mechanics
Leila@
ynt__assignment_yardimi_1.zipMidterm 1 & Solutions.pdf
(1) Fourier-expand the following periodic input force:
2π ω
π ω
3π ω
4π ω
0− π ω
A
−A
f (t)
t
f (t) = a0
2 + a1 cos(ωt)+ a2 cos(2ωt)+ ...
+ b1 sin(ωt)+ b2 sin(2ωt)+ ...
= a0
2 + an cos(nωt)+ bn sin(nωt)( )
n=1
∞
∑
Find the coefficients a0, an , and bn
an = ω π
f (t)cos(nωt)dt 0
2π ω∫ = ω
π Aω π t cos(nωt)dt
0
π ω∫ + ω
π −A( )cos(nωt)dtπ
ω
2π ω∫
= Aω 2
π 2 t cos(nωt)dt 0
π ω∫ − Aω
π cos(nωt)dtπ
ω
2π ω∫
= Aω 2
π 2 1 nω
t sin nωt( ) + 1 n2ω 2 cos(nωt)⎡
⎣⎢ ⎤ ⎦⎥ 0
π ω − Aω
π 1 nω
sin nωt( )⎡ ⎣⎢
⎤ ⎦⎥ π ω
2π ω
= Aω 2
π 2 0 + 1 n2ω 2 cos(nπ )⎛
⎝⎜ ⎞ ⎠⎟ − 0 + 1
n2ω 2 cos(0)⎛ ⎝⎜
⎞ ⎠⎟
⎡ ⎣⎢
⎤ ⎦⎥ − Aω
π 0 − 0[ ]
= Aω 2
π 2 1
n2ω 2 (−1)n − 1 n2ω 2
⎛ ⎝⎜
⎞ ⎠⎟
= A n2π 2 −1( )n −1⎡⎣ ⎤⎦
= − 2A n2π 2 when n∈odd
0 when n∈even
⎧ ⎨ ⎪
⎩⎪
a0 = ω π
f (t)dt 0
2π ω
∫
= ω π
Aω π t dt
0
π ω
∫ + ω π
−A( )dt π ω
2π ω
∫ = Aω 2
π 2 t dt 0
π ω
∫ − Aω π
1⋅dt π ω
2π ω
∫
= Aω 2
π 2 1 2 t 2⎛
⎝⎜ ⎞ ⎠⎟ 0
π ω − Aω
π t( ) π
ω
2π ω = Aω 2
2π 2 π 2
ω 2 − 0 ⎛ ⎝⎜
⎞ ⎠⎟ − Aω
π 2π ω
− π ω
⎛ ⎝⎜
⎞ ⎠⎟ =
A 2 − A
= − A 2
Solu:on
bn = ω π
f (t)sin(nωt)dt 0
2π ω∫ = ω
π Aω π t sin(nωt)dt
0
π ω∫ + ω
π −A( )sin(nωt)dtπ
ω
2π ω∫
= Aω 2
π 2 t sin(nωt)dt 0
π ω∫ − Aω
π sin(nωt)dtπ
ω
2π ω∫
= Aω 2
π 2 − 1 nω
t cos nωt( ) + 1 n2ω 2 sin(nωt)⎡
⎣⎢ ⎤ ⎦⎥ 0
π ω + Aω
π 1 nω
cos nωt( )⎡ ⎣⎢
⎤ ⎦⎥ π
ω
2π ω
= Aω 2
π 2 − π nω 2 cos nπ( ) + 0⎛
⎝⎜ ⎞ ⎠⎟ − 0 + 0( )⎡
⎣⎢ ⎤ ⎦⎥ + A nπ
cos 2nπ( )− cos nπ( )⎡⎣ ⎤⎦
= − A nπ
−1( )n + A nπ
1− −1( )n⎡⎣ ⎤⎦
= A nπ
1− 2(−1)n⎡⎣ ⎤⎦
= 3A
nπ when n∈odd
− A nπ
when n∈even
⎧
⎨ ⎪⎪
⎩ ⎪ ⎪
(2) Given a standard mass-spring-damper mechanical system:
!!x + 2 !x + 5x = f (t), 0 < ς <1
f (t) is a periodic force with Fourier expansion:
f (t) = 2 π
1 kk=1
∞
∑ cos kωt( )
Calculate the steady state response of x, which is xp , the particular soluiton for !!x + 2 !x + 5x = f (t).
