UNIT 5: STRUCTURAL DESIGN & DRAWING (RCC-I) - EXAM-FOCUSED NOTES

1.0 FUNDAMENTALS OF LIMIT STATE DESIGN (THEORETICAL FOUNDATION)

1.1 Principles of Limit State Design (LSD)

• Concept: Structure designed to satisfy two limit states:

  • Ultimate Limit State (ULS): Safety against collapse (strength, stability). Loads are factored loads ($$\displaystyle \gamma_f \times \text{characteristic load} $$).

  • Serviceability Limit State (SLS): Comfort & utility under normal use (deflection, cracking). Loads are characteristic loads.

• Partial Safety Factors ($$\displaystyle \gamma_m, \gamma_f $$):

  • $$\displaystyle \gamma_m $$: For materials (concrete $$\displaystyle \gamma_c = 1.5 $$, steel $$\displaystyle \gamma_s = 1.15 $$ for LSM).

  • $$\displaystyle \gamma_f $$: For loads (Dead $$\displaystyle \gamma_d = 1.5 $$, Live $$\displaystyle \gamma_l = 1.5 $$, Wind/Seismic $$\displaystyle \gamma_w = 1.2 $$ or $1.5$).

• Design Equation: $$\displaystyle S_d \leq S_{allowable} $$, where $$\displaystyle S_d = \gamma_f \times S_{characteristic} $$ (load effect) and $$\displaystyle S_{allowable} = \frac{S_{characteristic}}{\gamma_m} $$ (material strength).

1.2 Key Definitions

• Characteristic Load ($$\displaystyle P_k $$): Load with 95% probability of not being exceeded in 50 years (from codes/statistics).

• Characteristic Strength ($$\displaystyle f_{ck}, f_y $$): Material strength below which not more than 5% of test results fall.

  • $$\displaystyle f_{ck} $$: Characteristic compressive strength of concrete (cube/cylinder).

  • $$\displaystyle f_y $$: Characteristic yield strength of steel.

• Partial Safety Factor ($$\displaystyle \gamma_m $$): Factor accounting for material variability, manufacturing, construction.

1.3 Balanced Section: WSM vs. LSM

1.4 Assumptions in Compression Member Design (IS 456)

1. Plane sections before bending remain plane after bending.

2. Maximum strain in concrete at outermost compression fiber = $0.0035$.

3. Stress in concrete = $$\displaystyle 0.45 f_{ck} $$ for $$\displaystyle f_{ck} \leq 50 $$ MPa (IS 456 Eq. 39).

4. Stress in steel: $$\displaystyle f_s = f_y $$ if $$\displaystyle \epsilon_s \geq \frac{f_y}{1.15E_s} + 0.002 $$; else linear.

5. Perfect bond between steel & concrete (no slip).

6. Stresses in concrete & steel derived from stress-strain curves.

[!TIP] Exam Focus: Differences between WSM & LSM balanced sections are a very common 5-7m question. Draw neat sketches showing stress & strain diagrams for both.

2.0 FLEXURAL DESIGN OF BEAMS (THEORY & NUMERICAL)

2.1 Stress-Strain Diagrams (IS 456)

• Concrete: Parabolic up to $0.002$ strain, then linear to $0.0035$ at peak. For LSM design, simplified as rectangular stress block of $$\displaystyle 0.45f_{ck} $$ over depth $$\displaystyle 0.9x_u $$.

• Steel: Elasto-plastic. Yield plateau after $$\displaystyle f_y $$. Strain hardening beyond.

2.2 Assumptions in Flexural Design (LSM)

1. Plane sections remain plane.

2. Strain in concrete $\leq 0.0035$.

3. Tensile strength of concrete ignored.

4. Stress in steel = $$\displaystyle f_y $$ (if yielded), in concrete = $$\displaystyle 0.36 f_{ck} $$ (for $$\displaystyle f_{ck} \leq 50 $$ MPa) over $$\displaystyle 0.48 x_u $$ (or $$\displaystyle 0.9x_u $$ for stress block centroid).

