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CE-601 · Structural Design & Drawing (RCC-I)/Quick Revision Short Notes

Structural Design & Drawing (RCC-I) (CE-601) - Unit 2 Short Notes

UNIT 2: RCC DESIGN & DETAILING - SHORT NOTES

(Aligned with CE-601 Past Papers 2022-2025)


1. FUNDAMENTALS OF LIMIT STATE DESIGN (LSD)

Philosophy: LSD is a probabilistic method ensuring structure does not reach limit states (collapse, serviceability) during design life. Contrasts with Working Stress Method (WSM) which uses elastic theory with constant safety factors.

Key Parameters:

  • Characteristic Loads ($$\displaystyle F_k $$): Loads with 95% survival probability (dead, live, wind, seismic).

  • Characteristic Strengths ($$\displaystyle f_{ck} $$, $$\displaystyle f_{yk} $$): Material strengths with 5% defect probability.

  • Partial Safety Factors:

    • Load factors ($$\displaystyle \gamma_f $$): 1.5 (DL+LL), 1.2 (DL only), 0.9 (earthquake).

    • Material factors ($$\displaystyle \gamma_m $$): 1.5 (concrete), 1.15 (steel).

Stress-Strain Curves (IS 456):

  • Concrete: Parabolic-rectangular for LSM. For WSM, modulus of elasticity $$\displaystyle E_c = 5000\sqrt{f_{ck}} $$.

  • Steel: Linear elastic up to yield, then plastic. $$\displaystyle E_s = 2 \times 10^5 $$ N/mm².

Assumptions in Flexure (IS 456):

  1. Plane sections remain plane.

  2. Tensile strength of concrete ignored.

  3. Perfect bond between steel & concrete.

  4. Strain in steel & concrete at same level is equal.

[!TIP] Exam Focus: Difference between balanced section in LSM (x_u = x_{u,lim}) vs. WSM (σ_c = σ_{c,max}, σ_st = f_s).


2. FLEXURAL DESIGN OF BEAMS

Singly Reinforced Rectangular Section (LSM)

Balanced Section: NA at limiting depth $$\displaystyle x_{u,lim} = 0.48d $$ (Fe415). Strain in steel = 0.87f_y/Es + 0.002. Moment Capacity:

$$M_u = 0.36 f_{ck} b x_u (d - 0.42 x_u) \quad \text{or} \quad M_u = 0.87 f_y A_{st} z$$

where $$\displaystyle z = d - 0.42 x_u $$ (lever arm).

Design Steps for Given $$\displaystyle M_u $$:

  1. Compute $$\displaystyle M_u = 1.5 \times M $$ (service load moment).

  2. Find $$\displaystyle R_u = \frac{M_u}{b d^2} $$.

  3. From $$\displaystyle R_u $$ chart, get $$\displaystyle p_t = \frac{A_{st}}{bd} \times 100 $$.

  4. Compute $$\displaystyle A_{st} = \frac{p_t}{100} b d $$.

  5. Check: $$\displaystyle x_u = \frac{0.87 f_y A_{st}}{0.36 f_{ck} b} < 0.48d $$.

Under/Over-Reinforced:

  • Under: $$\displaystyle A_{st} < A_{st,lim} $$ → $$\displaystyle x_u < x_{u,lim} $$ (ductile failure).

  • Over: $$\displaystyle A_{st} > A_{st,lim} $$ → $$\displaystyle x_u > x_{u,lim} $$ (brittle, avoid).

Doubly Reinforced Beams

Necessity: When $$\displaystyle M_u > M_{u,lim} $$ for given $b,d$. Design:

  1. Compute $$\displaystyle M_{u1} = M_{u,lim} $$ for section with $$\displaystyle A_{st,lim} $$.

  2. Additional moment $$\displaystyle M_{u2} = M_u - M_{u1} $$.

  3. Additional steel $$\displaystyle A_{sc} = \frac{M_{u2}}{0.87 f_y (d - d')} $$.

  4. Total $$\displaystyle A_{st} = A_{st,lim} + A_{sc} $$ (since both steels at same stress).

T-Beams

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

$$b_f = b_w + \frac{l_o}{6} \quad \text{or} \quad b_f = b_w + 4D_f \quad \text{or} \quad b_f = \text{spacing of beams}$$

Take least.

Analysis:

  • If $$\displaystyle x_u \leq D_f $$: NA in flange → treat as rectangular with $$\displaystyle b = b_f $$.

