Analyses a straight beam in any material over any number of spans and cantilevers: the reactions, bending moments, shear forces and deflections, with haunches, settlements, springs and hinges, and the imposed loads in their worst arrangement. Every step shows its formula.

What it checks

Indeterminacy: r − (2 + h), with r the reactions, springs included, and h the hinges. Above zero, the forces depend on the stiffness and a settlement produces forces.
Loads: the total as entered ΣF, the position of its resultant x_F, and ΣF_var for the loads in the worst arrangement, all present.
Worst arrangement: a variable load may be present or absent on each segment it covers. All N_arr = 2ᵏ arrangements of the k pieces are covered, with only k + 1 analyses, since the analysis is linear. The results become an envelope.
Reactions at every support: R under the loads as entered, R_max and R_min, with the equilibrium check ΣR = ΣF.
At the points: M_max, M_min, the shear force left and right of each point, and w_max and w_min.
Per segment: the largest sagging and hogging moments and where they occur, the largest shear force, the largest deflection, the deflection relative to the supports w_rel and l/|w_rel|.
Deflection, optional: |w_rel| ≤ l/n_lim for every segment; for a cantilever l is its own length.
Sections you choose, up to four: bending moment, shear on each side, deflection and rotation.

What you enter and what you get

You enter the number of segments and the modulus E of the material. A span between two supports is one segment, a cantilever is one segment, and a span with a hinge in it is two. For each segment: its length, its second moment of area, and whether the section is constant or has straight or parabolic haunches at its ends. For each point: the support (pinned or roller, fixed, free, guided up and down, guided left and right, or springs), a settlement, and a continuous or hinged joint. Loads are placed anywhere along the beam: distributed loads, uniform or trapezoidal, concentrated forces and concentrated moments, each as entered or in the worst arrangement. Then the deflection limit and the sections you want reported.

You get the diagrams of moment, shear and deflection and every result above, with a pass or fail for each deflection check.

Assumptions and limits

A straight beam that carries no axial force. One modulus E for the whole length; only E·I enters the forces on rigid supports.
In a haunch the width is constant, so I grows with the cube of the depth.
Linear analysis: characteristic or design values, as you need them. Forces downward and moments clockwise are positive; enter a negative value for the opposite direction.
A concentrated force in the worst arrangement stays at its entered position, present or absent.
Settlements are included with the loads as entered.
An analysis only: no section is verified. The deflection limit comes from the project or the code you design to.

Worked example

A beam of four segments, 6.0 m, 7.5 m with parabolic haunches of 1.5 m at both ends, 6.0 m and a cantilever of 1.8 m, with E = 32,000 N/mm², a settlement of 5 mm at point 2, 22 kN/m as entered and 12 kN/m in the worst arrangement over the whole length, 40 kN at the end of the cantilever, and a limit of l/300. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Number of segments n	4
Modulus of elasticity E	32000 N/mm²
Length l₁	6.0 m
Second moment of area I₁	5400 10⁶ mm⁴
Section along the segment	Constant
Length l₂	7.5 m
Second moment of area I₂	5400 10⁶ mm⁴
Section along the segment	Parabolic haunches
Length of the haunch at the left end a_A,2	1.5 m
Second moment of area at the left end I_A,2	10800 10⁶ mm⁴
Length of the haunch at the right end a_B,2	1.5 m
Second moment of area at the right end I_B,2	10800 10⁶ mm⁴
Length l₃	6.0 m
Second moment of area I₃	5400 10⁶ mm⁴
Section along the segment	Constant
Length l₄	1.8 m
Second moment of area I₄	5400 10⁶ mm⁴
Section along the segment	Constant
Support	Pinned
Settlement of the support Δ₀	0 mm
Support	Roller
Settlement of the support Δ₁	0 mm
Joint	Continuous
Support	Roller
Settlement of the support Δ₂	5 mm
Joint	Continuous
Support	Roller
Settlement of the support Δ₃	0 mm
Joint	Continuous
Support	Free
Number of distributed loads	2
Start x_a,1	0 m
End x_b,1	21.3 m
Intensity at the start q_a,1	22 kN/m
Intensity at the end q_b,1	22 kN/m
Arrangement	As entered
Start x_a,2	0 m
End x_b,2	21.3 m
Intensity at the start q_a,2	12 kN/m
Intensity at the end q_b,2	12 kN/m
Arrangement	Worst case
Number of concentrated forces	1
Position x_P,1	21.3 m
Force (downward positive) P₁	40 kN
Arrangement	As entered
Number of concentrated moments	0
Check the deflections against a limit	Yes
Limit: the segment length divided by n_lim	300
Number of sections for their own results	1
Position of section 1 x_S1	9.75 m

Calculation: The beam
Step	Value	Ref.
Length of the beamΣ l_i	l_tot = 21.30 m
Reactions against equations of equilibriumr = 5 reactions (4 vertical and rotational, springs included; 1 horizontal); 3 equations of equilibrium + 1 for each of the h = 0 hingesstatically indeterminate, degree 2; for bending, degree 2: the moments depend on the stiffness, and settlements produce forces	r − (3 + h) = 2

Step	1Segment 10.00 → 6.00 m	2Segment 26.00 → 13.50 m	3Segment 313.50 → 19.50 m	4Segment 419.50 → 21.30 m
Length l m	6.00	7.50	6.00	1.80
Bending stiffness E·I MN·m²E·I between the haunches	172.8	172.8	172.8	172.8
Bending stiffness at the left end E·I_A MN·m²	·	345.6	·	·
Bending stiffness at the right end E·I_B MN·m²	·	345.6	·	·
Haunch shape	·	parabolic	·	·

