Design Aids for Reinforced Concrete to IS 456:1978

IS SP Part 16: Scope - Key Specifications & Formulas

Scope covers:

• Working stress and limit state design methods for reinforced concrete.
• Flexure, shear, development length, deflection, and slender compression members.
• Material properties and stress-strain relationships.

Key Parameters & Symbols

Important Formulas

1. Moment of Inertia (Gross Section)

[ I_g = \frac{b d^3}{12,000} ]

2. Moment of Inertia (Cracked Section)

[ I_r = \text{Values tabulated for } \frac{d'}{d} = 0.05, 0.10, 0.15, 0.20 ]

3. Depth of Neutral Axis (Elastic Theory)

[ x_u = \text{Values tabulated for } \frac{d'}{d} = 0.05, 0.10, 0.15, 0.20 ]

4. Development Length for Deformed Bars

[ l_d = \frac{f_y \times \phi}{4 \times \tau_{bd}} ]

• (f_y) = yield strength of steel
• (\phi) = bar diameter
• (\tau_{bd}) = design bond stress (from tables)

5. **Perm

IS SP Part 16 — Design of Flexural Members: Key Formulas & Tables

1. Flexural Design Basics (Working Stress Method)

• Moment of Resistance (M) for singly reinforced rectangular beam:
  [ M = 0.36 f_{ck} b d^2 ] where,

  • (f_{ck}) = characteristic compressive strength of concrete (N/mm²)
  • (b) = width of beam (mm)
  • (d) = effective depth (mm)

• Limiting moment of resistance factor ( \frac{M_u}{b d^2 f_{ck}} ) for T-beams and doubly reinforced sections available in tables (Clause 25.0).

2. Reinforcement Percentages for Doubly Reinforced Sections

• Provided for different (f_y) (250, 415, 500 N/mm²) and (f_{ck}) values.
• Example: For (f_y = 415) N/mm², refer Table 58 (Page 94).

3. Shear Design

• Permissible shear stress ( \tau_c ) in concrete, vertical stirrups, and bent-up bars detailed in Tables 80-83 (Pages 207-208).

4. Development Length

• For plain and deformed bars, based on (f_y) and bar diameter, see Tables 83-86 (Pages 208-209).

5. Moment of Inertia

• Values for gross and cracked sections (I_r) given for various (d'/d) ratios (Tables 86-90, Pages 220-224).

Summary Table Snippet:

IS SP Part 16: Design of Compression Members - Key Points

1. Axially Loaded Compression Members (Clause 3.1 & 6.2)

• Design Strength, P_u is governed by: [ P_u = 0.4 f_{ck} A_c + 0.67 f_y A_s ] where:

  • ( f_{ck} ) = characteristic compressive strength of concrete (N/mm²)
  • ( f_y ) = yield strength of steel (N/mm²)
  • ( A_c ) = area of concrete section (mm²)
  • ( A_s ) = area of steel reinforcement (mm²)

• Slenderness Ratio (\lambda = \frac{l_{eff}}{r}) controls buckling capacity.

2. Slender Compression Members (Clause 10.0, Table 60)

• Reduction factor ( \phi ) depends on slenderness and reinforcement ratio ( \rho ).
• Table excerpt for Rectangular Sections with equal reinforcement on four sides (for ( f_y = 415 ) N/mm²):

• For circular and rectangular sections, values vary; use corresponding table values.

3. Design Aids (Charts 78-85)

• Provide interaction curves for Tension with Bending for rectangular sections with reinforcement distributed equally on four sides.
• These charts help determine required reinforcement under combined axial load and bending moments.

Summary Diagram of Compression Member Design Flow

flowchart TD
    A[Determine Loads] --> B[Calculate Axial Load \(P_u\)]
    B --> C[Check Slenderness Ratio \(\lambda\)]
    C -->|Short| D[Use Axial Load Formula]
    C -->|Slender| E[Apply Reduction Factor \(\phi\) from Table 60]
    D --> F[Design Reinforcement]
    E --> F
    F --> G[Check Interaction with Bending (

IS SP Part 16: Shear and Torsion Reinforcement Key Points

1. Shear Reinforcement

• Permissible shear stress in concrete (Table 80) depends on concrete grade (fck) and percentage tensile reinforcement (Pt).
• Use Table 62 for vertical stirrup design (two-legged stirrups):

• Example: For fy=250 N/mm², 8 mm dia stirrups at 14 cm spacing provide Vus/d = 1.562 kN/cm.

• Bent-up bars capacity (Table 63) depends on bar diameter and angle (45° or 60°).

