Free worked example · complete calculation

Gable Portal Frame Analysis Worked Example

See the complete reactions and force diagrams first, then follow the full nine-DOF direct stiffness calculation for this fixed-base gable frame.

8 m span 4 m eaves 6 m apex Fixed column bases 4 kN/m on both rafters 18 kN·m apex moment

One worked loading condition First-order linear elastic analysis 9 active joint DOFs

Open in 2D Calculator See the solved answer

An 8 metre wide fixed-base rigid gable frame with 4 metre eaves, a 6 metre apex, 4 kilonewtons per metre on both inclined rafters and an 18 kilonewton-metre apex moment. — 8 m span4 m eaves6 m apex

Answer at a glance

Apex horizontal movement: 0.229 mm
Apex vertical movement: -0.919 mm
Left / right vertical reactions: 18.550 / 13.450 kN
Governing bending moment: 13.217 kN·m
Maximum resultant movement: 1.115 mm

Gable portal frame fundamentals

What is gable portal frame analysis?

A gable portal frame is a rigid structural frame with columns supporting two inclined rafters that meet at a roof apex. Moment-resisting eave and apex joints make the columns and rafters act continuously, so roof loading produces axial force, shear force and bending moment throughout the frame.

Gable portal frame analysis calculates horizontal, vertical and moment support reactions; eave and apex translations and rotations; and the axial-force, shear-force and bending-moment response of the columns and inclined rafters. Because the rigid fixed-base frame is statically indeterminate, the direct stiffness method is used to satisfy equilibrium and displacement compatibility.

Geometry, joints and load path

How to recognize a gable portal frame

Two columns and two inclined rafters form a pitched or gable roof profile.
Rigid eave and apex joints transfer moment as well as axial force and shear.
Each inclined rafter has a local axis, so a local-transverse roof load has both global horizontal and vertical components.
Fixed bases develop horizontal reactions, vertical reactions and base moments.

Matrix-analysis workflow

How to analyze a gable portal frame

Define the column, eave and apex geometry, member properties, rigid joints, fixed supports and roof loads.
Number the three global DOFs at each joint and retain the nine free DOFs at N2, N3 and N4.
Write the 6 × 6 local axial-and-bending matrix for each column and rafter.
Use the rafter direction cosines to transform stiffness and consistent load vectors into global axes.
Assemble and solve the 9 × 9 reduced stiffness equation for eave and apex movements.
Recover all six base reactions, member end actions and the axial, shear, moment and deflected-shape diagrams.

How is a gable portal frame analyzed?

Each column and inclined rafter is represented by a 2D frame element. Local stiffness matrices and consistent rafter-load vectors are transformed into global axes, assembled at the shared joints and reduced by the support restraints. Joint displacements are solved first, followed by reactions and member end actions.

How are loads applied to inclined rafters?

The load direction must match the intended physical loading. In this worked model, 4 kN/m acts in each rafter’s local negative-y direction, normal to the rafter. The transformation matrix converts those local actions into global horizontal, vertical and moment components.

What reactions does a fixed-base gable frame have?

Each fixed base can develop horizontal reaction Rx, vertical reaction Ry and moment reaction Mz. Both bases therefore contribute six reaction components, all of which are reported in the solved table and reaction diagram.

Does a symmetric gable frame always have symmetric reactions?

Only when the geometry, stiffness, supports and loading are symmetric. The counter-clockwise apex moment in this example breaks load symmetry, so the two base reactions and rafter actions differ even though the frame geometry is symmetric.

01 · Model inputs

The worked gable portal-frame model

A symmetric gable portal frame spans 8 m, has 4 m eaves and a 6 m apex. Both column bases are fixed and every column-to-rafter and apex connection is rigid. Each 4.472 m rafter carries a 4 kN/m load in its local negative-y direction, and node N3 carries an 18 kN·m counter-clockwise moment.

