Analyses any frame in a vertical plane, node by node or from its bays and storeys: the reactions, bending moments, shear and normal forces and displacements, with sloping and haunched members, hinges, springs, settlements, temperature changes and the variable loads in their worst arrangement. Every step shows its formula.

What it checks

Indeterminacy: 3m + r − (3j + c), with m the members, r the reactions (springs included), j the nodes and c the conditions at the hinges. Above zero, the forces depend on the stiffness, and settlements and temperature produce forces.
Loads: the totals as entered ΣF_x and ΣF_y, and ΣF_x,var / ΣF_y,var for the loads in the worst arrangement, all present.
Worst arrangement: each of the k variable loads may be present or absent. All N_arr = 2ᵏ arrangements are covered with k + 1 analyses, and the results become an envelope.
Support reactions: R_x, R_y and M_R from the loads as entered, with their largest and smallest values, and the equilibrium checks ΣR_x = −ΣF_x and ΣR_y = ΣF_y.
At the nodes and member ends: the displacements and rotations, and the normal force (tension positive), shear force and bending moment (tension on the right face positive), largest and smallest.
Along each member: M_max and M_min with their positions, the largest shear force, N_max and N_min, the deflection w_rel from the line through the displaced ends, and l/|w_rel|.
Checks, optional: |w_rel| ≤ l/n_lim for every member, and |u_x| ≤ h/n_h for every node above the lowest support.
Chosen sections, up to four, with the envelope.

What you enter and what you get

You describe the frame node by node, or by bays and storeys with one section for the columns and one for the beams. You enter E, the same for every member, and whether the axial shortening of the members is included. For each node: its coordinates x and y, the support (free, pinned, fixed, two kinds of roller, two kinds of guide, or springs in x, y and against turning) and its settlements. For each member: its start and end nodes, I, A, a constant section or straight or parabolic haunches, and rigid or hinged ends.

Loads: up to eight distributed loads, downward per metre of member or in plan, horizontal per metre of height, or perpendicular to the member, uniform or varying; up to eight forces on members; up to six loads at nodes; up to four temperature loads, uniform or different on the two faces. Each is as entered or in the worst case. Then the limits and the chosen sections.

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

Assumptions and limits

Straight members in a vertical plane. Linear first-order analysis: characteristic or design values, as you need them. Downward and to the right are positive, moments at nodes clockwise.
For members of different materials, enter one E and scale I and A of the others by the ratio of the moduli.
In a haunch the width is constant, so I grows with the cube of the depth; the area is taken as constant.
With the axial shortening neglected, no area is asked and no uniform temperature change can be applied.
A temperature difference between the faces gives the free curvature α_T·ΔT_d/h, h the depth of the member.
To join several members by one pin, give each of them a hinge at that node, or all but one.
An analysis only: no member is verified.

