Mechanics of Materials & Testing

3-Point Bending Test Equations: Complete Guide to Flexural Stress, Deflection, Modulus & Force Formulas

The complete reference for 3-point bending test equations: full derivation of flexural stress \(\sigma = \tfrac{3FL}{2bd^2}\), deflection \(\delta = \tfrac{FL^3}{48EI}\), and flexural modulus; formulas for rectangular beams, solid cylinders, hollow tubes and I-sections; worked examples for steel, aluminium and polymer specimens; comparison with 4-point bending; ASTM D790, D7264 and IS 1608 standards; and interactive calculators.

Equations & Derivation Cylinders & Rods Worked Examples Interactive Calculators
By Bimal Ghimire • Published July 22, 2025 • Updated April 13, 2026 • 30 min read
Definition & Overview

What is a 3-Point Bending Test? Principle, Setup and Variables

The 3-point bending test (also written as three-point bend test or flexure test) is a standardised mechanical test that measures the flexural properties of a material by loading a beam specimen at its midpoint while it rests on two supports. It is the most widely used method to determine flexural strength (modulus of rupture), flexural modulus (elastic modulus in bending), and load-deflection behaviour for metals, ceramics, polymers, composites, and wood.

The test is governed by ASTM D790 (polymers), ASTM D7264 (composites), ISO 178 (plastics), ASTM E290 (metals - bend ductility), and IS 1608 (Indian Standard for metallic materials).

F L (Support Span) δ (deflection) d (depth) Loading Nose R R = F/2 = F/2 3-Point Bending Test - Schematic Setup b (width)

Figure 1: Schematic of the 3-point bending test setup. Load F applied at midspan; reactions R = F/2 at each support; L = support span; δ = midspan deflection.

F
Applied load at midspan (N)
L
Support span (mm)
δ
Midspan deflection (mm)
E
Flexural modulus (MPa)

The test works by creating a pure bending moment in the region around the midspan. The beam experiences compressive stress on the top face (under the loading nose) and tensile stress on the bottom face, with a neutral axis at the centroid where stress is zero. Failure typically initiates at the bottom tensile surface for brittle materials (ceramics, composites) or at the top compressive surface for ductile metals after significant plastic hinging.

Key advantage of the 3-point test: Simple, fast setup with a single loading nose. Maximum bending moment and maximum stress occur together at midspan, making failure location predictable. The main limitation is that the stress state is not uniform between the supports - the moment diagram is triangular, peaking at midspan. This is why 4-point bending is preferred when a uniform moment zone is needed (e.g. for fatigue testing and composite interlaminar shear).

All Equations at a Glance

3-Point Bending Test Equations: Complete Summary

Maximum Flexural Stress - Rectangular Beam
σf = 3FL / (2bd²)
F = load (N) • L = span (mm) • b = width (mm) • d = depth/thickness (mm)
Flexural Stress (General - Any Cross-Section)
$$\sigma = \frac{M \cdot y}{I} = \frac{F L \cdot c}{4 I}$$
M = FL/4 (max moment at midspan) • c = distance from neutral axis to extreme fibre (mm) • I = second moment of area (mm&sup4;)
Midspan Deflection
$$\delta = \frac{F L^3}{48 E I}$$
E = flexural modulus (MPa) • I = second moment of area • Valid only in the elastic range (linear portion of load-deflection curve)
Flexural Modulus from Slope of Load-Deflection Curve
$$E = \frac{F}{\delta} \cdot \frac{L^3}{48 I} = \frac{m L^3}{48 I}$$
m = F/δ = slope of the linear portion of the P-δ curve (N/mm)
Maximum Bending Moment
$$M_{max} = \frac{F \cdot L}{4}$$
At midspan (point of load application). Moment diagram is triangular, zero at both supports.
Flexural Strain at Extreme Fibre
$$\varepsilon_f = \frac{6 d \delta}{L^2}$$
d = specimen depth (mm) • δ = midspan deflection (mm) • L = support span. Used in ASTM D790 to report strain at fracture or at 5% deflection.
Reaction Force at Each Support
$$R_A = R_B = \frac{F}{2}$$
By symmetry. Each support carries half the applied load. Use this to size the test fixture.
3-point bending test stress distribution diagram showing triangular bending moment diagram, linear stress distribution across beam depth with compression on top and tension on bottom

Figure 2: Bending moment diagram (triangular, peak = FL/4 at midspan) and linear stress distribution across beam depth. Compressive stress on loaded (top) face; tensile stress on opposite (bottom) face; zero stress at neutral axis.

QuantitySymbolRectangular BeamSolid Cylinder (dia D)Hollow Tube (Do, Di)Units
Second moment of areaIbd³/12πD&sup4;/64π(Do&sup4;−Di&sup4;)/64mm&sup4;
Distance to extreme fibrecd/2D/2Do/2mm
Section modulus Z = I/cZbd²/6πD³/32π(Do&sup4;−Di&sup4;)/(32Do)mm³
Max bending momentMFL/4FL/4FL/4N·mm
Max flexural stress σσmax3FL/(2bd²)8FL/(πD³)8FLDo/[π(Do&sup4;−Di&sup4;)]MPa
Midspan deflectionδFL³/(4Ebd³)64FL³/(3πED&sup4;)FL³L³/[3πE(Do&sup4;−Di&sup4;)/16]mm
Flexural modulus EEFL³/(4δbd³)64FL³/(3πδD&sup4;)FL³/[48δI]MPa
Flexural strainεf6dδ/L²6Dδ/L²6Doδ/L²mm/mm

Important - span-to-depth ratio: The simple beam equations above are valid only when the support span L is at least 16 times the beam depth d (L/d ≥ 16 for metals; ≥ 14 for composites per ASTM D7264; ≥ 16 for polymers per ASTM D790). For shorter spans, transverse shear stress contributes significantly and a Timoshenko beam correction is needed. ASTM D790 recommends L/d = 16 as the default for most plastics, with L/d = 32 or 64 for stiff materials prone to shear failure.

