Structural Design & Drawing (RCC-I) (CE-601) - Unit 4 Short Notes

UNIT 4: Comprehensive Design and Detailing of RCC Members

1.0 Limit State Method & Fundamental Concepts

Principles of Limit State Design

• Structure designed for ultimate limit state (ULS) (strength, stability) and serviceability limit state (SLS) (deflection, cracking).

• Partial safety factors (γ_f for loads, γ_m for materials) applied to characteristic values to obtain design values.

• Characteristic load (P_k): Load with 95% probability of not being exceeded during service life.

• Characteristic strength (f_ck, f_yk): Material strength with 5% probability of being inferior.

Partial Safety Factors (IS 456:2000)

Balanced, Under-Reinforced, Over-Reinforced Sections (LST)

• Balanced Section: Strain in steel (ε_s) = yield strain (ε_y) and strain in extreme concrete fibre (ε_cu) = 0.0035 simultaneously. x_u = x_u,lim.

• Under-Reinforced: Steel yields before concrete crushes. ε_s ≥ ε_y, ε_cu reached later. Ductile failure (desired).

• Over-Reinforced: Concrete crushes before steel yields. ε_s < ε_y, ε_cu reached first. Brittle failure (avoid).

[!TIP] Exam Focus: Always sketch strain & stress diagrams for balanced section. State that under-reinforced is safer due to ductility.

Comparison: Balanced Section in WSM vs LST

2.0 Flexural Members: Beams

2.1 Singly Reinforced Rectangular Beams (LST)

Design Steps (LST):

1. Compute design moment M_u = γ_f × factored load effect.

2. For given b, d, f_ck, f_y, find M_u,lim for balanced section:

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

where $$\displaystyle x_{u,lim} = 0.48d $$ for Fe415 (0.46d for Fe500).

3. If M_u ≤ M_u,lim → Singly reinforced.

   Find x_u from:

$$M_u = 0.36 f_{ck} b x_u \left( d - 0.42 x_u \right)$$

4. Area of steel A_st:

$$A_{st} = \frac{0.36 f_{ck} b x_u}{0.87 f_y}$$

5. Provide A_st ≥ min. steel (0.85bd/f_y).

Moment of Resistance (M_R):

$$M_R = T_s \times \text{lever arm} = 0.87 f_y A_{st} \left( d - 0.42 x_u \right)$$

2.2 Doubly Reinforced Beams (LST)

When M_u > M_u,lim:

1. Additional moment M_u2 = M_u - M_u,lim.

2. Compression steel A_sc for M_u2:

$$A_{sc} = \frac{M_{u2}}{0.87 f_y \left( d - d' \right)}$$

(d' = effective cover to compression steel).

3. Total A_st = A_st1 (for M_u,lim) + A_sc.

4. Check: A_sc ≤ 0.04bD (max. compression steel).

2.3 T-Beams (LST)

Effective Flange Width (b_f) (IS 456 Cl. 23.1.2):

$$b_f = \begin{cases} b_w + \frac{l_0}{6} & \text{for simply supported} \\ b_w + \frac{l_0}{10} & \text{for continuous} \\ b_w + \frac{l_0}{12} & \text{for cantilever} \end{cases}$$

where l₀ = distance between points of contraflexure.

Design:

1. Assume NA within flange (x_u ≤ D_f). Check:

$$\frac{M_u}{0.36 f_{ck} b_w D_f} \leq 0.45 f_{ck} D_f (d - 0.42 D_f)$$

2. If true → Flanged beam (NA in flange). Use b_f in place of b.

3. If false → Rectangular beam (NA in web). Solve for x_u using b_w.

2.4 Shear in Beams

Critical Sections for Shear (IS 456):

• For simply supported beam: at face of support.

• For continuous beam: at 0.5d from face of support (inner face of support).

[!TIP] Diagram Required: Always sketch beam with supports, show critical sections at support face (simply supported) and at 0.5d from face (continuous).

Design for Shear:

• Design shear force V_u = factored shear at critical section.

• Nominal shear stress τ_v = V_u / (b d).

• Permissible shear stress τ_c (Table 19, IS 456) for given f_ck & p_t%.

• If τ_v ≤ τ_c → No shear reinforcement needed (min. stirrups @ 0.75% provided).

• If τ_v > τ_c,max (Table 20) → Redesign section.

