Frame element, end to end

Results.

Example	Reference	Records	Largest deviation
Three-hinged parabolic arch, load over half the span	Closed form	2	0%
Cantilever, point load at the tip	Published source	1	0%
Cantilever, uniformly distributed load	Published source	1	0%
Simply supported beam, distributed load	Published source	1	0%
Beam fixed at both ends, distributed load	Published source	2	0%
Beam fixed at both ends, support settlement	Published source	1	0%

Problem. The model is built, the system is solved and the result is compared with the closed form. Cantilever, simply supported and fixed-fixed beams, a support settlement and a three-hinged arch built of inclined members are covered.

Reference document. The closed-form solutions of Timoshenko, Strength of Materials, are used. For the support settlement the reference is the displacement method, and for the arch statics.

The tip deflection of a cantilever is given by these relations for a point load at the tip and for a uniformly distributed load.

\delta_{uc} = \dfrac{PL^3}{3EI},\qquad \delta_{uc} = \dfrac{wL^4}{8EI}

In a fully fixed beam the settlement of one end produces this end moment.

M = \dfrac{6\,EI\,\Delta}{L^2}

Numerical model. The beams are modelled with four to eight elements. All six freedoms are held at the cantilever support and at fixed ends. At a simple support $u_y$ , $u_z$ and $r_x$ are held at both ends, and $u_x$ is also held at the left end. The arch is modelled with sixteen straight frame elements whose nodes lie on its axis.

Three-hinged parabolic arch, load over half the span

Parameters. L = 24 · f = 6 · segments = 16

Closed form. Statics, with the maximum moment at a quarter of the span

H = wL^2/(16f),\qquad M_{max} = wL^2/64

Quantity	Reference	Computed	Deviation
Horizontal thrust $H$	120	120	0%
Maximum bending moment	180	180	0%

Note. The moments in the table are nodal moments. The small bulge at the middle of each segment on the loaded half comes from representing the curved axis with straight segments, and its size is $wa^2/8$ . Modelling curved members explains it.

Cantilever, point load at the tip

Parameters. element count = 4 · L = 6

Published source. Timoshenko, Strength of Materials

Quantity	Reference	Computed	Deviation
Tip deflection $PL^3/3EI$	-0.1125	-0.1125	0%

Cantilever, uniformly distributed load

Parameters. element count = 6 · L = 6

Published source. Timoshenko, Strength of Materials

Quantity	Reference	Computed	Deviation
Tip deflection $wL^4/8EI$	-0.081	-0.081	0%

Simply supported beam, distributed load

Parameters. element count = 8 · L = 6

Published source. Timoshenko, Strength of Materials

Quantity	Reference	Computed	Deviation
Mid-span deflection $5wL^4/384EI$	-0.0105469	-0.0105469	0%

Beam fixed at both ends, distributed load

Parameters. element count = 8 · L = 6

Published source. Timoshenko, Strength of Materials

Quantity	Reference	Computed	Deviation
Mid-span deflection $wL^4/384EI$	-0.00210938	-0.00210938	0%
End moment $wL^2/12$	30	30	0%

Beam fixed at both ends, support settlement

Parameters. δ = -0.01 · L = 6

Published source. The classical displacement method

Quantity	Reference	Computed	Deviation
End moment $6EI\delta/L^2$	26.6667	26.6667	0%

Three-hinged parabolic arch, load over half the span#

Cantilever, point load at the tip#

Cantilever, uniformly distributed load#

Simply supported beam, distributed load#

Beam fixed at both ends, distributed load#

Beam fixed at both ends, support settlement#

Three-hinged parabolic arch, load over half the span

Cantilever, point load at the tip

Cantilever, uniformly distributed load

Simply supported beam, distributed load

Beam fixed at both ends, distributed load

Beam fixed at both ends, support settlement