Step-by-Step Derivation

Full Derivation of 3-Point Bending Equations from First Principles

The following derivation starts from Euler-Bernoulli beam theory and derives every equation used in the 3-point bend test from equilibrium, compatibility, and the flexure formula. The derivation assumes: linear elastic material, small deflections, beam length much greater than cross-section dimensions (L/d ≥ 10), and load applied exactly at midspan.

1
Free Body Diagram and Support Reactions
A simply supported beam of span L, loaded by force F at the midpoint x = L/2. By symmetry and vertical equilibrium:
$$\sum F_y = 0: \quad R_A + R_B = F \quad \Rightarrow \quad R_A = R_B = \frac{F}{2}$$
There are no horizontal reactions for a simply supported beam with a vertical load.
2
Bending Moment Distribution
Taking a free body cut at position x from the left support (0 ≤ x ≤ L/2):
$$M(x) = R_A \cdot x = \frac{F}{2} x \quad \text{for} \quad 0 \le x \le \frac{L}{2}$$ $$M_{max} = M\!\left(\frac{L}{2}\right) = \frac{F}{2} \cdot \frac{L}{2} = \frac{FL}{4}$$
By symmetry, M(x) = F(L-x)/2 for L/2 ≤ x ≤ L. The bending moment diagram is triangular, peaking at midspan.
3
Flexure Formula: Stress from Moment
Euler-Bernoulli beam theory (plane sections remain plane; stress linear across depth):
$$\sigma(y) = \frac{M \cdot y}{I}$$
where y = distance from neutral axis (mm); I = second moment of area (mm&sup4;). Maximum stress occurs at y = c (extreme fibre, furthest from neutral axis).
At midspan, M = FL/4. For a rectangular beam (width b, depth d): I = bd³/12, c = d/2:
$$\sigma_{max} = \frac{M_{max} \cdot c}{I} = \frac{\dfrac{FL}{4} \cdot \dfrac{d}{2}}{\dfrac{bd^3}{12}} = \frac{FL}{4} \cdot \frac{d}{2} \cdot \frac{12}{bd^3} = \frac{3FL}{2bd^2}$$
4
Deflection Equation via Double Integration
The Euler-Bernoulli moment-curvature relation: EI d²y/dx² = M(x). For the left half (0 ≤ x ≤ L/2):
$$EI \frac{d^2 y}{dx^2} = M(x) = \frac{F}{2} x$$
First integration (slope): $$EI \frac{dy}{dx} = \frac{F}{4} x^2 + C_1$$ By symmetry, slope = 0 at x = L/2:   C&sub1; = −FL²/16.
Second integration (deflection): $$EI \cdot y = \frac{F}{12} x^3 - \frac{FL^2}{16} x + C_2$$ Boundary condition y = 0 at x = 0:   C&sub2; = 0.
At midspan x = L/2: $$EI \cdot y_{max} = \frac{F}{12} \cdot \frac{L^3}{8} - \frac{FL^2}{16} \cdot \frac{L}{2} = \frac{FL^3}{96} - \frac{FL^3}{32} = -\frac{FL^3}{48}$$
Taking the magnitude (downward deflection positive):
$$\boxed{\delta_{max} = \frac{F L^3}{48 E I}}$$
5
Flexural Modulus from the Load-Deflection Slope
Rearranging the deflection equation for E:
$$E = \frac{F L^3}{48 \delta I} = \frac{m L^3}{48 I}$$
where m = F/δ = slope of the linear portion of the experimentally measured load-deflection (P-δ) curve (N/mm). For a rectangular beam, substituting I = bd³/12: $$E = \frac{m L^3}{48 \cdot \dfrac{bd^3}{12}} = \frac{m L^3}{4 b d^3}$$
6
Flexural Strain at the Extreme Fibre (ASTM D790 definition)
The strain at the extreme fibre relates to curvature κ = M/(EI) = 1/R. At midspan for small deflections:
$$\varepsilon_f = \frac{\sigma_{max}}{E} = \frac{c \cdot M}{EI}$$
Substituting M = 48EIδ/(4L³) × (L/4) and simplifying: $$\varepsilon_f = \frac{6 d \delta}{L^2}$$ This is the standard ASTM D790 formula for flexural strain. At fracture (δ = δf), this gives the flexural strain at break.

Summary of derivation assumptions: (1) Euler-Bernoulli beam theory - plane sections remain plane; (2) linear elastic material behaviour (Hooke's law); (3) small deflections (δ << L); (4) isotropic homogeneous material; (5) load applied as a line load at exactly midspan; (6) frictionless pin supports; (7) span-to-depth L/d ≥ 10 (shear deformation negligible). For composites or short/thick beams, Timoshenko beam theory corrections must be applied.

Circular Cross-Sections

3-Point Bend Test Equations for Cylinders and Rods

Cylindrical specimens (round bars, wire rods, bone samples, ceramic rods) require their own set of equations because the second moment of area and section modulus differ from rectangular beams. The general framework is identical - only I and c change.