Span8 mSingle gable bay

Eave / apex4 m / 6 m2 m roof rise

SupportsFixed + fixedSix restrained base DOFs

ConnectionsRigidNo member-end releases

Section5000 mm²I = 200 × 10⁶ mm⁴

MaterialSteelE = 200 GPa

Input table

Nodes and restraints

Coordinates in metres

Node	X	Y	Support	Restrained DOFs
N1	0.0	0.0	Fixed	UX, UY, RZ
N2	0.0	4.0	Free	—
N3	4.0	6.0	Free	—
N4	8.0	4.0	Free	—
N5	8.0	0.0	Fixed	UX, UY, RZ

Input table

Applied loads

One worked loading condition

Load	Target	Definition
L1	N3	18.000 kN·m global MZ
L2	M2	-4.000 kN/m to -4.000 kN/m in LOCAL Y
L3	M3	-4.000 kN/m to -4.000 kN/m in LOCAL Y

02 · Solved answer

Complete support reactions and member-force diagrams

The 8 kN/m roof load and 12 kN horizontal load act together as the one worked loading condition. The diagrams below are exported by the Optimal Beam calculator from this exact model.

Optimal Beam support reaction diagram showing horizontal, vertical and moment reactions at both fixed bases

All six fixed-base components

Support reactions

+X right · +Y up · +Mz counter-clockwise

Support	Horizontal Rx (kN)	Vertical Ry (kN)	Moment Mz (kN·m)
N1	5.198 kN	18.550 kN	-7.574 kN·m
N5	-5.198 kN	13.450 kN	9.972 kN·m

Optimal Beam axial-force diagram for the worked gable portal frame — Tension positive · compression negative

Optimal Beam shear-force diagram for the worked gable portal frame — Member local axes

Optimal Beam bending-moment diagram for the worked gable portal frame — Sagging positive

Optimal Beam deflected shape for the worked gable portal frame — Values remain unamplified

Calculated joint response

Roof-joint movement

Translations and rotation

Node	UX (mm)	UY (mm)	RZ (mrad)
N2	-0.129	-0.074	-0.282
N3	0.229	-0.919	0.316
N4	0.608	-0.054	0.042

Calculated member response

Member end actions

Local i-to-j axes

Member	Axial (kN)	Vi / Vj (kN)	Mi / Mj (kN·m)
M1	-18.550	-5.198 / -5.198	7.574 / -13.217
M2	-12.945	14.267 / -3.622	-13.217 / 10.587
M3	-10.664	8.183 / -9.706	-7.413 / -10.819
M4	-13.450	5.198 / 5.198	-10.819 / 9.972

03 · Hand calculation

Direct stiffness calculation, step by step

This is the numerical path from the applied loads to the six reactions. The same model values and member axes are used by the calculator.

Nine unknown joint DOFs, assembled from four full frame elements

The two fixed bases set six translations and rotations to zero. The eaves and apex leave nine unknown global degrees of freedom. Every numerical matrix below is generated from the published model values.

Calculation units are kN, m and radians. Global +X is right, +Y is up and +RZ is counter-clockwise.

Step 1

Number the global degrees of freedom

Start with every joint translation and rotation, then identify which values are known.

{d} = {u₁, v₁, θ₁, u₂, v₂, θ₂, u₃, v₃, θ₃, u₄, v₄, θ₄, u₅, v₅, θ₅}ᵀ {dc} = {u₁, v₁, θ₁, u₅, v₅, θ₅}ᵀ = {0, 0, 0, 0, 0, 0}ᵀ {df} = {u₂, v₂, θ₂, u₃, v₃, θ₃, u₄, v₄, θ₄}ᵀ

N1 and N5 are fixed. N2, N3 and N4 each retain horizontal translation, vertical translation and rotation.

Joint	Global DOFs	Status
N1	u₁, v₁, θ₁	Restrained
N2	u₂, v₂, θ₂	Free
N3	u₃, v₃, θ₃	Free
N4	u₄, v₄, θ₄	Free
N5	u₅, v₅, θ₅	Restrained

Static-indeterminacy check Ds = 3m + r − 3j → Ds = 3(4) + 6 − 3(5) = 3

The rigid frame is externally and internally indeterminate to degree 3.

Step 2

Write the local 2D frame-element stiffness matrix

This is the complete axial-and-bending matrix used for each unreleased member.

