← All tools

Tools · Structural analysis

SFD and BMD solver

Build a beam, load it, and watch the shear force, bending moment and deflected shape redraw instantly, with the reactions and equations worked out for you.

Standard cases with their textbook formulas: see why a point load gives a triangle and a UDL gives a parabola.

Choose a case

Values

Standard results

ReactionsRA = RB = W2= 202 = 10 kN
Max SFVmax = W2= 202 = 10 kN
Max BMMmax = WL4 (at centre)= 20 × 64 = 30 kN·m
Max deflectionδ = WL³48EI= 20 × 6³48 × 16000 = 5.625 mm

With only point loads, SF is constant between loads and jumps by the load at each one, so the BMD is made of straight lines (dM/dx = V). The peak BM is where the SF changes sign.

20 kNRA = 10 kNRB = 10 kN33Shear force diagram (SFD) (kN, left-up +)1010-10-10Bending moment diagram (BMD) (kN·m, sagging +)30Deflected shape (mm)-5.63
Move over the diagrams to read values at any section.
Max sagging BM
30.00kN·m
at x = 3 m
Max hogging BM
0.00kN·m
none
Max shear
10.00kN
at x = 0 m
Max deflection
5.625mm
at x = 3 m
Points of contraflexure
none

Deflection, step by step (derivation of the formula)

  1. Flexural rigidity EIE in kN/m², I in m⁴

    E = 200 GPa = 200 × 10⁶ kN/m²; I = 8000 cm⁴ = 8000 × 10⁻⁸ m⁴

    EI = 2 × 10⁸ × 8 × 10⁻⁵ = 16000 kN·m²

  2. Bending moment at a section x in the last segment (Macaulay form)Write every load in one expression; a bracket [x − a] counts only when x > a

    M(x) = 10x − 20[x − 3] + 10[x − 6]

    Reaction RA10x
    Point load 20 kN− 20[x − 3]
    Reaction RB10[x − 6]
  3. Differential equation of the elastic curveEI·d²ydx² = M(x)

    EI·d²ydx² = 10x − 20[x − 3] + 10[x − 6]

  4. Integrate once: slope equation∫[x − a]ⁿ dx = [x − a]ⁿ⁺¹n + 1; integrate brackets as a whole

    EI·dydx = 10x²2 − 20[x − 3]²2 + 10[x − 6]²2 + C1

  5. Integrate again: deflection equation

    EI·y = 10x³6 − 20[x − 3]³6 + 10[x − 6]³6 + C1·x + C2

  6. Boundary conditionsFind the constants C1 and C2

    At support A (x = 0 m), deflection y = 0: 0 = 0 + C1 × 0 + C2

    At support B (x = 6 m), deflection y = 0: 0 = 270 + C1 × 6 + C2

    C1 = −45 kN·m²

    C2 = 0 kN·m³

  7. Deflection at x = 3 mPut x in the deflection equation

    EI·y = 10(3)³6 − 45 × 3 + 0

    (2 bracket terms are dropped because [x − a] is zero or negative there)

    EI·y = −90 kN·m³

    y = −9016000 = −0.005625 m = −5.625 mm (downward)

    Slope: EI·dydx = 0 → θ = −1.33 × 10⁻¹¹ rad

  8. Maximum deflectionWhere dydx = 0 (or at a free end)

    Solving EI·dydx = 0 numerically gives x = 3 m

    y max = −5.625 mm (downward)

Sign convention: x from the left end, y upward positive, so a sag comes out negative; sagging moment positive. Results agree with the stiffness solution shown in the diagrams (-5.625 mm there).