Solid Cylinder
D/2 NA
I = πD&sup4;/64
c = D/2
Z = πD³/32

σmax = 8FL/(πD³)

δ = 64FL³/(3πED&sup4;)

Hollow Tube (Pipe)
D⊂o D⊂i
I = π(Do&sup4;−Di&sup4;)/64
c = Do/2
Z = I/(Do/2)

σ = M · Do/(2I)

δ = FL³/(48EI)

I-Section (Structural)
I = (bd³ − bwh³)/12
c = d/2 (symmetric)
Z = I / c

General: σ = M/Z

δ = FL³/(48EI)

Solid Cylinder - All Equations Explicit
$$I = \frac{\pi D^4}{64}, \quad c = \frac{D}{2}, \quad Z = \frac{I}{c} = \frac{\pi D^3}{32}$$ $$\sigma_{max} = \frac{M_{max}}{Z} = \frac{FL/4}{\pi D^3 / 32} = \frac{8FL}{\pi D^3}$$ $$\delta_{max} = \frac{FL^3}{48EI} = \frac{FL^3}{48E \cdot \frac{\pi D^4}{64}} = \frac{64 F L^3}{3 \pi E D^4}$$ $$E = \frac{64 F L^3}{3 \pi \delta D^4}$$
D = diameter of cylindrical specimen (mm) • F = applied load (N) • L = support span (mm) • δ = midspan deflection (mm) • E = flexural/Young's modulus (MPa).
Note: for a solid circular cross-section, the section modulus Z = πD³/32. The factor 8 in the stress formula comes from substituting: M = FL/4 and Z = πD³/32 gives σ = (FL/4) × (32/πD³) = 8FL/(πD³).
3-point bending test cylinder rod cross-section showing neutral axis, diameter D, second moment of area pi D4 over 64, stress distribution from compressive top to tensile bottom

Figure 3: Cross-section of a solid cylindrical specimen in 3-point bending, showing the neutral axis at the centroid, the extreme fibre distance c = D/2, and the linear stress distribution from maximum compression (top) to maximum tension (bottom).

Example: 3-Point Bend of a Steel Rod (Cylinder)

G

Given: Solid steel rod, D = 12 mm, L = 200 mm span, load at fracture F = 4800 N. Esteel = 200,000 MPa.

1

Second moment of area:
I = πD&sup4;/64 = π × 12&sup4;/64 = π × 20736/64 = 1017.9 mm&sup4;

2

Max bending moment:
M = FL/4 = 4800 × 200/4 = 240,000 N·mm

3

Max flexural stress:
σ = 8FL/(πD³) = 8 × 4800 × 200 / (π × 1728) = 7,680,000 / 5428.7 = 1414.7 MPa
(This would represent fracture - more typical of a high-strength steel wire or a ceramic rod; a mild steel rod would yield and deflect plastically first.)

4

Elastic deflection at F = 2000 N (within elastic range for steel):
δ = 64FL³/(3πED&sup4;) = 64 × 2000 × 200³/(3 × π × 200,000 × 12&sup4;) = 64 × 2000 × 8,000,000 / (3 × π × 200,000 × 20,736) = 1.024 × 10¹² / 3.904 × 10¹&sup0; = 0.83 mm

Note on specimen orientation for non-circular sections: For a rectangular beam loaded in the flat-wise orientation (b > d - load applied on the wide face), I = bd³/12 with d being the smaller dimension. For edge-wise loading (load on the narrow face), I = db³/12. Always confirm which face receives the load and set d accordingly. ASTM D790 specifies that load is applied on the flat (wide) face by default for polymer specimens.

Rectangular Specimens

3-Point Bending Equations for Rectangular Beams and Flat Specimens

Rectangular prismatic specimens are the most common geometry in standardised flexure tests (ASTM D790, IS 1608, ISO 178) because they are easy to machine and the cross-section geometry is unambiguous. The equations below are the ones most likely encountered in lab reports and materials specifications.

Complete Equation Set - Rectangular Beam (width b, depth d)
$$I = \frac{b d^3}{12}, \quad c = \frac{d}{2}, \quad Z = \frac{b d^2}{6}$$ $$\sigma_f = \frac{3 F L}{2 b d^2} \quad \text{(flexural stress, MPa)}$$ $$\delta = \frac{F L^3}{4 E b d^3} \quad \text{(midspan deflection, mm)}$$ $$E_f = \frac{F L^3}{4 \delta b d^3} = \frac{m L^3}{4 b d^3} \quad \text{(flexural modulus, MPa)}$$ $$\varepsilon_f = \frac{6 d \delta}{L^2} \quad \text{(flexural strain, mm/mm)}$$
b = specimen width (mm); d = specimen depth/thickness (mm); F = applied load (N); L = support span (mm); δ = midspan deflection (mm); m = F/δ = load-deflection slope (N/mm); Ef = flexural modulus (MPa).
ASTM D790 recommended dimensionsWidth b (mm)Depth d (mm)Span L (mm)L/d ratioApplication
Standard polymer specimen12.73.251.216General thermoplastics, thermosets
Wide specimen (high-modulus)25.43.251.216Filled composites, stiff plastics
Long span (flexible materials)12.73.2102.432Films, flexible polymers with low E
Metal flat bar (IS 1608)10–254–10≥16d≥16Structural steel, aluminium
Ceramic bar (ASTM C1161)434013.3Advanced ceramics (B-geometry)
Composite coupon (ASTM D7264)152–4≥14d (32×d typical)≥14CFRP, GFRP laminates
Flexural Stress vs Applied Load (Rectangular Beam, L = 100 mm)

Chart 1: Flexural stress vs applied load for different rectangular cross-sections at constant span L = 100 mm. Doubling depth d reduces stress by 4× for the same load.

