Structural Design & Drawing (RCC-I) (CE-601) - Unit 3 Short Notes

UNIT 3: STRUCTURAL DESIGN & DRAWING (RCC-I) - EXAM-FOCUSED NOTES

1.0 FUNDAMENTALS OF LIMIT STATE DESIGN (THEORETICAL FOUNDATION)

1.1 Principles of Limit State Method (LSM)

• Philosophy: Structure designed for ultimate limit state (ULS) (safety against collapse) and serviceability limit state (SLS) (comfort, durability).

• Design Equation: Design Load (P_u) = γ_f * Characteristic Load (P_k) and Design Strength (f_d) = f_k / γ_m

  • γ_f = Partial safety factor for loads (Table 18, IS 456).

  • γ_m = Partial safety factor for materials (Concrete: 1.5, Steel: 1.15).

• Characteristic Load (P_k): Load not exceeded by 95% of probability (e.g., dead load, live load from code).

• Characteristic Strength (f_k): Strength below which not more than 5% of test results fall (e.g., f_ck for concrete, f_y for steel).

1.2 Stress-Strain Curves (IS 456)

• Concrete (M grades): Parabolic up to 0.002 strain, then linear to ultimate strain ε_cu = 0.0035 (for f_ck ≤ 50 MPa). Stress block simplified for design.

• Steel (Fe grades): Perfectly plastic beyond yield strain ε_y = f_y / 1.15E. No defined ultimate strain (ductile).

1.3 Balanced, Under-Reinforced, Over-Reinforced Sections

[!TIP] Exam Focus: Sketch strain diagrams & compare moment capacities. Under-reinforced is ductile & preferred.

• x_{u,lim} depends on f_ck and steel grade. For Fe415 & M25, x_{u,lim}/d ≈ 0.48.

• Significance: LSM aims for under-reinforced sections for warning before failure.

1.4 Nominal vs. Design Stresses/Loads

• Nominal Stress/Load: Characteristic value (f_ck, P_k).

• Design Stress/Load: Value after applying partial safety factors (f_d = f_ck/γ_m, P_u = γ_f P_k).

1.5 Comparison: WSM vs. LSM (Flexure)

2.0 DESIGN OF FLEXURAL MEMBERS (BEAMS & SLABS)

2.1 Singly Reinforced Rectangular Section (LSM)

Given M_u, find p_t (A_st / bd)

1. Compute M_u / (b d^2 f_ck).

2. From Table 2, IS 456 (or formula), find p_t and x_u/d.

3. A_st = (p_t / 100) * b * d.

4. Check: x_u ≤ x_{u,lim} (under-reinforced). If not, go for doubly reinforced.

Key Formula:

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

2.2 Doubly Reinforced Rectangular Section

Necessity: When M_u > M_{u,lim} for singly reinforced section. Strain Diagram: Compression steel yields (ε_sc ≥ ε_y) if d' is adequate. Design Steps:

1. M_u1 = M_{u,lim} for singly reinforced part.

2. M_u2 = M_u - M_u1 to be resisted by compression steel.

3. A_sc = M_u2 / [0.87 f_y (d - d')]

4. A_st = A_st1 + A_sc (where A_st1 from M_u1).

2.3 T-Beams & L-Beams

Effective Flange Width (b_f):

$$b_f = \left[ \frac{l_0}{6} + b_w + 6 D_f \right] \quad \text{or} \quad b_f = b_w + \frac{l_0}{10} \quad \text{or} \quad b_f = b_w + \text{adjacent spacing}$$

(whichever is least, l_0 = effective span, D_f = flange thickness).

Design Procedure:

1. Assume NA lies in flange (x_u ≤ D_f). Check if M_u ≤ 0.36 f_ck b_w D_f (d - 0.42 D_f).

2. If YES, treat as rectangular with width b_w.

3. If NO, NA in web. Solve for x_u from:

$$M_u = 0.36 f_{ck} \left[ b_w x_u (d - 0.42 x_u) + (b_f - b_w) D_f (d - 0.42 x_u) \right]$$

(for `x_u > D_f`). Then find `A_st`.