• If τ_c < τ_v ≤ τ_c,max → Provide shear reinforcement:

$$A_{sv} = \frac{V_u - V_c}{0.87 f_y d} \times \frac{s_v}{\text{no. of legs}}$$

where V<sub>c</sub> = τ<sub>c</sub> b d.

Shear Transfer Mechanism at Flexural-Shear Crack

[!TIP] High Frequency: Explain with force components diagram.

2.5 Torsion in Beams

Design for Combined Bending, Shear & Torsion:

1. Compute equivalent shear force V_e:

$$V_e = V_u + \frac{T_u}{D}$$

where D = overall depth.

2. Compute equivalent bending moment M_e:

$$M_e = M_u + \frac{T_u}{2} \left(1 + \frac{D}{b}\right)$$

3. Design for V_e (as shear) and M_e (as bending) simultaneously.

4. Provide torsional reinforcement:

   • Longitudinal: 2A_st (total) in tension zone + 2A_st in compression zone (for closed ties).

   • Transverse (stirrups): 4-legged closed ties @ s_v:

$$A_{st, \text{torsion}} = \frac{T_u \times s_v}{0.87 f_y \times 2A_o}$$

    where A<sub>o</sub> = area enclosed by centreline of closed tie.

2.6 Sketches (High Frequency)

Cross-section of Singly Reinforced Beam (LST):

Reinforcement Details:

• Singly reinforced: Tension steel only.

• Doubly reinforced: Tension + compression steel (2-4 bars).

• T-beam: Flange reinforcement (distribution steel) + rib reinforcement.

• With shear/torsion: Stirrups (2/4-legged) closed for torsion.

3.0 Slabs

One-way vs Two-way Slabs:

Design of Simply Supported One-way Slab (LST):

1. Thickness: For simply supported, l/d ≈ 20 (initial guess). Check deflection later.

2. Effective span l = clear span + support width (min.).

3. Factored load: w_u = 1.5 (DL + SL + FF).

4. Maximum B.M. (mid-span): M_u = w_ul²/8.

5. Depth of NA: x_u = 0.48d (Fe415). M_u,lim per unit width.

6. A_st = M_u / (0.87 f_y (d - 0.42x_u)).

7. Provide A_st ≥ 0.15% of gross area (Fe415).

Safe Superimposed Load (SSL) Calculation:

Given M_R (from provided steel), find SSL:

$$SSL = \frac{8 M_R}{1.5 \times l^2} - DL - FF$$

where DL = self-weight (γ_c × thickness).

Reinforcement Layout:

• Main bars: Along short span (for one-way) or both spans (two-way), spaced ≤ 3d or 300 mm.

• Distribution bars: Along long span, spaced ≤ 5d or 450 mm.

• Curtailment: Main bars curtailed at 0.7l from support (for simply supported).

• Development length: At supports, bars must extend beyond critical section.

4.0 Columns

Assumptions in Design (IS 456):

1. Plane sections remain plane (linear strain).

2. Maximum strain in concrete = 0.0035.

3. Stress in concrete = 0.45f_ck at NA (for tied columns).

4. Steel stress = f_y (if yielded) or (f_y/1.15) (if not).

5. Ignore tensile strength of concrete.

Short Columns under Axial Load:

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

Check: p = (A_sc/A_g) × 100% ≤ 4% (max).

Columns with Uni-axial Bending (Reinforcement on Two/Four Sides):

• Two sides (e.g., about major axis): Steel only on two faces parallel to bending axis.

• Four sides: Steel on all faces (for bi-axial or heavy loads).

• Design: Use interaction diagram (Cl. 39.4, IS 456) or approximate formula:

$$\frac{P_u}{p_{u,lim}} + \frac{M_u}{m_{u,lim}} \leq 1$$

where p<sub>u,lim</sub> and m<sub>u,lim</sub> are for balanced section.

Columns with Bi-axial Bending (High Frequency): Approximate Method (IS 456 Cl. 39.6):

$$\frac{M_{ux}}{M_{ux,lim}} + \frac{M_{uy}}{M_{uy,lim}} \leq 1$$

where M_ux,lim and M_uy,lim are limiting moments for axial load P_u about respective axes (from interaction diagram for uni-axial case).

Reinforcement Details:

• Longitudinal: Min. 0.8% (for 4 bars), max. 6% (including overlaps). Dia ≤ 1/6th least lateral dimension.

