This patch test applies constant tensile stress to a domain.

Nine Q4 elements used.

Parameter	Symbol	Value
Young's Modulus	E	2540 kPa
Poisson's Ratio	nu	0.3
Applied Traction	t	3.495 kPa

Plane stress conditions applied

N/A

Nodes 1-10 are restrained with their displacements according to the exact solution:

Horizontal displacement: (1 - nu)* t / E * x

Vertical displacement: (1 - nu)* t / E * y

The displacements are prescribed on the degrees of freedom related to the nodes 1-10 so as to establish tensile stress (3.495 kPa) in the horizontal direction of the domain. The total horizontal reaction force is 8.0385 N/m. The total vertical reaction force is 5.06775 N/m. The stress components are in the x- and y-directions are 3.495 kPa and there is no shear stress.

Displacements are computed at the nodes 11-15, and the exact displacements shown in the below table are produced at these nodes.

Node	x (m)	y (m)	Force X (N)	Force Y (N)	Displacement x (m)	Displacement y (m)
1	0	0	-1.33975	-1.26694	0.00000	0.00000
2	1.45	0	1.33975	-1.26694	0.00140	0.00000
3	1.45	2.3	1.33975	1.26694	0.00140	0.00222
4	0	2.3	-1.33975	1.26694	0.00000	0.00222
5	0.725	0	0	-2.53388	0.00070	0.00000
6	1.45	0.7667	2.6795	0	0.00140	0.00074
7	1.45	1.5333	2.6795	0	0.00140	0.00148
8	0.725	2.3	0	2.53388	0.00070	0.00222
9	0	1.5333	-2.6795	0	0.00000	0.00148
10	0	0.7667	-2.6795	0	0.00000	0.00074
11	0.7866	1.1018	0	0	0.00076	0.00106
12	0.4513	1.6262	0	0	0.00043	0.00157
13	0.8677	0.5081	0	0	0.00084	0.00049
14	1.0304	1.6240	0	0	0.00099	0.00156
15	0.5993	0.4069	0	0	0.00058	0.00039