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CivilSolve

Statics & Mechanics of Materials

Beam Deflection by Superposition

Calculate beam deflection and slope for standard simply supported and cantilever load cases. Combine closed-form solutions and find peak deflection.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
loadsApplied loads from the superposition case table; effects are summed
loads 1
eval atOptional positions at which to report deflection and slope

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SS midspan point load: P=10 kN, L=6 m, E=200 GPa, I=50e6 mm⁴

Given

configuration
simply-supported
span
6 m
modulus
200 GPa
moment of inertia
50000000 mm^4
loads
  1. 1.kind pointmagnitude 10 kNposition 3 mdirection down

Assumptions

  • Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
  • Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
  • Simply supported: pin and roller at the two ends of the span; deflection is zero at both supports.

Solution steps

  1. Flexural rigidity

    All handbook deflection formulas share the beam stiffness EI in the denominator.

    flexural rigidity EI = E × I

    EI = 10000 kN·m^2

  2. Simply supported beam with a point load

    Handbook closed-form case; its deflection adds to the other loads by superposition.

    y = −Pbx(L² − b² − x²)/(6·L·EI) for x ≤ a (mirror form beyond the load)

    Contribution at x = 3 m: y = -4.5 mm

  3. Maximum deflection

    For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

    for a load right of center (a ≥ b): xmax = √((L² − b²)/3), on the longer segment

    δmax = -4.5 mm at x = 3 m (downward)

    = -4.5 mm

  4. End slopes

    The support rotations bound the elastic curve; useful for checking serviceability and for slope-deflection methods.

    θA and θB from the slope function

    θA = -0.00225 rad; θB = 0.00225 rad

Results

Maximum deflection (signed; negative = downward)

-4.5mm

Location of maximum deflection

3m

Slope at the left support (radians)

-0.00225rad

Slope at the right support (radians)

0.00225rad

Where this answer was checked
source
δmax = PL³/48EI, θ = PL²/16EI (NCEES FE Reference Handbook) — simply supported, concentrated load at midspan
verified by
hand-recomputed
derivation
EI = 200e9·5e-5 = 1.0e7 N·m². δmax = 10e3·216/(48·1e7) = 2.16e6/4.8e8 = 4.5e-3 m = 4.5 mm DOWN → −4.5 mm at x = 3 m. θA = −PL²/16EI = −10e3·36/(16·1e7) = −2.25e-3 rad; θB = +2.25e-3 rad.

Example 2

US SS full UDL: w=2 kip/ft, L=20 ft, E=29000 ksi, I=600 in⁴

Given

configuration
simply-supported
span
20 ft
modulus
29000 ksi
moment of inertia
600 in^4
loads
  1. 1.kind udlmagnitude 2 kip/ft

Assumptions

  • Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
  • Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
  • Simply supported: pin and roller at the two ends of the span; deflection is zero at both supports.

Solution steps

  1. Flexural rigidity

    All handbook deflection formulas share the beam stiffness EI in the denominator.

    flexural rigidity EI = E × I

    EI = 120800 kip·ft^2

  2. Simply supported beam with a full-span uniform load

    Handbook closed-form case; its deflection adds to the other loads by superposition.

    y = −wx(L³ − 2Lx² + x³)/(24·EI); δmax = 5wL⁴/(384·EI) at midspan

    Contribution at x = 10 ft: y = -0.4138 in

  3. Maximum deflection

    For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

    slope = 0 at the extremum (or at a free end)

    δmax = -0.4138 in at x = 10 ft (downward)

    = -0.4138 in

  4. End slopes

    The support rotations bound the elastic curve; useful for checking serviceability and for slope-deflection methods.

    θA and θB from the slope function

    θA = -0.005517 rad; θB = 0.005517 rad

Results

Maximum deflection (signed; negative = downward)

-0.4138in

Location of maximum deflection

10ft

Slope at the left support (radians)

-0.005517rad

Slope at the right support (radians)

0.005517rad

Where this answer was checked
source
δmax = 5wL⁴/384EI, θ = wL³/24EI (NCEES FE Reference Handbook) — simply supported, uniform load over the whole span
verified by
hand-recomputed
derivation
In inches: w = 2000/12 = 166.667 lbf/in, L = 240 in. δmax = 5·166.667·240⁴/(384·29e6·600) = 2.76480e12/6.6816e12 = 0.413793 in DOWN → −0.41379 in at midspan (10 ft). θA = −wL³/24EI = −166.667·1.38240e7/(24·1.74e10) = −5.51724e-3 rad; θB = +5.51724e-3.

Example 3

cantilever tip load: P=5 kN, L=2 m, E=70 GPa, I=40e6 mm⁴

Given

configuration
cantilever
fixed end
left
span
2 m
modulus
70 GPa
moment of inertia
40000000 mm^4
loads
  1. 1.kind pointmagnitude 5 kNposition 2 mdirection down

Assumptions

  • Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
  • Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
  • Cantilever fixed at the left end, free at the other.

Solution steps

  1. Flexural rigidity

    All handbook deflection formulas share the beam stiffness EI in the denominator.

    flexural rigidity EI = E × I

    EI = 2800 kN·m^2

  2. Cantilever with a point load

    Handbook closed-form case; its deflection adds to the other loads by superposition.

    y = −Px²(3a − x)/(6·EI) for x ≤ a; beyond the load the beam is straight; δtip = −Pa²(3L − a)/(6·EI)

    Contribution at x = 2 m: y = -4.762 mm

  3. Maximum deflection

    For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

    slope = 0 at the extremum (or at a free end)

    δmax = -4.762 mm at x = 2 m (downward)

    = -4.762 mm

  4. Free-end deflection and slope

    The cantilever's tip carries the accumulated deflection and rotation.

    tip deflection and tip slope

    δtip = -4.762 mm; θtip = -0.003571 rad

Results

Maximum deflection (signed; negative = downward)

-4.762mm

Location of maximum deflection

2m

Deflection at the free end (signed; negative = downward)

-4.762mm

Slope at the free end (radians)

-0.003571rad

Where this answer was checked
source
δtip = PL³/3EI, θtip = PL²/2EI (NCEES FE Reference Handbook) — cantilever fixed left, load at the free end
verified by
hand-recomputed
derivation
EI = 70e9·4e-5 = 2.8e6 N·m². δtip = 5e3·8/(3·2.8e6) = 4.7619e-3 m DOWN → −4.7619 mm at x = 2 m. θtip = −PL²/2EI = −5e3·4/(2·2.8e6) = −3.5714e-3 rad.