How do you calculate the deflection of a beam?

Beam deflection depends on the load, span, support conditions and the beam stiffness EI (Young's modulus E times the second moment of area I). For standard cases there are closed-form formulas: a simply supported beam under a uniform load w deflects 5wL⁴/384EI at mid-span, while a cantilever under an end point load P deflects PL³/3EI at the tip. This calculator applies the correct formula for each beam case automatically and also gives the deflection at any position along the span.

What is the difference between a shear force diagram (SFD) and a bending moment diagram (BMD)?

The shear force diagram (SFD) plots the internal shear force along the beam - the net transverse force on the section - while the bending moment diagram (BMD) plots the internal bending moment, which causes the beam to bend. They are related: the slope of the bending moment diagram at any point equals the shear force there, and the bending moment is a maximum (or minimum) where the shear force passes through zero. This tool draws both diagrams for every beam case.

Where is the maximum bending moment in a simply supported beam?

For a simply supported beam under a uniform load, the maximum bending moment is at mid-span and equals wL²/8. For a single central point load it is also at mid-span and equals PL/4. For an off-centre point load at distance a from one support (with b = L − a), the maximum moment occurs under the load and equals Pab/L. In general the maximum moment is where the shear force changes sign.

What is the maximum bending moment in a cantilever beam?

A cantilever carries its maximum bending moment at the fixed support (the wall), not at the free end. Under a uniform load w it equals wL²/2; under an end point load P it equals PL; under an end moment M₀ the moment is constant and equals M₀. The moment is hogging (tension on the top fibre), which this calculator shows as a negative value.

What is the difference between a fixed (built-in) beam and a simply supported beam?

A simply supported beam rests on a pin and a roller, so the supports carry vertical reactions but no moment - it is statically determinate. A fixed (built-in) beam is rigidly clamped at both ends, so the supports also resist rotation and develop fixing moments - it is statically indeterminate. Fixing the ends roughly halves the mid-span moment and reduces deflection to about one-fifth of the simply supported value, but introduces hogging moments at the supports.

Why is a propped cantilever statically indeterminate?

A propped cantilever is fixed at one end and propped (a roller) at the other, giving more reaction components than the three equilibrium equations can solve. The extra (redundant) reaction is found from a compatibility condition - that the deflection at the prop is zero. For a uniform load this gives reactions of 5wL/8 at the fixed end and 3wL/8 at the prop, a support moment of −wL²/8, and a maximum sagging moment of 9wL²/128.

What deflection limit should a beam satisfy?

Serviceability deflection limits are commonly span/360 for brittle finishes under imposed load, span/250 for total load, and span/200 for cantilevers, though the governing code and project specification take precedence. This calculator reports the elastic deflection in millimetres so it can be compared against the relevant span/n limit for your design.

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Beam Analysis Theory - Shear, Moment & Deflection

The theory behind this beam calculator: how support reactions, shear force, bending moment and deflection are derived for the standard beam types - simply supported, cantilever, fixed-end, propped cantilever, overhanging and continuous beams - and the closed-form formulas for each.

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A beam is a structural member that carries transverse loads primarily by bending. Analysing a beam means finding four things: the support reactions, the internal shear force $V(x)$ , the internal bending moment $M(x)$ , and the deflection $\delta(x)$ . This reference derives all four for the standard textbook beams - simply supported, cantilever, fixed-end (built-in), propped cantilever, overhanging and continuous - under point loads, uniformly distributed loads (UDL), triangular (linearly varying) loads and applied moments, and gives the closed-form formula for each. These are the same results the calculator evaluates.

Sign conventions and the quantities

Consistent sign conventions are essential for reading the diagrams correctly. Throughout this tool:

Loads - downward point loads $P$ (kN) and distributed loads $w$ (kN/m) are positive.
Shear force $V(x)$ (kN) - the algebraic sum of the transverse forces to the left of the section. A positive shear tends to push the left part up relative to the right.
Bending moment $M(x)$ (kN·m) - positive for sagging (concave up, tension on the bottom fibre) and negative for hogging (concave down, tension on top). Support moments at fixed ends are hogging, so they print as negative values.
Deflection $\delta(x)$ (mm) - downward positive. Slope $\theta = d\delta/dx$ is in radians.
Flexural rigidity $EI$ - Young's modulus $E$ times the second moment of area $I$ . It is the beam's bending stiffness; deflection is inversely proportional to it.

The load–shear–moment–deflection relationships

Shear, moment and deflection are linked by differential relationships that underpin every beam diagram. Working from the distributed load intensity $w(x)$ downward:

\frac{dV}{dx} = -w(x), \qquad \frac{dM}{dx} = V(x), \qquad EI\,\frac{d^{2}\delta}{dx^{2}} = M(x)

Three consequences follow directly, and they let you sketch any diagram by hand:

The bending moment is a maximum or minimum exactly where the shear force is zero (since $dM/dx = V = 0$ ). This is the single most useful rule for locating $M_{\max}$ .
Over an unloaded segment the shear is constant and the moment varies linearly; under a UDL the shear varies linearly and the moment is parabolic; under a triangular load the moment is cubic.
A concentrated load causes a step in the shear diagram; an applied moment causes a step in the moment diagram.

