BÀI TẬP LẦN 2 NGUYỄN THANH PHONG 1

Sinh viên: Nguyễn Thanh Phong
MSSV: G1002398
Lớp: GT10HK

KHÍ ĐÀN HỒI
Giảng viên: TS. Trần Tiến Anh
Ngày nộp: 06/03/2014

BÀI TẬP 1-4
Introduction to Structural Dynamics and Aeroelasticity
Problem 1:
Show that the equation of motion for longitudinal vibration pf a uniform beam is the same ass
that for a string, viz.

EA
Solution:

∂ 2u
∂ 2u
=
m
∂x 2
∂t 2

y

x

ds
T+
u

v

u+

∂u
dx
∂x

∂T
dx
∂x

dx
We have:

With β = 0
So :

∂
∂ 2u
T =m 2
∂x
∂t

&

0=m

∂ 2v

∂t 2

Such as the force of the rod vertically then u turn stretches over time, and v is constant (the
derivative with time 0)
Equation (2.19) T = EAε

Let us presuppose the existence of a static-equilibrium solution of the string deflection so that


whereT0 and ε o are constants and δ=l−l0 is the change in the length of the string between its
stretched and unstretched states.
If the steady-state tension T0 is sufficiently high, the perturbation deflections about the staticequilibrium solution are very small. Thus, we can assume:

we find that the equations of motion can be reduced to two linear partial differential
equations:

∂ 2uˆ
∂ 2uˆ
EA 2 = m 2
∂x
∂t
Problem 2:
Show that the equation of motion for longitudinal vibration pf a uniform beam is the same ass
that for a string, viz.
l

2

 ∂u 
P = ∫ EA  ÷ dx

 ∂x 
0
Solution:
To solve problems involving the forced response of strings using Lagrange’s equation, we
need an expression for the strain energy, which is caused by extension of
the string, viz.

where, as before

and the original length is0. To pick up all of the linear terms in Lagrange’s equations, we
must include all terms in the energy up through the second power of the unknowns. Taking
the pertinent unknowns to be perturbations relative to the stretched but undeflected string, we
can again write


ForEAequal to a constant, the strain energy is:

Because
:
Strain energy simplifies to

the strain energy becomes

We assuming that v is const and ε o << 1 so we have:
l

2

 ∂u 
P = ∫ EA  ÷ dx

 ∂x 
0
Problem 3:
Show that the equation of motion for longitudinal vibration pf a uniform beam is the same ass
that for a string, viz.
l

2

 ∂u 
K = ∫m 
÷ dx
∂
t


0
Solution:
To solve problems involving the forced response of strings using Lagrange’s equation, we
also need the kinetic energy. The kinetic energy for a differential length of
string is

With:

∂v
= 0 (v is const) so we have:
∂t
2
l
 ∂u  2 

 ∂u 
dK = m  ÷  dx ⇒ K = ∫ m  ÷ dx
 ∂t 
 ∂t  
0