2014 - In ETABS, shell or area element has two types of stiffnesses i.e. inplane stiffness refers as f11, f22 and f12 and out-of-plane stiffness refers ass modifiersFull description
Stiffness of soil
Document provides the equivalent material for different codes. For carbon steel Forging and and casting material included. SS material equivalent is also included.Full description
Full description
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can compare diffirent standard
leg training
Excel sheet for pile design
Excel sheet for pile design
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3 d stiffness matrix
Equivalent Material Grades
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Descripción: A handbook for matlab -scilab equivalent command, useful for those who use matlab or scilab (can be downloaded free from its official web) for mechanical or thermal problems of materials or any oth...
58
.6
C H AP A P TE TE R
U nd n d am am p e
F re re e V ib ib ra ra t o n
Figure 3.9 Cantileve Cantileve beam with tip load.
.-
.. ..---.--1
~--------L--------~
where
(3.53)
T A B L E 3.
Equi Equiva vale le
Spri Spring ng
on onst st nt
Case 1.
Equivalen Equivalen Sprin Constant Constant
axial axial spring spring in parall parallel el '/'/'/'/'
'/'N'/' k2
'/'N'/' k3
Ir"""",,{J 2.
axia axia spri spring ng in seri series es
(continued)
E qu q u iv iv a
3.3
Case ar
St
ss
Equi Equiva vale lent nt Spri Spring ng Cons Consta tant nt
klk3
k2k3
k,
igid
-,
...-
~&;"""""";j~ -
AE
cros crosss-se sect ctio iona na area area elas elasti ti modu modulus lus (continued)
59
CHAP CHAPTE TE
60
damp damped ed Free Free Vibr Vibrat atio io
Case
xi
he
Equi Equiva vale lent nt Spri Spring ng Cons Consta tant nt
ri
elas elasti ti
shea shea modu modulu lu
GJ
tors tors on
nsta nsta
of cr ss
ctio ctio
(7Td /32)
Di L[Dbl
pi
rs on
Da
ring ring
(DbIDa)3]
EI
Youn Young' g'
modu modulu lu
3EI EI
--- --(continued)
3.3
E qu q u iv iv a
St
61
Equi Equiva vale lent nt Spri Spring ng Cons Consta tant nt
Case
48EI
IF
ElL
192EI
-----
~~
~_J 3EI(a
b)
r~-_~_
+_-_-~{---
~------L------~
768EI
iF
.............
_._;,-----
24EI
-------lJ;"
~I~a~ (continued)
C H AP A P TE TE R
U nd n d am am pe p e d F re re e V ib ib ra ra t o n
Equi Equiva vale lent nt Spri Spring ng Cons Consta tant nt
Case
3EI (L+a)a
----------...
11--__
.-
1W iI
aj
EI
EI
lilll / . R i . · g . i d . 1 J = i EI
le ti rm elem elemen ents ts in para parall llel el (ke)p,
lati lati
ri
th ti (ke)s, an then then appr approp opri ri
(3.54) or (3.55)
Figu Figure re 3.10 3.10
where
stem stem it prin prings gs in aral aralle le an in seri series es
(3.56) and (3.57)
.3
E qu q u iv iv a
S t n es es s
(ke)p and (ke)s
he
ke is
(3.58)
E XA M L E
and (b) mass m. Solution
Springs k, and k2 and Let
k3
k,
3.49 3.49
111
(1)
or ' k k3
k2
-x
m(k,
k3
(2)
(3)
k3)X
(5)
E XA M L E
he al on in flex flexur ural al rigi rigidi dity ty EI an leng length th