{d′e} = {u′ᵢ, v′ᵢ, θᵢ, u′ⱼ, v′ⱼ, θⱼ}ᵀ [k′e] = local axial-and-bending stiffness matrix

	uᵢ	vᵢ	θᵢ	uⱼ	vⱼ	θⱼ
uᵢ	EA/L	0	0	−EA/L	0	0
vᵢ	0	12EI/L³	6EI/L²	0	−12EI/L³	6EI/L²
θᵢ	0	6EI/L²	4EI/L	0	−6EI/L²	2EI/L
uⱼ	−EA/L	0	0	EA/L	0	0
vⱼ	0	−12EI/L³	−6EI/L²	0	12EI/L³	−6EI/L²
θⱼ	0	6EI/L²	2EI/L	0	−6EI/L²	4EI/L

Rows and columns follow {u′i, v′i, θi, u′j, v′j, θj}. Translational and rotational terms carry the compatible kN–m units.

Section properties in calculation units

E = 200 GPa, A = 5000 mm², I = 200 × 10⁶ mm⁴

E = 200,000,000 kN/m²; A = 0.005 m²; I = 0.0002 m⁴; EA = 1,000,000 kN; EI = 40,000 kN·m²

Coefficient	Definition	Purpose
EA/L	axial rigidity ÷ length	Axial stiffness
12EI/L³	12 × flexural rigidity ÷ L³	Transverse stiffness
6EI/L²	6 × flexural rigidity ÷ L²	Shear-rotation coupling
4EI/L	4 × flexural rigidity ÷ L	Near-end rotational stiffness
2EI/L	2 × flexural rigidity ÷ L	Far-end rotational coupling

Term	4 m column	4.472 m rafter
EA/L	250,000.000 kN/m	223,606.798 kN/m
12EI/L³	7,500.000 kN/m	5,366.563 kN/m
6EI/L²	15,000.000 kN	12,000.000 kN
4EI/L	40,000.000 kN·m	35,777.088 kN·m
2EI/L	20,000.000 kN·m	17,888.544 kN·m

Numerical [k′e]

Column matrix

M1 and M4 · L = 4.000 m

	uᵢ	vᵢ	θᵢ	uⱼ	vⱼ	θⱼ
uᵢ	250,000.000	0.000	0.000	-250,000.000	0.000	0.000
vᵢ	0.000	7,500.000	15,000.000	0.000	-7,500.000	15,000.000
θᵢ	0.000	15,000.000	40,000.000	0.000	-15,000.000	20,000.000
uⱼ	-250,000.000	0.000	0.000	250,000.000	0.000	0.000
vⱼ	0.000	-7,500.000	-15,000.000	0.000	7,500.000	-15,000.000
θⱼ	0.000	15,000.000	20,000.000	0.000	-15,000.000	40,000.000

Numerical [k′e]

Rafter matrix

M2 and M3 · L = √20 = 4.472136 m

	uᵢ	vᵢ	θᵢ	uⱼ	vⱼ	θⱼ
uᵢ	223,606.798	0.000	0.000	-223,606.798	0.000	0.000
vᵢ	0.000	5,366.563	12,000.000	0.000	-5,366.563	12,000.000
θᵢ	0.000	12,000.000	35,777.088	0.000	-12,000.000	17,888.544
uⱼ	-223,606.798	0.000	0.000	223,606.798	0.000	0.000
vⱼ	0.000	-5,366.563	-12,000.000	0.000	5,366.563	-12,000.000
θⱼ	0.000	12,000.000	17,888.544	0.000	-12,000.000	35,777.088

Step 3

Rotate each member matrix into global X/Y

Direction cosines connect the member’s local axial/transverse directions to the frame axes.

[ke] = [Te]ᵀ[k′e][Te]

Columns and rafters use the same local matrix form. Their direction cosines rotate local axial and transverse behavior into global X and Y.

