To determine the stress–strain state of a symmetric wedge of unit thickness subjected to bending by a uniformly distributed load.
Demidov S.P., Theory of Elasticity, Moscow, Vysshaya Shkola, 1979.
To determine the stress σrr in the wedge at a distance of r = 5 m from the apex for the case of wedge bending under a uniformly distributed load, as well as the angle θ0 at which σrr=0.
A uniformly distributed load of intensity q is applied to the upper face of the wedge of unit thickness.
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Wedge thickness h = 1 m
Radius defining the wedge domain R = 15 m
Wedge apex angle 2α = 30°
Modulus of elasticity Е = 3 * 107 kPa
Poisson's ratio ν = 0.2
Restraints in all degrees of freedom are applied along the arc boundary of the wedge.
Uniformly distributed load: q = 10 kN/m
The problem is solved in a 3D formulation (model type 5).
The model is generated with FE type 44 – arbitrary quadrilateral FE of shell and FE type 42 – arbitrary triangular FE of shell.
The finite element mesh consists of 60 elements along the radius and 40 elements along the circumference.
The local axes of plates for the results are aligned in such a way that each local Y1-axis is directed towards the apex of the wedge.
Nodes: 2461. Elements: 2400.
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K = tgα-α
σrr = q(2α-2θ-(1-cos2θ)*tgα)-sin2θ)/2K
σθθ = q(2α-2θ-(1+cos2θ)*tgα)+sin2θ)/2K
σrθ = q(1-tgα*sin2θ-cos2θ)/2K
Without additional side nodes:
| Point | The unknown | Analytical solution | LIRA-FEM | Error, % |
| r = 5 m when θ = 0 | Ny, kN/m2 | 98 | 94,635 | 3,4337 |
| r = 5 m when θ = α | Ny, kN/m2 | -107 | -104,63 | 2,215 |
| σrr = 0 | θ0, ° | 13,4 | 14,25 | 5,9649 |
With additional side nodes:
| Point | The unknown | Analytical solution | LIRA-FEM | Error, % |
| r = 5 m when θ = 0 | Ny, kN/m2 | 98 | 94,974 | 3,0878 |
| r = 5 m when θ = α | Ny, kN/m2 | -107 | -104,97 | 1,8972 |
| σrr = 0 | θ0, ° | 13,4 | 14,25 | 5,9649 |
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