  • If $$\displaystyle x_u > D_f $$: NA in rib → consider flange contribution as $$\displaystyle C_{f} = 0.45 f_{ck} (b_f - b_w) D_f $$.


3. SHEAR AND TORSION IN BEAMS

Shear Design

Critical Sections: At support face & at $d$ from support face for simply supported beams. Nominal Shear Stress:

$$\tau_v = \frac{V_u}{b d}$$

Design Shear Strength: $$\displaystyle \tau_{c} $$ from Table 19 (IS 456) based on $$\displaystyle p_t $$ and $$\displaystyle f_{ck} $$.

Shear Reinforcement:

  • If $$\displaystyle \tau_v > \tau_{c,max} $$: Redesign section.

  • If $$\displaystyle \tau_c < \tau_v < \tau_{c,max} $$: Provide stirrups.

$$A_{sv} = \frac{(\tau_v - \tau_c) b s}{0.87 f_y} \quad \text{or} \quad \frac{V_{us}}{0.87 f_y d}$$

where $s$ = spacing.

Maximum Spacing: $$\displaystyle s_{max} = 0.75d $$ (stirrups), $d$ (bent-up bars).

Torsion

Ultimate Torsional Moment $$\displaystyle T_u $$:

  • Longitudinal steel: $$\displaystyle A_{st} = \frac{T_u}{f_y} \times \frac{u}{2p_t d} $$ (approx).

  • Transverse (stirrups): $$\displaystyle A_{sv} \times \frac{s_v}{p_{hov}} = \frac{T_u}{f_y} $$.

  • Reinforcement Detailing: Closed ties along entire length, additional longitudinal bars at corners.

[!TIP] Common Pitfall: For torsion, provide both longitudinal & transverse reinforcement. Ignoring transverse leads to failure.


4. COMPRESSION MEMBERS (COLUMNS)

Short Columns (Uni-axial Bending)

Design using Interaction Diagram (IS 456):

  1. Compute $$\displaystyle P_u / (f_{ck} b D) $$ and $$\displaystyle M_u / (f_{ck} b D^2) $$.

  2. Get $$\displaystyle p_t $$ from chart.

  3. Calculate $$\displaystyle A_{sc} = \frac{p_t}{100} b D $$.

  4. Provide bars along column face parallel to bending (2 sides) or all four sides.

Limits: $$\displaystyle p_{t,min} = 0.8\% $$, $$\displaystyle p_{t,max} = 4\% $$ (ties), $6\%$ (spiral).

Bi-axial Bending

Use IS 456 Eq. 39:

$$\frac{M_{ux}}{M_{ux1}} + \frac{M_{uy}}{M_{uy1}} \leq 1$$

where $$\displaystyle M_{ux1}, M_{uy1} $$ are moments for same $$\displaystyle P_u $$ from interaction diagram about x & y axes.

Bar Arrangement: Place bars at corners for equal moment resistance about both axes. Minimum 4 bars for rectangular, 6 for circular.

Circular Columns

  • Lateral Ties: Diameter ≥ 8 mm, spacing ≤ least lateral dimension, ≥ 300 mm.

  • Helical Reinforcement: Pitch ≤ 75 mm, ≥ 25 mm, ≥ 6 times core dia. Helical confinement increases load capacity by 10% if pitch criteria met.


5. FOOTINGS

Isolated Rectangular Footing

Design Steps:

  1. Load: $$\displaystyle P_u = 1.5 \times (P_{col} + self-weight) $$.

  2. Base area: $$\displaystyle A = \frac{P_u}{q_{safe}} $$ (consider eccentricity if any).

  3. Thickness: For shear, $$\displaystyle h \geq \frac{V_u}{\tau_c b} $$; for moment, $$\displaystyle M_u = \frac{P_u e}{2} $$.

  4. Critical Sections:

    • Shear: At $d$ from face of column (one-way) & at $d$ from face in both directions (two-way).

    • Bending: At face of column.

  5. Reinforcement: $$\displaystyle M_u = 0.87 f_y A_{st} (d - a/2) $$.

Stepped Footing

  • For heavy loads, provide steps to reduce punching shear.

  • Each step designed as individual footing.

  • Total depth = sum of step heights + top slab thickness.