M1 · N1–N2

θ = 90°

c = 0.000000 · s = 1.000000

0.000000	1.000000	0.000000	0.000000	0.000000	0.000000
-1.000000	0.000000	0.000000	0.000000	0.000000	0.000000
0.000000	0.000000	1.000000	0.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.000000	1.000000	0.000000
0.000000	0.000000	0.000000	-1.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.000000	0.000000	1.000000

M2 · N2–N3

θ = 26.565051°

c = 0.894427 · s = 0.447214

0.894427	0.447214	0.000000	0.000000	0.000000	0.000000
-0.447214	0.894427	0.000000	0.000000	0.000000	0.000000
0.000000	0.000000	1.000000	0.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.894427	0.447214	0.000000
0.000000	0.000000	0.000000	-0.447214	0.894427	0.000000
0.000000	0.000000	0.000000	0.000000	0.000000	1.000000

M3 · N3–N4

θ = -26.565051°

c = 0.894427 · s = -0.447214

0.894427	-0.447214	0.000000	0.000000	0.000000	0.000000
0.447214	0.894427	0.000000	0.000000	0.000000	0.000000
0.000000	0.000000	1.000000	0.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.894427	-0.447214	0.000000
0.000000	0.000000	0.000000	0.447214	0.894427	0.000000
0.000000	0.000000	0.000000	0.000000	0.000000	1.000000

M4 · N4–N5

θ = -90°

c = 0.000000 · s = -1.000000

0.000000	-1.000000	0.000000	0.000000	0.000000	0.000000
1.000000	0.000000	0.000000	0.000000	0.000000	0.000000
0.000000	0.000000	1.000000	0.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.000000	-1.000000	0.000000
0.000000	0.000000	0.000000	1.000000	0.000000	0.000000
0.000000	0.000000	0.000000	0.000000	0.000000	1.000000

Step 4

Transform both rafter UDLs into consistent global nodal actions

The rafter loads enter the global equation together with the 18 kN·m apex moment.

{p′e} = {0, qL/2, qL²/12, 0, qL/2, −qL²/12}ᵀ q = −4 kN/m; L = √20 = 4.472136 m {p′M2} = {0, -8.944272, -6.666667, 0, -8.944272, 6.666667}ᵀ {Ff} = {4, -8, -6.666667, 0, -16, 18, -4, -8, 6.666667}ᵀ

M2 and M3 have opposite roof slopes, so their local-y forces transform differently in global X and Y. Their apex force components combine with the applied +18 kN·m nodal moment.

Free DOF	Applied action
u₂	4.000 kN
v₂	-8.000 kN
θ₂	-6.667 kN·m
u₃	0.000 kN
v₃	-16.000 kN
θ₃	18.000 kN·m
u₄	-4.000 kN
v₄	-8.000 kN
θ₄	6.667 kN·m

Step 5

Assemble the four member contributions

Common joint DOFs land in the same rows and columns, so their stiffness terms add directly.

[K] = Σ [Ae]ᵀ[Te]ᵀ[k′e][Te][Ae] [Kff] = [Kff]M1 + [Kff]M2 + [Kff]M3 + [Kff]M4

Each transformed 6 × 6 element matrix is placed into the 15 × 15 global matrix at its two node locations. Selecting the nine free DOFs produces [Kff].

Member	Element DOFs	Global placement
M1	1–6	u₁, v₁, θ₁, u₂, v₂, θ₂
M2	1–6	u₂, v₂, θ₂, u₃, v₃, θ₃
M3	1–6	u₃, v₃, θ₃, u₄, v₄, θ₄
M4	1–6	u₄, v₄, θ₄, u₅, v₅, θ₅

[Kff]M1

N1–N2 contribution to the nine free joint DOFs

	u₂	v₂	θ₂
u₂	7,500.000	0.000	15,000.000
v₂	0.000	250,000.000	0.000
θ₂	15,000.000	0.000	40,000.000
u₃	0.000	0.000	0.000
v₃	0.000	0.000	0.000
θ₃	0.000	0.000	0.000
u₄	0.000	0.000	0.000
v₄	0.000	0.000	0.000
θ₄	0.000	0.000	0.000