[!TIP] Common Pitfall: Forgetting to check NA position first. Always verify if NA is within flange.

2.4 Shear in Beams

Critical Section for Shear: At d from face of support (for simply supported) or at face of support (for continuous).

Shear Stress Computation:

$$τ_v = \frac{V_u}{b d}$$

Permissible Shear Stress (τ_c): From Table 19, IS 456 based on f_ck and p_z (percentage of tension steel).

Design of Shear Reinforcement:

1. If τ_v < τ_c,max (Table 20), no shear reinforcement needed if τ_v ≤ τ_c.

2. If τ_c < τ_v < τ_c,max, provide stirrups:

   • V_{us} = V_u - τ_c * b * d

   • Spacing s_v ≤ min(0.75d, 300mm).

   • For V_{us}, use A_{sv} / s_v * 0.87 f_y * d (vertical stirrups).

Shear Transfer Mechanism: At flexural-shear crack, force transferred by:

1. Shear in concrete (aggregate interlock).

2. Doweling action of tension steel.

3. Stirrups crossing the crack.

2.5 Torsion in Beams (Combined Bending & Torsion)

• Design: For T_u (ultimate torsion), provide longitudinal and transverse reinforcement.

• Transverse: Closed ties (stirrups) along length. A_{st} / s_v for torsion added to shear requirement.

• Longitudinal: Additional bars along corners of section (4 bars, one at each corner). Area from:

$$A_{st} = \frac{T_u}{s_v} * \frac{1}{0.87 f_y} * \frac{u_1}{2.5 d}$$

(`u_1` = perimeter of centerline of torsion reinforcement).

• Sketch: Show main bars + corner bars + closed ties.
  DiagramCANVAS: Beam cross-section with torsion reinforcement: 4 corner bars and closed stirrups

2.6 Detailing of Reinforcement in Beams

• Curtailment: Tension bars cut where moment reduces. Lapped with development length.

• Development Length (L_d):

$$L_d = \frac{ϕ σ_{st}}{4 τ_{bd}}$$

(`τ_{bd}` from Table 66, IS 456; depends on `f_ck` & steel grade).

• Sketch Types:

  • Simply Supported: Top bars at supports, bottom bars at midspan.

  • Continuous: Top bars over supports, bottom bars at midspans.

  • Fixed: Top & bottom bars at supports.

2.7 Design of One-Way & Two-Way Slabs

Two-Way Slab (Restrained Corners):

1. Effective Span: l_x, l_y (clear span + width of support).

2. Type: If l_y / l_x < 2 → Two-way.

3. Moments (for square/rectangular panels, corners down):

   • Negative moments at edges (continuous).

   • Positive moments at mid-spans.

   • Use Table 27, IS 456 for coefficients (depends on edge conditions).

4. Moment Calculation:

$$M_x = α_x w l_x^2, \quad M_y = α_y w l_y^2$$

(`w` = design load/unit area, `α` = coefficient).

5. Reinforcement:

   • Shorter span direction: Provide main reinforcement for M_x.

   • Longer span direction: Provide distribution steel (min. 0.15% for Fe415) for M_y if M_y > 0.5 M_x.

6. Checks:

   • A_st from M_u formula.

   • x_u < 0.45d (for Fe415).

   • Minimum reinforcement (Table 26).

7. Detailing: Distribution steel at top in shorter span? No. Provide corner bars (top) in both directions if corners are restrained.

[!TIP] Two-Way Slab Formulation: Always write: i) Effective span, ii) Type check (l_y/l_x), iii) Load calculation, iv) Moment coefficients from Table 27, v) M_x, M_y calculation, vi) Reinforcement for both spans, vii) Checks.

2.8 Control of Deflection

IS 456 Limits (L/d ratio):

• For simply supported: L/d ≤ 20 (for Fe415) or L/d ≤ 20 * (250 / σ_{st}) (WSM).

• For continuous: L/d ≤ 26.

• For cantilever: L/d ≤ 7.

