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MANE 7100: Mechanical Engineering Foundations II
Chet VanGaasbeek
Homework 2
9/29/2012
Problem 1
In two-dimensional elasticity theory, the stress function φ(x, y) defined by the relationships:
σxx =
∂2φ
∂y 2
(1)
σyy =
∂2φ
∂x2
(2)
σxy = σyx = −
∂2φ
∂x∂y
(3)
Substitute the above expressions into the equilibrium equations and obtain the stress function
equation. Take into account also the equations of compatibility.
The equilibrium equations are as follows:
∂σxx ∂σxy ∂σxz
+
+
+ Xx = 0
∂x
∂y
∂z
(4)
∂σyx ∂σyy ∂σyz
+
+
+ Xy = 0
∂x
∂y
∂z
(5)
∂σzx ∂σzy ∂σzz
+
+
+ Xz = 0
∂x
∂y
∂z
(6)
with body forces Xx , Xy , Xz . Applying the assumption of plane stress (given this is a two
dimensional elastic problem, equations 4, 5, 6 become:
∂σxx ∂σxy
+
+ Xx = 0
∂x
∂y
(7)
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MANE 7100: Mechanical Engineering Foundations II
Chet VanGaasbeek
∂σyx ∂σyy
+
+ Xy = 0
∂x
∂y
(8)
In the absence of body forces or assuming body forces are small with respect to the problem,
the plane stress approximation becomes:
∂σxx ∂σxy
+
=0
∂x
∂y
(9)
∂σyx ∂σyy
+
=0
∂x
∂y
(10)
Applying the condition of compatibility1 :
2
∂2
∂
(σxx + σyy ) = 0
+
∂x2 ∂y 2
(11)
yields the equation:
∂4φ
∂4φ
∂4φ
+
2
+
=0
∂x4
∂x2 ∂y 2 ∂y 4
(12)
which is the biharmonic equation∇4 φ = 0. Solutions for the stress field satisfy equation 12.
A common approach is to assume a polynomial φ less than degree 4, which satisfies Equation
12.
Problem 2
A simply supported beam has length L = 1, breadth b = 1, and height h = 0.1. The elastic
modulus of the material is E = 1011 and its Poisson’s ratio is ν = 0.3.
A downward load of P = 105 N is applied at the midpoint of the cross section along the upper
surface of the beam. Use the finite element method to determine approximate solutions to
this problem. Repeat with a distributed load across the entire cross section Q = 105 N/m.
Compare both results to those obtained from elementary beam theory.
Point Load, Midpoint of the Cross Section
For a simply supported beam with a central load (fixed at x = 0, roller support at x = L),
the loading condition is as follows:
P
x=0
x=L
1
see http://homepages.engineering.auckland.ac.nz/p̃kel015/SolidMechanicsBooks/Part II/03
ElasticityRectangular/ElasticityRectangular 02 StressFunction.pdf
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MANE 7100: Mechanical Engineering Foundations II
Chet VanGaasbeek
x=L
x = L/2
x=0
The maximum deflection for this plate is:
P L3
48EI
where E is the modulus of elasticity, L is 1, P is 10000 N, and I is:
umax =
I=
b ∗ h3
(1m)(0.1m)3
=
= 8.333 ∗ 10−5 m4
12
12
The maximum deflection is:
100000N(1m)3
P L3
=
= 2.5 ∗ 10−4 m
umax =
11
−5
4
48EI
48 ∗ (10 P a)(8.333 ∗ 10 m )
Using Simulia’s Abaqus 6.10-1 FEA software, a model is constructed as follows:
Part & Materials
The part is a one dimensional line with length = 1 meter.
A steel material is constructed with E = 101 1 Pa, ν = 0.3 in accordance with the problem
statement. A homogeneous beam section is created and assigned a rectangular cross section
(b = 1 meter, h = 0.1 meter). The beam section is orientated along the beam to accurately
represent the geometric conditions of the problem.
Step Definition, Loads, and Boundary Conditions
The analysis is performed in a linear static load step, taking place over one second of application time. No nonlinear effects are considered2 . A point partition divides the part into
two separate sections to define the load.
The load is applied over the course of the step using a concentrated point load at the
node midway along the span of 100,000 N.
Boundary Conditions for simply supported beam:
• The x = 0 node is fixed such that there is no displacement or rotation with the exception
of the z axis rotation, which is allowed to freely vary.
• The x = L node is fixed such that there is no displacement or rotation with the
exception of the x direction which is allowed to slide and the z axis rotation, which is
allowed to freely vary.
2
Second order terms in the solver are neglected.
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MANE 7100: Mechanical Engineering Foundations II
Chet VanGaasbeek
Meshing & Results
Initially the beam is meshed using 10 B31 2-node linear shear-flexible beam elements with
no bias. The resulting maximum deflection is 0.0002571 meters, which is within 3% of the
expected analytical result.
A second run of the analysis is run using 10,000 B32 2 node quadratic beam elements.
The resulting deflection produced is 0.0002576 m, which is not appreciably different from
the initial run.
Figure 1: 10 element beam analysis. Point load, midspan of the beam.
Y
X
Z
Line (Distributed) Load, Mid-Line of the Cross Section.
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