Worked example

A two-bay frame 16 m wide and 6 m high, with E = 32,000 N/mm²: columns fixed at x = 0 and 16 m and pinned at 10 m, where the base settles 8 mm; a 10 m beam with parabolic haunches 2 m long at both ends and a 6 m beam; the right column hinged at its top. The beams carry 30 kN/m as entered and 20 kN/m in the worst arrangement; also in the worst arrangement, 4.5 kN/m of wind on the left column, 15 kN horizontally at its top, and on the long beam a warming of 15 °C with 10 °C between its faces. Limits l/300 and h/300. Every check passes. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Description of the frame	Node by node
Number of nodes j	6
Number of members m	5
Modulus of elasticity E	32000 N/mm²
Axial shortening of the members	Included
Horizontal coordinate x₁	0 m
Vertical coordinate y₁	0 m
Support	Fixed
Horizontal settlement Δ_x,1	0 mm
Vertical settlement Δ_y,1	0 mm
Horizontal coordinate x₂	0 m
Vertical coordinate y₂	6 m
Support	Free (a joint)
Horizontal coordinate x₃	10 m
Vertical coordinate y₃	6 m
Support	Free (a joint)
Horizontal coordinate x₄	10 m
Vertical coordinate y₄	0 m
Support	Pinned
Horizontal settlement Δ_x,4	0 mm
Vertical settlement Δ_y,4	8 mm
Horizontal coordinate x₅	16 m
Vertical coordinate y₅	6 m
Support	Free (a joint)
Horizontal coordinate x₆	16 m
Vertical coordinate y₆	0 m
Support	Fixed
Horizontal settlement Δ_x,6	0 mm
Vertical settlement Δ_y,6	0 mm
Start node	Node 1
End node	Node 2
Second moment of area I₁	7200 10⁶ mm⁴
Cross-section area A₁	240000 mm²
Section along the member	Constant
Connection at its ends	Rigid at both ends
Start node	Node 2
End node	Node 3
Second moment of area I₂	24300 10⁶ mm⁴
Cross-section area A₂	360000 mm²
Section along the member	Parabolic haunches
Length of the haunch at the start a_A,2	2 m
Second moment of area at the start I_A,2	48600 10⁶ mm⁴
Length of the haunch at the end a_B,2	2 m
Second moment of area at the end I_B,2	48600 10⁶ mm⁴
Connection at its ends	Rigid at both ends
Start node	Node 4
End node	Node 3
Second moment of area I₃	7200 10⁶ mm⁴
Cross-section area A₃	240000 mm²
Section along the member	Constant
Connection at its ends	Rigid at both ends
Start node	Node 3
End node	Node 5
Second moment of area I₄	24300 10⁶ mm⁴
Cross-section area A₄	360000 mm²
Section along the member	Constant
Connection at its ends	Rigid at both ends
Start node	Node 6
End node	Node 5
Second moment of area I₅	7200 10⁶ mm⁴
Cross-section area A₅	240000 mm²
Section along the member	Constant
Connection at its ends	Hinge at the end
Number of distributed loads	5
On	Member 2
Direction	Downward, per metre of member
Extent	The whole member
Intensity at the start q_a,1	30 kN/m
Intensity at the end q_b,1	30 kN/m
Arrangement	As entered
On	Member 4
Direction	Downward, per metre of member
Extent	The whole member
Intensity at the start q_a,2	30 kN/m
Intensity at the end q_b,2	30 kN/m
Arrangement	As entered
On	Member 2
Direction	Downward, per metre of member
Extent	The whole member
Intensity at the start q_a,3	20 kN/m
Intensity at the end q_b,3	20 kN/m
Arrangement	Worst case
On	Member 4
Direction	Downward, per metre of member
Extent	The whole member
Intensity at the start q_a,4	20 kN/m
Intensity at the end q_b,4	20 kN/m
Arrangement	Worst case
On	Member 1
Direction	Horizontal (to the right), per metre of height
Extent	The whole member
Intensity at the start q_a,5	4.5 kN/m
Intensity at the end q_b,5	4.5 kN/m
Arrangement	Worst case
Number of forces on members	0
Number of loads at nodes	1
At	Node 2
Horizontal force (to the right positive) F_x,1	15 kN
Vertical force (downward positive) F_y,1	0 kN
Moment (clockwise positive) M₁	0 kN·m
Arrangement	Worst case
Number of temperature loads	1
Coefficient of thermal expansion α_T	10 10⁻⁶/°C
On	Member 2
Uniform change ΔT₁	15 °C
Difference, right face minus left face ΔT_d,1	10 °C
Depth of the member h₁	900 mm
Arrangement	Worst case
Check the deflection of the members against a limit	Yes
Limit: the member length divided by n_lim	300
Check the horizontal displacement of the nodes	Yes
Limit: the height divided by n_h	300
Number of sections for their own results	1
On	Member 2
Position, from the start node a_S1	5 m

Calculation: The frame
Step	Value	Ref.
Overall width and heightfrom the node coordinates	B × H = 16.00 × 6.00 m
Unknowns against equations of equilibriumm = 5 members, r = 8 reactions (springs included), j = 6 nodes, c = 1 conditions at the hingesstatically indeterminate, degree 4: the forces depend on the stiffness; settlements and temperature produce forces	3m + r − (3j + c) = 4

Step	1Member 1node 1 → 2	2Member 2node 2 → 3	3Member 3node 4 → 3	4Member 4node 3 → 5
Length l m	6.00	10.00	6.00	6.00
Slope from the horizontal α °anticlockwise from the x axis, start to end	90.0	0.0	90.0	0.0
Bending stiffness E·I MN·m²between the haunches	230.4	777.6	230.4	777.6
Axial stiffness E·A MN	8	12	8	12
Haunches, start / end a_A / a_B m	·	2.00 / 2.00	·	·
Hinges	—	—	—	—

Step	5Member 5node 6 → 5	Ref.
Length l m	6.00
Slope from the horizontal α °anticlockwise from the x axis, start to end	90.0
Bending stiffness E·I MN·m²between the haunches	230.4
Axial stiffness E·A MN	8
Haunches, start / end a_A / a_B m	·
Hinges	end

Calculation: The loads
Step	Value	Ref.
Total horizontal load as enteredto the right positivewithout the loads in the worst arrangement	ΣF_x = 0.0 kN
Total vertical load as entereddownward positive	ΣF_y = 480.0 kN
Loads in the worst arrangement, all present	ΣF_x,var / ΣF_y,var = 42.0 / 320.0 kN