Factors Affecting Deflection: Span L, support condition, load intensity, stiffness (I), steel percentage, creep, shrinkage.

Measures to Reduce Deflection:

1. Increase effective depth d.

2. Use higher grade steel (Fe500 vs Fe415).

3. Increase compression reinforcement (top steel).

4. Reduce span (add supports).

5. Use lighter aggregates, control cracking.

3.0 DESIGN OF COLUMNS (COMPRESSION MEMBERS)

3.1 Assumptions in Design (IS 456)

1. Maximum compression strain in concrete = 0.002.

2. Strain in steel ≤ yield strain (ε_y).

3. Plane sections remain plane.

4. Perfect bond between steel & concrete.

5. Stress in steel = 0.87 f_y (for design).

6. Stress in concrete = 0.45 f_ck (for design).

3.2 Short Columns

Axially Loaded: P_u ≤ 0.4 f_ck A_c + 0.67 f_y A_sc Uni-axial Bending: Use Interaction Diagram (or formula from SP 16). For rectangular sections:

$$ \frac{P_u}{f_{ck} b D} = \frac{1}{f_{ck}} \left[ 0.45 f_{ck} A_c + 0.87 f_y A_{sc} \right] \text{ vs } \frac{M_u}{f_{ck} b D^2} $$

Bi-axial Bending (Very High Priority):

IS 456 Eq. (39):

$$ \frac{P_{uz}}{P_{uz}} + \frac{M_{ux}}{M_{ux1}} + \frac{M_{uy}}{M_{uy1}} \leq 1.0 \quad \text{for short columns}$$

Where:

• P_{uz} = axial load capacity under concentric compression.

• M_{ux1}, M_{uy1} = moment capacities about x & y axes for P_u = 0.

• M_{ux}, M_{uy} = applied moments.

• For rectangular sections with uni-axial moment only: M_{uy1} replaced by (M_{uy} / M_{uy1}) * (1 - P_u / P_{uz}) term.

Design Procedure for Bi-axial:

1. Assume p_t (%), compute P_{uz}, M_{ux1}, M_{uy1}.

2. Check Eq. (39). If not satisfied, increase p_t.

3. Alternatively, use Bresler's Load Contour (approximate):

$$ \left( \frac{P_u}{P_{uz}} \right)^{1.5} + \left( \frac{M_{ux}}{M_{ux1}} \right)^{1.5} + \left( \frac{M_{uy}}{M_{uy1}} \right)^{1.5} \leq 1.0 $$

3.3 Longitudinal Reinforcement

• Min: 0.8% of gross area (A_g).

• Max: 6% of A_g (for columns).

• Arrangement: For b/D ratio, bars on 2 sides (if b/D > 0.65) or 4 sides (if b/D ≤ 0.65). Minimum 4 bars for rectangular, 6 for circular.

3.4 Transverse Reinforcement

• Lateral Ties:

  • Diameter: ≥ ϕ_main / 4 and ≥ 6 mm.

  • Spacing: ≤ least of (i) smallest lateral dimension, (ii) 16 × ϕ_main, (iii) 300 mm.

• Helical Reinforcement:

  • Diameter: ≥ ϕ_main / 6 and ≥ 8 mm.

  • Pitch: ≤ least of (i) 75 mm, (ii) 1/4 × core diameter, (iii) 3 × helical bar diameter.

  • Difference: Helical provides continuous lateral support, increases buckling load by ~20% (core size effect). Ties provide discrete support.

3.5 Slender Columns (Brief)

• Effective Length (l_eff): Depends on end conditions (Table 28, IS 456).

• Slenderness Ratio (λ): λ = l_eff / r (r = radius of gyration).

• If λ > 12 for unbraced, or λ > 40 for braced → ** slender**. Design load increased by moment magnification.

3.6 Detailing of Column Reinforcement (Sketches)

• Show longitudinal bars with clear cover (from Table 16, IS 456).

• Show ties/helix with spacing.

• Lap in longitudinal bars: stagger, provide additional ties.

• Development at base/top.

