w(x,t) x = 0 x = l...or, p j+ω j 2p j=fˆ jsinΩt the particular solutions for these...
TRANSCRIPT
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Example A4.19
Given: Consider the bending beam shown below with a concentrated harmonic force acting at themid-length point C on the beam.
Find: For this problem:a) Use the modal uncoupling approach to determine the particular solution wP (t) = W (x)sin⌦t
of the beam‘s EOM.b) Make a hand sketch of |W (x = L/2)| vs. ⌦ for a range of values of ⌦ covering up through the
first four resonances of the response. Provide details on how you arrived at this hand sketch.
w(x,t)
x = 0 x = L
EI , ρA
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ME 563 – Fall 2014
Hom
ework Problem
8.4 C
onsider the bending beam show
n below w
ith a concentrated harmonic force acting at
the mid-length point C
on the beam.
a) U
se the modal uncoupling approach to determ
ine the particular solution wP (x,t)=
W(x)sinΩ
t of the beam’s EO
M. Feel free to use natural frequency and
modal function results from
a previous homew
ork assignment in arriving at your
solution here.
b) M
ake a hand sketch of W(x
=L/2) vs. Ω
for a range of values of Ω covering
up through the first four resonances of the response. Provide details on how you
arrived at this hand sketch.
c) R
epeat b) above for W(x
=L)
.
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or,
pj +ω j2pj = f̂ j sinΩt
The particular solutions for these modally-uncoupled EOMs are given by:
pjP (t) =f̂ j
ω j2 −Ω2
sinΩt
Therefore, the beam response is:
wP (x, t) = Wj( )(x)pjP (t)
j=1
∞
∑
= W j( )(x)f̂ j
ω j2 −Ω2
⎛
⎝⎜
⎞
⎠⎟
j=1
∞
∑⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪sinΩt
=W (x)sinΩt
From this, we have:
W L2
⎛⎝⎜
⎞⎠⎟ = W
j( ) L2
⎛⎝⎜
⎞⎠⎟
f̂ jω j2 −Ω2
⎛
⎝⎜
⎞
⎠⎟
j=1
∞
∑
= f0W j( ) L
2⎛⎝⎜
⎞⎠⎟
⎡⎣⎢
⎤⎦⎥2
ρA W ( j)2 dx0
L
∫⎛
⎝⎜
⎞
⎠⎟
1ω j2 −Ω2
⎛
⎝⎜
⎞
⎠⎟
j=1
∞
∑
Comments:
• Resonances will occur for the jth mode when Ω =ω j , unless W( j)(L / 2) = 0 .
• Anti-resonances will occur between every pair of resonances since the coefficient of 1/ ω j
2 −Ω2( ) in the above sum is positive for all modes.