2. Torsion Reinforcement

• Design by equivalent shear force (Ve) and equivalent bending moment (Me):

[ V_e = V + 1.6 \times \frac{T}{u} ]

Where:

• (V) = Shear force

• (T) = Torsional moment

• (u) = Perimeter of the section

• Use shear design charts/tables for Ve to find required shear reinforcement.

• Example: For a 30x60 cm beam, M15 concrete, and steel fy=415 N/mm²:

[ V_e = 95 + 1.6 \times \frac{45 \times 10^3}{0.3} = 335 \text{ kN} ]

3. Design Shear Strength of Concrete (To') — Table 61 (N/mm²)

IS SP Part 16: Control of Deflection – Key Points

1. Deflection Limits and Factors (Clause 2.5.3)

• For simply supported rectangular members, spans ≤ 10 m: Use values directly from deflection charts.
• For spans > 10 m: Multiply chart values by
  [ \text{Factor} = \frac{10}{\text{Span (m)}} ]
• For continuous spans, multiply by 1.3.
• For cantilevers, multiply by 0.35.
• Cantilevers > 10 m: Calculate deflection explicitly to ensure limits are met.

2. Flat Slabs (Clause 2.5.6)

• Use the longer span for deflection checks.
• If drop panels are not provided, multiply span-to-effective depth ratio by 0.9.

3. Span-to-Effective Depth Ratio (l/d) Control

• Use charts/tables for permissible l/d ratios.
• Modify for continuous spans and cantilevers as above.

Summary Table: Multiplying Factors for Deflection Control

Example Formula for Deflection Check

[ \frac{l}{d_{eff}} \leq \text{Permissible ratio from charts} \times \text{Multiplying factor} ]

Where:

• ( l ) = span length
• ( d_{eff} ) = effective depth of member

flowchart TD
    A[Start: Identify Member Type] --> B{Span ≤ 10 m?}
    B -- Yes --> C[Use Chart Values Directly]
    B -- No --> D[Multiply Chart Values by 10/span]
    C --> E{Member Type?}
    D --> E
    E -- Continuous --> F[Multiply by 1.3]
    E -- Cantilever --> G[Multiply by 0.35]
    E

Slenderness Effects & Additional Moments (IS SP Part 16)

1. Additional Moments due to Slenderness (Clause 38.7.1)

• Additional moments are added to initial moments after modification.
• Use multiplying factor k to reduce additional moments:

[ k = \frac{P_{uz} - P_u}{P_{uz} - P_b} < 1 ]

Where:

• ( P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_s ) (ultimate axial load capacity)
• ( P_u ) = applied axial load
• ( P_b ) = axial load at maximum concrete compressive strain (0.0035) and steel tensile strain (0.002)

Use Table 60 for coefficients to calculate ( P_b ).

2. Additional Moments as Eccentricities

[ M_{ux} = P_u \times e_{x} = P_u \times \frac{l_{ex}^2}{2000D} ]

[ M_{uy} = P_u \times e_{y} = P_u \times \frac{l_{ey}^2}{2000b} ]

Where:

• ( l_{ex}, l_{ey} ) = effective lengths about major and minor axes
• ( D, b ) = cross-section dimensions

3. Additional Eccentricity Values (Table for ( e_x/D ), ( e_y/b ))

Key Formulas, Tables & Specifications for Working Stress Method (IS SP Part 16)

1. Development Length and Anchorage

• Use Table 67 for anchorage values of standard hooks and bends applicable to Working Stress Method.
• Development length ( l_d ) depends on bar type and grade (see Tables 83-85).

2. Flexure - Moment of Resistance Factor (Table 68, 71)

• For Singly Reinforced Sections: [ M = m \times b d^2 \quad \text{or} \quad M = M/bd^2 \times b d^2 ] where ( m ) = moment of resistance factor (N/mm²), ( b ) = width, ( d ) = effective depth.
• Values vary with steel stress ( f_{st} ) (130, 140, 190, 230, 275 N/mm²) and concrete stress ( f_{cb} ) (5.0 to 10.0 N/mm²).

3. Reinforcement Percentages and Moment Factors

• Tables provide ( P_t = \frac{A_s}{bd} \times 100 ) vs. moment factors for various steel grades.
• Example extract:
  | ( P_t ) (%) | ( f_{st} = 230 ) N/mm² | Moment Factor ( M/bd^2 ) (N/mm²) | |---------------|--------------------------|------------------------------------| | 0.3 | 0.638 | 1.01 | | 0.5 | 1.042 | 1.11 |

4. Shear Design

• Permissible shear stress ( \tau_c ) for concrete from Table 80.
• Shear reinforcement details in Tables 81 (vertical stirrups) and 82 (bent-up bars).

5. Moment of Inertia

• Gross section: ( I_g = \frac{b d^3}{12,000} ) (Table 86).
• Cracked section inertia ( I_r ) depends on ( d'/d ) ratio (Tables 87-90).