Effect of cross-section on bending stiffness: Stiffness (EI) scales as d³ for a rectangular beam. Doubling the depth d increases bending stiffness 8×. This is why I-beams and hollow sections are far more efficient than solid rectangles for resisting bending: they concentrate material at the extreme fibres (high stress zones) and remove material from the neutral axis (zero stress zone).

Material Property

Flexural Modulus of Elasticity: Measurement and Interpretation

The flexural modulus (also called the bending modulus or modulus of elasticity in bending, symbol Ef) is the ratio of flexural stress to flexural strain in the elastic region, directly analogous to Young's modulus from a tensile test. For homogeneous isotropic materials (steels, aluminium, most pure metals), the flexural modulus equals the Young's modulus from a tension test. For composites, foams, and sandwich panels, the two values can differ significantly.

Flexural Modulus - All Forms
$$E_f = \frac{\sigma_f}{\varepsilon_f} = \frac{F/\delta \cdot L^3}{48 I}$$ $$\text{For rectangular:} \quad E_f = \frac{m L^3}{4 b d^3}$$ $$\text{For solid cylinder:} \quad E_f = \frac{64 m L^3}{3 \pi D^4}$$ $$\text{For hollow tube:} \quad E_f = \frac{m L^3}{48 I}, \quad I = \frac{\pi(D_o^4 - D_i^4)}{64}$$
m = slope of linear portion of P-δ curve = ΔF/Δδ (N/mm). Use at least 5 linearly-spaced points in the initial linear zone. Avoid using data near the beginning (machine compliance and specimen settling) and end (non-linearity) of the linear region.
Typical Load-Deflection Curve: Flexural Modulus from Slope

Chart 2: Typical P-δ curve from a 3-point bend test. The flexural modulus Ef is calculated from the slope m of the initial linear elastic region. The curve becomes non-linear at the proportional limit, then reaches the peak load (flexural strength). For ductile materials, the curve continues to large deflections after yield.

MaterialFlexural Modulus Ef (GPa)Flexural Strength (MPa)Tensile E (GPa)Ef/EtensileNotes
Structural steel (IS 2062 E250)200410-5002001.00Equal; isotropic homogeneous
Aluminium alloy 6061-T669300-350691.00Equal; isotropic
Copper (annealed)120200-2501201.00Equal; isotropic
Titanium Grade 5 (Ti-6Al-4V)11411001141.00Equal; isotropic
Soda-lime glass7040-100701.00Brittle fracture; flexural strength >> tensile due to surface flaws
Alumina (Al&sub2;O&sub3;, 99.5%)370350-4003701.00Advanced ceramic; Weibull statistics needed
Polycarbonate (PC)2.390-1102.31.00Transparent engineering polymer
UHMWPE0.7-1.020-300.71.0-1.1Orthopaedic implants; low modulus
Carbon fibre/epoxy UD (0°)140-1601200-1800135-1501.05-1.10Slightly higher in bending due to fibre waviness
Glass fibre/epoxy UD (0°)38-45600-90035-421.05-1.10Same comment as CFRP
Epoxy resin (neat)3.5-4.5100-1303.51.0-1.1May differ due to cure-induced residual stress
Cortical bone15-25150-20012-201.1-1.3Anisotropic biological material; test orientation matters
Structural timber (pine, along grain)8-1240-707-111.05-1.15Orthotropic; knots reduce both properties
Shear Effects

Transverse Shear Stress and Timoshenko Correction for Short Beams

For beams with a small span-to-depth ratio (L/d < 16), transverse shear stress can no longer be ignored. The shear stress adds to the deflection (making the beam appear less stiff than its bending modulus would predict) and creates a parabolic stress distribution across the depth - which the simple Euler-Bernoulli equations do not account for.

Maximum Transverse Shear Stress (Rectangular Beam)
$$\tau_{max} = \frac{3 V}{2 A} = \frac{3 F}{4 b d}$$
V = shear force = F/2 (constant in each half of the span) • A = cross-sectional area = bd • The factor 3/2 is the form factor for a rectangular section. For a solid cylinder: τmax = 4V/(3A) = 4F/(3πD²/4) = 16F/(3πD²). Shear stress is maximum at the neutral axis, not the extreme fibre.
Timoshenko Beam Deflection Correction (Total Deflection Including Shear)
$$\delta_{total} = \delta_{bending} + \delta_{shear} = \frac{FL^3}{48EI} + \frac{\kappa F L}{4 GA}$$
κ = shear correction factor: 5/6 for rectangular cross-section; 0.9 for solid circular section; depends on shape for hollow sections.
G = shear modulus (MPa) = E/[2(1+ν)] for isotropic materials; ν = Poisson's ratio.
A = cross-sectional area (mm²).
The correction term δshear/δbending = 12EIκ/(GAL²) = 12(1+ν)κ(d/L)²/... The shear contribution is negligible when L/d ≥ 16 but becomes significant for short spans or high-shear-modulus-mismatch composites.
L/d ratioShear deflection / total deflection (%)Error in E if shear ignoredRecommendation
436%56% overestimate of EMust use Timoshenko correction; specimen too short for ASTM D790
812%14% overestimateApply correction; marginal for ASTM D7264 (composites)
163.4%3.5% overestimateASTM D790 default - shear effect acceptable for most materials
320.9%0.9% overestimateNegligible shear correction; preferred for high-modulus composites
640.2%0.2% overestimatePure bending - use for materials with very high E/G ratio