Deflection is obtained by integrating $M(x)/EI$ twice (the double-integration method) and applying the boundary conditions - zero deflection at supports, zero slope at fixed ends. For statically indeterminate beams (fixed-end, propped cantilever, continuous) the redundant reactions are first found from compatibility - typically that a deflection or rotation is zero at a redundant support - using methods such as the three-moment theorem, moment-area method or unit-load method.

Determinate vs. indeterminate beams

A beam is statically determinate when the reactions can be found from the three equations of planar equilibrium alone ( $\sum F_x = 0,\ \sum F_y = 0,\ \sum M = 0$ ) - this covers simply supported, cantilever and overhanging beams. It is statically indeterminate when there are more reaction components than equilibrium equations - fixed-end (two extra), propped cantilever (one extra) and continuous beams. The extra unknowns (the redundants) require compatibility of deformation to solve. Indeterminate beams are stiffer and distribute moment to the supports, which usually makes them more material- efficient than determinate ones.

Simply supported beam

A simply supported beam rests on a pin at one end and a roller at the other over a span $L$ . It is statically determinate. It develops a sagging moment that peaks where the shear is zero and has the largest deflection of the common single-span arrangements, so deflection often governs design. Key results:

Load case	Reactions	Max bending moment	Max deflection
Central point load P	$R_A = R_B = \tfrac{P}{2}$	$M_{\max} = \tfrac{PL}{4}$	$\delta_{\max} = \tfrac{PL^{3}}{48EI}$
Point load P at a (b = L−a)	$R_A = \tfrac{Pb}{L},\ R_B = \tfrac{Pa}{L}$	$M_{\max} = \tfrac{Pab}{L}$	$\text{under load, by integration}$
UDL w over span	$R_A = R_B = \tfrac{wL}{2}$	$M_{\max} = \tfrac{wL^{2}}{8}$	$\delta_{\max} = \tfrac{5wL^{4}}{384EI}$
Triangular load (0 → w₀)	$R_A = \tfrac{w_0 L}{6},\ R_B = \tfrac{w_0 L}{3}$	$M_{\max} = \tfrac{w_0 L^{2}}{9\sqrt3}$	$\approx 0.00652\tfrac{w_0 L^{4}}{EI}$
Applied moment M₀ at a	$R_A = \tfrac{M_0}{L},\ R_B = -\tfrac{M_0}{L}$	$M = \tfrac{M_0 a}{L}\ \text{(jump at a)}$	$\text{by integration}$

For the UDL case the shear varies linearly from $+wL/2$ at the left support to $-wL/2$ at the right, crossing zero at mid-span where the parabolic moment peaks at $wL^{2}/8$ . The triangular-load maximum moment occurs at $x = L/\sqrt3 \approx 0.577L$ , not mid-span.

Cantilever beam

A cantilever is fixed (encastré) at one end and free at the other. The fixed support supplies both a vertical reaction and a moment reaction. The bending moment is hogging and largest at the fixed end; the deflection is largest at the free tip. Cantilevers are comparatively flexible, so tip deflection frequently controls.

Load case	Reactions	Max bending moment	Max deflection
Point load P at free end	$R = P,\ M_A = -PL$	$M_{\max} = PL\ \text{(fixed)}$	$\delta_{\max} = \tfrac{PL^{3}}{3EI}$
Point load P at a	$R = P,\ M_A = -Pa$	$M_{\max} = Pa$	$\delta_{tip} = \tfrac{Pa^{2}}{6EI}(3L-a)$
UDL w over length	$R = wL,\ M_A = -\tfrac{wL^{2}}{2}$	$M_{\max} = \tfrac{wL^{2}}{2}$	$\delta_{\max} = \tfrac{wL^{4}}{8EI}$
Triangular (max w₀ at fixed)	$R = \tfrac{w_0 L}{2},\ M_A = -\tfrac{w_0 L^{2}}{6}$	$M_{\max} = \tfrac{w_0 L^{2}}{6}$	$\delta_{\max} = \tfrac{w_0 L^{4}}{30EI}$
Moment M₀ at free end	$M_A = M_0,\ V = 0$	$M = M_0\ \text{(constant)}$	$\delta_{\max} = \tfrac{M_0 L^{2}}{2EI}$

Fixed-end (built-in) beam

A fixed-end beam is rigidly clamped at both ends, so each support resists rotation and develops a hogging fixing moment. It is statically indeterminate to the second degree. The fixing moments reduce the mid-span moment and stiffen the beam dramatically - for a uniform load the central deflection is only $wL^{4}/384EI$ , one-fifth of the simply supported value, and the peak moment $wL^{2}/12$ (at the supports) is two-thirds of the simply supported $wL^{2}/8$ .