Calculation: The worst arrangement
Step	Value	Ref.
Loads that may each be present or absentdistributed load 3, distributed load 4, distributed load 5, load at a node 1, temperature load 1	k = 5
Arrangements coveredN_arr = 2ᵏ; max = base + Σ (positive contributions), min = base + Σ (negative contributions)each load is analysed once; at every section the loads that increase the value are added for the largest result, those that decrease it for the smallest — the same as all 2ᵏ combinations, with k + 1 analyses	N_arr = 32

Calculation: Support reactions
Step	Value	Ref.
Equilibrium: sum of the horizontal reactionsequals −ΣF_x = -0.00 kNa check of the analysis (loads as entered)	ΣR_x = 0.00 kN
Equilibrium: sum of the vertical reactionsequals ΣF_y = 480.00 kN	ΣR_y = 480.00 kN

Step	1Node 1fixed	4Node 4pinned	6Node 6fixed
Horizontal reaction, from the loads as entered R_x kNto the right positive	30.67	-20.86	-9.80
largest R_x,max kN	51.94	-18.81	-9.80
smallest R_x,min kN	-2.80	-40.81	-21.72
Vertical reaction, from the loads as entered R_y kNupward positive	170.07	192.66	117.27
largest R_y,max kN	258.88	412.23	170.89
smallest R_y,min kN	155.19	192.66	90.15
Reaction moment, from the loads as entered M_R kN·manticlockwise positive	-21.71	·	58.82
largest M_R,max kN·m	53.90	·	130.30
smallest M_R,min kN·m	-57.58	·	58.82

Calculation: At the nodes
Step	1Node 1(0.00; 0.00)	2Node 2(0.00; 6.00)	3Node 3(10.00; 6.00)	4Node 4(10.00; 0.00)
Largest horizontal displacement u_x,max mmto the right positive	0.00	5.54	6.80	0.00
Smallest horizontal displacement u_x,min mm	0.00	2.90	3.07	0.00
Largest vertical displacement w,max mmdownward positive	0.00	0.20	8.32	8.00
Smallest vertical displacement w,min mm	0.00	0.12	8.15	8.00
Largest rotation θ,max mradanticlockwise positive; none at a node joining only hinged ends	0.000	-1.796	1.016	-1.009
Smallest rotation θ,min mrad	0.000	-2.651	0.445	-2.188

Step	5Node 5(16.00; 6.00)	6Node 6(16.00; 0.00)
Largest horizontal displacement u_x,max mmto the right positive	6.79	0.00
Smallest horizontal displacement u_x,min mm	3.06	0.00
Largest vertical displacement w,max mmdownward positive	0.13	0.00
Smallest vertical displacement w,min mm	0.07	0.00
Largest rotation θ,max mradanticlockwise positive; none at a node joining only hinged ends	2.086	0.000
Smallest rotation θ,min mrad	1.714	0.000

Calculation: At the member ends
Step	1Member 1node 1 → 2	2Member 2node 2 → 3	3Member 3node 4 → 3	4Member 4node 3 → 5
Largest normal force at the start N_i,max kNtension positive	-155.2	-28.8	-192.7	-9.8
Smallest normal force at the start N_i,min kN	-258.9	-62.4	-412.2	-21.7
Largest normal force at the end N_j,max kN	-155.2	-28.8	-192.7	-9.8
Smallest normal force at the end N_j,min kN	-258.9	-62.4	-412.2	-21.7
Largest shear force at the start V_i,max kNV = dM/dx	2.8	258.9	40.8	157.3
Smallest shear force at the start V_i,min kN	-51.9	155.2	18.8	61.7
Largest shear force at the end V_j,max kN	-19.5	-129.9	40.8	-90.1
Smallest shear force at the end V_j,min kN	-56.6	-256.0	18.8	-170.9
Largest bending moment at the start M_i,max kN·mtension on the right face positive	57.6	-118.1	0.0	169.9
Smallest bending moment at the start M_i,min kN·m	-53.9	-254.0	0.0	-43.6
Largest bending moment at the end M_j,max kN·m	-118.1	38.4	244.8	0.0
Smallest bending moment at the end M_j,min kN·m	-254.0	-269.9	112.9	0.0