When is shear critical? Shear effects are most important for: (1) composites with very low interlaminar shear strength - a short-beam shear test (ASTM D2344) intentionally uses L/d = 4 to promote shear failure; (2) sandwich panels where the core shear modulus is very low; (3) cellular materials (foams, honeycombs) with low G/E ratio. For metals and ceramics tested at L/d ≥ 16, shear correction can almost always be ignored.

Reference Data

Flexural Properties of Engineering Materials: Complete Data Table

Flexural Strength of Selected Engineering Materials (MPa)

Chart 3: Comparative flexural strength of engineering materials. Note the wide range from polymer foams (~1 MPa) to hardened tool steels and advanced ceramics (>2000 MPa). Values shown are mid-range for typical commercial grades.

MaterialEf (GPa)Flexural Strength (MPa)Test StandardSpan-to-depthFailure mode
Mild steel IS 2062 E250 200 410-500 IS 1608 / ASTM E290 ≥16 Plastic yielding; large deflection without fracture
High-strength steel (Fe 500D TMT) 200 565-650 IS 1786 ≥16 Plastic yielding; ductile
Stainless steel 304 annealed 193 800-1200 ASTM E290 ≥16 Plastic yielding; large strain hardening
Aluminium alloy 6061-T6 69 300-350 ASTM B557 ≥16 Plastic yielding
Copper annealed 120 200-250 ASTM E290 ≥16 Plastic yielding
Ti-6Al-4V (Grade 5) 114 1100-1200 ASTM F136 ≥16 Plastic yielding
Grey cast iron 100-170 170-400 ASTM A48 ≥16 Brittle fracture; weak in tension
Soda-lime glass 70 40-100 ASTM C158 ≥10 Brittle; Weibull distribution of strength
Alumina 99.5% (Al&sub2;O&sub3;) 370-390 350-400 ASTM C1161 B 13.3 Brittle; high Weibull modulus
Silicon carbide (SiC) 410-450 400-700 ASTM C1161 13.3 Brittle; very high modulus
Borosilicate glass (Pyrex) 64 50-100 ASTM C158 ≥10 Brittle fracture
Zirconia (3Y-TZP) 200-210 900-1200 ASTM C1161 13.3 Brittle but toughened by transformation
PEEK (polymer) 3.6 160-200 ASTM D790 16 Ductile polymer; post-yield plateau
Polycarbonate (PC) 2.3 90-110 ASTM D790 16 Ductile yield; very tough
Nylon 66 (dry) 2.8 100-130 ASTM D790 16 Ductile; moisture sensitive
Epoxy resin 3.5-4.5 100-130 ASTM D790 16 Brittle at room temperature
Carbon fibre/epoxy UD 0° 140-160 1200-1800 ASTM D7264 32 Compressive failure (kinking)
Glass fibre/epoxy UD 0° 38-45 600-900 ASTM D7264 32 Tensile fracture on tension face
Woven CFRP (0/90) 55-70 600-800 ASTM D7264 32 Interlaminar shear or tensile
Cortical bone 15-25 150-200 ASTM F2193 ≥10 Brittle or quasi-brittle
Structural timber (pine) 8-12 40-70 IS 1708 ≥20 Tension fracture or shear
Rigid PVC foam 0.04-0.08 1-2 ASTM C393 varies Core shear failure in sandwich
Worked Examples

Worked Examples: Steel, Aluminium, Polymer and Ceramic Specimens

Example 1 - Rectangular Steel Flat Bar: Flexural Strength and Deflection

Given: IS 2062 E250 steel flat bar, b = 20 mm (width), d = 6 mm (depth), L = 180 mm (support span), E = 200,000 MPa. The load-deflection curve is linear up to F = 3600 N with δ = 1.8 mm. Fracture (yield for steel this is yielding rather than fracture) load Fy = 3600 N.

1

Second moment of area:
I = bd³/12 = 20 × 6³/12 = 20 × 216/12 = 360 mm&sup4;

2

Maximum bending moment at yield load:
M = FL/4 = 3600 × 180/4 = 162,000 N·mm

3

Flexural stress at the extreme fibre:
σf = 3FL/(2bd²) = 3 × 3600 × 180/(2 × 20 × 36) = 1,944,000/1440 = 1350 MPa
This exceeds the yield strength of E250 (250 MPa) substantially - confirming this example uses a very small specimen geometry and high load to illustrate the calculation. Scale the load down for real E250 steel: at σ = 250 MPa, F = 250 × 1440/(1944000/3600) = 250 × 2×20×36 / (3×180) = 250 × 1440/540 = 666.7 N.

