Find beam deflection and slope by double integration, including partial distributed loads and overhangs. See the elastic curve and peak deflection.
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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.type pinposition 0 m
2.type rollerposition 6 m
loads
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
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
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
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
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
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
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.type pinposition 0 ft
2.type rollerposition 20 ft
loads
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
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
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
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
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
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
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.type fixedposition 0 m
loads
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
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
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
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
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
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
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.