linearly varying perimeter shear stress, rather than only the code fraction (1 − α). (IS 456:2000, Clause 31.6.2.2) Also neglect the favourable moment of the soil reaction inside the perimeter.

J_p = d[2B_p(L_p2)² + L_p³6] = 0.910451 m⁴. (Derived by integration over the four shear faces)
τ_moment = M_u(L_p2)J_p × 1000 = 0.1072 N/mm². (Linear shear distribution, IS 456:2000, Clause 31.6.2.2)
τ_max = 0.5108 + 0.1072 = 0.6180 N/mm².
k_s = min[1, 0.5 + 400600] = 1. (IS 456:2000, Clause 31.6.3.1)
Permissible k_sτ_c = 1(0.25√25) = 1.25 N/mm². (IS 456:2000, Clause 31.6.3.1)
0.6180 < 1.25: punching shear passes without shear reinforcement.

The expression for J_p is obtained by integrating r²dA over the four vertical shear faces. It is not an SP 16 chart or an empirical amplification factor.

6. Footing-Bar Development and Edge Thickness

For M25 deformed bars in tension: τ_bd = 1.4(1.6) = 2.24 N/mm². (IS 456:2000, Clause 26.2.1.1)
L_d(16 mm) = 16(361.05)4 × 2.24 = 644.73 mm. (IS 456:2000, Clause 26.2.1)
Available from column face to bar end: x, 1300 − 75 = 1225 mm; y, 1400 − 75 = 1325 mm. (Critical development section: IS 456:2000, Clause 34.2.4.3)
Both exceed 644.73 mm. Continue all bars to the outer cover line; no curtailment is required.
Uniform edge thickness 800 mm > 150 mm minimum. (IS 456:2000, Clause 34.1.2)

7. Column–Footing Joint: Bearing and Reinforcement

Loaded area A₂ = 400 × 600 = 240,000 mm².
At the footing, even a concentric geometrically similar supporting area 800 × 1200 mm fits wholly within the pad, giving √(A₁A₂) = 2, the maximum permitted enhancement. (IS 456:2000, Clause 34.4)
Footing-side permissible bearing = 0.45(25)(2) = 22.5 N/mm². (IS 456:2000, Clause 34.4)
Column-side basic permissible bearing = 0.45(25) = 11.25 N/mm². (IS 456:2000, Clause 34.4)
Corresponding axial-only bearing capacities on the gross contact area are 5400 kN and 2700 kN, both above 2000 kN.

That axial-only comparison is not the whole joint check. A gross-concrete linear stress estimate, with the kPa-to-N/mm² conversion shown explicitly, is:

σ = 20000.24 × 1000 ± 150(0.4 × 0.6²6) × 1000 = 14.583 and 2.083 N/mm². (Elastic axial-plus-bending stress estimate)

Its maximum exceeds the column-side 11.25 limit. Therefore explicitly account for the supplied continued reinforcement. (IS 456:2000, Clause 34.4.1)

Arrange the 10×20 mm column bars symmetrically: four corners, one middle bar on each 400 mm face, and two intermediate bars on each 600 mm face. With nominal cover 40 mm and 10 mm column ties, the bar-centre cover is 60 mm. Measured from the highly compressed 600 mm extreme face, the layers are 3 bars at z = 60 mm, 2 at 220 mm, 2 at 380 mm and 3 at 540 mm.

An explicit strain-compatible check of this joint section makes the load sharing reproducible:

ε(z) = ε₀(1 − zx). (Linear strain distribution, IS 456:2000, Clauses 38.1 and 39.1)
In the applicable parabolic range,
σ_c = 0.67f_ck1.5[2(ε0.002) − (ε0.002)²]. (IS 456:2000, Clause 38.1, Figure 21)
All steel strains here are elastic, so σ_s = 200,000ε. (IS 456:2000, Figure 23A, elastic branch)

P_u = 400∫₀⁶⁰⁰σ_c(z) dz + Σ[σ_si − σ_ci]A_si,
M_u = 400∫₀⁶⁰⁰σ_c(z)(300 − z) dz + Σ[σ_si − σ_ci]A_si(300 − z_i). (Section equilibrium using IS 456:2000, Clause 39.1)

A solution is ε₀ = 0.001369124 and x = 663.935 mm, outside the section.
Concrete stress at the highly compressed edge = 10.056 N/mm²; at the other edge = 1.424 N/mm².
Both edges are in compression and the maximum is below 11.25 N/mm², hence also below the footing-side 22.5 N/mm².

Net concrete force, after subtracting displaced concrete = 1528.457 kN.
Steel compression force = 471.543 kN.
Total = 2000.000 kN.
Net concrete moment = 101.904 kN·m; steel moment = 48.096 kN·m.
Total = 150.000 kN·m.

Thus there is an equilibrium load-transfer state with compatible strains, acceptable concrete bearing and acceptable bar stresses. This check uses IS 456 material relations, not a guessed steel/concrete split. Subramanian §15.5.4, printed pp. 597–598 / PDF pp. 611–612, explains why bearing and the continued bars must be considered together.

8. Continue and Develop the Column Bars

Minimum extended steel or dowels = 0.005(400)(600) = 1200 mm², with at least four bars. (IS 456:2000, Clause 34.4.3)
Provided continuing column bars = 10π420² = 3141.59 mm².
L_dc(20 mm) = 20(361.05)4(2.24)(1.25) = 644.73 mm. (IS 456:2000, Clauses 26.2.1 and 26.2.1.1, compression bond enhancement)
L_dt(20 mm) = 20(361.05)4(2.24) = 805.92 mm. (IS 456:2000, Clause 26.2.1)

Continue all ten bars into the footing with adequate development. (IS 456:2000, Clause 34.4.2) A workable detail has each bar's horizontal 90° return centreline 150 mm above the footing underside, giving a 650 mm vertical projection below the joint. This exceeds L_dc. With a standard bend having a 50 mm centreline radius, the straight vertical length to the bend tangent is 600 mm; a 300 mm straight tail after the bend gives tension anchorage 600 + 8(20) + 300 = 1060 mm > 805.92 mm. (90°-bend credit: IS 456:2000, Clause 26.2.2.1(b)(1)) The bend must conform to the referenced standard bend requirements; do not credit 8d_b as additional compression length. (IS 456:2000, Clause 26.2.2.2)

The returns lie above the two footing-mesh layers: upper mesh top = 75 + 16 + 16 = 107 mm from the underside; return underside = 150 − 10 = 140 mm, leaving 33 mm clear. Orient/stagger tails within the generous plan projections so that they do not collide. The 800 mm thickness follows from at least 644.73 + 150 = 794.73 mm needed for the selected compression anchorage detail, in addition to the satisfied strength checks. Maintain ordinary column ties through the embedded cage; this question does not specify a seismic-frame joint.

Provide a centred 3.2 m × 3.2 m square footing, 800 mm uniform thickness, 75 mm bottom/side cover, and 17×16 mm bars each way at 189.625 mm centres (not more than 200 mm). Continue all 10×20 mm column bars with the checked anchorage. Service soil pressure, both flexural directions, one-way shear, eccentric punching, minimum steel, development and joint load transfer pass under the stated assumptions.