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# Validation Case: Statically Indeterminate Reaction Force Analysis

In this case, a bar assembly with three parts is fixed at two ends and two force loads are applied at the two cross-sectional areas in the middle. This type of system is deemed to be statically indeterminate as the reaction forces at the fixtures cannot be determined by the equations of statics alone and the deformation in the system must be taken into consideration.

The FEA model and the resulting comparison of the simulation results against the analytical calculations validates the use of following conditions for linear mechanical static analysis in Solve:

• Restraints
• Reaction Forces (outputs)

## Geometry • Geometry Dimensions:
• Part Cross Sectional Area = 0.1m by 0.1 m
• Part 1 Length (L1) = 0.4 m
• Part 2 Length (L2) = 0.3 m
• Part 3 Length (L3) = 0.3 m

## Material

• Structural Steel (for all three parts)
• Young’s Modulus (E) = 200 GPa
• Poisson’s Ratio (ν) = 0.3
• Density (ρ) = 7856 kg/m3

## Physics

• Restraints
• Fixture 1
• Part 1, Face 5
• Fixture 2
• Part 3, Face 4
• Force 1
• 1000 N
• Part 2, Face 4
• Force 2
• 500 N
• Part 1, Face 4

## Meshing:

OnScale Solve automatically generates a 3D second-order tetrahedral mesh. The meshing statistics are:

Mesh Quality: Medium

## Reference Solution

The reference solution can be found in S.Timoshenko’s Strength of Materials 2nd Edition – Tenth Printing, page 21 and 25. The load P will be in equilibrium with the reactions R and R1 at the ends: Using Hooke’s equation for describing the linear relationship between force and the elongation of the prismatic bar: Where is total elongation, P is the force applied, A is the cross sectional area and E is Young’s Modulus of the bar.

As shortening in one part is equal to the elongation in the other, we can obtain: Cancelling the cross sectional area and Young’s Modulus term, the forces R and R1 are therefore shown to be inversely proportional to the distances of their points of application from the loaded cross section m.n: With these equations, we can rearrange for R and R1 in terms of P, a and b: For our geometry with a constant cross section, we can couple 2 parts together and treat each Force application as a separate case and sum the total reaction forces.

Outputs Analytical OnScale Solve OnScale Solve Error (%)
Reaction Force R 600 N 599.2 N -0.001
Reaction Force R1 900 N 900.8 N 0.001

Note simulated reaction forces are extracted using the Jupyter Notebook functionality from the kpi output (.json) file. 