[!TIP] Soil Pressure: For eccentric loading, $$\displaystyle q_{max/min} = \frac{P}{A} \pm \frac{M}{Z} $$. Check $$\displaystyle q_{max} \leq SBC $$.


6. SLABS

One-Way Slab

  • $$\displaystyle l_y / l_x > 2 $$.

  • Moment Coefficients (IS 456): Simply supported: $+0.083$, $-0.083$; Continuous: $+0.063$, $-0.075$ (end), $-0.05$ (int).

  • Design: $$\displaystyle M_u = \text{coeff} \times w l_x^2 $$. Provide main bars along short span, distribution along long span.

Two-Way Slab

  • $$\displaystyle l_y / l_x \leq 2 $$.

  • Coefficients (Table 26, IS 456): For restrained corners, $+0.076$, $-0.075$ (edges), $-0.063$ (corners).

  • Thickness: $l/d \leq 35$ (simply supported), 32 (continuous) for Fe415.

  • Critical Sections: At support faces for negative moments, mid-span for positive.

Flat Slabs

  • Drop panel depth ≥ 1/4 slab thickness.

  • Column head diameter ≤ 2/3 column size.

  • Use direct design method (not in syllabus focus).


7. STAIRCASES

Dog-Legged Stair:

  • Waist slab acts as simply supported beam over flights.

  • Effective Span: Horizontal distance between landings.

  • Loads: Dead (self-weight), live (4 kN/m² office), finishes (1 kN/m²).

  • Design Moment: $$\displaystyle M_u = \frac{w l^2}{8} $$ (if simply supported).

  • Reinforcement: Main bars (longitudinal) along slope, distribution bars perpendicular.

  • Landing: Cantilevered or supported on walls/beams.

Design Steps:

  1. Determine number of risers $$\displaystyle N = H/R $$, treads $N-1$.

  2. Horizontal length $$\displaystyle L = (N-1) \times T $$.

  3. Thickness $t \geq 100$ mm, $l/d \leq 25$.

  4. Calculate $$\displaystyle A_{st} $$ for $$\displaystyle M_u $$, check shear.

[!TIP] Sketch: Show waist slab as inclined beam, reinforcement in both directions, landing support details.


8. BOND, DEVELOPMENT LENGTH & LAP SPLICES

Bond Failure Mechanisms

  1. Pull-out: Bar slips out (good bond).

  2. Splitting: Concrete splits along bar (poor bond, high confinement needed).

Factors Affecting Bond:

  • Concrete strength, bar surface (deformed > plain), bar diameter, confinement, loading type.

Development Length ($$\displaystyle L_d $$)

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

where $$\displaystyle \tau_{bd} = k \sqrt{f_{ck}} $$ (k=1.6 for mild steel, 1.4 for HYSD), $$\displaystyle \sigma_{st} = 0.87 f_y $$.

Modifications:

  • For compression: $$\displaystyle L_d $$ reduced by 25%.

  • For epoxy-coated bars: increase by 50%.

  • For >12 mm bars: increase by 20%.

Lap Splices

  • Tension Lap Length: $$\displaystyle L_{lap} = L_d \times \frac{\text{area of larger bar}}{\text{area of smaller bar}} $$ (but ≥ $$\displaystyle L_d $$ of larger bar).

  • Staggering: Lap length ≥ $75\phi$ or $600$ mm, whichever more. Stagger by ≥ $75\phi$ or $600$ mm.

  • Compression Lap: $$\displaystyle 0.5 L_d $$ (but ≥ $24\phi$).

[!TIP] Common Error: Lap length for different bar sizes—use larger bar's $$\displaystyle L_d $$ as base, multiply by area ratio.


9. DEFLECTION & CRACKING CONTROL

Span/Effective Depth Ratio (IS 456 Table 19)

  • Cantilever: 7

  • Simply supported: 20

  • Continuous: 26

  • For Fe415, modify by factor $$\displaystyle k_t = \frac{0.25}{0.25 + \frac{M_t}{M}} \leq 1.5 $$.

Measures to Reduce Deflection:

  1. Increase effective depth.

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

  3. Add compression steel (doubly reinforced).

  4. Use smaller bar spacing.

Cracking Control:

  • Minimum Reinforcement: $$\displaystyle A_{st,min} = 0.12\% bD $$ (Fe415) for slabs.

  • Bar Spacing: ≤ 3D or 300 mm (whichever less) for slabs; ≤ 300 mm for beams.