  DiagramCANVAS: Cross-section of rectangular column with 4-bar arrangement and lateral ties at spacing

4.0 DESIGN OF FOOTINGS & FOUNDATIONS

4.1 Types of Footings (with Sketches)

1. Isolated: For single column.

   DiagramSEARCH: isolated square footing plan and section

2. Combined: For two columns.

   DiagramSEARCH: combined rectangular footing

3. Strip: For walls.

   DiagramSEARCH: strip footing section

4. Raft: For poor soil, many columns.

   DiagramSEARCH: raft foundation plan

5. Stepped: For depth > 1.5m, economical.

   DiagramSEARCH: stepped footing section

4.2 Design of Isolated Square/Rectangular Footing

Steps:

1. Load: P_u = 1.5 * (P_column + self-weight).

2. Size: A = P_u / SBC. For square: B = √A. For rectangular: assume B = D ± 0.5m.

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

   • Two-way shear (punching): Critical at d/2 from column face. V_{u2} = P_u - [ (B*D) - ( (B-2*0.5d)*(D-2*0.5d) ) ] * SBC.

   • τ_{v2} = V_{u2} / ( (B-0.5d)*(D-0.5d) ) ≤ 0.25 √f_ck (Table 19).

   • Solve for d from shear. h = d + clear cover (≥ 50mm).

4. Bending Moment: Critical at column face (for one-way). M_u = SBC * (B/2 - b/2) * (D/2)^2 (for square col).

5. Reinforcement: A_st = M_u / (0.87 f_y * (d - 0.416 x_u)) (approx). Provide in both directions.

6. Checks: Development length, minimum steel (Table 26), bearing pressure (service).

4.3 Design of Stepped Footing

• Steps for Economy: When depth required > 1.5m. Steps of 0.3–0.5m.

• Reinforcement: Each step acts as a cantilever slab. Design each step for its moment.

• Detailing: Vertical bars from lower step to upper step, distribution steel.

  DiagramCANVAS: Stepped footing section with reinforcement in each step

5.0 DESIGN OF STAIRCASES

5.1 Types

Dog-legged, Open-well, Cantilever, Straight, Spiral.

5.2 Design of Dog-Legged Staircase (Very High Priority)

Given: Floor height H, Riser R, Tread T, Waist thickness t, Width w.

Steps:

1. Number of Risers: N_r = H / R (round up). Number of Treads = N_r - 1.

2. Staircase Length: L_s = (N_r / 2 - 1) * T (for dog-leg, 2 flights).

3. Effective Span: For flight: l_eff = L_s + w (if supported on walls). For landing: l_eff = clear span.

4. Loading (on plan):

   • Dead load: w_d = (R + t) * w / T (kN/m run).

   • Live load: as per IS 875 (2.5–5 kN/m² for office/residential).

   • Total UDL on flight: w_u = 1.5 * (w_d + LL * w).

5. Design Waist Slab as Simply Supported/Continuous Slab:

   • M_{max} = w_u * l_eff^2 / 8 (for simply supported flight).

   • Compute A_st for M_u (one-way slab if w < 1m, else two-way? Usually one-way).

6. Landing Slab: Design similarly, effective span = clear span between supports.

7. Reinforcement Detailing:

   • Flight: Main bars (tension) along slope, distribution bars perpendicular.

   • Landing: Main bars in shorter span if two-way, else along length.

   • Sketch: Show flight & landing with main bars, top bars at landing support, curtailment.

     DiagramCANVAS: Dog-legged staircase plan and elevation with reinforcement

6.0 BOND, DEVELOPMENT LENGTH & SPLICING

6.1 Bond Failure Mechanisms

1. Failure in Concrete: Splitting (radial cracks) or crushing (near bar surface).

2. Failure in Steel: Slippage (if bond stress < adhesion).

6.2 Development Length (L_d)

• Concept: Length required to develop full stress in bar.

• Formula (IS 456):

$$L_d = \frac{ϕ σ_{st}}{4 τ_{bd}}$$

Where:

*   `σ_{st}` = design stress in bar = 0.87 f_y.

