Axial bar element

Results.

Example	Reference	Records	Largest deviation
Symmetric two-bar truss, load at the apex	Closed form	1	0%
Tension-only X bracing, single-storey panel	Closed form	1	< 0.001%
Compression-only X bracing, single-storey panel	Closed form	1	< 0.001%
Ten-storey frame, tension-only X bracing in every panel	Closed form	2	< 0.001%
Gap (contact) element, the gap closes	Closed form	4	0%
Bar with an initial force, top end on a spring	Closed form	1	0%
Pre-tensioned cable slackening under support settlement	Closed form	1	0%

Problem. Four behaviours of the axial bar element are tested. A tension-only bar drops out in compression and a compression-only bar drops out in tension. A gap element carries no force until the gap closes. The $N_0$ of a bar with an initial force is the force it would carry if neither end moved.

Reference document. The references come from statics. The linear case is also compared with a frame element pinned at both ends.

The axial force in the acting diagonal of an X braced panel is found from the storey shear by this relation.

N = \dfrac{V}{\cos\theta}

Once the gap has closed, the translation of the gap element is computed from this relation.

u = -\dfrac{P + k\,g}{k_{spring} + k_{bar}},\qquad \lvert u \rvert > g

The remaining force in a bar with an initial force and the elongation at which the bar goes slack are given by these relations.

N = N_0 + k\,e,\qquad e = -\dfrac{N_0}{k}

Numerical model. In the truss model the nodal rotations are held, because the bar element gives no rotational stiffness. With tension-only and compression-only bars and the gap element the solution is nonlinear. Which elements carry load is found by equilibrium iteration, and the drawing shows the final converged solution.

Symmetric two-bar truss, load at the apex

Parameters. P = 90 · sin θ = 0.6

Closed form. Nodal equilibrium

N = P/(2\sin\theta)

Quantity	Reference	Computed	Deviation
Axial force in the bar	-75	-75	0%

Tension-only X bracing, single-storey panel

Parameters. H = 120 · span = 4 · storey = 3

Closed form. The slack diagonal carries nothing, so the whole shear is the horizontal component of the acting diagonal

N = V/\cos\theta

Quantity	Reference	Computed	Deviation
Axial force in the acting diagonal	150	150	< 0.001%

Compression-only X bracing, single-storey panel

Parameters. H = 120 · span = 4 · storey = 3

Closed form. The slack diagonal carries nothing, so the whole shear is the horizontal component of the acting diagonal

N = V/\cos\theta

Quantity	Reference	Computed	Deviation
Axial force in the acting diagonal	-150	-150	< 0.001%

Ten-storey frame, tension-only X bracing in every panel

Parameters. storey = 10 · span = 4 · braces = 80

Closed form. Statics, the shear in every storey is the sum of the horizontal components of the working braces

V_k = \sum_i N_i \cos\theta

Quantity	Reference	Computed	Deviation
Largest relative difference between storey shear and the shear the braces carry	0	$5.773 \times 10^{-15}$	0%
Largest force in the braces that go into compression, over the largest tension	0	0.00000110338	< 0.001%

Note. Which of the eighty braces work is found together in the same equilibrium iteration, and the solution converges in four iterations. The lateral load acts the other way at the third, sixth and ninth storeys, so the working brace changes from storey to storey.

Gap (contact) element, the gap closes

Parameters. gap = 0.002 · P = -400 · spring stiffness = 5,000 · bar stiffness EA/L = 50,000

Closed form. The closed form, valid once the gap has closed

u = -\dfrac{P + k\,g}{k_{spring} + k_{bar}},\qquad \lvert u\rvert > g

Quantity	Reference	Computed	Deviation
Nodal translation	-0.00909091	-0.00909091	0%
Force in the gap element	-354.545	-354.545	0%
Force in the spring	-45.4545	-45.4545	0%
Sum of the gap element and spring forces	-400	-400	0%

Note. The top node is tied both to the gap element and to a spring in parallel with it. Until the gap closes the spring alone carries the load, taking 10 kN by then. The remaining 390 kN is shared between the two in proportion to their stiffness. The two forces add up to the applied load.

Bar with an initial force, top end on a spring

Parameters. N₀ = 25 · spring stiffness = 4,000

Closed form. The equilibrium condition

N_0 + k_{cable}\,u = -k_{spring}\,u

Quantity	Reference	Computed	Deviation
Remaining tension in the cable	3.44828	3.44828	0%

Note. $N_0$ is the tension the bar would have if neither end moved, equivalent to the bar being made $N_0/k$ = 1 mm too short. The bottom end is fixed and the top end sits on a spring, so the tension pulls the top end down by 0.86 mm and the bar shortens by that much. The tension left in the bar equals the spring's upward push on the node. There is no external load, so the system is in equilibrium on its own.

Pre-tensioned cable slackening under support settlement

Parameters. N₀ = 25 · threshold settlement = 0.001

Closed form. The cable force and the slack threshold

N = N_0 + k\,e,\qquad e = -N_0/k

Quantity	Reference	Computed	Deviation
Force before the threshold, at half the settlement	12.5	12.5	0%

Symmetric two-bar truss, load at the apex#

Tension-only X bracing, single-storey panel#

Compression-only X bracing, single-storey panel#

Ten-storey frame, tension-only X bracing in every panel#

Gap (contact) element, the gap closes#

Bar with an initial force, top end on a spring#

Pre-tensioned cable slackening under support settlement#