Results Verification Has the model been correctly implemented?

Checklist for Results Verification Data check Sum of reactions = 0.0 Restraints - no deformations at restrained freedoms Symmetry - check symmetry for symmetric structures with symmetrical loading. Check local equilibrium Form of results - internal forces Form of results - deformations Checking Model - internal forces Checking Model - deformations

Results Verification 1. Check the data

2. Check the sum of reactions 10 kN

Results Verification 2. Check the sum of reactions Compare the sum of the reactions on the structure in one direction (normally vertical) with the sum of the loads applied in that direction. It is best to use the sum of the loads calculated separately rather than the sum calculated by the software.

3. Check the restraints Where displacements are intended to be zero, they are zero. No nodes are restrained when they should be free. Look at the values of displacement Look at the supports on the model view

4. Check for Symmetry -1 Model 5 significant digits Results are symmetric but loading is not symmetric

4. Check for Symmetry -2 Model 15 significant digits Results are not symmetric

4. Check for Symmetry -3 Model Results are symmetric to 15 significant digits 15 significant digits

Results Verification 4. Check for Symmetry - 4 For a symmetrical structure with symmetrical loading check any pair of symmetrical values (only one pair need be checked). The two values should correspond to a high number of significant digits. Small differences may not be significant but do indicate that is an error somewhere. For example the sectional axes of a column may be wrongly defined.

Results Verification 5. Check local equilibrium Finding an error from an equilibrium check from the output of a FE program very unlikely. The value of doing an equilibrium check is that it helps in understanding how the forces in the system interact,

5. Check local equilibrium In LUSAS select: Utilities/Print Results Wizard/(Force/Moment) then select ‘Component and ‘Element Nodal’ then OK This gives a table like this

LUSAS sign convention for internal forces

Model of simply supported beam

Check equilibrium at node

Check equilibrium at node 57

Check equilibrium at node

Check equilibrium at node 53 Note that the load is on the node and is the statical equivalent of the sum of the member forces.

7. Look at the form of the form (i.e. the shape) of the results Model Deformed mesh Red line is straight?

Form of the results Deformed mesh 1 Bending moments for mesh 2 Deformed mesh 2 The displacements for mesh 2 seem to be reasonable but the bending moments are wrong at the base

Form of the results Error in support for mesh 2 BM for correct support Bending moments for mesh 2

Bridge truss model Central point load Deflection of beam with central point load As above but with very low shear stiffness Bending mode Shear mode Combination of Bending mode and shear modes

Form of the results Shear mode deflection of a beam Bending mode deflection of a beam

Form of the results Bending mode deformation

Uniformly distributed load. Deflection decreases towards the top Top point load. Horizontal deflection is fairly straight Form of the results Typical of shear mode deformation

7. Look at the form (i.e. the shape) of the results Bending moments Bending mode - Big columns

Results Verification 8. Check the values of the results - use a ‘checking model’ A checking model can be: A ‘back of an envelope’ hand calculation, i.e. a quick check. An alternative simplified computer model. A repeat of the main model carried out independently.

Sleipner Platform before collapse No deaths but very high financial loss

Sleipner - Plan

Sleipner - Mesh at Tricell

Sleipner - Detail of Mesh at Tricell Junction There was shear failure of the wall of the tricell when the tricell was full of water (67m) and the main cells were empty. This was an abnormal but realistic loading condition

Plan of tricell

Calculation (checking model) for shear stress in tricell wall of Sleipner Platform Operating Conditions: Span of Wall m Effective depth mm, Pressure head 67.0 m Pressure at 67 m depth: p =  gh = 1000 x 9.81 x 67 = 66x 10 4 N/m = p x area =66x 10 4 x 4.5 x 1.0 = 3000 x 10 3 N Load on 1.0 m strip at 67 m depth: W Max shear: V = W/2 = 3000 x 10 3 /2 = 1500 x 10 3 N Shear stress in concrete: v c = V/(bd) = 1500x10 3 /(500x 1000) = 3 N/mm 2 Maximum design shear stress (BS 8110 for unreinforced section): v c (BS8110) = 0.91 N/mm 2

Sleipner - Back of an envelope check Loss > $700m

Sleipner Collapse The collapse was due to the choice of an inadequate mesh of 3D elements for the walls of the tricells. The estimate of shear stress in the wall was about 1/3 of a realistic value. There was an error in model validation

Equivalent beam models For full description - see: Mech/Beam-deflection.pdf