📐 IS 1893 (Part 1) : 2016

Equivalent Diagonal Strut
Method for URM Infill Walls

A focused technical learning module on Clause 7.9.2.2 — modelling unreinforced masonry infill walls as diagonal compression struts in RC frame buildings under seismic loads.

Clause 7.9.2.2 URM Infill Walls Seismic Analysis RC MRF Buildings In-plane Stiffness Strut Width Formula
📖 Background

Why Do Infill Walls Matter in Seismic Design?

In India, most multi-storey RC frame buildings have their structural bays filled with unreinforced masonry (URM) walls — made of brick or block masonry. During an earthquake, these walls do not just sit passively inside the frame; they interact with the surrounding RC columns and beams, fundamentally changing the building's stiffness, strength, and seismic behaviour.


For decades, structural engineers conservatively ignored infill walls in analysis (treating buildings as "bare frames"). IS 1893:2016 now mandates that you account for their in-plane stiffness and strength, especially in Seismic Zones III, IV, and V.

⚠️
The Problem with Ignoring Infills
  • ❌ Leads to soft-storey irregularity when one storey has fewer infills (e.g., ground floor parking)
  • ❌ Underestimates building stiffness → incorrect natural period
  • ❌ Incorrect distribution of seismic forces between floors
  • ❌ Columns attract unexpected shear forces ("short column" effect)
✅
What IS 1893:2016 Requires
  • ✅ Model each URM infill as an equivalent diagonal strut
  • ✅ Analyse both bare frame and infilled frame
  • ✅ Design RC members for governing combination
  • ✅ Check irregularity per Table 6 for every storey
IS 1893 (Part 1):2016 — Clause 7.9.1

"In RC buildings with moment resisting frames and unreinforced masonry (URM) infill walls, variation of storey stiffness and storey strength shall be examined along the height of the building considering in-plane stiffness and strength of URM infill walls. If storey stiffness and strength variations render it to be irregular as per Table 6, the irregularity shall be considered especially in Seismic Zones III, IV and V."

Fig. 7 from IS 1893:2016
URM Infill Wall (t) w θ PIN PIN l (clear length of infill) h (clear height) L (diagonal length) Column Ic Equivalent Diagonal Strut — IS 1893:2016, Fig. 7

The URM infill wall (dashed outline) is replaced by a diagonal compression strut (shaded band) of width w and thickness t. Ends are pin-jointed to the RC frame.

💡
Key Physical Insight: During lateral loading (earthquake), the URM infill wall acts like a diagonal brace. The wall separates from the frame along two of its edges, and contact is maintained only in two corners. This diagonal compression path is captured by the equivalent strut model.
📐 Clause 7.9.2.2

The Equivalent Diagonal Strut Formulas

1️⃣
Step 1 — Modulus of Elasticity of Masonry (Clause 7.9.2.1)
Masonry Elastic Modulus
Em = 550 × fm
where Em and fm are in MPa

If you don't have direct test data for the masonry prism strength fm, calculate it from individual brick and mortar strengths:

Masonry Prism Compressive Strength (IS 1905)
fm = 0.433 × fb0.64 × fmo0.36
fb = brick unit compressive strength (MPa)  |  fmo = mortar compressive strength (MPa)
SymbolMeaningTypical Range
fmMasonry prism compressive strength2 – 10 MPa
fbBrick unit compressive strength3.5 – 25 MPa (IS 1077)
fmoMortar compressive strength2.5 – 10 MPa
EmModulus of elasticity of masonry1100 – 5500 MPa
2️⃣
Step 2 — Relative Stiffness Parameter αh

This dimensionless parameter captures the relative stiffness between the masonry infill and the surrounding RC frame columns. It is the core of the method.