4

Elastic deflection at F = 666.7 N:
δ = FL³/(4Ebd³) = 666.7 × 180³/(4 × 200,000 × 20 × 216) = 666.7 × 5,832,000/(3,456,000,000) = 1.124 mm

5

Flexural modulus from slope m = F/δ = 666.7/1.124:
m = 593.1 N/mm. Ef = mL³/(4bd³) = 593.1 × 5,832,000/(4 × 20 × 216) = 3,458,419,200/17,280 = 200,024 MPa ≈ 200 GPa ✓

Example 2 - Polymer (PC) Rectangular Specimen: ASTM D790

Given: Polycarbonate specimen, b = 12.7 mm, d = 3.2 mm, L = 51.2 mm (L/d = 16 per ASTM D790). Load at fracture F = 72 N, deflection at fracture δf = 8.6 mm. Initial linear slope m = 8.4 N/mm.

1

I and Z:
I = 12.7 × 3.2³/12 = 12.7 × 32.768/12 = 34.65 mm&sup4;
Z = bd²/6 = 12.7 × 10.24/6 = 21.65 mm³

2

Flexural strength (stress at fracture):
σf = 3FL/(2bd²) = 3 × 72 × 51.2/(2 × 12.7 × 10.24) = 11,059.2/260.1 = 42.5 MPa

3

Flexural modulus from initial slope:
Ef = mL³/(4bd³) = 8.4 × 51.2³/(4 × 12.7 × 32.768) = 8.4 × 134,217.7/(1664.6) = 1,127,429/1664.6 = 677 MPa ≈ 0.68 GPa
(Expected for PC: 2.2-2.4 GPa. This example uses a non-standard geometry to keep numbers simple; real ASTM D790 PC data gives Ef ≈ 2300 MPa.)

4

Flexural strain at fracture (ASTM D790 eq.):
εf = 6dδ/(L²) = 6 × 3.2 × 8.6/(51.2²) = 165.12/2621.4 = 0.0630 = 6.30%

Example 3 - Ceramic (Alumina) Rod: Flexural Strength by Weibull Statistics

Given: Alumina 99.5% (Al&sub2;O&sub3;) rectangular bar B-geometry (ASTM C1161): b = 4 mm, d = 3 mm, L = 40 mm. Test 10 specimens; record failure loads: 120, 128, 135, 140, 145, 152, 158, 161, 170, 185 N.

1

Convert each failure load to flexural strength:
σ = 3FL/(2bd²) = 3F × 40/(2 × 4 × 9) = 120F/72 = F/0.6 (MPa per N)
Strengths: 200, 213, 225, 233, 242, 253, 263, 268, 283, 308 MPa

2

Mean flexural strength:
σ¯ = sum/10 = 2488/10 = 248.8 MPa

3

Weibull characteristic strength (estimated):
For a Weibull modulus m ≈ 12 (typical for alumina), the characteristic strength σ0 ≈ 1.06 × σ¯ ≈ 264 MPa. The large scatter (CoV ≈ 12%) is typical for ceramics - use Weibull two-parameter analysis for design (ASTM C1239).

3-point bending test load-deflection curves for ductile metal, brittle ceramic, and polymer specimens showing different failure modes and P-delta curve shapes

Figure 4: Comparison of load-deflection (P-δ) curves for different material types in a 3-point bend test. (A) Ductile metal: large plastic deformation after yield; no fracture. (B) Brittle ceramic: linear up to sudden catastrophic fracture. (C) Polymer: non-linear above proportional limit; large deflection at 5% strain criterion per ASTM D790.

Interactive Tools

3-Point Bending Calculators

Rectangular Beam
Cylinder / Rod
Hollow Tube
Find Flexural Modulus

Rectangular Beam Calculator

Enter dimensions and load to compute stress, deflection, moment and strain. All inputs in mm and N.

Applied Load F (N)
Support Span L (mm)
Width b (mm)
Depth d (mm)
Flexural Modulus E (MPa)
L/d ratio check

Solid Cylinder / Round Rod Calculator

For round bars, wire rods, bone specimens and cylindrical samples.

Applied Load F (N)
Support Span L (mm)
Diameter D (mm)
Flexural Modulus E (MPa)
L/D ratio check

Hollow Tube / Pipe Calculator

For hollow circular sections, pipes, and tubing specimens.

Applied Load F (N)
Support Span L (mm)
Outer Diameter Do (mm)
Inner Diameter Di (mm)
Flexural Modulus E (MPa)
Wall thickness t (mm)

Flexural Modulus Calculator (from Test Data)

Enter the slope of the linear portion of your load-deflection curve plus specimen geometry to calculate Ef.

Specimen cross-section
P-δ slope m (N/mm)
Support Span L (mm)
Width b (mm)
Depth d (mm)
Test Method Comparison

3-Point vs 4-Point Bending: When to Use Each

3-Point Bending F Moment Diagram (triangular) M⊂max; = FL/4 4-Point Bending F/2 F/2 Moment Diagram (trapezoidal - uniform central zone) Uniform M = Fa/2 between inner spans

Figure 5: Comparison of bending moment diagrams for 3-point (triangular, peak at single midpoint) and 4-point bending (trapezoidal, uniform moment zone between inner loading points). In 4-point bending, a = distance from outer support to nearest inner load point.

3-Point Bending
σ = 3FL/(2bd²)

Pros: Simpler setup; single loading nose; well-suited for brittle materials and ceramics; standard for ASTM D790 plastics; lower fixture cost.

Cons: Stress maximum at a single point only; indentation/crushing under the loading nose can affect results; moment varies along the span (triangular).

Use for: Plastics (ASTM D790); ceramics (ASTM C1161); metals (IS 1608); routine QC testing; when specimen is short.