Load case	Reactions	Max bending moment	Max deflection
Central point load P	$R_A = R_B = \tfrac{P}{2}$	$M_{end} = -\tfrac{PL}{8},\ M_{mid}=+\tfrac{PL}{8}$	$\delta_{\max} = \tfrac{PL^{3}}{192EI}$
Point load P at a (b = L−a)	$R_A = \tfrac{Pb^{2}(3a+b)}{L^{3}}$	$M_A = -\tfrac{Pab^{2}}{L^{2}},\ M_B=-\tfrac{Pa^{2} b}{L^{2}}$	$\text{by integration}$
UDL w over span	$R_A = R_B = \tfrac{wL}{2}$	$M_{end} = -\tfrac{wL^{2}}{12},\ M_{mid}=+\tfrac{wL^{2}}{24}$	$\delta_{\max} = \tfrac{wL^{4}}{384EI}$

Propped cantilever

A propped cantilever is fixed at one end and simply propped (a roller) at the other - indeterminate to the first degree. The redundant prop reaction follows from the compatibility condition that the deflection at the prop is zero. For a uniform load the fixed end takes $5wL/8$ and the prop $3wL/8$ , the hogging moment at the fixed end is $-wL^{2}/8$ , and the maximum sagging moment is $9wL^{2}/128$ at $x = 5L/8$ from the fixed end.

Load case	Reactions	Max bending moment	Max deflection
UDL w over span	$R_A = \tfrac{5wL}{8},\ R_B = \tfrac{3wL}{8}$	$M_A = -\tfrac{wL^{2}}{8},\ M^+ = \tfrac{9wL^{2}}{128}$	$\delta_{\max} \approx \tfrac{wL^{4}}{185EI}$
Central point load P	$R_A = \tfrac{11P}{16},\ R_B = \tfrac{5P}{16}$	$M_A = -\tfrac{3PL}{16},\ M_{load}=\tfrac{5PL}{32}$	$\approx \tfrac{7PL^{3}}{768EI}$

Overhanging beam

An overhanging beam is a simply supported span with one or both ends cantilevering past a support. It is statically determinate. The load on the overhang induces a hogging moment over the support that partly cancels the sagging moment in the span - a deliberately efficient arrangement used in balconies, canopies and bridge approach spans. With a balanced overhang the positive and negative moments can be made roughly equal, minimising the peak. For a single overhang $c$ carrying UDL $w$ over the whole length, the moment over the support is $-wc^{2}/2$ ; a point load $P$ at the tip gives $-Pc$ there. A long enough overhang can lift the far support into uplift (a negative reaction) - which this tool reports.

Continuous beam

A continuous beam runs over three or more supports as a single member. It is statically indeterminate, with one redundant per interior support, solved classically by the three-moment (Clapeyron) theorem or moment distribution. Continuity develops hogging moments over the interior supports that relieve the span moments, so a continuous beam is stiffer and carries more load than the equivalent series of simply supported spans. For two equal spans $L$ under a uniform load, the interior support takes $5wL/4$ , the end supports $3wL/8$ each, the hogging moment over the middle support is $-wL^{2}/8$ , and the maximum span sagging moment is $9wL^{2}/128$ .

Load case	Reactions	Max bending moment	Max deflection
Two equal spans, UDL w	$R_A = R_C = \tfrac{3wL}{8},\ R_B = \tfrac{5wL}{4}$	$M_B = -\tfrac{wL^{2}}{8},\ M^+ = \tfrac{9wL^{2}}{128}$	$\text{(by integration)}$
Two equal spans, central P each	$R_A = R_C = \tfrac{5P}{16},\ R_B = \tfrac{11P}{8}$	$M_B = -\tfrac{3PL}{16},\ M_{span}=\tfrac{5PL}{32}$	$\text{(by integration)}$

Deflection limits - serviceability

Beyond strength, beams are checked for deflection at the serviceability limit state to control sag, cracking of finishes and user comfort. Typical span/ $n$ limits (the governing code and project specification always take precedence) are:

Condition	Typical limit
Beams with brittle finishes (imposed load)	$L/360$
General beams (total load)	$L/250$
Cantilevers	$L/180$ to $L/250$
Beams supporting machinery / sensitive cladding	$L/500$ or stricter

The calculator reports the elastic deflection in millimetres so it can be compared directly against the relevant span/ $n$ limit. Remember that deflection scales with $1/EI$ and with the fourth powerof span for distributed loads - doubling the span increases UDL deflection sixteen-fold, which is why long-span beams are deflection-controlled.

How to use this calculator

Pick a beam type and load case from the left, enter the span, load and the section stiffness $E$ (GPa) and $I$ (cm⁴), and the tool returns the reactions, the maximum shear, moment and deflection with their locations, the value of $V$ , $M$ and $\delta$ at any position you choose, and the shear-force, bending-moment and deflected- shape diagrams. Hover over any diagram to read off the value at that point.

These are standard linear-elastic results assuming constant $EI$ , small deflections and material within the elastic range. They neglect shear deformation (Euler–Bernoulli theory), which is adequate for ordinary slender beams. Always design to the governing code and project specification.

Frequently asked questions

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