Relative Stiffness Parameter (Dimensionless)
αh = h × Em · t · sin 2θ 4 Ec Ic h 0.25
αh = h × [ (Em × t × sin2θ) / (4 × Ec × Ic × h) ]^0.25
SymbolMeaningUnit
hClear height of URM infill between top beam soffit and bottom slab topmm
lClear length of URM infill between faces of vertical RC elements (columns)mm
tThickness of infill wall (same as strut thickness)mm
θAngle of diagonal strut with horizontal = arctan(h/l)degrees
EmModulus of elasticity of masonry (= 550 fm)MPa
EcModulus of elasticity of RC frame material (concrete)MPa
IcMoment of inertia of the adjoining column cross-section (about bending axis)mm⁴
LDiagonal length of strut = √(h² + l²)mm
3️⃣
Step 3 — Width of Equivalent Diagonal Strut

This is the main formula of Clause 7.9.2.2. Once αh is known, the strut width is calculated as:

Main Formula — Clause 7.9.2.2(b)
w = 0.175 × αh−0.4 × L
w = width of equivalent diagonal strut (mm)  |  L = diagonal length of infill = √(h² + l²) (mm)
📌
Strut Thickness: The thickness of the equivalent strut equals the thickness t of the original URM infill wall. This is valid provided:
h/l < 12   and   l/t < 12
🪟
Walls with Openings (Cl. 7.9.2.2c): For URM infill walls that have doors or windows, no reduction in strut width is required as per IS 1893:2016. The same formula applies.
4️⃣
Step 4 — Modelling the Strut in Structural Software

  1. Pin-jointed connections

    The ends of the diagonal strut shall be pin-jointed to the RC frame nodes (beam-column junctions). No moment is transferred — the strut is purely axial.

  2. Assign cross-section

    Create a rectangular section: width = w (calculated), depth = t (infill thickness). Assign material as masonry with E = Em.

  3. Compression only

    Masonry cannot take tension. The diagonal strut should be defined as a compression-only member (tension stiffness = 0). Under reversed loading, the opposite diagonal activates.

  4. Two struts per panel

    In practice, for dynamic (bidirectional) analysis, both diagonal struts may be modelled simultaneously to capture both directions of loading.

  5. Two analyses required (Cl. 7.9.1.1)

    Analyse (a) the bare frame and (b) the frame with URM infills. RC members shall be designed for the governing combination of stress resultants from both analyses.

📊 Worked Example

Step-by-Step Solved Problem

📝 Problem Statement

A typical bay of a G+4 RC MRF building has an infill wall panel with the following data. Determine the equivalent diagonal strut dimensions as per IS 1893:2016 Cl. 7.9.2.2.

Clear height of infill (h)3000 mm
Clear length of infill (l)4500 mm
Thickness of infill wall (t)230 mm
Brick compressive strength (fb)10 MPa
Mortar compressive strength (fmo)7.5 MPa
Concrete grade of RC frameM25 (Ec = 25000 MPa)
Column cross-section (b × d)350 mm × 450 mm
Bending axis of columnAbout 350 mm dimension (strong axis)
Step 1: Calculate masonry prism strength fm
fm = 0.433 × fb0.64 × fmo0.36
= 0.433 × 100.64 × 7.50.36
= 0.433 × 4.365 × 2.127 = 4.02 MPa
Modulus of elasticity of masonry:
Em = 550 × fm = 550 × 4.02 = 2211 MPa
Step 2: Calculate geometric parameters
θ = arctan(h/l) = arctan(3000/4500) = 33.69°
sin 2θ = sin(2 × 33.69°) = sin(67.38°) = 0.9231
L = √(h² + l²) = √(3000² + 4500²) = √(9×10⁶ + 20.25×10⁶) = 5408.3 mm
Column moment of inertia (bending about strong axis):
Ic = (b × d³)/12 = (350 × 450³)/12 = 2.672 × 10⁹ mm⁴
📐
Note: The moment of inertia Ic is taken about the bending axis perpendicular to the plane of the infill wall — typically the strong axis if the infill spans in-plane with the frame.
Step 3: Calculate relative stiffness parameter αh
αh = h × [ (Em × t × sin2θ) / (4 × Ec × Ic × h) ]0.25

Substituting values:

= 3000 × [ (2211 × 230 × 0.9231) / (4 × 25000 × 2.672×10⁹ × 3000) ]0.25
Numerator = 2211 × 230 × 0.9231 = 4,692,460
Denominator = 4 × 25000 × 2.672×10⁹ × 3000 = 8.016 × 10¹⁷
Ratio = 4,692,460 / (8.016 × 10¹⁷) = 5.854 × 10⁻¹²
αh = 3000 × (5.854 × 10⁻¹²)0.25 = 3000 × 0.04908 / 3000 × …
αh ≈ 4.86 (dimensionless)
💡
Interpretation: A lower αh means the infill is flexible relative to the column (the column is stiff). A higher αh means the infill is stiffer relative to the column. Typical values range from 2 to 10.
Step 4: Calculate strut width w
w = 0.175 × αh−0.4 × L
= 0.175 × (4.86)−0.4 × 5408.3
(4.86)−0.4 = e(-0.4 × ln4.86) = e(-0.4×1.581) = e-0.632 = 0.5316
w = 0.175 × 0.5316 × 5408.3 = 502.6 mm
Strut thickness = t = 230 mm
✅
Result: The equivalent diagonal strut has a width of ~503 mm and thickness of 230 mm. Its material has Em = 2211 MPa and it is pin-jointed at both ends.
Step 5: Applicability Checks (IS 1893:2016 Cl. 7.9.2.2)

Check 1: h/l < 12 3000/4500 = 0.67 ✓ PASS
Check 2: l/t < 12 4500/230 = 19.6 ✓ PASS

Final Summary
ParameterValueUnit
fm (Masonry prism strength)4.02MPa
Em (Masonry modulus)2211MPa
θ (Strut angle)33.69degrees
L (Diagonal length)5408.3mm
αh (Relative stiffness)4.86—
w (Strut width)502.6mm
t (Strut thickness)230mm
🧮 Interactive Calculator

Diagonal Strut Width Calculator

Enter your infill wall and frame parameters below. All calculations follow IS 1893 (Part 1):2016, Clauses 7.9.2.1 and 7.9.2.2.

Equivalent Diagonal Strut — IS 1893:2016

Cl. 7.9.2.1 + 7.9.2.2
📏 Infill Wall Geometry
Between top beam soffit and bottom slab top
Between faces of adjacent columns
Nominal wall thickness (e.g. 230 for 9", 115 for 4.5")
Cl. 7.9.2.2(c): No reduction needed for openings
🧱 Masonry Material Properties
Typical: 3.5 (low) to 25 (high) MPa
M5=5MPa, M7.5=7.5MPa, M10=10MPa
🏗️ RC Frame Properties
IS 456: Ec = 5000√fck MPa
Ic = bd³/12 (bending about strong axis)
📋 Project Information (for Report)
📊 Calculation Results
🔢 Step-by-Step Calculation
StepParameterFormula / ExpressionValue
🎯 Summary

Key Takeaways for Students

Core Formula
w = 0.175·αh⁻⁰·⁴·L
Width of equivalent diagonal strut (Cl. 7.9.2.2b)
Thickness Condition
t (infill)
h/l < 12 and l/t < 12 must be satisfied
End Condition
Pin-jointed
No moment transfer — axial compression only
📌
Complete Clause Summary
ClauseContent
7.9.1RC MRF buildings with URM infills must examine storey stiffness & strength variations. Irregularities per Table 6 must be addressed.
7.9.1.1RC members to be designed for governing combination from (a) bare frame analysis, and (b) infilled frame analysis.
7.9.2.1Em = 550 fm (MPa); fm = 0.433 fb⁰·⁶⁴ fmo⁰·³⁶
7.9.2.2(a)Ends of diagonal struts shall be pin-jointed to RC frame.
7.9.2.2(b)Width w = 0.175 αh⁻⁰·⁴ L, where αh = h[(Em·t·sin2θ)/(4EcIch)]⁰·²⁵
7.9.2.2(c)For walls with openings: no reduction in strut width required.
7.9.2.2 (last para)Strut thickness = t, provided h/l < 12 and l/t < 12.
⚠️
Common Student Mistakes:
  • Using the full frame height H instead of clear infill height h
  • Using frame span L instead of clear infill length l for computing θ
  • Taking Ic about the wrong axis (must be about the column axis parallel to the infill plane)
  • Using diagonal length L in metres instead of mm (keep consistent units throughout)
  • Forgetting that two analyses (bare frame + infilled frame) are required