Analyses the deck of a box-girder bridge as a continuous beam: any number of spans and cantilevers, haunched segments, settlements, elastic supports and hinges, with the variable loads in their worst arrangement. Every step shows its formula.

What it checks

The system: the deck is a chain of segments between numbered points, one at every support, hinge and change of section. The degree of indeterminacy is r − (2 + h), with r the reactions, springs included, and h the hinges. In an indeterminate deck, settlements produce forces.
The loads: the total as entered ΣF, its resultant at x_F, and ΣF_var for the loads in the worst arrangement.
The worst arrangement: each piece of a variable load, segment by segment, may be present or absent. With k pieces, all N_arr = 2ᵏ arrangements are covered with k + 1 analyses, and the results become an envelope.
Support reactions: R from the loads as entered, R_max and R_min, and the equilibrium check ΣR = ΣF.
At every point: M_max and M_min, the shear on each side (V_max,l, V_min,l, V_max,r, V_min,r) and the deflections w_max and w_min.
In each segment: the largest sagging and hogging moments with their positions, the largest shear force, the largest deflection, the largest deflection relative to the supports w_rel and l/|w_rel|.
Deflection check, optional: |w_rel| ≤ l/n_lim for every segment.
Chosen sections, up to four: the moment, the shear on each side, the deflection and the rotation, with the envelope.

What you enter and what you get

You enter the number of segments, the modulus E, and for each segment its length and its second moment of area, constant or with straight or parabolic haunches at either end. At each point you choose the support (pinned or roller, fixed, free, guided up and down, guided left and right, or elastic with a vertical and a rotational spring), its settlement and whether the deck is continuous or hinged. Then up to eight distributed loads, uniform or trapezoidal, up to eight concentrated forces and up to four concentrated moments, each as entered or in the worst arrangement, the deflection limit and the chosen sections.

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

Assumptions and limits

A straight beam with no axial force. E is the same along the whole length.
In a haunch the width is taken as constant, so I grows with the cube of the depth.
The analysis is linear: enter characteristic or design values as you need them. Forces downward and moments clockwise are positive.
A force in the worst arrangement is present or absent at the position you enter; it is not moved along the deck.
Settlements are included with the loads as entered.
The deflection of a segment is measured from the line through its supported ends: a settlement moves that line, it does not bend the segment.

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

How is the traffic placed on the spans?

Enter it as a load in the worst arrangement. It is then present or absent on each segment it covers, and the results become the envelope over all the arrangements.

Why does a settlement of a support change the moments?

A continuous deck is statically indeterminate: a support that settles relative to the others bends it. In the worked example the degree is 2 and point 2 settles 5 mm.

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

Box-girder bridge deck, longitudinal analysis

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