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The BST binders · Beams binder · Solution procedure — the order of steps · updated 27.9.2026

Beam solution procedureThe order of steps for solving a statics beam

The solution procedure is the recipe. 9 fixed steps — whatever exercise is in front of you.

Reactions → N → S → Z → M → Checks

In every step: goal, formula, order of work, common mistake — and the flagship exercise solved next to it.

  • Signs A clockwise moment is +. A clockwise concentrated moment → M jumps by +M₀. Positive M is drawn below the axis.
  • Supports Triangle = pinned support (H, V) · triangle on wheels = roller support (V)
  • Letters A, B — the supports. C, D, E… — the points where the loads start and end, from left to right, as in BST
1 t/m1 t1 t45°1 t·mABCFDEG113221
The flagship exercise: every type of force in one exercise — a point force, a distributed load, an inclined force and a concentrated moment · scroll the drawing sideways
  1. 01

    Sketch

    Goal identify the supports, loads and dimensions

    Supports
    Roller support1 unknown (V)
    Pinned support2 unknowns (H, V)
    Fixed support3 unknowns (H, V, M)
    Loads
    PPoint force
    QDistributed load
    M0Concentrated moment

    Order of work

    1. Mark the type of every support.
    2. Mark every load and angle.
    3. Count the unknowns → must be 3.

    In the flagship exercise

    • A — pinned support (H, V) · B — roller support (V)
    • Count of unknowns2 + 1 = 3 ✓

    ⚠ Common mistake Confusing a roller support with a pinned support.

  2. 02

    Resolving inclined forces

    Goal resolve every inclined force into its components

    Fx = F · cos α
    Fy = F · sin α
    F
    Magnitude of the force
    α
    Angle given in the question
    Fx
    Horizontal component
    Fy
    Vertical component

    Order of work

    1. Read α from the question.
    2. Substitute into the formula and get Fx, Fy.
    3. Mark the direction of the components on the sketch.
    4. Erase the original force from the sketch.

    Direction of the components

    Down-rightFx → rightFy ↓ down
    Down-leftFx ← leftFy ↓ down
    Up-leftFx ← leftFy ↑ up
    Up-rightFx → rightFy ↑ up

    In the flagship exercise

    • Point loadFC = 1 t
    • Point loadFF = 1 t
    • Vertical componentFFy = FF · sin α = 1 · sin 45° = 0.707 t
    • Horizontal componentFFx = FF · cos α = 1 · cos 45° = 0.707 t

    ⚠ Common mistake A mistake in the direction of the components. The direction of the original force sets the sign of Fx and Fy.

  3. 03

    Resultant of a distributed load

    Goal to convert a distributed load into a resultant force

    Presultant = Q · L
    Q
    Load intensity
    L
    Segment length

    Order of work

    1. Measure L of the loaded segment.
    2. Calculate the resultant P = Q · L.
    3. Place the resultant exactly at the middle of the distributed-load segment.
    4. Use the resultant — for reactions only.

    In the flagship exercise

    • Distributed load — resultantPDE = q · L = 1 · 3 = 3 t

    ⚠ Common mistake Using the resultant in the S and M diagrams.

  4. 04

    Reactions

    Goal to find the support reactions

    ΣMA = 0
    ΣMB = 0
    ΣFy = 0 (check)
    A, B
    The two supports
    ΣM
    Sum of moments about the axis
    lever arm
    Horizontal distance to the force

    Order of work

    1. Choose a support about which the check will be made.
    2. Check the moments about this point: force · distance from the support axis.
    3. Solve the equation → the reaction of the second support.
    4. Repeat the operation — sum of moments about the second support → the first reaction.
    5. ΣFy = 0 serves only as a check that we made no mistake.

    In the flagship exercise

    ΣFx = 0 — horizontal equilibrium

    • −(0.707) + Ax = 0
    • → to the rightAx = 0.707 t

    ΣMA = 0 — moment equilibrium about A

    • (1 · 1) + (0.707 · 7) + (3 · 3.5) + (1) − (By · 10) = 0
    • ↑ upwardBy = 1.745 t

    ΣMB = 0 — moment equilibrium about B

    • (Ay · 10) − (1 · 9) − (0.707 · 3) − (3 · 6.5) + (1) = 0
    • ↑ upwardAy = 2.962 t

    ΣFy = 0 — check

    • Ay − FCy − FFy − PDE + By = 0
    • 2.962 − 1 − 0.707 − 3 + 1.745 = 0
    • Sum of the vertical forcesΣFy = 0 ✓

    ⚠ Common mistake A concentrated moment M₀ enters ΣM with its full value, without multiplying by a distance.