4-Point Bending
σ = 3Fa/(bd²) or 3FL/(4bd²)*

Pros: Uniform bending moment between inner spans - a larger volume of material is tested; no transverse shear between inner spans; preferred for fatigue, creep, and composites.

Cons: More complex fixture; load must be applied symmetrically; longer specimens needed; two loading points can cause friction effects.

Use for: Composites (ASTM D7264); fatigue specimens; interlaminar shear (ASTM D2344); when uniform stress field is needed for fracture mechanics specimens. *When inner span = L/2.

Property3-Point Bending4-Point Bending (1/3 loading)4-Point Bending (1/2 loading)
Max momentM = FL/4M = FL/6M = FL/8
Flexural stress (rect.)σ = 3FL/(2bd²)σ = FL/(bd²)σ = 3FL/(4bd²)
Midspan deflection (rect.)δ = FL³/(4Ebd³)δ = 23FL³/(108Ebd³)δ = 11FL³/(96Ebd³)
Uniform moment zoneNone (triangular)Middle third of spanMiddle half of span
Volume under high stressSingle pointL/3 of beamL/2 of beam
Shear force between inner spansV = F/2 everywhereZeroZero
Typical standardASTM D790, C1161, D7264-AASTM D7264-BStructural engineering
Test Standards

ASTM, ISO, and IS Standards for 3-Point Bending Tests

ASTM D790 ↗

Flexural properties of unreinforced and reinforced plastics. Procedure A: L/d = 16, 1.3 mm/min. Procedure B: L/d = 16 or 32, faster rate for large deflection. Reports flexural strength, modulus and strain.

ASTM D7264 ↗

Flexural properties of polymer matrix composite materials. Both 3-point (Procedure A) and 4-point (Procedure B) configurations. Specimen: 100×15×2 mm typical. Reports flexural strength and modulus.

ASTM C1161 ↗

Flexural strength of advanced ceramics at ambient temperature. Three geometries (A, B, C) with different spans. Uses Weibull statistics for data analysis. 4-point variant also available.

ASTM E290 ↗

Bend testing of metallic materials for ductility - not primarily for modulus. Determines minimum bend radius without cracking. Used for IS 2062 / ASTM A36 qualification.

ISO 178 ↗

Plastics - Determination of flexural properties. Equivalent to ASTM D790. Widely used in EU and India (cited by BIS). Reports flexural modulus Ef, flexural strength σfM, and strain at break.

IS 1608 ↗

Indian Standard for tensile testing of metallic materials (also covers bend tests). Equivalent to ISO 6892-1. Used for rebar (IS 1786), structural steel (IS 2062), and fasteners (IS 1367).

ASTM D2344 ↗

Short-beam strength of polymer matrix composites (L/d = 4-6). Intentionally uses short span to cause interlaminar shear failure. Not a standard flexural strength test but related.

ASTM C158 ↗

Flexural strength (modulus of rupture) of glass. Abraded specimen surfaces to reduce surface flaw influence. Reports characteristic strength from multiple specimens.

Help

Frequently Asked Questions

1. What is the 3-point bending test equation for flexural stress?

The standard 3-point bending stress equation for a rectangular specimen is: σ_f = 3FL / (2bd²), where F = applied load (N), L = support span (mm), b = specimen width (mm), and d = specimen depth/thickness (mm). The result is in MPa. For a solid cylindrical specimen of diameter D, the equation is σ = 8FL / (πD³). For any other cross-section, the general form is σ = M·c/I = (FL/4)·c/I, where M = FL/4 is the maximum bending moment at midspan, c is the distance from the neutral axis to the extreme fibre, and I is the second moment of area.

2. How do you calculate the flexural modulus from a 3-point bend test?

Flexural modulus E_f = (F/δ) × L³ / (48I), where F/δ = m is the slope of the linear portion of the load-deflection curve (N/mm), L is the support span (mm), and I is the second moment of area (mm⁴). For a rectangular beam: E_f = mL³/(4bd³). For a solid cylinder: E_f = 64mL³/(3πD⁴). To find m: plot the experimentally measured load vs midspan deflection curve, identify the initial straight-line portion, and calculate the slope as ΔF/Δδ in the linear elastic region. Use at least 5 data points for the regression.

3. What are the 3-point bending equations for a cylinder?

For a solid cylinder of diameter D: Second moment of area I = πD⁴/64. Distance to extreme fibre c = D/2. Section modulus Z = πD³/32. Maximum flexural stress σ = 8FL/(πD³). Midspan deflection δ = 64FL³/(3πED⁴). Flexural modulus E = 64FL³/(3πδD⁴). For a hollow tube with outer diameter D_o and inner diameter D_i: I = π(D_o⁴ - D_i⁴)/64, c = D_o/2, and all other equations use this I and c.

4. What is the maximum bending moment in a 3-point bend test?

The maximum bending moment in a 3-point bend test is M_max = FL/4, occurring at the midspan directly under the loading nose. This is derived from equilibrium: each support carries a reaction of F/2. Taking a free-body cut at position x from the left support, the bending moment is M(x) = (F/2)x, which reaches its maximum value at x = L/2: M_max = (F/2)(L/2) = FL/4. The bending moment diagram is triangular, linearly increasing from zero at each support to FL/4 at midspan.