Step	5Member 5node 6 → 5	Ref.
Largest normal force at the start N_i,max kNtension positive	-90.1
Smallest normal force at the start N_i,min kN	-170.9
Largest normal force at the end N_j,max kN	-90.1
Smallest normal force at the end N_j,min kN	-170.9
Largest shear force at the start V_i,max kNV = dM/dx	21.7
Smallest shear force at the start V_i,min kN	9.8
Largest shear force at the end V_j,max kN	21.7
Smallest shear force at the end V_j,min kN	9.8
Largest bending moment at the start M_i,max kN·mtension on the right face positive	-58.8
Smallest bending moment at the start M_i,min kN·m	-130.3
Largest bending moment at the end M_j,max kN·m	0.0
Smallest bending moment at the end M_j,min kN·m	0.0

Calculation: Along each member
Step	1Member 1node 1 → 2	2Member 2node 2 → 3	3Member 3node 4 → 3	4Member 4node 3 → 5
Largest positive moment M_max kN·m	57.6	462.4	244.8	292.0
its position a mfrom the start node	0.00	5.00	6.00	2.63
Largest negative moment M_min kN·m	-254.0	-269.9	·	-43.6
its position a m	6.00	10.00	·	0.00
Largest shear force \|V\|_max kN	56.6	258.9	40.8	170.9
Largest tension N_max kN	·	·	·	·
Largest compression N_min kN	-258.9	-62.4	-412.2	-21.7
Largest deflection across the member w_rel mmfrom the line through its displaced ends (a cantilever: from its supported end), towards the right face positive	-2.05	6.12	2.45	1.45
its position a m	3.63	5.00	3.50	2.88
Relative deflection over the length l/\|w_rel\|	2,928	1,634	2,445	4,138

Step	5Member 5node 6 → 5	Ref.
Largest positive moment M_max kN·m	·
its position a mfrom the start node	·
Largest negative moment M_min kN·m	-130.3
its position a m	0.00
Largest shear force \|V\|_max kN	21.7
Largest tension N_max kN	·
Largest compression N_min kN	-170.9
Largest deflection across the member w_rel mmfrom the line through its displaced ends (a cantilever: from its supported end), towards the right face positive	-1.31
its position a m	2.50
Relative deflection over the length l/\|w_rel\|	4,595

Calculation: Check — deflection of every member
Step	Value	Ref.
Deflection of member 1\|w_rel\|_max,1 ≤ l₁/n_lim	η = 0.102 · Pass10 % of the capacity used
Deflection of member 2\|w_rel\|_max,2 ≤ l₂/n_lim	η = 0.184 · Pass18 % of the capacity used
Deflection of member 3\|w_rel\|_max,3 ≤ l₃/n_lim	η = 0.123 · Pass12 % of the capacity used
Deflection of member 4\|w_rel\|_max,4 ≤ l₄/n_lim	η = 0.072 · Pass7 % of the capacity used
Deflection of member 5\|w_rel\|_max,5 ≤ l₅/n_lim	η = 0.065 · Pass7 % of the capacity used

Calculation: Check — horizontal displacement of the nodes
Step	Value	Ref.
Node 2, 6.00 m above the lowest support\|u_x,2\| ≤ h/n_h	η = 0.277 · Pass28 % of the capacity used
Node 3, 6.00 m above the lowest support\|u_x,3\| ≤ h/n_h	η = 0.340 · Pass34 % of the capacity used
Node 5, 6.00 m above the lowest support\|u_x,5\| ≤ h/n_h	η = 0.339 · Pass34 % of the capacity used

Calculation: At the chosen sections
Step	1Section 1member 2, a = 5.00 m	Ref.
Largest bending moment M_max kN·mtension on the right face positive	462.4
Smallest bending moment M_min kN·m	235.8
Largest shear force, before the section V_max,l kNon the side of the start node	20.1
Smallest shear force, before the section V_min,l kN	-6.0
Largest shear force, after the section V_max,r kN	20.1
Smallest shear force, after the section V_min,r kN	-6.0
Largest normal force N_max kNtension positive	-28.8
Smallest normal force N_min kN	-62.4
Largest horizontal displacement u_x,max mmto the right positive	6.07
Smallest horizontal displacement u_x,min mm	3.08
Largest vertical displacement w_max mmdownward positive	10.35
Smallest vertical displacement w_min mm	7.98

How this module is verified

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

Questions

What does the degree of indeterminacy tell me?

Above zero, the forces depend on the stiffness of the members, and settlements and temperature changes produce forces. The frame of the worked example is indeterminate to degree 4: 5 members, 8 reactions, 6 nodes and 1 condition at the hinge.

Should the axial shortening be included?

Included, each member lengthens and shortens under its normal force, E·A/l: the exact first-order result. Neglected, the members keep their length, as in the classical hand methods.

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: 14 November 2026

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