Engineering guide · Beam fundamentals · Reviewed July 30, 2026

Beam Deflection Formulas & Equations

Compare 12 verified beam formulas for reactions, shear force, bending moment, slope and deflection across simply supported, cantilever, fixed, overhanging and continuous beams.

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Beam formula library

Positive moment is sagging. Downward deflection is reported as positive unless a case explicitly identifies lift or signed displacement.

12 formula cases shown

Simply supported

Center point load

Point load P at x = L/2.

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PLAB
RA = RB
= P / 2
Mmax
= PL / 4 at midspan
δmax
= PL3 / 48EI at midspan
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Shear, left half V(x) = +P/2 0 < x < L/2
Shear, right half V(x) = −P/2 L/2 < x < L
Moment, left half M(x) = Px/2 0 ≤ x ≤ L/2
Moment, right half M(x) = P(L−x)/2 L/2 ≤ x ≤ L
Deflection, left half y(x) = Px(3L2−4x2)/(48EI) 0 ≤ x ≤ L/2
End slope magnitude |θA| = |θB| = PL2/(16EI)

Engineering note: Symmetry places equal reactions, zero shear immediately across the load, and maximum bending and deflection at midspan.

Read the complete simply supported guide

Point load at any position

P at x = a; use b = L − a.

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PabAB
RA
= Pb / L
RB
= Pa / L
M at P
= Pab / L at x = a
δ at P
= Pa2b2 / 3EIL not always the maximum
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Left-span shear V(x) = Pb/L 0 < x < a
Right-span shear V(x) = −Pa/L a < x < L
Left-span moment M(x) = Pbx/L 0 ≤ x ≤ a
Right-span moment M(x) = Pa(L−x)/L a ≤ x ≤ L
Left-span deflection y(x) = Pbx(L2−b2−x2)/(6LEI) 0 ≤ x ≤ a
Right-span deflection y(x) = Pa(L−x)[L2−a2−(L−x)2]/(6LEI) a ≤ x ≤ L

Engineering note: The maximum moment occurs under the point load. The deflection directly under P is convenient to calculate, but the true maximum deflection generally shifts away from the load unless a = b.

Read the complete simply supported guide

Full-span UDL

Uniform load w over the complete span L.

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wLAB
RA = RB
= wL / 2
Mmax
= wL2 / 8 at midspan
δmax
= 5wL4 / 384EI at midspan
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Shear V(x) = w(L/2−x) 0 ≤ x ≤ L
Moment M(x) = wx(L−x)/2 0 ≤ x ≤ L
Deflection y(x) = wx(L3−2Lx2+x3)/(24EI) 0 ≤ x ≤ L
End slope magnitude |θA| = |θB| = wL3/(24EI)

Engineering note: The shear is linear, the moment is parabolic and both the zero-shear point and maximum deflection occur at midspan.

Read the complete simply supported guide

Cantilever

End point load

Point load P at the free tip.

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PLA
RA
= P
|MA|
= PL at fixed end
δmax
= PL3 / 3EI at tip
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Shear V(x) = −P 0 < x < L
Moment M(x) = −P(L−x) 0 ≤ x ≤ L
Slope magnitude θ(x) = Px(2L−x)/(2EI) 0 ≤ x ≤ L
Downward deflection y(x) = Px2(3L−x)/(6EI) 0 ≤ x ≤ L
Tip slope θL = PL2/(2EI)

Engineering note: The fixed end carries the full vertical reaction and the peak hogging moment. Slope and deflection both reach their largest magnitude at the free tip.

Read the complete cantilever guide

Point load at any position

P at x = a from the fixed end; free tip at L.

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PaA
RA
= P
|MA|
= Pa at fixed end
δ at P
= Pa3 / 3EI at x = a
δtip
= Pa2(3L−a) / 6EI at x = L
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Moment before load M(x) = −P(a−x) 0 ≤ x ≤ a
Moment after load M(x) = 0 a ≤ x ≤ L
Deflection before load y(x) = Px2(3a−x)/(6EI) 0 ≤ x ≤ a
Deflection after load y(x) = Pa2(3x−a)/(6EI) a ≤ x ≤ L
Slope after load θ(x) = Pa2/(2EI) a ≤ x ≤ L

Engineering note: The unloaded segment beyond P has zero shear and moment but continues as a straight tangent, so the free tip moves farther than the point of load.

Read the complete cantilever guide

Full-span UDL

Uniform load w over the complete cantilever L.

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wLA
RA
= wL
|MA|
= wL2 / 2 at fixed end
δmax
= wL4 / 8EI at tip
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Shear V(x) = −w(L−x) 0 ≤ x ≤ L
Moment M(x) = −w(L−x)2/2 0 ≤ x ≤ L
Slope magnitude θ(x) = wx(3L2−3Lx+x2)/(6EI) 0 ≤ x ≤ L
Downward deflection y(x) = wx2(6L2−4Lx+x2)/(24EI) 0 ≤ x ≤ L
Tip slope θL = wL3/(6EI)

Engineering note: The shear varies linearly and the hogging moment varies quadratically from the fixed end to zero at the free tip.