Hint : xp = xp,k k=0
∞
∑ = α k cos(kωt)+ βk sin(kωt) k=0
∞
∑ ,
Calculate α k & βk
!!x + 2 !x + 5x = f (t) = 2 π
1 kn=0
∞
∑ cos kωt( )
xp,k =α k cos(kωt)+ βk sin(kωt) ⇒ !xp,k = −α kkω sin(kωt)+ βkkω cos(kωt)
!!xp,k = −α kk 2ω 2 cos(kωt)− βkk
2ω 2 sin(kωt) ⇒
−α kk 2ω 2 cos(kωt)− βkk
2ω 2 sin(kωt)+ 2 −α kkω sin(kωt)+ βkkω cos(kωt)[ ]+ 5 α k cos(kωt)+ βk sin(kωt)[ ] = 2 kπ cos kωt( )
= −k2ω 2α k + 2βkkω + 5α k( )cos(kωt)+ −βkk 2ω 2 − 2α kkω + 5βk( )sin(kωt)
= 5 − k2ω 2( )α k + 2kω ⋅βk⎡⎣ ⎤⎦cos(kωt)+ −2kω ⋅α k + 5 − k2ω 2( ) ⋅βk⎡⎣ ⎤⎦sin(kωt)
⇒
5 − k2ω 2( )α k + 2kω ⋅βk = 2 kπ
−2kω ⋅α k + 5 − k2ω 2( ) ⋅βk = 0
⎧ ⎨ ⎪
⎩⎪
⇒
α k
βk
⎡
⎣ ⎢
⎤
⎦ ⎥ =
5 − k2ω 2 2kω −2kω 5 − k2ω 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
−1 2 kπ 0
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥ = 1 5 − k2ω 2( )2 + 2kω( )2
5 − k2ω 2 −2kω 2kω 5 − k2ω 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
2 kπ 0
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
= 1 5 − k2ω 2( )2 + 2kω( )2
2 kπ
5 − k2ω 2( ) 4ω π
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥
Solu:on
⇒
xp = 1
5 − k2ω 2( )2 + 2kω( )2 2 kπ
5 − k2ω 2( )cos kωt( ) + 4ω π sin kωt( )⎡
⎣⎢ ⎤ ⎦⎥k=0
∞
∑
= 2 kπ
⋅ 1 ω n
2 − k2ω 2( )2 + 2kω( )2 5 − k2ω 2( )cos kωt( ) + 2kω sin kωt( )⎡⎣ ⎤⎦
k=0
∞
∑
m J
y = 2sin 5t( )
x b
k
(3) A block with mass m is suspended inside a box via a spring (k) and a damper (b). A roller with roatry inertia J and radius r is hinged to the box. The block and the roller are coupled by a set of gear and gear rack, such that x = rθ . The box is shaked up and down following a function y = 2sin 5t( ).
r
θ
This system can be described by a standard mass-spring-damper-force format: M!!x + B!x + Kx = F Calculate M , B, K , and F.
Hint: You can use either of the following methods to solve this problem: (1) Calculate the equivalent mass with respect to displacement x. (2) Assume there is a tangent force between the block and the roller, layout two dynamics equations using
F∑ = m!!x and F ⋅r∑ = J !!θ , and finally combine all the equatoin into one using x = rθ .
m!!x = −k x − y( )− b !x − !y( )−T J !!θ = Tr x = rθ ⇒
T = J !!θ r = J !!x r2
m!!x = −k x − y( )− b !x − !y( )− J r2 !!x
m + J r2
⎛ ⎝⎜
⎞ ⎠⎟ !!x + b !x + kx = ky + b!y
y = 2sin 5t( ) !y = 10cos 5t( ) ⇒
m + J r2
⎛ ⎝⎜
⎞ ⎠⎟ !!x + b !x + kx = 2k sin 5t( ) +10bcos 5t( )
Solu:on
m J r
T T b !x − !y( )
k x − y( )
Midterm 2 & Solutions.pdf
Midterm II & Solu/ons
m1 m2
F(t)(1) (60%)
Given the above mass-and-spring system. F(t) = sin(3t) is applied to mass m1. The system has zero initial conditions, x1(0)= x2 (0) = !x1(0) = !x2 (0) = 0
(a) Derive the dynamics equations for displacements x1 and x2, using the following numbers:
m1 = 1, m2 = 1, k1 = 2 − 2, k2 = 2, k3 = 3− 2, F(t) = sin(3t). Identify matrices m[ ] and k[ ] (b) Calculate the natural frequencies ω1 and ω 2. (c) Calculate the eigenvectors X (1) and X (2)of the system. Make sure to normalize the eigenvetors with respect to m[ ]. (d) Convert the dynamics equations into equations of generalized displacement q1 and q2, where x[ ] = X[ ] q[ ].
