Homework: 7
Figure 1 illustrates a one element plane stress mesh. A traction t = [200 0]T is applied on the
right hand side. The thickness of the part is 1. The material has a modulus E=1000 and Poiso
ME/EMech 461. HW3
Figure 1 illustrates a rope under a gravitational led b = 10. A concentrated force P = 1000
is applied the free end. The cross section is A = .01. The modulus of Elasticity is E = 10
ME/EMCH 461 HW: 1
Part 1
Derive the material matrix D for plane stress conditions by enforcing that z = 0 in the 3D
stress-strain relations.
Part 2
The 2D square domain of Figure 1 is subjected to the
Final Project
Option 1: Propose a project
If you have a some problem you would like to work on, please, send me an email (by 11/17/11
5:00pm) with a short description so that I can gure out if its app
ME461
Homework 2
For the system in the figure below determine the following:
a) Derive the stiffness matrix, displacement vector, and load vector using Newtons
Law.
b) Derive the above by minimizing t
2/16/08 7:35 PM
\mnelabs02\users\dxs961\desktop\Dilip_ME461_HW3b.m
% Program which uses 5 quadratic elements to solve for displacement and
% stress fields.
clear all;
nnode=11; % Number of nodes
nelem
2/16/08 7:35 PM
\mnelabs02\users\dxs961\desktop\Dilip_ME461_HW3a.m
% Program which uses 10 linear elements to solve for displacement and
% stress fields.
clear all;
nnode=11; % Number of nodes
nelem=1
ME461
Homework 6
A quadrilateral element is shown in the figure below. The nodal displacement vector is
q=[.1 .1 0 0 0 0 0 0]^T. Plane strain conditions are assumed with E=1000 and =.3
Determine the l
ME/EMech 461. HW4
Figure 1 illustrates a rope under a gravitational led b = 10. A concentrated force P = 1000
is applied the free end. The cross section is A = .01. The modulus of Elasticity is E = 10
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ME 461
Homework Assignment 5
Problem 1.
Complete this ANONYMOUS survey: https:/goo.gl/forms/ErKfnMViQMYSFb133
Problem 2.
Redo the 2-element problem we did in class (image shown below), but with the p
Consider the bar shown in g. P3.1.Cross-seconal area A: = 1.2 in.2.andYoimg's modu-
lusE = 3|] X 105 psi.qu = cfw_102 1.5.qu = 0.025 in.,deterntine thefollowingfbyhand
calculation):
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h- P i|-
ME 461 Fall 2016 HW_1
* Note: this and future homeworks refer to the 4th edition of the Textbook
1) P1.3, P1.7
2) P1.12 using the method of minimum potential energy. After determining the
displacement
Problem 4.7
Input File:
Next line is problem title
< 2D TRUSS ANALYSIS >
Problem 4.7
NN NE NM NDIM NEN NDN
4 3 1 2 2 2
ND NL NMPC
6 1 0
Node# X Y
1
0 0
2 -0.45 0.6
3 0.45 0.6
4
0.8 0.6
Elem# N1 N2 Mat