[Kff]M2

N2–N3 contribution to the nine free joint DOFs

	u₂	v₂	θ₂	u₃	v₃	θ₃
u₂	179,958.751	87,296.094	-5,366.563	-179,958.751	-87,296.094	-5,366.563
v₂	87,296.094	49,014.610	10,733.126	-87,296.094	-49,014.610	10,733.126
θ₂	-5,366.563	10,733.126	35,777.088	5,366.563	-10,733.126	17,888.544
u₃	-179,958.751	-87,296.094	5,366.563	179,958.751	87,296.094	5,366.563
v₃	-87,296.094	-49,014.610	-10,733.126	87,296.094	49,014.610	-10,733.126
θ₃	-5,366.563	10,733.126	17,888.544	5,366.563	-10,733.126	35,777.088
u₄	0.000	0.000	0.000	0.000	0.000	0.000
v₄	0.000	0.000	0.000	0.000	0.000	0.000
θ₄	0.000	0.000	0.000	0.000	0.000	0.000

[Kff]M3

N3–N4 contribution to the nine free joint DOFs

	u₃	v₃	θ₃	u₄	v₄	θ₄
u₂	0.000	0.000	0.000	0.000	0.000	0.000
v₂	0.000	0.000	0.000	0.000	0.000	0.000
θ₂	0.000	0.000	0.000	0.000	0.000	0.000
u₃	179,958.751	-87,296.094	5,366.563	-179,958.751	87,296.094	5,366.563
v₃	-87,296.094	49,014.610	10,733.126	87,296.094	-49,014.610	10,733.126
θ₃	5,366.563	10,733.126	35,777.088	-5,366.563	-10,733.126	17,888.544
u₄	-179,958.751	87,296.094	-5,366.563	179,958.751	-87,296.094	-5,366.563
v₄	87,296.094	-49,014.610	-10,733.126	-87,296.094	49,014.610	-10,733.126
θ₄	5,366.563	10,733.126	17,888.544	-5,366.563	-10,733.126	35,777.088

[Kff]M4

N4–N5 contribution to the nine free joint DOFs

	u₄	v₄	θ₄
u₂	0.000	0.000	0.000
v₂	0.000	0.000	0.000
θ₂	0.000	0.000	0.000
u₃	0.000	0.000	0.000
v₃	0.000	0.000	0.000
θ₃	0.000	0.000	0.000
u₄	7,500.000	0.000	15,000.000
v₄	0.000	250,000.000	0.000
θ₄	15,000.000	0.000	40,000.000

Assembled result

Reduced global stiffness matrix [Kff]

DOF order: u₂, v₂, θ₂, u₃, v₃, θ₃, u₄, v₄, θ₄

DOF	u₂	v₂	θ₂	u₃	v₃	θ₃	u₄	v₄	θ₄
u₂	187,458.751	87,296.094	9,633.437	-179,958.751	-87,296.094	-5,366.563	0.000	0.000	0.000
v₂	87,296.094	299,014.610	10,733.126	-87,296.094	-49,014.610	10,733.126	0.000	0.000	0.000
θ₂	9,633.437	10,733.126	75,777.088	5,366.563	-10,733.126	17,888.544	0.000	0.000	0.000
u₃	-179,958.751	-87,296.094	5,366.563	359,917.502	0.000	10,733.126	-179,958.751	87,296.094	5,366.563
v₃	-87,296.094	-49,014.610	-10,733.126	0.000	98,029.220	0.000	87,296.094	-49,014.610	10,733.126
θ₃	-5,366.563	10,733.126	17,888.544	10,733.126	0.000	71,554.175	-5,366.563	-10,733.126	17,888.544
u₄	0.000	0.000	0.000	-179,958.751	87,296.094	-5,366.563	187,458.751	-87,296.094	9,633.437
v₄	0.000	0.000	0.000	87,296.094	-49,014.610	-10,733.126	-87,296.094	299,014.610	-10,733.126
θ₄	0.000	0.000	0.000	5,366.563	10,733.126	17,888.544	9,633.437	-10,733.126	75,777.088

Step 6

Apply the fixed-base boundary conditions and solve

Partitioning removes the six known zero displacements without discarding their reaction rows.

[[Kcc, Kcf], [Kfc, Kff]] {dc, df}ᵀ = {Fc, Ff}ᵀ {dc} = 0 ⇒ [Kff]{df} = {Ff} [Kff]{df} = {Ff}

Applying the six base restraints leaves the 9 × 9 equation for the eave and apex translations and rotations.