Read the complete cantilever guide

Fixed beam

Center point load

Point load P at midspan with both ends fixed.

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PLAB
RA = RB
= P / 2
MA = MB
= −PL / 8 at supports
Mmid
= +PL / 8 at midspan
δmax
= PL3 / 192EI at midspan
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Moment, left half M(x) = P(x/2−L/8) 0 ≤ x ≤ L/2
Deflection, left half y(x) = Px2(3L−4x)/(48EI) 0 ≤ x ≤ L/2
Slope, left half θ(x) = Px(L−2x)/(8EI) 0 ≤ x ≤ L/2
Zero-moment locations x = L/4 and 3L/4
Boundary conditions y(0) = y(L) = θ(0) = θ(L) = 0

Engineering note: End restraint creates negative support moments. The moment changes sign at the quarter points and the midspan deflection is one quarter of the corresponding simple-span value.

Read the complete fixed beam guide

Full-span UDL

Uniform load w over a fixed-fixed span L.

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wLAB
RA = RB
= wL / 2
MA = MB
= −wL2 / 12 at supports
Mmid
= +wL2 / 24 at midspan
δmax
= wL4 / 384EI at midspan
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Shear V(x) = w(L/2−x) 0 ≤ x ≤ L
Moment M(x) = −wL2/12 + wLx/2 − wx2/2 0 ≤ x ≤ L
Downward deflection y(x) = wx2(L−x)2/(24EI) 0 ≤ x ≤ L
Slope θ(x) = wx(L−x)(L−2x)/(12EI) 0 ≤ x ≤ L
Zero-moment locations x = L(1 ± 1/√3)/2 0.2113L and 0.7887L

Engineering note: The end slopes are zero. Negative support moments reduce the positive midspan moment and make the ideal beam five times stiffer in deflection than the comparable simple span.

Read the complete fixed beam guide

Overhanging

Tip point load

P at the free tip; span AB = L and overhang = a.

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PLaAB
RA
= −Pa / L downward
RB
= P(1 + a / L)
MB
= −Pa at roller
δtip
= Pa2(L + a) / 3EI downward
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Moment in AB M(x) = −Pax/L 0 ≤ x ≤ L
Moment on overhang M(x) = −P(L+a−x) L ≤ x ≤ L+a
Center of AB δAB,center = −PaL2/(16EI) at x = L/2
Largest lift within AB δAB,max = −PaL2/(9√3 EI) at x = L/√3
Reaction check RA + RB = P

Engineering note: The negative reaction at A represents uplift and requires a physical hold-down. The supported span lifts while the free tip moves downward.

Read the complete overhanging guide

Full-length UDL

w over the support span L and overhang a.

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wLaAB
RA
= w(L2 − a2) / 2L
RB
= w(L + a)2 / 2L
MB
= −wa2 / 2 at roller
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Positive-moment location x = RA/w when RA > 0
Positive moment there M+ = RA2/(2w) compare with |MB|
Center of AB δAB,center = wL2(5L2−12a2)/(384EI) at x = L/2
Free-tip deflection δtip = wa(−L3+4La2+3a3)/(24EI) signed; positive downward
Reaction check RA + RB = w(L+a)

Engineering note: The center-of-span and free-tip values are references, not guaranteed global maxima. The complete deflected curve can reverse direction as the overhang ratio changes.

Read the complete overhanging guide

Continuous

Two equal spans, UDL on both

Two spans ℓ with the same w and constant EI.

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wℓℓABC
RA = RC
= 3wℓ / 8
RB
= 5wℓ / 4
MB
= −wℓ2 / 8 interior support
M+max
= 9wℓ2 / 128 at 3ℓ/8
δmax
= wℓ4 / 184.63EI approximately
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Total length 2ℓ two equal spans
Deflection location x = 0.4215ℓ from A or C
Reaction check 3wℓ/8 + 5wℓ/4 + 3wℓ/8 = 2wℓ
Compatibility condition yB = 0 with one continuous member through B

Engineering note: The interior support carries the largest reaction and develops hogging moment. Use ℓ for one span, not the total beam length 2ℓ.

Read the complete continuous guide

Point load on one span

P at the center of span AB; span BC is unloaded.

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PℓℓABC
RA
= 13P / 32
RB
= 11P / 16
RC
= −3P / 32 uplift
MB
= −3Pℓ / 32 interior support
M+max
= 13Pℓ / 64 under P
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Maximum displacement δmax ≈ 0.015012Pℓ3/EI in span AB
Total length 2ℓ two equal spans
Reaction check 13P/32 + 11P/16 − 3P/32 = P
Compatibility condition yB = 0 with constant EI

Engineering note: Pattern loading breaks symmetry and can lift the remote support C. Real continuous beams should be checked for multiple realistic load arrangements.

Read the complete continuous guide

Formula notation

Symbols, sign convention and assumptions

Use this technical basis when checking a selected case. Values labelled as magnitudes use absolute-value notation.

