solution

ME 360 – H01
Name _________________________
1) Norton text problem 4-10. Additionally - write a MATLAB m-file to plot vertical deflection
of the beam as a function of position along the beam. The plot should be fully annotated with
meaningful labels, titles, and legends as appropriate. See Section 4.9 and Appendix B in Norton
for help with beam analysis. The MATLAB code provided below shows two ways to code beam
functions. You must attach copy of your MATLAB code.
% h01_demo.m - demonstration code for ME 360 H01
% HJSIII, 14.09.03
% constants
L = 10;
a = 4;
% fill column vector from 0 to L with step 0.5
x = ( 0 : 0.5 : L )';
% number of elements in x
n = length( x );
% beam <x-a> function using for-loop and if-statement
for i = 1 : n,
if x(i) >= a,
beam1(i) = 1 + 2*x(i) - 2.5*(x(i)-a);
else
beam1(i) = 1 + 2*x(i);
end
end
% beam <x-a> function using conditional-test and dot-multiply
beam2 = 1 + 2*x - 2.5*(x-a).*(x>=a);
% plot both
figure( 1 )
clf
plot( x,beam1,'r-', x,beam2,'go' )
xlabel( 'x [inches]' )
ylabel( 'beam1 and beam2 [inches]' )
title( 'Sample plot for beam <x-a> function' )
legend( 'if-else', 'conditional-test' )
% bottom of h01_demo
yMAX = -128 mm
from Appendix D
ME 360 – H01
Name _________________________
P = (100 kg) (9.81 m/sec2) = 981 N
w = 305 mm
MMAX = P (1.3 m) = 1.2753x106 N.mm
(
t = 32 mm
E = 10.3 GPa
I = w t3 / 12 = 8.329x105 mm4
c = t/2 = 16 mm
)
M c 1.2753x10 6 N.mm (16 mm )
N
σ=
=
= 24.50
= 24.50 MPa
5
4
I
mm 2
8.329 x10 mm
use MPa units for E in deflection equation
yMAX = -128.8 mm
from MATLAB code
N
(MPa ) mm 4
(
N
[
mm ] =
= mm
(MPa )(mm )
)
3
ME 360 – H01
Appendix B Fig B-3a Norton
Name _________________________
ME 360 – H01
Name _________________________
Norton problem 4-10, HJSIII
20
Vertical deflection of beam [mm]
0
-20
-40
-60
-80
-100
-120
-140
0
200
400
600
800 1000 1200 1400
Position along beam [mm]
% h01.m - ME 360 Norton problem 4-10
% HJSIII, 14.09.03
%
a
b
L
F
E
I
constant values
= 2000;
% mm
= 700;
% mm
= 2000;
% mm
= 981;
% N
= 10.3e3;
% MPa
= 8.329e5; % mm^4
% values along beam
n = 100;
dx = L / n;
x = ( 0 : n )' * dx;
% beam deflection function
y_bracket = ((b-a)/b)*(x.^3).*(x>=0) + (a/b)*((x-b).^3).*(x>=b) ...
- ((x-a).^3).*(x>=a)
+ b*(a-b)*x ;
y = F * y_bracket /6 /E /I;
max_deflection = min(y)
% graph
figure( 1 )
clf
plot( x, y )
xlabel( 'Position along beam [mm]' )
ylabel( 'Vertical deflection of beam [mm]' )
title( 'Norton problem 4-10, HJSIII' )
% bottom of h01
1600
1800
2000
ME 360 – H01
Name _________________________
2) Norton text problem 4-19. Pin length is not required. See Section 4.8 in Norton for help with
direct shear. Be certain to round up to nearest standard size.
dROD = 0.75 inch
tension
AROD = π dROD2 / 4 = 0.4418 in2
σROD = FROD / AROD = 30,000 psi
FROD = σROD AROD = 13,253.6 lbf
P = 10 FROD = 132,536 lbf
APIN = π dPIN2 / 4
direct shear
APIN = 3.3134 in2
ASHEAR = 2 APIN
τ = P / ASHEAR = 20,000 psi
dPIN = 2.054 in
ASHEAR = P / τ = 6.6268 in2
round up to 2.125 inch DIA
ME 360 – H01
Name _________________________
3) Build a SolidWorks model of a stepped shaft that is 0.75 inch DIA by 1 inch long reduced to
0.5 inch DIA by 0.75 inches long with a 0.05 in fillet radius at the step. Attach the other end of
the larger section to a 1.5 inch by 1.5 inch by 0.25 inch thick plate for future FEA studies.
Include a generous fillet between the plate and larger shaft. Provide a hardcopy image of your
model.

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