• Transverse (ties/helix):

  • Ties: φ ≥ 6 mm, spacing ≤ least of: (i) 16×longitudinal bar dia, (ii) 48×tie dia, (iii) least section dimension.

  • Helix: φ ≥ 6 mm, pitch ≤ 25 mm, 75 mm, or D_c/6 (min.).

Sketches:

5.0 Footings

Types of Footings (High Frequency):

• Isolated Square/Rectangular: For single column.

• Stepped: When depth is large (economical).

• Combined: For two or more columns.

• Continuous/Strip: For row of columns.

• Mat/Raft: For heavy loads/weak soil.

Design of Isolated Footing (Flexure & Bond):

1. Size (Plan): Based on bearing capacity:

$$A_{req} = \frac{P_u}{SBC}$$

Provide square/rectangular footing with dimensions ≥ A<sub>req</sub>.

Check: **Maximum pressure** q<sub>max</sub> = P/A (1 + e×6/(bL²)) ≤ SBC.

2. Thickness (h): Check for shear (one-way & two-way) and bond.

   • Two-way shear (punching shear): Critical at d/2 from column face.

$$V_{u, \text{punch}} = P_u \left[1 - \frac{(a+2d)(b+2d)}{A_{foot}}\right]$$

  Check τ<sub>v,punch</sub> = V<sub>u,punch</sub> / (perimeter × d) ≤ τ<sub>c,max</sub>.

- **One-way shear**: Check along critical section at d from column face.

3. Reinforcement (Flexure): Design as one-way/two-way slab.

   • For rectangular footing: B.M. about both axes.

   • Provide steel in both directions (mesh).

4. Bond: Check development length L_d for footing bars at critical section.

Sketches (to scale):

6.0 Staircases

Dog-legged Staircase (High Frequency):

• Two flights with mid-landing (180° turn).

• Waist slab thickness (t): 1/6 to 1/8 of going (T).

• Riser (R) = 150-180 mm, Tread (T) = 250-300 mm.

• Effective span l = 2√(R² + T²) (for simply supported on walls).

Loads on Waist Slab:

1. Dead Load (DL):

   • Self-weight = γ_c × t × 1 m width.

   • Step weight = γ_c × R × (T - thickness of step finish) × 1 m.

2. Live Load (LL): As per IS 875 (2.5-5 kN/m² for offices).

3. Floor Finish: 0.5-1 kN/m².

4. Total UDL w_u = 1.5 (DL + LL + FF).

Design of Waist Slab (as Simply Supported/Continuous Slab):

• Effective span l = clear distance between supports.

• Maximum B.M. (mid-span for simply supported): M_u = w_ul²/8.

• Thickness: Initially, l/d ≈ 20 (simply supported).

• Reinforcement: Main bars along inclined length (along flight), distribution bars perpendicular.

• Curtailment: Main bars curtailed at 0.7l from landing support.

Support Conditions:

• Supported on walls: Landing slab rests on walls → simply supported.

• Stringer beams: Waist slab acts as T-beam with stringer as web.

Sketches:

7.0 Bond, Development Length & Splices

Bond Failure Mechanisms (High Frequency):

1. Pull-out Failure: Steel pulled out from concrete (smooth bars, low confinement).

2. Cover Splitting: Concrete cover splits along bars (high bond stress, small cover).

3. Bar Yielding: Steel yields before bond fails (rare, high steel stress).

4. Cone Failure: For anchorage devices.

[!TIP] Exam Tip: Pull-out = bond stress < tensile strength of concrete. Cover splitting = bond stress > tensile strength of concrete → cracks.

Development Length (L_d):

• Minimum length for steel to develop yield stress.

• Formula (Tension):

$$L_d = \frac{0.87 f_y}{4 \tau_{bd}}$$

where τ<sub>bd</sub> = design bond stress (Table 66, IS 456) for given f<sub>ck</sub> & steel type.

• For Compression: L_d (compression) = L_d (tension) × 0.5.

• Factors: Increase L_d by 1.5 for: (i) epoxy-coated bars, (ii) bars in compression with > 12 mm dia, (iii) bars at shallow depth.

Lap Splices (High Frequency):

• When bars > 1m apart or stress > 0.8f_y → mechanical splices/welding.

• Lap Length (L_lap):

$$L_{lap} = L_d \times \text{multiplier}$$

Multiplier = 1.0 (if % steel ≤ 25%), 1.2 (if 25% < % steel ≤ 50%), 1.5 (if > 50%).