P point load w load per length L span or beam length a, b load-position distances x position from the left datum E Young’s modulus I second moment of area EI flexural rigidity

Sign convention

Reactions, V, M, θ and y

Coordinates
+x runs from the left datum. +y and positive deflection are downward.
Loads & reactions
P and w are downward load magnitudes. +R is upward; a negative reaction denotes uplift.
Shear V
Positive shear is downward on the right cut face of the left beam segment.
Moment M
+M is sagging. −M is hogging.
Slope θ
θ = dy/dx; positive slope descends to the right.
Deflection δ
+δ is downward; negative displacement is upward.

Model assumptions

When these formulas apply

  • Small-displacement Euler–Bernoulli bending and linear-elastic material response.
  • Straight, prismatic member with constant E and I.
  • Ideal supports without settlement, slip or support flexibility.
  • Static point or uniform loads at the positions and lengths shown.
  • Consistent units for loads, geometry and EI.
  • No shear deformation, yielding, cracking, buckling, vibration, second-order effects or code checks.

For partial or mixed loads, varying EI or flexible supports, use a complete analysis model.

Build a custom beam Find E, I and section properties

Coefficient comparison

Two load families, two explicit baselines

Compare supports inside one card only. The point-load and uniform-load coefficients have different dimensional forms and are never compared directly.

Point-load family

Hold P, L, E and I constant

δmax = CPPL3/EI
Baseline

Simply supported + P at midspan is defined as 1.00× for this card.

SupportLoadCritical momentCPvs baseline
Simply supportedBaseline P at midspan PL/4 1/48 1.00×
Fixed-fixed P at midspan PL/8 1/192 0.25×
Cantilever P at free tip PL 1/3 16.00×

Uniform-load family

Hold w, L, E and I constant

δmax = CwwL4/EI
Baseline

Simply supported + full-span UDL is defined as 1.00× for this card.

SupportLoadCritical momentCwvs baseline
Simply supportedBaseline Full-span UDL w wL2/8 5/384 1.00×
Fixed-fixed Full-span UDL w wL2/12 at ends 1/384 0.20×
Cantilever Full-span UDL w wL2/2 1/8 9.60×

Reading the ratio: 0.25× means one quarter of that card’s simply supported baseline deflection; 16.00× means sixteen times that baseline for equal load magnitude, length and EI.

Common questions

Beam formula FAQ

What is the general beam deflection equation?

Euler-Bernoulli theory relates bending moment to curvature through EIκ(x) = M(x). With this page’s sagging-positive and downward-deflection-positive convention, EI times the second derivative of y(x) equals −M(x). Integrating the curvature and applying the beam boundary conditions gives slope and deflection.

Which beam deflection formula should I use?

Match the support arrangement, load type and load position exactly. A simply supported center-load formula, for example, is not valid for an off-center load, fixed beam or cantilever.

What do E and I mean in beam formulas?

E is Young’s modulus of the material and I is the second moment of area of the cross-section about the bending axis. Their product EI is the flexural rigidity.

Why does beam deflection depend on length cubed or length to the fourth power?

Point-load deflection commonly scales with L cubed, while distributed-load deflection commonly scales with L to the fourth power. Span therefore has a very strong effect on stiffness checks.

Is the deflection under an off-center point load always the maximum?

No. For an off-center load on a simply supported beam, the deflection at the load is easy to calculate, but the true maximum generally occurs at another position unless the load is centered.

Can I superimpose these beam formulas?

Yes, when the beam remains linear elastic and the support conditions and EI are unchanged. Add signed reactions, shear, moment, slope and deflection from compatible load cases.

When should I use the full beam calculator instead of a formula?

Use the full solver for mixed or partial loads, arbitrary support coordinates, more spans, varying EI, spring supports, settlements or when you need complete diagrams rather than a closed-form benchmark.

Do these equations include shear deformation?

No. The listed elastic-curve equations use Euler-Bernoulli bending and neglect transverse shear deformation. Deep beams and short spans may require a theory such as Timoshenko beam theory.

Engineering authorship

Prepared and reviewed by

Formula scope, boundary conditions, sign conventions, worked substitutions and modeling limitations were technically reviewed for the exact beam cases presented on this page. Last technical review: July 30, 2026.

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Sources and verification

Reference basis and checking process

The equations are classical, closed-form Euler–Bernoulli beam solutions. Coefficients and boundary-condition behavior for the 12 launch cases spanning simply supported, cantilever, fixed, overhanging and two-span continuous beams were checked against the references below before the worked values were published.

How the examples were checked

Each expected value was recalculated from the displayed equation and checked against force and moment equilibrium where applicable. The worked-example links carry the same geometry, stiffness and loading into the calculator so the response diagrams can be compared with the hand check.

Review the published solver verification methodology

These checks verify the stated idealized analysis cases; they do not constitute design certification or replace project-specific review of loading, restraints, stability, connections and governing requirements.