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Beam problems (part1- simply supported beam with UDL and point load)

 Beam problem

Problem1 Description:

A simply supported Beam of 4000 mm length as shown in fig. A point load of 500 N and UDL load of 0.4 N/mm will be applied to a solid steel beam with a rectangular cross section. The cross section of the beam is 200mm*300mm while the modulus of elasticity of the steel is 200GPa and Poisson’s ratio is 0.3. Find the maximum deflection, Shear forces, Bending moment, and Reaction forces.

problem figure

Results:

Discipline: Structural

Analysis Type: Static

Element type: Beam, 2 node 188

Material model: Linear elastic isotropic.

Section: common section of beam – rectangular c/s, B= 200, H=300.

Key points: 4 key points are taken at x= 0, 2000, 3000, 4000. (*all dimensions are in mm)

Meshing: for all 3 elements, the number of element divisions are taken as 30. 

Displacement: at key-point1 (x=0), due to hinged support constraints are chosen in translation in all directions and rotation in x and y direction. Because the beam can only rotate in z direction.

Load: at key-point 2 (x=2000), -500N point load is applied in y direction, and between key-point 1 (x=0) and key-point 3 (x=3000) UDL is taken of magnitude 0.4 N/mm. 

Max deflection and Nodal displacement:

 


  • As per result, we get max displacement of 0.045507 mm. as per nodal displacement graph, displacement is max near the middle section of the beam.

Shear force:

 

  • After selecting SMIS6 and SMIS19 in line element result, we get shear force result.

  • As per graph, we get maximum shear force of 700N (+ve y-direction) at key-point4 and minimum shear force of -986.668N (-ve y-direction) at key-point 1.

  • Between key-point 1 and 3, due to uniformly distributed load (0° curve), we get a line in SFD (1° curve).

  • At key-point 2, due to point load, SFD changes its direction. 

 

Bending moment:

 

  • After selecting SMIS3 and SMIS16 in line element result, we get a bending force result.

  • In result, maximum bending force of 0.119E+07 N.mm occurs at element 31 (near to mid span) and minimum bending force of 11666.7 N.mm occurs at element 90.

  • Here between key-point 1 and 3, due to uniformly distributed load (0° curve) we get curvature (2° curve -parabola) in BMD and in the remaining part, the curve is line.

  • After performing practical, it can be observed from fig that SFD is 1° higher than load diagram and BMD is 1° higher than SFD.


 

Reaction forces:

 

  • There is load in Y-direction only, hence reaction force in x and z direction are zero. In y direction, 

    • at node 1, reaction force = 1000 N 

    • at node 62, reaction force = 700 N, both reactions are in +ve y direction.

  • Due to the simply supported joint at both ends of the beam (at node 1 and node 62), there is no moment in the z-axis direction (2D case). But at node 1 and 62, in x and y direction, moments are present to restrict deformation. 

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