10. DETAILING & DRAWING

Cross-Sections (Sketches Required):

  • Singly/doubly reinforced beam (show $$\displaystyle A_{st} $$, $$\displaystyle A_{sc} $$, cover, stirrups).

  • T-beam (flange width, NA position).

  • Column (uni-axial: bars on two faces; bi-axial: bars at corners).

  • Isolated footing (plan: column location, bars; section: depth, bars, cover).

Critical Sections Summary:

Element Flexure Shear
Beam At mid-span (pos), supports (neg) At $d$ from support face
Slab Mid-span (pos), supports (neg) At $d$ from support (one-way)
Footing At column face At $d$ from column face (both dir.)
Column At top/bottom for moments Not critical (axial)

Bar Bending Schedule (BBS) Items:

  • Bar mark, diameter, cutting length, number, total length.

  • Cutting length = clear span + 2 × hook length - bends (4d per bend).

Cover (IS 456):

  • Moderate exposure: 20 mm (beams), 30 mm (slabs).

  • Severe: 45 mm (beams), 30 mm (slabs).


11. COMPARATIVE STUDIES

LSM vs. WSM

Feature LSM WSM
Approach Probabilistic, ultimate load Deterministic, working load
Safety Partial factors on loads & materials Single factor of safety
Balanced Section $$\displaystyle x_u = 0.48d $$ (Fe415) $$\displaystyle \sigma_c = \sigma_{c,max} $$, $$\displaystyle \sigma_{st} = f_s $$
Stress Block Parabolic-rectangular (0.36f_ck) Linear (σ_c = (σ_{c,max}/x) × depth)
Moment Capacity $$\displaystyle M_u = 0.36 f_{ck} b x_u (d - 0.42 x_u) $$ $$\displaystyle M = (σ_c b x/2)(d - x/3) $$

WSM Example:

Given $$\displaystyle σ_c $$, $$\displaystyle σ_{st} $$, $m$:

  1. $$\displaystyle C = T $$ → $$\displaystyle \frac{σ_c}{m} b x = σ_{st} A_{st} $$.

  2. $$\displaystyle M = \frac{σ_c}{m} b x \left(d - \frac{x}{3}\right) $$.


12. SPECIAL TOPICS FROM PAST PAPERS

Composite Beam-Slab (Hall Design)

  • Effective Flange Width: $$\displaystyle b_f = b_w + \frac{l_o}{6} $$ or $$\displaystyle b_f = b_w + 4D_f $$ (min).

  • Design: Treat as T-beam. Compute $$\displaystyle x_u $$: if $$\displaystyle x_u \leq D_f $$, use $$\displaystyle b_f $$; else, subtract flange contribution.

  • Safe Superimposed Load: Find $$\displaystyle M_{u,allow} $$ for given steel, then $$\displaystyle w_{allow} = \frac{8M_{allow}}{l^2} $$ minus self-weight.

Two-Way Slab Design Steps (Complete)

  1. Check $$\displaystyle l_y/l_x \leq 2 $$.

  2. Determine slab thickness: $l/d \leq 32$ (continuous, Fe415).

  3. Compute total load $$\displaystyle w = DL + LL + finish $$.

  4. Find $$\displaystyle M_u = \text{coeff} \times w l_x^2 $$.

  5. Calculate $$\displaystyle A_{st} = \frac{M_u}{0.87 f_y z} $$ (z=0.9d).

  6. Check for shear: $$\displaystyle \tau_v = \frac{V_u}{b d} < \tau_c $$.

  7. Provide distribution steel (min 0.12%).

  8. Detailing: main bars in both directions, crank at supports.

Types of Footings (with Sketches)

  • Isolated: For single column.

  • Combined: For two columns.

  • Stepped: For heavy loads, in stages.

  • Raft: For poor soil, covers entire area.

  • Mat: Similar to raft, with beams.


Final Exam Strategy:

  • Numericals: Start with $$\displaystyle M_u = 1.5M $$, check $$\displaystyle x_u $$ limit, use IS 456 tables for $$\displaystyle \tau_c $$, $$\displaystyle p_t $$.

  • Theoretical: Define terms, draw strain diagrams, list assumptions.

  • Detailing: Always show clear cover, bar labels, stirrup spacing.

\boxed{\text{Master LSM fundamentals, T-beam flange width, shear critical sections, and column interaction diagrams for high marks.}}

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