Civil Engineering

Bending Moment Calculator

Find the maximum bending moment, shear force and deflection in a beam. Choose one of five standard load cases — simply supported or cantilever under uniformly distributed or point load, or a fixed-end beam — enter the span, load and cross-section, and this calculator returns the moment and shear diagrams' peak values together with the second moment of area, section modulus, radius of gyration, bending stress and the span-to-deflection ratio. Concrete sections also get a long-term creep deflection.

Elastic beam theory · IS 456 Cl.23.2 creep · L/δ ≥ 250 serviceability

📏 Beam, Load and Section

Inputs convert when you switch. Results are always reported in SI — see the note under the results.

Load Case

Span L (m)

Distributed Load w (kN/m)

Cross-Section

Flange Width b (mm)

Overall Depth d (mm)

Web Thickness t_w (mm)

Flange Thickness t_f (mm)

Elastic Modulus E (GPa)

200 steel, 25–35 concrete, 70 aluminium, 10–12 timber.

Yield / Permissible Stress f_y (MPa)

📊 Beam Results

Press Calculate Beam to analyse the beam.

ƒ Governing Formulae

Load cases

Case	Max moment M	Max shear V	Max deflection δ	Location of M_max
Simply supported, UDL	wL²/8	wL/2	5wL⁴/384EI	Midspan
Simply supported, point load	PL/4	P/2	PL³/48EI	Midspan
Cantilever, UDL	wL²/2	wL	wL⁴/8EI	Fixed end
Cantilever, point load	PL	P	PL³/3EI	Fixed end
Fixed both ends, UDL	wL²/12	wL/2	wL⁴/384EI	Supports (midspan is wL²/24)

Section properties

Rectangular  I = bd³/12  A = bd
Circular  I = πD⁴/64  A = πD²/4
Hollow box  I = (bd³ − b_id_i³)/12, b_i = b − 2t_w, d_i = d − 2t_f
I-section  I = [bd³ − (b − t_w)h_w³]/12, h_w = d − 2t_f

Stress, stiffness and serviceability

Z = I / y y = d/2 (D/2 for circular)
σ = M / Z
r = √(I / A)
δ_long-term = δ × θ θ = 2.5 for concrete, 1.0 otherwise
serviceability limit: L/δ ≥ 250

w — uniformly distributed load, kN per metre run

P — concentrated point load, kN

L — span, or projection for a cantilever

I — second moment of area about the bending axis

Z — elastic section modulus

y — distance from neutral axis to extreme fibre

r — radius of gyration, used for buckling checks elsewhere

E — elastic modulus, entered in GPa

f_y — yield or permissible bending stress

θ — creep multiplier for sustained load on concrete

The stress check compares σ directly against f_y, with no factor of safety applied. That is a working-stress comparison against the yield value, so a beam reported as passing at σ = 249 MPa against f_y = 250 MPa has no margin whatsoever. If you want a design check, either enter a permissible stress that already includes your safety factor — 0.66f_y for a compact section under allowable stress design, for example — or read the reported σ and apply your own factor. This is the single most important thing to understand about the output on this page.

Other assumptions. The theory is elastic and small-deflection throughout, so no plastic redistribution and no second-order effects. Sections are treated as symmetric about the bending axis, which means y is taken as half the depth. Self-weight of the beam is not added to the load — include it in w yourself. There is no check on lateral-torsional buckling, web crippling, shear stress in the web, or local flange buckling, any of which can govern before bending stress does in a slender steel section. Concrete is treated as a homogeneous elastic material, so the section properties take no account of reinforcement or of cracking, and the results for a concrete beam are indicative of stiffness trends rather than a code-compliant analysis.

📝 Worked Example 1 — Simply Supported Steel I-Beam

Given

A simply supported steel I-beam spanning 6.0 m carrying a uniformly distributed load of 20 kN/m. The section is 150 mm wide by 300 mm deep with an 8 mm web and 12 mm flanges. Steel with E = 200 GPa and f_y = 250 MPa.

Step 1 — second moment of area

h_w = 300 − 2 × 12 = 276 mm
I = [150 × 300³ − (150 − 8) × 276³] / 12
I = [4.050 × 10⁹ − 2.985 × 10⁹] / 12 = 88.709 × 10⁶ mm⁴

Step 2 — section modulus and radius of gyration

y = 300/2 = 150 mm, so Z = 88.709 × 10⁶ / 150 = 591.39 × 10³ mm³
A = 2 × 150 × 12 + 276 × 8 = 5808 mm²
r = √(88.709 × 10⁶ / 5808) = 123.59 mm

Step 3 — moment and shear

M = wL²/8 = 20 × 6² / 8 = 90.000 kN·m at midspan
V = wL/2 = 20 × 6 / 2 = 60.000 kN at each support

Step 4 — bending stress

σ = M/Z = 90 × 10⁶ N·mm / 591 394 mm³ = 152.18 MPa
Against f_y = 250 MPa that is 61 percent utilisation — but remember there is no safety factor in that comparison.