5. Perfect bond.

2.3 Types of Sections (with Sketches)

• Balanced: $$\displaystyle x_u = x_{u,lim} $$. Both materials reach strain limits simultaneously.

  • $$\displaystyle x_{u,lim} = \frac{0.0035}{0.005 + f_y / E_s} $$ (for Fe415, $$\displaystyle x_{u,lim} \approx 0.48d $$).

• Under-Reinforced: $$\displaystyle x_u < x_{u,lim} $$. Steel yields first → ductile failure (warning). Preferred design.

• Over-Reinforced: $$\displaystyle x_u > x_{u,lim} $$. Concrete crushes first → brittle failure. Avoided by limiting $$\displaystyle p_t \leq p_{t,lim} $$.

[!DIAGRAM: CANVAS: Draw strain diagram (linear) and stress diagram (rectangular block for concrete, constant $$\displaystyle f_y $$ for steel) for balanced, under, over sections. Label $$\displaystyle x_u $$, $d$, $0.0035$, $$\displaystyle f_y $$, $$\displaystyle 0.36f_{ck} $$.]

2.4 Design Formulas (Singly Reinforced Rectangular)

Depth of Neutral Axis:

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

For balanced section (steel just yielded, $$\displaystyle f_s = f_y $$):

$$x_{u,lim} = \frac{0.0035}{0.005 + f_y / E_s} \cdot d \quad \text{(IS 456 Eq. 38)}$$

Lever Arm:

$$z = d - 0.42 x_u \quad (\text{for } x_u \leq x_{u,lim})$$

Moment of Resistance ($$\displaystyle M_u $$):

$$M_u = 0.36 f_{ck} b x_u (d - 0.42 x_u) = T \cdot z = f_s A_{st} \cdot z$$

Limiting Moment ($$\displaystyle M_{u,lim} $$):

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

Area of Steel ($$\displaystyle A_{st} $$) for given $$\displaystyle M_u $$:

1. If $$\displaystyle M_u \leq M_{u,lim} $$ (Under-reinforced):

$$A_{st} = \frac{M_u}{0.87 f_y z} \quad \text{where } z = 0.87d \text{ (approx)}$$

More accurately, solve quadratic from $$\displaystyle M_u = 0.36 f_{ck} b x_u (d - 0.42 x_u) $$ for $$\displaystyle x_u $$, then $$\displaystyle A_{st} = \frac{0.36 f_{ck} b x_u}{0.87 f_y} $$.

2. If $$\displaystyle M_u > M_{u,lim} $$: Design as doubly reinforced.

2.5 Doubly Reinforced Rectangular Beam

• When used: $$\displaystyle M_u > M_{u,lim} $$ or architectural depth constraint.

• Analysis:

  1. $$\displaystyle M_{u1} = M_{u,lim} $$ (carried by $$\displaystyle A_{st1} $$ & concrete).

  2. $$\displaystyle M_{u2} = M_u - M_{u,lim} $$ (carried by $$\displaystyle A_{sc} $$ & additional $$\displaystyle A_{st2} $$).

  3. For $$\displaystyle M_{u2} $$: $$\displaystyle A_{sc} = A_{st2} = \frac{M_{u2}}{(f_{sc} - 0.45 f_{ck})(d - d')} $$.

     • $$\displaystyle f_{sc} $$ from IS 456 Table B-2 (strain in compression steel $$\displaystyle \epsilon_{sc} = 0.0035 - \frac{0.0035 x_{u,lim}}{d'} $$).

  4. Total: $$\displaystyle A_{st} = A_{st1} + A_{st2} $$, $$\displaystyle A_{sc} $$ as above.

• Key Check: Ensure $$\displaystyle A_{sc} $$ & $$\displaystyle A_{st2} $$ provided are equal and symmetrical for equilibrium.

