sinmecanik
TRANSCRIPT
-
8/17/2019 sinmecanik
1/6
-
8/17/2019 sinmecanik
2/6
◦
-
8/17/2019 sinmecanik
3/6
-
8/17/2019 sinmecanik
4/6
◦
-
8/17/2019 sinmecanik
5/6
0 = 0
0 = 0
E = 30X 106 lbin2
-
8/17/2019 sinmecanik
6/6
◦
m = w32,17 ft
s2
= 150lb32,17 ft
s2
= 0,385 lbs2
in
K = 193EI L3
I = 112bh3 = 1
12(20in)(0,5in)3 = 0,2083in4
K = ( 193(30
X106lb
in2 )(0,2083in4)
(100in)3 = 1200 lb
in
wn =
km
=
1200lb
in
0,385 lbs2
in
= 55,57 rads
S eq =
F 0
k =
50lb
1200 lbin = 0,04166in
X S eq
= 11−( w
wn )2
X = S eq1−(
62,832 rads
55,57 rads)2
= −0,15in
dla/tales/documentos/lmt/ramirez ro/capitulo4.pdf