6. Symbols & Definitions

• ( A_s ): Area of tensile steel
• ( A

Neutral Axis Depth (xu) and Strain Distribution - IS SP Part 16

1. Maximum Depth of Neutral Axis (Clause 2.2)

• Formula for xu,max:

[ x_{u, max} = \frac{0.005}{0.005 + 0.87 \times f_y / E_s} \times d ]

• Where:

  • (d) = effective depth
  • (f_y) = yield strength of steel
  • (E_s) = modulus of elasticity of steel

• Table B (Clause 2.2) provides values of (x_{u,max}/d) for different steel grades.

2. Neutral Axis Depth for Under-Reinforced Section (Clause 6.1.2)

• General expression for neutral axis depth:

[ k = \frac{x_u}{d} = \frac{-100 + \sqrt{10000 + P_m^2}}{2 P_m} ]

where (P_m = \frac{P_t}{f_{ck} b d}) (reinforcement ratio factor)

• Moment of resistance:

[ M = f_{ck} b d^2 \times \text{function of } k ]

Tables 68-71 tabulate (M / (f_{ck} b d^2)) vs. (P_t) for various concrete grades.

3. Stress Block Parameters When Neutral Axis Lies Outside Section (Clause 3.2.2)

Minimum Eccentricity for Columns (IS SP Part 16, Clause 24.4)

• Minimum eccentricity, ( e_{min} ), to consider in design: [ e_{min} = \max \left( \frac{500}{D} + 20, , 2, \text{cm} \right) ] where (D) = lateral dimension of the column in mm.

• Check: ( e_{min} \leq 0.05 \times D ) (Clause 38.3)
  If true, simplified design methods apply.

Additional Eccentricity for Slender Columns (Clause 24.1.2 & Table I)

For slender columns, additional eccentricities ( e_{x} ) and ( e_{y} ) are considered:

• Additional moments: [ M_{ux} = P_u \times e_x, \quad M_{uy} = P_u \times e_y

IS SP Part 16: Development Length and Anchorage

1. Development Length (ld) for Reinforcement Bars

• Depends on:

  • Bar diameter (d)
  • Grade of steel (fy)
  • Grade of concrete (fck)
  • Whether bars are in tension or compression

• Tables for ld (in cm):

(Refer to Tables 64-66 for full range and other concrete grades)

2. Anchorage Value of Hooks and Bends (Table 67)

• Standard Hook ≈ 1.64 × bar diameter
• 90° Bend ≈ 0.8 × bar diameter

3. Minimum Development Length Factor (k)

• Mild steel: k = 2
• Cold worked steel: k = 4

Summary Formula for Development Length (approximate):

[ l_d = k \times \frac{f_y \times d}{4 \times \sqrt{f_{ck}}} ]

where:

• (

Helical and Lateral Tie Reinforcement Design (IS SP Part 16)

1. Strength Adjustment (Clause 38.4):

• Strength of compression member with helical reinforcement = 1.05 × strength with lateral ties.
• For design, divide the applied load and moment by 1.05 before using lateral tie charts.

2. Shear Reinforcement (Clause 12.5 & Tables):

• Typical shear reinforcement: 10 mm two-legged stirrups @ 12.5 cm spacing satisfy codal requirements for common grades.
• Use Table 61 for concrete shear strength To' (N/mm²) based on % steel (Pt) and concrete grade (fck).
• Use Table 62 for vertical stirrup capacity Vus/d (kN/cm) based on stirrup diameter, spacing, and yield strength (fy).
• Use Table 63 for bent-up bars shear capacity Vus (kN) for different bar diameters and angles.

3. Design Formula for Helical Reinforcement Strength:

[ P_{helical} = 1.05 \times P_{lateral} ]

Where:

• ( P_{helical} ) = design axial load capacity with helical reinforcement
• ( P_{lateral} ) = design axial load capacity with lateral ties

Summary Table for Shear Strength To' (Excerpt from Table 61)

Practical Notes:

• Use helical reinforcement for better confinement and strength (5% increase).

Key Formulas & Tables for Reinforcement Design (IS SP Part 16 & IS 456:1978)

1. Design of Singly Reinforced Beam

• Moment of Resistance, Mu = 0.87 * fy * Ast * d * (1 - (Ast * fy) / (fck * b * d))
• Where:
  • fy = yield strength of steel (N/mm²)
  • Ast = area of tensile steel (mm²)
  • d = effective depth (mm)
  • fck = characteristic compressive strength of concrete (N/mm²)
  • b = width of beam (mm)

2. Design of Doubly Reinforced Beam

• Use Clause 10.0 Tables (Page 199-206) for various steel grades and concrete strengths.
• Design moment, Mu = Moment resisted by tension steel + Moment resisted by compression steel.
• Refer to Working Stress Percentages and Reinforcement Tables for steel grades (195 - 230 N/mm²).