Solved vector in calculation units

DOF	Value	Unit
u₂	-0.000128733158	m
v₂	-0.000074199008	m
θ₂	-0.000282145744	rad
u₃	0.000229000698	m
v₃	-0.000919113240	m
θ₃	0.000316187894	rad
u₄	0.000608336400	m
v₄	-0.000053800992	m
θ₄	0.000042344123	rad

Same answer in readable units

DOF	Value	Unit
u2	-0.128733	mm
v2	-0.074199	mm
θ2	-0.282146	mrad
u3	0.229001	mm
v3	-0.919113	mm
θ3	0.316188	mrad
u4	0.608336	mm
v4	-0.053801	mm
θ4	0.042344	mrad

Two rows written out

187458.751(-0.000128733158) + 87296.094(-0.000074199008) + 9633.437(-0.000282145744) − 179958.751(0.000229000698) − 87296.094(-0.00091911324) − 5366.563(0.000316187894) = 4 kN (u₂)

-179958.751(-0.000128733158) − 87296.094(-0.000074199008) + 5366.563(-0.000282145744) + 359917.502(0.000229000698) + 10733.126(0.000316187894) − 179958.751(0.0006083364) + 87296.094(-0.000053800992) + 5366.563(0.000042344123) = 0 kN (u₃)

Row	[Kff]{df}	Applied Ff
u₂	4.000000 kN	4.000000 kN
v₂	-8.000000 kN	-8.000000 kN
θ₂	-6.666667 kN·m	-6.666667 kN·m
u₃	0.000000 kN	0.000000 kN
v₃	-16.000000 kN	-16.000000 kN
θ₃	18.000000 kN·m	18.000000 kN·m
u₄	-4.000000 kN	-4.000000 kN
v₄	-8.000000 kN	-8.000000 kN
θ₄	6.666667 kN·m	6.666667 kN·m

Step 7

Recover the horizontal, vertical and moment reactions

The restrained rows were not solved for displacement; they are now used to recover all six base actions.

{Rc} = [Kcf]{df} − {Fc}

The full 6 × 9 coupling block maps all nine free joint movements into the six restrained base rows; the transformed rafter-load actions at restrained DOFs are then subtracted.

Restrained-to-free coupling block

Reaction recovery matrix [Kcf]

Maps the complete free-joint vector into the fixed-base reaction rows

-7,500.000	0.000	-15,000.000	0.000	0.000	0.000
0.000	-250,000.000	0.000	0.000	0.000	0.000
15,000.000	0.000	20,000.000	0.000	0.000	0.000
0.000	0.000	0.000	-7,500.000	0.000	-15,000.000
0.000	0.000	0.000	0.000	-250,000.000	0.000
0.000	0.000	0.000	15,000.000	0.000	20,000.000

N1 and N5

[R] = [Kcf] {-0.000128733158, -0.000074199008, -0.000282145744, 0.000229000698, -0.000919113240, 0.000316187894, 0.000608336400, -0.000053800992, 0.000042344123}ᵀ

{5.197685, 18.549752, -7.573912, -5.197685, 13.450248, 9.971928}ᵀ Reaction order follows N1 [Rx, Ry, Mz], then the second fixed base [Rx, Ry, Mz]

Support	Rx (kN)	Ry (kN)	Mz (kN·m)
N1	5.197685	18.549752	-7.573912
N5	-5.197685	13.450248	9.971928

Step 8

Recover member end actions and close the equilibrium checks

The selected member displacement vector is returned to its local stiffness equation, including its consistent load actions.

{f′M2} = [k′M2]{d′M2} − {p′M2} {f′M2} = {12.944652, 14.266927, 13.216827, -12.944652, 3.621617, 10.586811}ᵀ

The N2–N3 global displacement vector is rotated into M2 local axes. The elastic end vector and the negative consistent load vector give the recovered left-rafter end actions.

