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Structural Analysis

Beam Calculator

Reactions, shear force, bending moment and deflection for simply supported, cantilever and fixed-fixed beams under point loads, UDLs, trapezoidal loads and applied moments.

Unit Reference

Lengthm

ForcekN

UDLkN/m

MomentkN·m

EGPa

Icm⁴

Deflectionmm

Areacm²

StressMPa

Section dimmm

Beam Type

Simply Supported

Pin + roller supports

Cantilever

Fixed at A, free at B

Fixed-Fixed

Fixed supports at A and B

Span & Section

Length L (m)Section

Modulus E (GPa)Moment of Inertia I (cm⁴)

Loads

LabelPosition (m)Force, ↓ (kN)Type

LabelStart (m)End (m)Load, ↓ (kN/m)Type

LabelStart (m)End (m)w start (kN/m)w end (kN/m)Type

No trapezoidal loads.

LabelPosition (m)Moment (kN·m)DirType

No applied moments.

Load Combinations

γ_G (dead) — 1.35γ_Q (live/wind/snow) — 1.50

Applied factors — G: 1.35, Q: 1.50, W: 1.50, S: 1.50.

Method & Assumptions

Reactions and internal forces follow classical elastic (Euler–Bernoulli) beam theory — statics for the simply supported and cantilever cases, and the fixed-end-moment / superposition method for the fixed-fixed case. Deflections assume a constant EI along the span. These results are independent of design code; the code toggle sets the default unit system and the bending-moment sign convention used below.

Beam Diagram

Reactions

Reaction R_A

55.50 kN

Reaction R_B

55.50 kN

Shear Force Diagram

V_max = 55.50 kN @ x=0.00mV_min = -55.50 kN @ x=6.00m

Bending Moment Diagram

M_max = 105.75 kN·m @ x=3.00mM_min = 0.00 kN·m @ x=0.00m

Sign convention: sagging (positive) moment is plotted below the axis — Eurocode convention, matching the deflected shape below.

Deflection

δ_max = 21.70 mm @ x=3.00m (downward)L / δ = 277

Key Results Table

x (m)	V (kN)	M (kN·m)	δ (mm)
0.00V_max · M_min	55.50	0.00	0.00
1.50	35.25	68.06	15.26
3.00M_max · δ_max	-15.00	105.75	21.70
4.50	-35.25	68.06	15.26
6.00V_min	-55.50	0.00	0.00

Highlighted rows mark the governing extremes: V_max/V_min, M_max/M_min and δ_max.

Hand Calculations

1. Applied Loads (after factoring)

ΣF (point loads) = 30.00 kN

ΣF (UDL + trapezoidal) = 81.00 kN

W = ΣF = 111.00 kN

2. Reactions (statics)

ΣM_A = 0 → R_B = [Σ(load × arm) + ΣM_applied] / L = 55.50 kN

ΣF_y = 0 → R_A = W − R_B = 111.00 − 55.50 = 55.50 kN

3. Internal Forces

V(x), M(x) obtained by integrating equilibrium along the span (see diagrams above)

V_max = 55.50 kN @ x = 0.00m, V_min = -55.50 kN @ x = 6.00m

M_max = 105.75 kN·m @ x = 3.00m, M_min = 0.00 kN·m @ x = 0.00m

4. Deflection

δ(x) = −∫∫ [M(x)/EI] dx dx, with E = 200.00 GPa, I = 8360.00 cm⁴

Boundary conditions: y(0) = 0, y(L) = 0

δ_max = 21.70 mm @ x = 3.00m (L / δ = 277)