Step 5 — deflection and serviceability

EI = 200 × 10⁶ kN/m² × 88.709 × 10⁻⁶ m⁴ = 17 742 kN·m²
δ = 5wL⁴/384EI = 5 × 20 × 6⁴ / (384 × 17 742) = 19.023 mm
L/δ = 6000 / 19.023 = 315, comfortably above the limit of 250

Maximum bending moment 90.000 kN·m at midspan

Maximum shear force 60.000 kN at the supports

Maximum deflection 19.023 mm, L/δ = 315

Bending stress 152.18 MPa against 250 MPa — PASS

Section properties I = 88.709 × 10⁶ mm⁴, Z = 591.39 × 10³ mm³, r = 123.59 mm

Overall verdict — PASS on both strength and serviceability

These are the calculator's default inputs. Press Calculate Beam without changing anything and you should get exactly these figures back. Try switching the load case to cantilever with the same span and load: the moment jumps from 90 to 360 kN·m and the deflection from 19 to 183 mm, which is the clearest demonstration there is of what a second support buys you.

⚠ Worked Example 2 — Strong Enough, Still Unacceptable

Given

A solid rectangular steel bar, 100 mm wide by 150 mm deep, spanning 8.0 m simply supported and carrying just 5 kN/m. E = 200 GPa, f_y = 250 MPa.

The strength check passes easily

I = 100 × 150³ / 12 = 28.125 × 10⁶ mm⁴, Z = 375.00 × 10³ mm³
M = 5 × 8² / 8 = 40.000 kN·m
σ = 40 × 10⁶ / 375 000 = 106.67 MPa against 250 MPa — only 43 percent utilised

And the beam is unusable

EI = 200 × 10⁶ × 28.125 × 10⁻⁶ = 5625 kN·m²
δ = 5 × 5 × 8⁴ / (384 × 5625) = 47.407 mm
L/δ = 8000 / 47.407 = 169, well short of the 250 limit

Bending stress 106.67 MPa against 250 — passes with 57 percent spare

Deflection 47.4 mm, L/δ = 169 against 250 required — FAILS

Verdict — not acceptable, and only the serviceability check says so

Why deflection governs long spans

Stress varies with L² but deflection varies with L⁴. Double the span and the stress goes up four times while the deflection goes up sixteen. That is why short heavily loaded members are usually governed by strength and long lightly loaded ones almost always by deflection, and why checking stress alone is not enough. The cure here is depth, not width: deflection depends on d³, so taking this section to 100 × 200 mm cuts the deflection by more than half while adding only a third to the material.

📐 Inputs, Units and Accepted Ranges

Input	SI unit	Imperial unit	Accepted	Notes
Load case	—	—	five standard cases	Sets whether the load is distributed or concentrated
Span L	m	ft	> 0	Clear span, or projection for a cantilever
Distributed load w	kN/m	kip/ft	≥ 0	UDL cases; self-weight not added automatically
Point load P	kN	kip	≥ 0	Point load cases; at midspan or the free end
Width b	mm	in	> 0	Flange width for I and box sections
Depth d	mm	in	> 0	Overall depth; must exceed 2t_f
Diameter	mm	in	> 0	Circular sections only
Web thickness t_w	mm	in	> 0, less than b	I and box sections; box requires 2t_w < b
Flange thickness t_f	mm	in	> 0, 2t_f < d	I and box sections
Elastic modulus E	GPa	GPa	> 0, flagged above 500	Enter GPa, not MPa — a factor of 1000 error is common
Yield stress f_y	MPa	MPa	> 0	Compared directly to σ with no safety factor

A value of E between 15 and 50 GPa is taken as concrete and triggers the 2.5× long-term creep multiplier. If you are analysing a different material whose modulus happens to fall in that band, the long-term figure will not apply to it — read the short-term deflection instead.