2.6 T-Beam Design

• Effective Flange Width ($$\displaystyle b_f $$):

$$b_f = b_w + \frac{l_o}{6} \quad \text{or} \quad b_w + \frac{l_o}{12} + b_w \quad \text{or} \quad b_w + \frac{l_o}{4} \quad (\text{whichever is least})$$

where $$\displaystyle l_o $$ = distance between points of zero contraflexure.

• Check if T-section valid: Compare $$\displaystyle D_f $$ (slab thickness) with $$\displaystyle x_u $$.

  • If $$\displaystyle x_u \leq D_f $$: Rectangular section of width $$\displaystyle b_f $$.

  • If $$\displaystyle x_u > D_f $$: Actual T-section. Stress block extends into rib.

$$M_u = 0.36 f_{ck} (b_f - b_w) D_f (d - 0.42 D_f) + 0.36 f_{ck} b_w x_u (d - 0.42 x_u)$$

    Solve for $$\displaystyle x_u $$ iteratively, then $$\displaystyle A_{st} $$.

[!TIP] Numerical Focus: T-beam problems (Jun 2025, May 2023) often require checking $$\displaystyle x_u $$ vs. $$\displaystyle D_f $$. Always compute $$\displaystyle b_f $$ first.

3.0 SHEAR AND TORSION IN BEAMS

3.1 Critical Sections for Shear (IS 456)

• At support face for simply supported beams.

• At face of support for continuous beams (where moment is zero).

• For beams with point loads, at $d$ from support face.

[!DIAGRAM: CANVAS: Sketch simply supported beam, continuous beam over two spans. Mark critical shear sections: at support for SSB, at $d$ from support face for point load, at support face for continuous.]

3.2 Shear Transfer Mechanism

At a flexural-shear crack, shear force $V$ is resisted by:

1. Aggregate Interlock: Roughness of crack faces.

2. Dowel Action: Tension steel crossing crack.

3. Stirrups (Shear Reinforcement): Primary resistance in LSM.

3.3 Design for Shear

1. Nominal Shear Stress: $$\displaystyle \tau_v = \frac{V_u}{b d} $$ (for rectangular sections).

2. Permissible Shear Stress ($$\displaystyle \tau_c $$): From IS 456 Table 19, based on $$\displaystyle f_{ck} $$ and $$\displaystyle p_t = \frac{A_{st}}{b d} \times 100\% $$.

3. Design Shear Reinforcement:

   • If $$\displaystyle \tau_v \leq \tau_{c,min} $$: Provide minimum stirrups ($$\displaystyle A_{sv} \geq 0.4 \frac{b s_v}{0.87 f_y} $$).

   • If $$\displaystyle \tau_{c,min} < \tau_v \leq \tau_c $$: Provide stirrups as per $$\displaystyle \tau_v - \tau_c $$.

$$V_{us} = V_u - \tau_c b d$$

$$A_{sv} = \frac{V_{us} s_v}{0.87 f_y d}$$

    Spacing $$\displaystyle s_v \leq 0.75d $$ (vertical) or $d$ (inclined).

*   If $$\displaystyle \tau_v > \tau_{c,max} $$: **Increase beam size** (cannot design).

4. Maximum Shear Stress: $$\displaystyle \tau_{v,max} = 0.5 \sqrt{f_{ck}} $$ (for $$\displaystyle f_{ck} \leq 40 $$ MPa).

3.4 Combined Bending & Torsion

• Design Torsional Moment ($$\displaystyle T_u $$): From loading.

• Equivalent Shear & Bending (IS 456):

$$V_{u,eq} = V_u + \frac{T_u}{D}$$

$$M_{u,eq} = M_u + \frac{T_u}{2} \left(1 + \frac{D}{b}\right) \text{ or } \frac{T_u}{2}\left(1 + \frac{b}{D}\right) \text{ (whichever gives larger } M_{u,eq})$$

• Reinforcement Detailing:

  • Longitudinal: Extra top/bottom bars at corners (torsion).