*   `τ_{bd}` = design bond stress (Table 66, IS 456) depends on `f_ck` and steel grade.

*   For bars in compression, `L_d` is 25% less.

• Factors: f_ck, f_y, bar diameter ϕ, coating, confinement.

6.3 Lap Splices

• Lap Length: L_lap = L_d (for bars in tension, same grade).

• For different bar diameters: L_lap based on larger diameter.

• Transverse Reinforcement: If A_st > 2% in tension zone, provide additional stirrups over lap length (spacing ≤ 150mm or 4×main bar dia).

• Staggering: Laps should be staggered (min. stagger = 75mm or 4×lap length). Avoid laps at points of max moment.

7.0 THEORETICAL CONCEPTS & DETAILED DRAWING

7.1 Strain & Stress Diagrams for Singly Reinforced Beam (Ultimate State)

[[DIAGRAM: CANVAS: Rectangular beam cross-section at ULS showing:

• Strain: Linear from 0 at NA to ε_cu=0.0035 at top, ε_s at steel.

• Stress: Concrete: parabolic to 0.36 f_ck at NA, then rectangular 0.45 f_ck? No, simplified rectangular stress block: 0.36 f_ck over depth 0.42 x_u.

• Steel: 0.87 f_y (if yielded).

• NA: At depth x_u.

• Lever arm: z = d - 0.42 x_u. ]]

7.2 Critical Sections (with Sketches)

• Beam:

  • Moment: At midspan (SS), or at face of support (continuous).

  • Shear: At d from face of support (SS), or at face (continuous).

  • Bond: At support (for simply supported).

• Slab:

  • Moment: Midspan (one-way), or at panels (two-way).

  • Shear: At d from support (for one-way).

• Footing:

  • Bending: At face of column.

  • One-way shear: At d from face of column (along length).

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

• DiagramCANVAS: Composite sketch showing critical sections for beam, slab, footing with dashed lines

7.3 Design of Beams for Shear Only

Given V_u, b, d, f_ck, f_y.

1. Compute τ_v = V_u / (b d).

2. Get τ_c from Table 19 (based on p_z). Assume p_z = 0.5% if unknown.

3. If τ_v > τ_c, provide shear reinforcement:

   • V_{us} = V_u - τ_c b d.

   • A_{sv} / s_v = V_{us} / (0.87 f_y d).

   • Choose ϕ for stirrups (2-legged: A_{sv} = πϕ^2), find s_v.

   • Check s_v ≤ min(0.75d, 300mm).

7.4 Calculation of Moment of Resistance (Given Section)

• Singly Reinforced: Find x_u from A_st = 0.87 f_y / (0.36 f_ck) * b x_u. If x_u > x_{u,lim}, use over-reinforced formula (not preferred). Else, M_u = 0.36 f_ck b x_u (d - 0.42 x_u).

• Doubly Reinforced:

  1. Find x_u from A_st - A_sc = 0.36 f_ck b x_u / (0.87 f_y).

  2. If x_u > D_f (for T-beam), use T-beam formula.

  3. M_u = 0.36 f_ck b x_u (d - 0.42 x_u) + (A_sc - A_{sc1}) * 0.87 f_y (d - d') (where A_{sc1} is compression steel in balanced section).

• T-Beam: Check if NA in flange (x_u ≤ D_f). If yes, M_u = 0.36 f_ck b_w x_u (d - 0.42 x_u) + 0.45 f_ck (b_f - b_w) D_f (d - 0.42 x_u).

7.5 Sketches of Reinforcement Details

• Beam: Show longitudinal bars (tension/compression), stirrups (spacing), development at supports, curtailment.

• Column: Longitudinal bars (arrangement), ties/helix (pitch), laps.

• Footing: Main bars (both directions), distribution bars, column dowels, footing thickness.

• Staircase: Flight bars (slope), landing bars, top bars, curtailment.

• Slab: Main bars (shorter span), distribution bars (longer span), corner bars (if restrained), edge beams.

\boxed{\text{Always use Limit State Method (LSM) for design unless specified. Check exposure condition for cover from Table 16, IS 456.}}