(e) Solve q1 and q2
( f ) Solve x1 and x2
Hint: You need this equation: sin(x)sin(y) = 1 2
cos(x − y)− cos(x + y)[ ]
x2x1
k1 k2 k3
(a)
!!x1 + 2x1 − 2x2 = sin(3t)
!!x2 − 2x1 + 3x2 = 0 ⇒ [m]= 1 0
0 1 ⎡
⎣ ⎢
⎤
⎦ ⎥, [k]= 2 − 2
− 2 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
(b)
−ω 2 1 0 0 1
⎡
⎣ ⎢
⎤
⎦ ⎥X + 2 − 2
− 2 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ X = 0 ⇒ 2 −ω 2 − 2
− 2 3−ω 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ X = 0
ω 4 − 5ω 2 + 4 = 0 ⇒ω 2 = 1 or 4 ⇒ω1 = 1, ω 2 = 2
(c)
ω1 = 1⇒ 1 − 2 − 2 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
X1 (1)
X2 (1)
⎡
⎣ ⎢
⎤
⎦ ⎥ = 0; let X1
(1) = a ⇒ X (1) = a
a / 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
Let X (1)T m[ ]X (1) = 1⇒ a2 1 1 / 2⎡ ⎣⎢
⎤ ⎦⎥
1 0 0 1
⎡
⎣ ⎢
⎤
⎦ ⎥
1 1/ 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ = 1⇒ a2 ⋅ 3
2 = 1⇒ a = 2
3 ⇒ X (1) = 2 / 3
1 / 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
ω 2 = 2 ⇒ −2 − 2 − 2 −1
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
X1 (2)
X2 (2)
⎡
⎣ ⎢
⎤
⎦ ⎥ = 0; let X1
(2) = b ⇒ X (2) = b
− 2b
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
Let X (2)T m[ ]X (2) = 1⇒ b2 1 − 2⎡ ⎣⎢
⎤ ⎦⎥
1 0 0 1
⎡
⎣ ⎢
⎤
⎦ ⎥
1 − 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ = 1⇒ b2 ⋅3= 1⇒ b = 1
3 ⇒ X (2) = 1/ 3
− 2 / 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
(d)
X[ ] = X (1) X (2)⎡ ⎣⎢
⎤ ⎦⎥ =
2 / 3 1 / 3 1 / 3 − 2 / 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ ,
Q1
Q2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ = X[ ]T F[ ] = 2 / 3 1 / 3
1 / 3 − 2 / 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
sin(3t) 0
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ =
2 / 3sin(3t)
1 / 3sin(3t)
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⇒General displacement equation: !!qj +ω j 2qj =Qj , j = 1,2
⇒ !!q1 + q1 = 2 / 3sin(3t), !!q2 + 4q2 = 1/ 3sin(3t)
(e)
1 ω j
sin(ωτ )sin(ω j (t −τ ))dτ o
t
∫ = 1 2ω j
cos(ωτ −ω jt +ω jτ ))− cos(ωτ +ω jt −ω jτ )⎡⎣ ⎤⎦dτ o
t
∫
= 1 2ω j
sin((ω +ω j )τ −ω jt)) ω +ω j
− sin((ω −ω j )τ +ω jt)
ω −ω j
⎡
⎣ ⎢
⎤
⎦ ⎥
t 0
= 1 2ω j
sin(ωt) ω +ω j
− sin(ωt) ω −ω j
⎡
⎣ ⎢
⎤
⎦ ⎥ −
sin(−ω jt) ω +ω j
− sin(ω jt) ω −ω j
⎡
⎣ ⎢
⎤
⎦ ⎥
⎧ ⎨ ⎪
⎩⎪
⎫ ⎬ ⎪
⎭⎪
= 1 2ω j
−2ω j sin(ωt) ω 2 −ω j
2 + 2ω sin(ω jt) ω 2 −ω j
2
⎧ ⎨ ⎪
⎩⎪
⎫ ⎬ ⎪
⎭⎪ = 1 ω 2 −ω j
2 ω ω j
sin(ω jt)− sin(ωt) ⎛
⎝⎜ ⎞
⎠⎟
⇒
!!q1 + q1 = 2 / 3sin(3t)→ω = 3, ω1 = 1
q1 = 1 ω1
2 / 3sin(3τ ) o
t
∫ sin(ω1(t −τ ))dτ = 2 / 3 1 9 −1
3 1
sin(t)− sin(3t)⎛ ⎝⎜
⎞ ⎠⎟ =
2 / 3 8
3sin(t)− sin(3t)( )
!!q2 + 4q2 = 1/ 3sin(3t)→ ω = 3, ω 2 = 2
q2 = 1 ω 2
1 / 3sin(3τ ) o
t
∫ sin(ω 2 (t −τ ))dτ = 1/ 3 1 9 − 4
3 2
sin(2t)− sin(3t)⎛ ⎝⎜
⎞ ⎠⎟ =
1/ 3 5
3 2
sin(t)− sin(3t)⎛ ⎝⎜
⎞ ⎠⎟
( f )
x1(t) x2 (t)
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ = X[ ] q1(t)
q2 (t)
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ = 2 / 3 1 / 3
1 / 3 − 2 / 3
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
2 / 3 8
3sin(t)− sin(3t)( )
1/ 3 5
3 2 sin(t)− sin(3t)⎛
⎝⎜ ⎞ ⎠⎟
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥
= 2 / 3 ⋅ 2 / 3
8 3sin(t)− sin(3t)( ) + 1/ 3 ⋅ 1/ 3
5 3 2 sin(t)− sin(3t)⎛
⎝⎜ ⎞ ⎠⎟
1/ 3 ⋅ 2 / 3 8
3sin(t)− sin(3t)( )− 2 / 3 ⋅ 1/ 3 5
3 2 sin(t)− sin(3t)⎛
⎝⎜ ⎞ ⎠⎟
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥
=
7 20 sin(t)− 3
20 sin(3t)
2 40 sin(t)+ 2
40 sin(3t)
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥
(2) (40%)
J0
m, J
o
See above. A round table has a slot, inside the slot is a block of mass m and rotary inertia J. The mass is coupled to the table by a spring of spring-constant k, and it slides along the slot without friction. The spring is fully relax when the block is at the center of the table. The round table has center of mass O and rotary inertia JO with respect to O. External force f is applied to the block in the y direction. In response, the table rotates around O by angle θ and the block slides off-center by x.