  • Transverse: Closed stirrups along entire length, spacing $$\displaystyle s_v \leq \min\left(\frac{D}{4}, 300\right) $$.

[!TIP] Common Pitfall: Forgetting to use equivalent shear $$\displaystyle V_{u,eq} $$ for stirrup design when torsion is present.

4.0 SLAB DESIGN (ONE-WAY & TWO-WAY)

4.1 Classification

• One-Way: $$\displaystyle l_y/l_x \geq 2 $$. Bends in short direction.

• Two-Way: $$\displaystyle l_y/l_x < 2 $$. Bends in both directions.

• Simply Supported, Continuous, Cantilever.

4.2 One-Way Slab (Simply Supported)

1. Load Calculation (UDL): $$\displaystyle w = \text{SW} + \text{LL} + \text{FF} + \text{Partitions} $$.

   • SW = $$\displaystyle 0.125 \times 25 = 3.125 \text{ kN/m}^2 $$ (for 125 mm thick).

2. Effective Span ($$\displaystyle l_{eff} $$): Clear span + effective depth (for simply supported).

3. Depth (for deflection): $$\displaystyle l_{eff}/d \leq \text{IS 456 Table C} $$ (based on steel % & span type). For Fe415, simply supported, $l/d \leq 20$ (initial), modify with $$\displaystyle A_{st} $$.

4. Bending Moment (Mid-span): $$\displaystyle M_u = \frac{w l_{eff}^2}{8} $$ (factored).

5. Reinforcement:

   • Main steel (along $$\displaystyle l_x $$): $$\displaystyle A_{st} = \frac{M_u}{0.87 f_y z} $$.

   • Distribution steel (along $$\displaystyle l_y $$): $$\displaystyle A_{st} = 0.0012 b D $$ (min).

6. Spacing: Main steel $\leq 3D$ or 300 mm; Distribution $\leq 5D$ or 450 mm.

4.3 Two-Way Slab (Restrained Corners)

• IS 456 Method (Coefficients):

  • For $$\displaystyle l_y/l_x = 1 $$: $$\displaystyle M_{ux} = \alpha_x w l_x^2 $$, $$\displaystyle M_{uy} = \alpha_y w l_y^2 $$.

  • $$\displaystyle \alpha_x, \alpha_y $$ from Table 26 (e.g., for corners held down, $$\displaystyle \alpha_x = 0.0625 $$ for $$\displaystyle l_y/l_x=1 $$).

• Reinforcement:

  • Two orthogonal sets. $$\displaystyle A_{stx} = M_{ux}/(0.87 f_y z_x) $$, $$\displaystyle A_{sty} = M_{uy}/(0.87 f_y z_y) $$.

  • Corner Reinforcement: Provide 4 bars (2 each way) at corners to hold down.

• Depth Check: $$\displaystyle l_{eff}/d \leq 28 $$ (for two-way, Fe415, continuous).

4.4 Numerical Steps

1. Compute loads (SW, LL, FF).

2. Find $$\displaystyle l_{eff} $$.

3. Assume depth $D$ (check deflection later), find $d$.

4. Calculate $$\displaystyle M_u $$ using coefficients.

5. Find $$\displaystyle A_{st} $$ for both directions.

6. Check $$\displaystyle p_t $$ for deflection (Table C).

7. Provide distribution steel.

5.0 COLUMN DESIGN (AXIAL LOAD WITH BENDING)

5.1 Short vs. Long Columns

• Short Column: $$\displaystyle l_{ex}/D \leq 12 $$ for braced, $\leq 5$ for unbraced (IS 456 Cl. 25.1.2). Failure by material crushing.

• Long/Slender Column: $$\displaystyle l_{ex}/D > \text{limit} $$. Failure by buckling. Requires increased $$\displaystyle P_u $$ for same $$\displaystyle M_u $$ (use interaction curves for short column with additional moment $$\displaystyle M_a = \frac{P_u}{1000} \left(\frac{l_{ex}}{D}\right)^2 $$).

