7/28/2019 nastran training
1/12
6 | 1
6.1 INTRODUCTION
FEMAP (Finite Element Modeling And Postprocessing) is an engineering analysis pre- and postprocessor
for the simulation of complex engineering problems using the finite element method. It runs on Microsoft
Windows and provides CAD import, modeling and meshing tools to create a finite element model, as well
as postprocessing functionality that allows structural engineers to interpret analysis results. The finite
element method allows engineers to virtually model components, assemblies, or systems to determine
behavior under a given set of boundaries loads conditions, and is typically used in the design process to
reduce costly prototyping and testing, evaluate differing designs and materials, and for structural
optimization to reduce weight.
Product simulation applications include basic strength analysis, frequency and transient dynamic
simulation, system-level performance evaluation and advanced response, fluid flow and multi-physics
engineering analysis for simulation of functional performance.
Femap is used by engineering organizations and consultants to model complex products, systems and
processes including satellites, aircraft, defense electronics, heavy construction equipment, lift cranes,
marine vessels and process equipment.
Using Femaps digital simulation capabilities we can:
Predict and improve product performance and reliability. Reduce time-consuming and costly physical prototyping and testing. Evaluate different designs and materials. Optimize our design and reduce material usage.
7/28/2019 nastran training
2/12
6 | 2
6.2 Case Studies
6.2.1 Beam Under bending
Consider the circular cantilever beam shown in Figure below. It is fixed at end A and loaded with force Fat the other end:
Where:
F= 100 N
L= 1 m
D= 0.08 m
E=29E6 Pa
= 0.32
6.2.1.1 Finite Element Model
Total number of element 10
Total number of nodes 11Total number of degree of freedom 60
7/28/2019 nastran training
3/12
6 | 3
6.2.1.2 Result
Total displacement (max) = 0.512 m Max comb stress= 1987945 Pa
Buckling factor of = 2.0251
7/28/2019 nastran training
4/12
6 | 4
6.2.2 PLATE UNDER INPLANE LOADING
It is required to perform plane-stress analysis for cantilever rectangular plate under distributed load asshown:
6.2.2.1 Problem Data
property value
Plate thickness 0.01 m
Poisson's ratio 0.32Young's modules of elasticity 29E6 PaDistributed load (in -Y direction) 100 PaThe dimensions 10 x 5 m
7/28/2019 nastran training
5/12
6 | 5
6.2.2.2 Finite Element Model
Total number of elements 100Total number of nodes 66Total number of degrees of freedom 396
6.2.2.3 Results
Von Misses stress with max= 10734.53 Pa Total displacement max = 0.0051 m
Buckling factor of = - 0.0024
7/28/2019 nastran training
6/12
6 | 6
6.2.3 BENDING PLATE0 (SQURE PLATE WITH A HOLE)
It is required to perform the stress and transverse displacement analysis for a square plate with a
circular hole. The plate is rigidly supported (CFCF) and subjected to distributed load in Y-direction as
shown:
6.2.3.1 Problem Data
property value
Plate thickness 0.01 mPoisson's ratio 0.32Young's modules of elasticity 10.3E6 Pa
Distributed load (in - Z-direction) 100 PaThe dimensions 1 x 1 m
7/28/2019 nastran training
7/12
6 | 7
6.2.3.2 Finite Element Model
Total number of elements 168Total number of nodes 108Total number of degrees of freedom 648
6.2.3.3 Results
Von Misses stress with max= 312533.612 Pa Total displacement (max) = 0.2825 m
Buckling factor of = - 84.213
7/28/2019 nastran training
8/12
6 | 8
6.2.4 SHELL STRUCTURE
It is required to perform the stress and transverse displacement analysis for a shell structure used as in
fuselage applications. The barrel is supported on rigid diaphragms, and loaded by uniform pressure.
6.2.4.1 Problem Data
property value
Plate thickness 0.01 mPoisson's ratio 0.32
Young's modules of elasticity 10.7E6 PaDistributed load (in X-direction) 100 Pa
The dimensions 1 x 1 mThe hole radius 0.2 m
7/28/2019 nastran training
9/12
6 | 9
6.2.4.2 Finite Element Model
Total number of elements 261Total number of nodes 161Total number of degrees of freedom 966
6.2.4.3 Results
Von Misses stress with max= 48416 Pa Total displacement (max) = 0.00652 m
Buckling factor of = -1.00965
7/28/2019 nastran training
10/12
6 | 10
7/28/2019 nastran training
11/12
6 | 11
7/28/2019 nastran training
12/12
6 | 12