Deflection of a Beam under a Uniformly Distributed Load

Let us consider a beam under a uniformly distributed load q. A length of the beam is L. The beam cross-section is a square. The length of the side of the square is a.

Sought quantity is the maximum deflection of the beam.
Let us use the following initial data: q = 3000 Pa, L = 0.5 m, a = 0.02 m.
Material characteristics: the Young's modulus E = 2.1E+011 Pa, Poisson's ratio ν = 0.28.

Deflection of a Beam under a Uniformly Distributed Load, the finite element model with applied loads and restraints

The finite element model with applied loads and restraints

The analytical solution is calculated by the formula:
, .

The maximal deflection of the beam obtains at x = L / 2 :
,
where J = a4 / 12 - the moment of inertia.

Thus, | w | = 1.7439E-005 m.

After carrying out calculation with the help of AutoFEM, the following results are obtained:

Table 1.Parameters of the finite element mesh

Finite Element Type	Number of Nodes	Number of Finite Elements
quadratic tetrahedron	220	322

Table 2. Result "Displacement OZ"

Numerical Solution Displacement OZ \| w* \|, m	Analytical Solution Displacement \| w \|, m	Error δ =100%* \|w* - w \| / \| w \|
1.7498E-005	1.7439E-005	0.34

Deflection of a Beam under a Uniformly Distributed Load, Result "Displacement, Z"

Conclusions:

The relative error of the numerical solution compared to the analytical solution is equal to 0.34% for quadratic finite elements.

*The results of numerical tests depend on the finite element mesh and may differ slightly from those given in the table.