5.2 Uni-axial Bending (IS 456 Interaction Charts)

• Given: $$\displaystyle b, D, f_{ck}, f_y, P_u, M_u $$ (about major axis).

• Steps:

  1. Compute $$\displaystyle p_t = \frac{A_{sc}}{b D} \times 100\% $$.

  2. Find $$\displaystyle M_{u,0} = 0.45 f_{ck} b D (1 - 0.045 p_t) $$ (for axial load only).

  3. Compute ratios: $$\displaystyle P_u / (b D f_{ck}) $$, $$\displaystyle M_u / (b D^2 f_{ck}) $$.

  4. Use IS 456 Chart 44 (for $$\displaystyle f_y = 415 $$) or Chart 45 (for $$\displaystyle f_y = 500 $$).

  5. Check if given $$\displaystyle (P_u, M_u) $$ lies below curve for assumed $$\displaystyle p_t $$. If not, increase $$\displaystyle p_t $$.

  6. Once safe, compute $$\displaystyle A_{sc} = p_t \times b D / 100 $$.

• Reinforcement Detailing:

  • Two-sided: Bars on 2 faces parallel to bending (for small $$\displaystyle M_u $$).

  • Four-sided: Bars on all faces (for larger $$\displaystyle M_u $$ or bi-axial).

  • Min: $0.8\% bD$, Max: $4\% bD$ (or $6\%$ with ties).

5.3 Bi-axial Bending (ISM/IS 456 Approximate Method)

• Given: $$\displaystyle P_u, M_{ux}, M_{uy} $$.

• Check:

$$\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha} \leq 1.0$$

where $$\displaystyle M_{ux1}, M_{uy1} $$ are uniaxial capacities for same $$\displaystyle P_u $$ and $$\displaystyle p_t $$, $$\displaystyle \alpha = 1.5 $$ to $2.0$ (use $1.5$ for preliminary).

• Design Iterative: Assume $$\displaystyle p_t \rightarrow $$ get $$\displaystyle M_{ux1}, M_{uy1} $$ from charts $$\displaystyle \rightarrow $$ check inequality $$\displaystyle \rightarrow $$ adjust $$\displaystyle p_t $$.

5.4 Transverse Reinforcement

• Lateral Ties:

  • Diameter $\geq 8$ mm or $$\displaystyle \frac{1}{4} $$ diameter of main bar (min).

  • Spacing $\leq$ least of: $D$, $16 \times$ tie diameter, $300$ mm.

  • Special Rule: Every 3rd or 4th tie should be continuous (wrap around).

• Helical Reinforcement (Circular):

  • Diameter $\geq 8$ mm.

  • Pitch $\leq$ least of: $D/6$, $75$ mm, $300$ mm.

  • Increased Capacity: $$\displaystyle P_{u,helical} = 1.05 \times P_{u,tied} $$ (if same steel).

5.5 Numerical Steps (Uni-axial)

1. Assume $$\displaystyle p_t = 1\% $$.

2. Find $$\displaystyle M_{u,0} $$ for zero moment.

3. Compute normalized ratios.

4. Read $$\displaystyle p_t $$ from chart.

5. If assumed $$\displaystyle p_t $$ ≠ read $$\displaystyle p_t $$, iterate.

6. Compute $$\displaystyle A_{sc} $$.

7. Design ties/helix.

6.0 FOOTING DESIGN (ISOLATED)

6.1 Design Steps (Isolated Rectangular Footing)

1. Loads: $$\displaystyle P_u $$ (factored column load) + self-weight of footing ($\approx 0.1 \times \text{volume}$).

2. Size of Footing:

   • Assume depth $h$, calculate $B \times L$ such that gross pressure $$\displaystyle q_g = \frac{P_u + \gamma_{c} B L h}{B L} \leq \text{SBC} $$.

   • Usually start with square footing: $$\displaystyle B = L = \sqrt{\frac{P_u}{\text{SBC}}} $$.