📚 Reference Tables

Elastic modulus and typical yield stress

Material	E, GPa	Typical f_y, MPa	Creep applied here?
Structural steel, mild	200	250	No
Structural steel, high yield	200	350 – 450	No
Stainless steel, austenitic	193	205 – 275	No
Concrete M25	25	—	Yes, θ = 2.5
Concrete M30	27.4	—	Yes, θ = 2.5
Concrete M40	31.6	—	Yes, θ = 2.5
Aluminium alloy 6061-T6	69	240	No
Timber, softwood C24	11	24 (bending)	No
Timber, hardwood D40	13	40 (bending)	No
Cast iron, grey	100	150 (compression)	No

Concrete modulus follows the IS 456 expression E = 5000√f_ck, so M25 gives 25 GPa and M40 gives 31.6 GPa.

Load case comparison — same span L, same total load, relative to the simply supported UDL case

Case	Moment ratio	Deflection ratio	Comment
Simply supported, UDL	1.00	1.00	Baseline
Simply supported, point load	2.00	1.60	Concentrating the load hurts moment most
Fixed both ends, UDL	0.67	0.20	End fixity is worth far more to stiffness than to strength
Cantilever, UDL	4.00	9.60	One support instead of two
Cantilever, point load	8.00	25.6	Worst case of the five by a wide margin

Deflection ratios assume the same total load W = wL applied in each case. The fixed-end row is the one worth remembering: fixing both ends cuts the moment by a third but the deflection by five times.

Section modulus of common shapes — showing how efficiently each uses its material

Section	Area, mm²	I, ×10⁶ mm⁴	Z, ×10³ mm³	Z per unit area
I-section 150 × 300, t_w 8, t_f 12	5808	88.71	591.4	101.8
Box 200 × 300, t 10	9600	120.72	804.8	83.8
Solid rectangle 200 × 400	80 000	1066.67	5333.3	66.7
Solid circle ø250	49 087	191.75	1534.0	31.3

The I-section delivers over three times the section modulus per unit of material that a solid circle does, because bending stress is proportional to distance from the neutral axis and an I-section puts its material where that distance is greatest.

Deflection limits in common use

Situation	Limit	At L = 6 m
General total deflection (checked here)	L/250	24.0 mm
Members supporting brittle finishes	L/350	17.1 mm
Deflection after finishes applied	L/350 or 20 mm, lesser	17.1 mm
Cantilevers, general	L/180	33.3 mm
Crane gantry girders, vertical	L/750 to L/1000	6.0 – 8.0 mm

❓ Frequently Asked Questions

How do you calculate the maximum bending moment in a simply supported beam?

For a uniformly distributed load the maximum bending moment occurs at midspan and equals the load per unit length times the span squared divided by eight. For a single point load at midspan it is the load times the span divided by four, again at midspan. Both assume the beam is simply supported, meaning it is free to rotate at each end and only one support resists horizontal movement.

Why does a cantilever have four times the moment of a simply supported beam?

Under the same uniformly distributed load the cantilever moment is the load times the span squared over two, against the span squared over eight for a simply supported beam, which is a factor of four. The cantilever has only one support, so the entire load has to be carried back to that single fixed end, and the moment builds all the way to it rather than being shared between two reactions. The deflection difference is even larger, at roughly nine and a half times.

What is the difference between section modulus and moment of inertia?

The second moment of area, commonly called the moment of inertia, measures how the area of a section is distributed about its neutral axis and governs deflection through the product of elastic modulus and that value. Section modulus is the moment of inertia divided by the distance from the neutral axis to the extreme fibre, and it governs stress because bending stress is the moment divided by the section modulus. In short, use moment of inertia for stiffness and section modulus for strength.

What deflection limit should a beam satisfy?

A span over deflection ratio of 250 is the common general limit for total deflection and is the value this calculator checks against. Members supporting brittle finishes such as plaster or masonry are often limited to span over 350 or span over 500 for the deflection occurring after the finish is applied. Deflection is a serviceability requirement rather than a strength one, so a beam can be perfectly safe against collapse and still be unacceptable because it sags, cracks finishes or feels bouncy underfoot.

Why is long-term deflection larger for a concrete beam?

Concrete creeps under sustained load, continuing to deform for years after the load is first applied, and it also shrinks as it dries. IS 456 accounts for this by multiplying the immediate elastic deflection by a creep factor, and this calculator applies a simplified factor of 2.5 when the elastic modulus entered falls in the range typical of concrete. A concrete beam that satisfies the deflection limit on its short-term value can easily fail it once the long-term multiplier is applied.

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Bending Moment Calculator — multicalci.com. Elastic small-deflection theory for symmetric sections. The stress check compares σ against f_y with no factor of safety. No allowance is made for self-weight, lateral-torsional buckling, web shear or crippling, or local buckling, and concrete sections are treated as homogeneous and uncracked. Results are indicative and must be verified by a qualified structural engineer.