  5. 05

    N diagram

    Goal to build the N diagram

    Sign convention

    • ←Horizontal force to the left — compresses the beam → N negative
    • →Horizontal force to the right — stretches the beam → N positive

    Order of work

    1. Look at the beam from the right, and start from the first horizontal force.
    2. The force points left → pushes into the beam → compression → N negative. Points right → pulls it → tension → N positive.
    3. Continue to the left: every horizontal force in the opposite direction — moves N in the opposite direction, by the magnitude of the force.
    4. Between horizontal forces — a straight line. At the end the graph returns to 0.

    In the flagship exercise

    • N1 = − 0.707 = −0.707 t
    • N2 = −0.707 + 0.707 = 0 t

    ⚠ Common mistake Putting a force that is not horizontal into N. Only forces along the X axis enter.

  6. 06

    S diagram

    Goal to build the S diagram

    Load type → graph movement

    Point loadSharp turn
    Distributed loadDiagonal line
    Triangular loadParabola

    Order of work

    1. Scan from left to right.
    2. In every segment — choose the graph movement by the type of load above it.
    3. The graph must close at 0 at the right end.

    In the flagship exercise

    SZ = 3.9622.962
    S diagram · scroll the drawing sideways
    • S1 = + 2.962 = 2.962 t
    • S2 = 2.962 − 1 = 1.962 t
    • S3 = 1.962 − (1 · 3) = −1.038 t
    • S4 = −1.038 − 0.707 = −1.745 t
    • S5 = −1.745 + 1.745 = 0 t
    • The zero point Z — the distances from it for calculating areas and Mmax: ZL = 1.962 , ZR = 1.038

    ⚠ Common mistake Starting from the left end instead of from the reaction. The graph starts at the point where the first force meets the beam.

  7. 07

    Point Z

    Goal the point where the shear graph passes from + to − linearly

    Z = SQ
    Z
    The 0 point of the shear
    S
    Shear height on the left side
    Q
    Load intensity in that zone

    In the flagship exercise

    • Distance of Z from the start of the segmentZ = SQ = 1.9621 = 1.962 m
    • Position of Z on the beamxZ = 2 + 1.962 = 3.962 m

    ⚠ Common mistake Using Q from another segment.

  8. 08

    M diagram

    Goal to build the M diagram by the area method

    In every segment — the type of area under the shear determines the shape of the movement in the moment and also the formula for calculating it.

    Shape in the shearAreaMovement in the moment
    rectangleb · hDiagonal line
    triangleb · h2Parabola
    trapezoid(a + b) · h2Parabola

    Order of work

    1. In every segment, identify the shape of the area under the shear.
    2. Calculate the area by the suitable formula.
    3. The value accumulates on the M diagram, in the suitable shape of movement.
    4. At a concentrated moment M₀ → a jump in M by its value.

    In the flagship exercise

    MZMmax = 6.849
    M diagram · scroll the drawing sideways
    • M1 = 0 = 0 t·m
    • M2 = 0 + (2.962 · 1) = 2.962 t·m
    • M3 = 2.962 + (1.962 · 1) = 4.924 t·m
    • M4 = 4.924 + (1.962 · 1.962) / 2 = 6.849 t·m
    • M5 = 6.849 + (−1.038 · 1.038) / 2 = 6.311 t·m
    • M6 = 6.311 + (−1.038 · 2) = 4.235 t·m
    • M7 = 4.235 + (−1.745 · 2) = 0.745 t·m
    • M8 = 0.745 + 1 = 1.745 t·m
    • M9 = 1.745 + (−1.745 · 1) = 0 t·m
    • Maximum positive moment — at a distance of 3.962 mMmax = 6.849 t·m

    ⚠ Common mistake Drawing a diagonal line under a trapezoid. Trapezoid = parabola.

  9. 09

    Final checks

    Goal to make sure that all the checks came to zero

    Allowed deviation ≤ 0.1
    in units of t / t·m

    Checks checklist

    1. ΣFy = 0 — the sum of all the vertical forces.
    2. The S graph closes at 0.
    3. The M graph closes at 0 at a free end or at a simple support.
    4. Mmax sits exactly on point Z.

    In the flagship exercise

    • Sum of the vertical forcesΣFy = 0 ✓
    • The S graph closes at 0 at the right endS(10) = 0 ✓
    • The M graph closes at 0M(10) = 0 ✓
    • The maximum moment sits on point ZxMmax = xZ = 3.962 m ✓

    ⚠ Common mistake Skipping the checks. S or M not closing points to a mistake in the reactions.

1 t/m1 t1 t45°1 t·mABAY = 2.962 tBY = 1.745 tAX = 0.707 tCFDEG113221
The flagship exercise solved: the reactions from the BST engine

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