Evaluate the natural frequency (radian/second) and natural period for the Example hetural systems shown below in figure. k = 45 kN/m L=4 m m = 10 kN/m E= 20000 MPa I= 1.5 x 10 m k = 45 kN/m L= 3.5 m (b) E = 25000 MPa k = 40 kN/m I = 1 x 10 m MADP m= 12 kN/m L= 5 m (c) PublicationS k = 30 kN/m E = 22000 MPa m = 30 kN/m I = 1.1 x 10 m m = 18 kN O 0.17 L 0.83 L L = 6 m E = 18000 MPa 4 I = 1.4 x 10 m*

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Evaluate the natural frequency (radian/second) and natural period for the
Example
k = 45 kN/m
L= 4 m
m = 10 kN/m
E= 20000 MPa
-4
I= 1.5 x 10m
k = 45 kN/m
L= 3.5 m
(b)
E = 25000 MPa
I = 1 x 10 m
k = 40 kN/m
-4
4
RAD
m D 12 kN/m
Publications
L= 5 m
(c)
k = 30 kN/m
E = 22000 MPa
4
m = 30 kN/m
-4
1 = 1.1 x 10 m*
m = 18 kN
0.17 L
0.83 L
(d)
L = 6 m
E = 18000 MPa
I= 1.4 x 10* m
Transcribed Image Text:Evaluate the natural frequency (radian/second) and natural period for the Example k = 45 kN/m L= 4 m m = 10 kN/m E= 20000 MPa -4 I= 1.5 x 10m k = 45 kN/m L= 3.5 m (b) E = 25000 MPa I = 1 x 10 m k = 40 kN/m -4 4 RAD m D 12 kN/m Publications L= 5 m (c) k = 30 kN/m E = 22000 MPa 4 m = 30 kN/m -4 1 = 1.1 x 10 m* m = 18 kN 0.17 L 0.83 L (d) L = 6 m E = 18000 MPa I= 1.4 x 10* m
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