f
x
θ
(a) Calculate the Kinetic energy T of the system. (b) Calculate the potential energy V of the system. (c) Using Lagrange formula with x and θ as the generalized displacements, drive the dynamics equations of the system when there is no external froce (free vibration), . (d) Continue (c), drive the dynamics equations of the system when froce f is applied (forced motion).
y
z
(a)
T = 1 2 J0 !θ 2 + 1
2 m xcosθ( )′2 + xsinθ( )′2⎡ ⎣⎢
⎤ ⎦⎥ + 1
2 J !θ 2
= 1 2 J0 + J( ) !θ 2 + 1
2 m !xcosθ − xsinθ ⋅ !θ( )2
+ !xsinθ + xcosθ ⋅ !θ( )2⎡ ⎣
⎤ ⎦
= 1 2 J0 + J( ) !θ 2 + 1
2 m !x2 cos2θ − 2x!x !θ sinθ cosθ + x2 sin2θ ⋅ !θ 2 + !x2 sin2θ + 2x!x !θ sinθ cosθ + x2 cos2θ ⋅ !θ 2( )
= 1 2 J0 + J( ) !θ 2 + 1
2 m !x2 + x2 !θ 2( )
(b)
V = 1 2 kx2
(c) d dt
∂T ∂ !x
⎛ ⎝⎜
⎞ ⎠⎟ −
∂T ∂x
+ ∂V ∂x
= 0
d dt
∂T ∂ !θ
⎛ ⎝⎜
⎞ ⎠⎟ −
∂T ∂θ
+ ∂V ∂θ
= 0
⇒
d dt
∂ ∂ !x
1 2 J0 + J( ) !θ 2 + 1
2 m !x2 + x2 !θ 2( )⎛
⎝⎜ ⎞ ⎠⎟
⎡ ⎣⎢
⎤ ⎦⎥ − ∂ ∂x
1 2 J0 + J( ) !θ 2 + 1
2 m !x2 + x2 !θ 2( )⎛
⎝⎜ ⎞ ⎠⎟ +
∂ ∂x
1 2 kx2⎛
⎝⎜ ⎞ ⎠⎟ = 0
d dt
∂ ∂ !θ
1 2 J0 + J( ) !θ 2 + 1
2 m !x2 + x2 !θ 2( )⎛
⎝⎜ ⎞ ⎠⎟
⎡ ⎣⎢
⎤ ⎦⎥ − ∂ ∂θ
1 2 J0 + J( ) !θ 2 + 1
2 m !x2 + x2 !θ 2( )⎛
⎝⎜ ⎞ ⎠⎟ +
∂ ∂θ
1 2 kx2⎛
⎝⎜ ⎞ ⎠⎟ = 0
⇒
m!!x −mx !θ 2 + kx = 0
J0 + J +mx 2( ) !!θ + 2mx!x !θ = 0
(d) External force f causes a displacement of xsinθ along the f 's direction
d dt
∂T ∂ !x
⎛ ⎝⎜
⎞ ⎠⎟ −
∂T ∂x
+ ∂V ∂x
= f ⋅ ∂ xsinθ( )
∂x d dt
∂T ∂ !θ
⎛ ⎝⎜
⎞ ⎠⎟ −
∂T ∂θ
+ ∂V ∂θ
= f ⋅ ∂ xsinθ( )
∂θ ⇒
m!!x −mx !θ 2 + kx = f ⋅sinθ
J0 + J +mx 2( ) !!θ + 2mx!x !θ = f ⋅ xcosθ