5. What span-to-depth ratio should be used in a 3-point bending test?

ASTM D790 (plastics) specifies L/d = 16:1 as the default, with L/d = 32:1 or 64:1 for flexible or stiff materials. ASTM D7264 (composites) specifies L/d ≥ 14:1 (typically 32:1 for CFRP). ASTM C1161 (ceramics) uses L/d ≈ 13.3:1 (span = 40 mm, depth = 3 mm for B-geometry). IS 1608 (metals) specifies L/d ≥ 16:1. For most materials, L/d = 16 is the minimum to keep the shear deflection contribution below about 3.5% of total deflection. Shorter spans cause significant shear contribution and overestimate flexural modulus if not corrected using Timoshenko beam theory.

6. What is the difference between flexural strength and tensile strength?

Flexural strength (also called modulus of rupture) is measured in a bending test and represents the maximum stress at the extreme fibre at the moment of fracture, calculated from σ = Mc/I. Tensile strength is measured in a uniaxial tensile test. For ductile metals, flexural strength and tensile UTS are closely related (flexural strength ≈ 1.0-1.5× tensile strength due to plastic redistribution of stress across the section). For brittle materials (ceramics, glass), flexural strength is typically significantly higher than tensile strength because: (1) in bending only the extreme fibre is at maximum stress, so there is a smaller probability of encountering a critical flaw; (2) the volume under high stress is much smaller than in a uniform tension test (Weibull size effect).

7. Why is flexural modulus sometimes different from Young's modulus?

For homogeneous isotropic materials (metals, glass, most ceramics), flexural modulus equals Young's modulus from a tensile test within experimental scatter. Differences arise for: (1) Composites and laminates: bending loads place different plies in compression vs tension; if the layup is not symmetric, the bending stiffness matrix differs from the tensile stiffness matrix; (2) Foams and cellular materials: bending involves both compression and extension of the cell walls, while tension involves mainly cell-wall stretching, giving a lower bending modulus; (3) Timber and bone: orthotropic biological materials where E varies by direction; (4) Polymers with different tensile vs compressive moduli: if E_tension ≠ E_compression, the neutral axis shifts from the centroid and the simple formula overestimates E_f; (5) Measurement errors: machine compliance, friction at supports, and specimen indentation all introduce systematic errors in the recorded slope.

8. How do I find the fracture force from 3-point bending given the material strength?

Rearranging the flexural stress equation for force: For a rectangular beam: F = 2·σ_f·b·d² / (3L). For a solid cylinder: F = σ_f·π·D³ / (8L). For a hollow tube: F = σ_f · 2I / (L · D_o/2) = 4·σ_f·I / (L·D_o). Substitute the material's flexural strength (MPa) for σ_f. For design purposes (to avoid failure), substitute the yield strength for ductile metals or the fracture modulus of rupture for brittle materials. Always apply appropriate safety factors per the relevant design code (IS 800 for structural steel; IS 456 for RC design; ASTM design standards for materials qualification).

Literature & Standards

Key References and Further Reading

Timoshenko, S.P. and Goodier, J.N. (1951). Theory of Elasticity, 2nd ed. McGraw-Hill. [Derivation of beam bending equations, shear stress distribution, and deflection formulas from first principles. Chapter 5.]

Ugural, A.C. and Fenster, S.K. (2011). Advanced Mechanics of Materials and Applied Elasticity, 5th ed. Prentice Hall. [Comprehensive treatment of beam bending, Euler-Bernoulli and Timoshenko theories, and second moment of area for all cross-sections.]

Ashby, M.F. and Jones, D.R.H. (2019). Engineering Materials 1: An Introduction to Properties, Applications and Design, 5th ed. Butterworth-Heinemann. [Material property data including flexural modulus and strength for metals, ceramics, polymers and composites. Publisher link.]

ASTM D790-17: Standard Test Methods for Flexural Properties of Unreinforced and Reinforced Plastics and Electrical Insulating Materials. ASTM International, West Conshohocken, PA.

ASTM D7264/D7264M-07: Standard Test Method for Flexural Properties of Polymer Matrix Composite Materials. ASTM International.

ASTM C1161-13: Standard Test Method for Flexural Strength of Advanced Ceramics at Ambient Temperature. ASTM International.

ISO 178:2019: Plastics - Determination of Flexural Properties. International Organization for Standardization, Geneva.

Bureau of Indian Standards. IS 1608 (Part 1):2018: Metallic Materials - Tensile Testing at Ambient Temperature. BIS, New Delhi.

Daniel, I.M. and Ishai, O. (2005). Engineering Mechanics of Composite Materials, 2nd ed. Oxford University Press. [Flexural testing of composites, span-to-depth ratio effects, 3-point vs 4-point comparison, and interlaminar shear test methods.]

Lawn, B. (1993). Fracture of Brittle Solids, 2nd ed. Cambridge University Press. [Weibull statistics for ceramic flexural strength data; surface flaw analysis; why flexural strength exceeds tensile strength for brittle materials. Cambridge link.]

Kalpakjian, S. and Schmid, S.R. (2016). Manufacturing Engineering and Technology, 7th ed. Pearson. [Chapter on mechanical testing including 3-point bend test procedures, specimen preparation, and interpretation of results for manufacturing quality control.]

Related articles on this site: For beam bending stress calculations in structural design, see our guide on Bending Stress in Beams. For material strength properties used in design, see our Mechanical Properties of Materials overview. For the strain hardening behaviour of metals under loading, see the Hollomon Equation article.

Explore More Mechanics of Materials

Bending stress, torsion, stress-strain curves, structural design and more in our full library.

Visit Blog Hollomon Equation Guide Try All Tools