M2 component	Elastic term	Fixed-end term	Recovered end action
Nᵢ	12.944652 kN	0.000000 kN	12.944652 kN
Vᵢ	5.322655 kN	8.944272 kN	14.266927 kN
Mᵢ	6.550160 kN·m	6.666667 kN·m	13.216827 kN·m
Nⱼ	-12.944652 kN	0.000000 kN	-12.944652 kN
Vⱼ	-5.322655 kN	8.944272 kN	3.621617 kN
Mⱼ	17.253478 kN·m	-6.666667 kN·m	10.586811 kN·m

ΣFx = 5.197685 − 5.197685 + transformed roof-load Fx = 0.000000 kNΣFy = 18.549752 + 13.450248 + transformed roof-load Fy = 0.000000 kNΣMN1 = base moments + N5 vertical reaction moment + transformed rafter-load moments + 18.000000 = 0.000000 kN·m

04 · Hand-versus-solver verification

The hand calculation matches the Optimal Beam result

Independent values from the calculation above are compared directly with the stored calculator result.

All declared checks passedVerified July 29, 2026

Check	Hand calculation	Optimal Beam	Difference	Status
Horizontal reaction at N1	5.197685 kN	5.197685 kN	0.000000%	Pass
Vertical reaction at N1	18.549752 kN	18.549752 kN	0.000000%	Pass
Base moment at N1	-7.573912 kN·m	-7.573912 kN·m	0.000000%	Pass
Horizontal reaction at N5	-5.197685 kN	-5.197685 kN	0.000000%	Pass
Vertical reaction at N5	13.450248 kN	13.450248 kN	0.000000%	Pass
Base moment at N5	9.971928 kN·m	9.971928 kN·m	0.000000%	Pass
Horizontal movement at N3	0.229001 mm	0.229001 mm	0.000000%	Pass
Vertical movement at N3	-0.919113 mm	-0.919113 mm	0.000000%	Pass
Rotation at N3	0.316188 mrad	0.316188 mrad	0.000000%	Pass
Left-rafter end moment at N3	10.586811 kN·m	10.586811 kN·m	0.000000%	Pass
Right-column axial force	-13.450248 kN	-13.450248 kN	0.000000%	Pass

05 · Analysis scope

What the worked model includes

This is a first-order structural analysis example, not a completed member or connection design.

Included in the model

Straight, prismatic 2D frame elements with axial and flexural stiffness.
Rigid, unreleased column-to-rafter and apex connections.
Fixed bases restraining horizontal translation, vertical translation and rotation.
A 4 kN/m local-transverse UDL on both rafters and an 18 kN·m apex moment.
First-order, small-displacement, linear-elastic response.
Member axial force, shear force, bending moment and deflected shape.

Outside the analysis scope

P-Delta effects, geometric nonlinearity or large-displacement response.
Member yielding, local or global buckling and plastic redistribution.
Connection, base plate, anchor, foundation or member capacity design.
Wind pressure generation, snow drift, load combinations or code-based load factors.
Out-of-plane behavior, purlins, bracing, diaphragm action or three-dimensional stability.
Temperature, settlement, dynamic, seismic or fatigue effects.

P-Delta boundary

Optimal Beam currently performs first-order linear-elastic analysis and does not include P-Delta effects. Use an appropriate second-order method when axial compression acting through sway may materially amplify moments or drift.

06 · Common modeling mistakes

Configuration errors that change the answer

Applying roof loading in global Y when the intended load is normal to each inclined rafter.

Using the same transformation sign for both rafters even though their slopes have opposite sine values.

Releasing the apex or eave rotations when the intended joints are moment-resisting.

Leaving either base rotation unrestrained when the idealization requires fixed bases.

Reporting only vertical reactions and omitting horizontal and moment reactions.

Treating first-order displacement as a P-Delta result or using analysis output as a completed code design.

Continue with the exact model

Open the free example or download the Optimal Beam report

The model, diagrams, result tables and calculation report come from this same calculator analysis.

Open free model in calculator Download Optimal Beam calculation report Download model JSON

Engineering authorship

Prepared and reviewed by Tamer Hijjawi, P.Eng.

Analysis scope: First-order linear elastic 2D frame analysis. Last model verification: July 29, 2026.

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