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transporto priemonių dinamika - Vilniaus Gedimino technikos ...

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Masių centro vektorius lygus:<br />

⎧xc<br />

⎫ ⎧x<br />

c b a i ⎫<br />

⎧a⎫<br />

⎪ ⎪ 1 ⎪ ⎪ 1 ⎪ ⎪<br />

{ Rc<br />

}= ⎨yc<br />

⎬ = ∫ ∫ ∫ ρ⎨yi<br />

⎬dxidyidzi<br />

= ⎨b⎬<br />

.<br />

⎪ m<br />

⎩z<br />

⎪ 0 0 0 ⎪<br />

c ⎭ ⎩z<br />

⎪ 2 ⎪<br />

i ⎭<br />

⎩c⎪<br />

⎭<br />

Stačiakampio gretasienio masių inercijos momentai kūno masių<br />

centro atžvilgiu lygūs:<br />

m ⎛ 2<br />

⎛ b⎞<br />

c<br />

Ixcxc = Ixx − m( yc + zc<br />

)= ( b + c )− m ⎜ ⎟<br />

⎝ ⎠<br />

+ ⎛ 2<br />

⎝ ⎜<br />

⎞ ⎞<br />

2 2 2 2<br />

⎜ ⎟<br />

3<br />

⎝<br />

2 2 ⎠ ⎟ =<br />

⎠<br />

m b<br />

2 c<br />

2<br />

( + )<br />

12<br />

;<br />

m ⎛ 2<br />

⎛ a⎞<br />

c<br />

Iycyc = Iyy − m( xc + zc<br />

)= ( a + c )− m ⎜ ⎟<br />

⎝ ⎠<br />

+ ⎛ 2<br />

⎝ ⎜<br />

⎞ ⎞<br />

2 2 2 2<br />

⎜ ⎟<br />

3<br />

⎝<br />

2 2 ⎠ ⎟ =<br />

⎠<br />

m a<br />

2 c<br />

2<br />

( + )<br />

12<br />

;<br />

m ⎛ 2<br />

⎛ a⎞<br />

b<br />

Izczc = Izz − m( xc + yc<br />

)= ( a + b )− m ⎜ ⎟<br />

⎝ ⎠<br />

+ ⎛ 2<br />

⎝ ⎜<br />

⎞ ⎞<br />

2 2 2 2<br />

⎜ ⎟<br />

3<br />

⎝<br />

2 2 ⎠ ⎟ =<br />

⎠<br />

m a<br />

2 b<br />

2<br />

( + )<br />

12<br />

;<br />

m m<br />

Ixcyc = Ixy + mxcyc<br />

=− ab + ab = 0 ;<br />

4 4<br />

m m<br />

Ixczc = Ixz + mxczc<br />

=− ac + ac = 0 ;<br />

4 4<br />

m m<br />

Iyczc = Iyz + myczc<br />

=− bc + bc = 0 .<br />

4 4<br />

Stačiakampio gretasienio masių inercijos tenzorius masių centro<br />

atžvilgiu lygus:<br />

( )<br />

⎡ m b<br />

2 c<br />

2<br />

⎤<br />

⎢ + 0 0<br />

12<br />

⎥<br />

⎢<br />

⎥<br />

m<br />

[ Icc<br />

]= ⎢<br />

( a<br />

2<br />

+ c<br />

2<br />

0<br />

⎢<br />

) 0 ⎥<br />

⎥<br />

.<br />

12<br />

⎢<br />

⎥<br />

⎢<br />

m<br />

( a<br />

2<br />

0 0<br />

+ b<br />

2)<br />

⎥<br />

⎣⎢<br />

12 ⎥ ⎦<br />

79

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