3. Thickness (for shear):

   • Punching Shear (at $d/2$ from column face):

$$\tau_{vp} = \frac{P_u}{4 (b_0 + d) d} \leq \tau_{cp}$$

    where $$\displaystyle b_0 $$ = perimeter of column at critical section.

*   **Beam Shear (at column face):** Check $$\displaystyle \tau_v \leq \tau_c $$ (usually governs for square footings).

*   Solve for $d$ from shear equation, then $$\displaystyle h = d + \text{cover} + \frac{1}{2} \text{bar dia} $$.

4. Bending Moment (Critical section at column face):

$$M_u = \frac{P_u}{2} \left( \frac{B - b}{2} \right) \text{ (per meter width)}$$

Design as **one-way slab** (if $$\displaystyle L > B $$) or **two-way slab** (if square) for bending.

5. Reinforcement:

   • Longitudinal (Bottom): $$\displaystyle A_{st} = M_u / (0.87 f_y z) $$ in both directions.

   • Distribution (Top): $$\displaystyle A_{st} = 0.0012 B L $$ (min, for shrinkage).

6. Checks:

   • Net pressure $$\displaystyle q_{net} = \frac{P_u}{B L} \leq \text{SBC} $$.

   • Gross pressure $$\displaystyle q_g \leq \text{SBC} $$.

   • Shear (punching & beam).

   • Bending.

6.2 Stepped Footing

• Steps provided for large depths to reduce steel.

• Each step acts as a cantilever slab.

• Design each step for bending & shear separately.

[!DIAGRAM: CANVAS: Plan of square footing showing column, critical sections for shear (at $d/2$ and column face) and bending (column face). Longitudinal section showing depth $h$, effective depth $d$, main bars (bottom), distribution bars (top).]

7.0 STAIRCASE DESIGN (DOG-LEGGED)

7.1 Key Parameters

• Riser (R): Vertical height (typically 150-180 mm).

• Tread (T): Horizontal width (typically 250-300 mm).

• Waist Slab Thickness (t): 100-150 mm.

• Flight Slope: $$\displaystyle \tan \theta = R/T $$.

• Number of Risers: $$\displaystyle N_R = \text{Total Height}/R $$.

• Number of Treads: $$\displaystyle N_T = N_R - 1 $$.

• Stair Width (W): Clear width (typically 1.0-1.5 m).

7.2 Design Steps (Waist Slab as Inclined Beam)

1. Loads on Waist Slab (per m² horizontal):

   • Self-weight = $t \times 25 \times \cos \theta$ (inclined).

   • Floor finish = $$\displaystyle 0.5-1.0 \text{ kN/m}^2 $$ (horizontal).

   • Live load = $$\displaystyle 3-5 \text{ kN/m}^2 $$ (office/public).

   • Total UDL ($w$): Sum of above $\times \cos \theta$ (to get along slope) or directly compute load per meter horizontal run.

2. Effective Span ($$\displaystyle l_{eff} $$):

   • Wall-supported: Clear distance between walls + $t$ (or bearing).

   • Stringer beam-supported: C/C distance between beams.

3. Bending Moment (Mid-flight): $$\displaystyle M_u = \frac{w l_{eff}^2}{8} $$ (simply supported).

   • Negative moment at landing support: Consider if landing is cantilevered.

4. Depth for Deflection: $$\displaystyle l_{eff}/d \leq 20 $$ (simply supported, Fe415).

5. Reinforcement:

   • Main Steel (along flight): $$\displaystyle A_{st} = M_u/(0.87 f_y z) $$ (bottom for mid-span, top at landing if cantilever).

   • Distribution Steel (across width): $$\displaystyle A_{st} = 0.0012 W t $$ (min).

6. Reinforcement Detailing:

   • Main bars run full length of flight + landing.

   • Provide extra bars at top of landing (if cantilevered).

   • Distribution bars across width.

[!TIP] Exam Focus: Dog-legged staircase is almost every paper. Remember: waist slab is an inclined beam. Calculate load along the slope or convert to horizontal UDL consistently.