Calculation: The loads
Step	Value	Ref.
Total load as enteredΣ (q_a + q_b)/2·(x_b − x_a) + Σ Pdownward positive; the loads placed in the worst arrangement are not included	ΣF = 508.6 kN
Position of its resultantΣ F·x / ΣF	x_F = 11.49 m
Loads in the worst arrangement, all presentdistributed loads present or absent segment by segment, forces and moments present or absent	ΣF_var = 255.6 kN

Calculation: The worst arrangement
Step	Value	Ref.
Load pieces that may each be present or absentload 2 on segment 1, load 2 on segment 2, load 2 on segment 3, load 2 on segment 4	k = 4
Arrangements coveredN_arr = 2ᵏ; max = base + Σ (positive contributions), min = base + Σ (negative contributions)the analysis is linear: each piece is analysed once, and at every section the pieces that increase the value are added for the largest result, those that decrease it for the smallest; this is the same as analysing all 2^k combinations, with only k + 1 analyses	N_arr = 16

Calculation: Support reactions
Step	Value	Ref.
Equilibrium: sum of the vertical reactionsequals ΣF of the loads as entereda check of the analysis: the reactions balance the loads	ΣR = 508.60 kN

Step	0Point 0pinned	1Point 1roller	2Point 2roller	3Point 3roller
Vertical reaction, from the loads as entered R kNupward positive	38.92	196.92	109.63	163.13
Largest vertical reaction R_max kN	71.62	294.73	206.35	221.50
Smallest vertical reaction R_min kN	32.38	190.80	98.58	156.84

Calculation: At the points
Step	0Point 0x = 0.00 m	1Point 1x = 6.00 m	2Point 2x = 13.50 m	3Point 3x = 19.50 m	4Point 4x = 21.30 m
Largest bending moment M_max kN·msagging positive	0.0	-154.5	10.6	-107.6	0.0
Smallest bending moment M_min kN·m	0.0	-229.5	-68.1	-127.1	0.0
Largest shear force, left of the point V_max,l kN	·	-91.7	-55.5	-77.2	40.0
Smallest shear force, left of the point V_min,l kN	·	-140.3	-110.9	-120.3	40.0
Largest shear force, right of the point V_max,r kN	71.6	154.5	95.4	101.2	·
Smallest shear force, right of the point V_min,r kN	32.4	99.1	43.1	79.6	·
Largest deflection w_max mmdownward positive	0.00	0.00	5.00	0.00	0.16
Smallest deflection w_min mm	0.00	0.00	5.00	0.00	-1.60

Calculation: In each segment
Step	1Segment 10.00 → 6.00 m	2Segment 26.00 → 13.50 m	3Segment 313.50 → 19.50 m	4Segment 419.50 → 21.30 m
Largest sagging moment M_max kN·m	75.4	127.9	91.0	·
its position x m	2.10	10.41	16.05	·
Largest hogging moment M_min kN·m	-229.5	-229.5	-127.1	-127.1
its position x m	6.00	6.00	19.50	19.50
Largest shear force \|V\|_max kN	140.3	154.5	120.3	101.2
Largest deflection w_max mmdownward positive, largest in magnitude	1.04	6.45	5.05	-1.60
its position x m	2.33	11.06	14.40	21.30
Largest deflection relative to the supports w_rel mmfrom the line through the supported ends; a cantilever: from its support	1.04	3.38	1.64	-1.60
its position x m	2.33	10.13	16.28	21.30
Relative deflection over the length l/\|w_rel\|	5,792	2,221	3,651	1,128

Calculation: Check — deflection of every segment
Step	Value	Ref.
Deflection of segment 1\|w_rel\|_max,1 ≤ l₁/n_lim	η = 0.052 · Pass5 % of the capacity used
Deflection of segment 2\|w_rel\|_max,2 ≤ l₂/n_lim	η = 0.135 · Pass14 % of the capacity used
Deflection of segment 3\|w_rel\|_max,3 ≤ l₃/n_lim	η = 0.082 · Pass8 % of the capacity used
Deflection of segment 4\|w_rel\|_max,4 ≤ l₄/n_lim	η = 0.266 · Pass27 % of the capacity used

Calculation: At the chosen sections
Step	1Section 1x = 9.75 m	Ref.
Largest bending moment M_max kN·msagging positive	120.6
Smallest bending moment M_min kN·m	52.4
Largest shear force, left of the section V_max,l kN	27.0
Smallest shear force, left of the section V_min,l kN	16.6
Largest shear force, right of the section V_max,r kN	27.0
Smallest shear force, right of the section V_min,r kN	16.6
Largest deflection w_max mmdownward positive	5.82
Smallest deflection w_min mm	3.62
Largest rotation θ_max mradanticlockwise positive	-0.882
Smallest rotation θ_min mrad	-1.017

How this module is verified

Verified against independent calculations used on real projects, before release.

Questions

Which loads go in the worst arrangement?

The variable loads, such as the imposed load. The permanent loads are entered as they are.

Does the material matter?

Only through E·I for the forces on rigid supports. The deflections, the springs and the settlements also need the value of E.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

Page updated: 8 November 2026

Continuous beam analysis

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