CivilSolve

Statics & Mechanics of Materials

Beam Deflection by Double Integration

Deflection and slope anywhere on a determinate single-span beam (simply supported with optional overhangs, or a cantilever) under any combination of point loads, partial or full uniform loads, triangular/trapezoidal loads, and applied moments — by integrating the exact bending-moment polynomials twice (EI y'' = M). Reports the maximum deflection with its location, support/tip slopes, values at requested points, and the deflected-shape diagram. Use for: deflections outside the standard handbook cases (partial UDL, overhangs, interior moments, triangular loads) or whenever the full elastic curve is wanted. Prefer beam-deflection-superposition for quick standard-case checks.

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

Inputs

Load a sample problem:
supportsSupports. Simply supported = pin + roller (overhangs allowed). Cantilever = one fixed support at x = 0 or x = span.
supports 1
loadsApplied loads; magnitudes positive with explicit direction/sense
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

span
6 m
supports
  1. 1.type pinposition 0 m
  2. 2.type rollerposition 6 m
loads
  1. 1.kind pointmagnitude 10 kNposition 3 mdirection down
modulus
200 GPa
moment of inertia
50000000 mm^4

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; moment positive = sagging.

Solution steps

  1. Support reactions

    Equilibrium gives the reactions, which fix the bending-moment diagram to be integrated.

    ΣFy = 0 and ΣM = 0

    RA = 5 kN; RB = 5 kN

  2. Integrate the moment diagram twice

    Within each segment the bending moment is an exact cubic polynomial, so slope and deflection integrate in closed form; slope and deflection are matched across every segment boundary.

    EI·y'' = M(x), integrated twice per polynomial segment with continuity at the breaks

    EI = 10000 kN·m^2; moment diagram has 2 polynomial segments

  3. Apply the boundary conditions

    The deflection vanishes at both supports; that pins down the two integration constants.

    y = 0 at each support

    Integration constants: C = -0.00225 rad (slope at the left end of the reference curve), D = 0 mm

  4. Maximum deflection

    Extreme deflection occurs where the slope crosses zero, or at a free end; every candidate is checked.

    slope = 0 at an interior extremum (free ends also checked)

    δmax = -4.5 mm at x = 3 m

    = -4.5 mm

  5. Slopes at the supports

    The support rotations bound the elastic curve between them.

    θA and θB from the slope function

    θA = -0.00225 rad; θB = 0.00225 rad

Deflected shapeDeflected shape0-4.5006x (m)y (mm)δ max (3, -4.5)

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 = 1.0e7 N·m². δmax = 10e3·216/(48·1e7) = 4.5e-3 m DOWN → −4.5 mm at x = 3 m. θA = −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

span
20 ft
supports
  1. 1.type pinposition 0 ft
  2. 2.type rollerposition 20 ft
loads
  1. 1.kind udlmagnitude 2 kip/ftstart 0 ftend 20 ft
modulus
29000 ksi
moment of inertia
600 in^4

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; moment positive = sagging.

Solution steps

  1. Support reactions

    Equilibrium gives the reactions, which fix the bending-moment diagram to be integrated.

    ΣFy = 0 and ΣM = 0

    RA = 20 kip; RB = 20 kip

  2. Integrate the moment diagram twice

    Within each segment the bending moment is an exact cubic polynomial, so slope and deflection integrate in closed form; slope and deflection are matched across every segment boundary.

    EI·y'' = M(x), integrated twice per polynomial segment with continuity at the breaks

    EI = 120800 kip·ft^2; moment diagram has 1 polynomial segments

  3. Apply the boundary conditions

    The deflection vanishes at both supports; that pins down the two integration constants.

    y = 0 at each support

    Integration constants: C = -0.005517 rad (slope at the left end of the reference curve), D = 0 in

  4. Maximum deflection

    Extreme deflection occurs where the slope crosses zero, or at a free end; every candidate is checked.

    slope = 0 at an interior extremum (free ends also checked)

    δmax = -0.4138 in at x = 10 ft

    = -0.4138 in

  5. Slopes at the supports

    The support rotations bound the elastic curve between them.

    θA and θB from the slope function

    θA = -0.005517 rad; θB = 0.005517 rad

Deflected shapeDeflected shape0-0.41380020x (ft)y (in)δ max (10, -0.4138)

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 = 166.667 lbf/in, L = 240 in, EI = 1.74e10 lbf·in². δmax = 5·166.667·3.31776e9/(384·1.74e10) = 0.413793 in DOWN → −0.41379 in at midspan. θA = −166.667·1.38240e7/(24·1.74e10) = −5.51724e-3 rad.

Example 3

cantilever full UDL: w=3 kN/m, L=3 m, E=200 GPa, I=30e6 mm⁴

Given

span
3 m
supports
  1. 1.type fixedposition 0 m
loads
  1. 1.kind udlmagnitude 3 kN/mstart 0 mend 3 m
modulus
200 GPa
moment of inertia
30000000 mm^4

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; moment positive = sagging.

Solution steps

  1. Support reactions

    Equilibrium gives the reactions, which fix the bending-moment diagram to be integrated.

    ΣFy = 0 and ΣM = 0

    R = 9 kN; wall moment M = -13.5 kN·m

  2. Integrate the moment diagram twice

    Within each segment the bending moment is an exact cubic polynomial, so slope and deflection integrate in closed form; slope and deflection are matched across every segment boundary.

    EI·y'' = M(x), integrated twice per polynomial segment with continuity at the breaks

    EI = 6000 kN·m^2; moment diagram has 1 polynomial segments

  3. Apply the boundary conditions

    At the fixed support both the deflection and the slope vanish; that pins down the two integration constants.

    y = 0 and y' = 0 at the wall

    Integration constants: C = 0 rad (slope at the left end of the reference curve), D = 0 mm

  4. Maximum deflection

    Extreme deflection occurs where the slope crosses zero, or at a free end; every candidate is checked.

    slope = 0 at an interior extremum (free ends also checked)

    δmax = -5.062 mm at x = 3 m

    = -5.062 mm

  5. Free-end deflection and slope

    The cantilever's free end carries the accumulated deflection and rotation.

    tip deflection and tip slope

    δtip = -5.062 mm; θtip = -0.00225 rad

Deflected shapeDeflected shape0-5.062003x (m)y (mm)δ max (3, -5.062)

Results

Maximum deflection (signed; negative = downward)

-5.062mm

Location of maximum deflection

3m

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

-5.062mm

Slope at the free end (radians)

-0.00225rad

Where this answer was checked
source
δtip = wL⁴/8EI, θtip = wL³/6EI (NCEES FE Reference Handbook) — cantilever fixed left, uniform load over the whole span
verified by
hand-recomputed
derivation
EI = 200e9·3e-5 = 6.0e6 N·m². δtip = 3e3·81/(8·6e6) = 5.0625e-3 m DOWN → −5.0625 mm at x = 3 m. θtip = −3e3·27/(6·6e6) = −2.25e-3 rad.