3. Reinforcement Percentage Limits (IS 456:1978, Clause 25.5.2)

4. Example: Singly Reinforced Beam

• Beam size: 300 mm x 600 mm
• Concrete: M15 (fck = 15 N/mm²)
• Steel: 415 N/mm²
• Factored moment, Mu = 170 kN-m
• Bar diameter: 25 mm, cover = 25 mm

Steps:

• Calculate effective depth, d = 600 - 25 - (25/2) = 588 mm
• Calculate Ast = Mu / (0.87 * fy * z), where z ≈ 0.95d
• Substitute values to find Ast.

5. Moment of Inertia & Neutral Axis Depth

• Refer to **Tables 86-94 (Pages 220-

Design Charts for Moment of Resistance (IS SP Part 16)

Key Points from Clause 2.3.1.1 & Related Clauses:

• Charts 1 to 18: Plotting depth (d) vs. percentage of reinforcement (pt) for different Mu/b values.
• Moment values are in kN·m per meter width.
• Charts cover 3 grades of steel and 2 concrete grades (M15, M20).
• Tables 1 to 4 extend this to 5 steel grades and 4 concrete grades up to M30.
• Tables tabulate pt vs. Mu/(b d²) for design convenience.

Important Formula for Moment of Resistance:

[ M_u = 0.87 f_y A_{st} (d - \frac{a}{2}) ]

Where:

• (M_u) = Ultimate moment of resistance (kN·m)
• (f_y) = Yield strength of steel (N/mm²)
• (A_{st}) = Area of tension reinforcement (mm²)
• (d) = Effective depth (mm)
• (a = \frac{A_{st} f_y}{0.36 f_{ck} b}) (depth of equivalent stress block)

Usage of Charts:

• Calculate Mu/b (moment per unit width).
• Refer to charts for given steel and concrete grades.
• Find corresponding pt and d values.
• Check if reinforcement and depth satisfy design requirements.

Quick Reference Table (Example):

Values are indicative; refer to IS SP Part 16 Tables 1-4 for exact data.

Notes:

• For singly reinforced sections, use the moment of resistance charts directly.
• For compression + bending, refer to Charts 27-50 (two/four-sided reinforcement).
• For tension + bending, refer to Charts 66-85.
• Charts show stress states at limit state of collapse

IS SP Part 16 — Reinforcement Percentages Key Details

1. Percentage of Reinforcement (pt)

• From Clause 1.79, example:
  [ p_t = 0.594% ]
• Used in design calculations for flexure.

2. Working Stress Method - Reinforcement Percentages Tables (Clause 10.0)

• Tables provide reinforcement percentages (pt) for various grades of steel and concrete strengths.
• Useful for doubly reinforced sections flexural design.

3. Sample Table Extract (Flexure - Doubly Reinforced Sections)

4. Design Formula for Percentage of Reinforcement

[ p_t = \frac{A_s}{b \times d} \times 100 ]

• (A_s) = area of tensile steel (mm²)
• (b) = width of beam (mm)
• (d) = effective depth (mm)

Summary

• Use Clause 10.0 tables

Deflection Control as per IS SP Part 16

Key Points from Clause 2.5.3 & 2.5.6

• Applicability:

  • Charts are for simply supported rectangular sections with spans ≤ 10 m.
  • For spans > 10 m (simply supported or continuous), multiply chart values by:
    [ \text{Factor} = \frac{10}{\text{Span (m)}} ]

• Modification factors for continuous spans and cantilevers:

  Type	Multiplying Factor
  Continuous spans	1.3
  Cantilevers	0.35

• For cantilevers > 10 m:
  Deflections must be explicitly calculated, not just scaled.

• Flat slabs (Clause 2.5.6):
  Use the longer span for deflection checks.
  If no drop panels, multiply span/effective depth ratio by 0.9.

Typical Deflection Control Formula (Span to Effective Depth Ratio)

[ \frac{L}{d} \leq \text{Value from charts} \times \text{Modification factors} ]

Where:

• (L) = span length
• (d) = effective depth of member

Summary Table for Deflection Factors

Practical Notes

• Use these factors to adjust span/effective depth ratios from deflection charts.
• For cantilevers longer than 10 m, perform detailed deflection calculations per IS 456.
• Ensure bar diameters and spacing comply with reinforcement limits to control cracking and deflection.

flowchart TD
    A[Read Deflection Chart Value] --> B{Span ≤ 10 m?}
    B -- Yes --> C[Use Chart Value