8.0 BOND, DEVELOPMENT LENGTH & LAP SPLICES

8.1 Bond Stress & Failure

• Bond Stress ($$\displaystyle \tau_{bd} $$): Average shear stress along steel-concrete interface.

$$\tau_{bd} = \frac{T_u}{n \times \Sigma (O \times L_d)}$$

where $$\displaystyle T_u $$ = tensile force, $n$ = number of bars, $O$ = perimeter, $$\displaystyle L_d $$ = development length.

• Failure Modes:

  1. Pull-out: Steel slips out (smooth bars).

  2. Splitting: Concrete splits along bar (common in tension).

  3. Anchorage: Steel yields before bond fails (good).

8.2 Development Length ($$\displaystyle L_d $$)

• Tension:

$$L_d = \frac{\phi \sigma_{st}}{4 \tau_{bd}}$$

where $$\displaystyle \tau_{bd} = 1.6 \frac{f_{ck}}{1.5} $$ (for plain bars) or $$\displaystyle 2.4 \frac{f_{ck}}{1.5} $$ (for deformed bars) (IS 456 Eq. 47).

Simplified: $$\displaystyle L_d = \frac{\phi f_y}{4 \tau_{bd}} $$ (with $$\displaystyle \tau_{bd} $$ from Table 66).

• Compression: $$\displaystyle L_d $$ is shorter (no splitting). $$\displaystyle L_d = \frac{\phi f_y}{4 \tau_{bd}} $$ but $$\displaystyle \tau_{bd} $$ is higher (use same formula, but bond is better in compression).

• Factors: $$\displaystyle L_d $$ increases if: bars are in tension, bars are in compression (less), bars are curtailed, bars are at top (reduce by 30% if $$\displaystyle > 12\phi $$ cover), laps.

8.3 Lap Splice

• Purpose: Join two bars when full length unavailable.

• Lap Length ($$\displaystyle L_{lap} $$):

  • Tension: $$\displaystyle L_{lap} = L_d \times \frac{\text{stress in shorter bar}}{\text{stress in longer bar}} $$ (if different diameters).

    • For same dia: $$\displaystyle L_{lap} = L_d $$ (if $\leq 4\phi$ diameter difference) or larger.

  • Compression: $$\displaystyle L_{lap} = \text{smaller of } (L_d \text{ in compression}, 24\phi) $$.

• Transverse Reinforcement: Provide stirrups (or helical) over lap length if:

  • $$\displaystyle A_{st} > 4\% bD $$ (compression), or

  • Lap length $$\displaystyle > 150 $$ mm (tension), or

  • Bars are large ($$\displaystyle > 36 $$ mm dia, direct lap not allowed, use welding or mechanical splices).

• Splicing Different Diameters: Stress in smaller bar governs. Use stress ratio method.

8.4 Standard Hooks & Bends

• Anchorage Value: A standard hook (180° bend + 4$\phi$ extension) provides $16\phi$ anchorage length (IS 456).

• Use: At simply supported beam supports, cantilever tops, etc.

[!TIP] Common Question: "Design lap splice for 12mm with 20mm bar." Solution: Compute $$\displaystyle L_d $$ for each bar (based on $\phi$), then $$\displaystyle L_{lap} = L_{d,12} \times \frac{f_{s,12}}{f_{s,20}} $$ where $$\displaystyle f_s $$ = stress in each bar at splice section (from moment calculation). Provide transverse reinforcement.

9.0 DEFLECTION & CRACK CONTROL

9.1 Approaches for Deflection Control (IS 456)

1. Limiting Span/Depth Ratio ($$\displaystyle l_{eff}/d $$): From Table C (based on steel % & support type).

2. Limiting Area of Steel ($$\displaystyle A_{st}/bd $$): Ensures under-reinforced section.

3. Use of Compression Reinforcement: Reduces long-term deflection.

9.2 Measures for Reducing Deflection

• Increase Beam Depth: Most effective ($l/d$ ratio).