• Lap for Different Diameter Bars (e.g., 12 mm with 20 mm):

  • Use larger L_d (for larger bar).

  • Lap length = larger L_d × multiplier.

  • Stagger laps by ≥ 75 mm or 0.3l_lap (min.).

8.0 Deflection and Cracking Control

Approaches for Deflection Control (IS 456 Cl. 23.2):

1. Limiting Span/Depth Ratio (Table 19):

   • For simply supported: l/d ≤ 20 (Cantilever: 7, Continuous: 26).

   • Modify by: k₁ (support condition), k₂ (steel %), k₃ (span > 10 m).

2. Limiting Deflection (Table 17):

   • For vertical deflection: l/250 (total), l/350 (immediate due to LL).

3. Check by Computation (Annex B): Calculate short-term & long-term deflection.

Measures for Reducing Deflection:

• Increase effective depth (d).

• Increase area of steel (but not beyond balanced %).

• Use higher grade steel (f_y).

• Provide compression steel (doubly reinforced).

• Reduce span (intermediate supports).

• Use prestressing.

9.0 Working Stress Method (WSM) Applications

Design of Singly Reinforced Beam (WSM):

1. Modular Ratio: m = E_s/E_c (≈ 10 for M20, 8.33 for M25).

2. Neutral Axis (x): From strain compatibility:

$$\frac{m \cdot \sigma_{st}}{f_c} = \frac{x}{d-x}$$

where σ<sub>st</sub> = permissible steel stress (Table B-2, IS 456).

3. Lever Arm (z): z = d - x/3.

4. Moment of Resistance:

$$M_R = \sigma_{st} A_{st} z = \frac{1}{2} f_c b x z$$

5. A_st = M_R / (σ_st z).

Doubly Reinforced Beam (WSM):

1. Find x from:

$$\frac{m \cdot \sigma_{sc}}{f_c} = \frac{x - d'}{x}$$

(σ<sub>sc</sub> = perm. stress in compression steel).

2. Moment from concrete: M_c = (1/2) f_c b x (d - x/3).

3. Additional moment M_u2 = M - M_c.

4. A_sc = M_u2 / [σ_sc (d - d')].

5. A_st = A_sc + (M_c / (σ_st z)).

10.0 Drawing and Detailing (Sketching Requirements)

General Rules for Sketches:

1. Show all dimensions (b, D, d, cover, bar dia, spacing).

2. Show reinforcement cage: Tension/compression bars, stirrups (type, spacing, dia).

3. Indicate clear cover (as per exposure condition, Table 16, IS 456).

4. Show bar bends (hooks, anchorage).

5. Label critical sections (shear, moment).

6. Use standard symbols for bars (e.g., 4#20 = 4 bars of 20 mm dia).

Essential Sketches from Past Papers:

1. Singly reinforced beam: Cross-section with strain & stress diagrams (as in 2.6).

2. Doubly reinforced beam: Cross-section showing compression & tension steel.

3. T-beam: Cross-section showing flange, rib, flange steel, stirrups.

4. Column (uni-axial/bi-axial): Cross-section with longitudinal bars & ties/helix.

5. Isolated footing: Plan & longitudinal section (as in 5.0).

6. Dog-legged staircase: Reinforcement in flight & landing (as in 6.0).

7. Shear reinforcement: Stirrups (open/closed) with spacing.

[!TIP] Exam Strategy: For 14m questions, always include step-by-step design calculations + neat sketch with all details. For 5m questions, focus on conceptual explanation with simple sketch.

Key Formulas Boxed:

Balanced steel ratio (LST):

\boxed{p_{t,lim} = \frac{0.36 f_{ck}}{0.87 f_y} \frac{x_{u,lim}}{d}}

Development length (tension):

\boxed{L_d = \frac{0.87 f_y}{4 \tau_{bd}}}

Effective flange width (T-beam):

\boxed{b_f = b_w + \frac{l_0}{6} \text{ (simply supported)}}

Punching shear stress (footing):

\boxed{\tau_{v,punch} = \frac{V_{u,punch}}{\text{perimeter} \times d}}

Equivalent shear & moment (torsion):

\boxed{V_e = V_u + \frac{T_u}{D}, \quad M_e = M_u + \frac{T_u}{2}\left(1 + \frac{D}{b}\right)}