• Use High-Strength Steel: Allows less steel for same $$\displaystyle M_u $$ → less steel % → higher $l/d$ limit.

• Add Compression Steel: Reduces $l/d$ limit factor.

• Control Cracking: Proper curing, min. cover, bar spacing.

9.3 Crack Width Control

• Limiting Bar Spacing: $s \leq \min(300 \text{ mm}, 3D)$ for flexure (IS 456 Cl. 26.3.2).

• Adequate Cover: Increases $$\displaystyle s_{max} $$.

• Use Smaller Diameter Bars: More bars at same area → less spacing.

10.0 REINFORCEMENT DETAILING & DRAWING (SKETCHES)

10.1 Singly Reinforced Beam Cross-Section

[!DIAGRAM: CANVAS: Cross-section of rectangular beam. Show:

• Strain diagram: linear from 0.0035 at top to $$\displaystyle \epsilon_s $$ at steel level.

• Stress diagram: rectangular block (0.36f_ck) over depth 0.9x_u, constant f_y in steel.

• Label: b, D, d, cover, x_u, z, A_st.

• Show compression zone (above NA) and tension zone.]

10.2 Reinforcement Details Sketches

• Simply Supported Beam: Main bars (bottom) full length, top bars (anchorage) at supports, stirrups (closer near supports).

• Continuous Beam: Top bars over supports (vega), bottom bars in spans. Stirrups throughout, denser near supports.

• Doubly Reinforced Beam: Top bars ($$\displaystyle A_{sc} $$) and bottom bars ($$\displaystyle A_{st} $$) both full length. Stirrups enclose both cages.

• T-Beam: Show flange width $$\displaystyle b_f $$, rib width $$\displaystyle b_w $$. Main bars in rib bottom. Flange may have nominal distribution steel.

• Square Column (2-sided): Bars on two faces parallel to bending moment.

• Square Column (4-sided): Bars on all four faces, ties with 135° hooks.

• Circular Column with Helix: Longitudinal bars around periphery, helical reinforcement with pitch.

• Isolated Footing:

  • Plan: Column at center, main bars (both ways) in bottom, distribution bars (top) grid.

  • Section: Show column, footing depth $h$, effective depth $d$, main bars (bottom), distribution bars (top), clear cover.

• Dog-Legged Staircase:

  • Flight: Waist slab with main bars (along flight) at bottom (mid-span) and top (at landing if cantilevered). Distribution bars across width.

  • Landing: Main bars in two directions (if simply supported on walls).

10.3 Standard Hook & Bend

• Hook: 180° bend + 4$\phi$ extension. Anchorage value = $16\phi$.

• Bend (for anchorage): 90° or 135° bend with $4\phi$ extension.

[!TIP] Drawing Questions (4-6m): Always draw clear cross-sections showing all bars, covers, and dimensions. Label $b, D, d, \phi, \text{spacing}$. For columns, show tie/helix details. For footings, show both plan and section.

SUMMARY OF HIGH-FREQUENCY TOPICS FOR EXAM:

1. Balanced/Under/Over Sections (Sketches mandatory).

2. Singly & Doubly Reinforced Beam Design (Find $$\displaystyle A_{st} $$ or $$\displaystyle M_u $$).

3. T-Beam Analysis ($$\displaystyle b_f $$, $$\displaystyle x_u $$ check).

4. Shear Design (Stirrups calculation).

5. One-Way & Two-Way Slab (Loads, $$\displaystyle M_u $$, $$\displaystyle A_{st} $$).

6. Column Design (Uni-axial & Bi-axial) using IS 456 charts.

7. Isolated Footing (Size, depth for shear, $$\displaystyle A_{st} $$).

8. Dog-Legged Staircase (Loads, $$\displaystyle M_u $$, reinforcement).

9. Lap Splice for Different Diameters (Calculation + transverse steel).

10. Reinforcement Detailing Sketches (Beam, Column, Footing, Stair).