MEM202 Engineering Mechanics - Statics
MEM
Chapter 5 Distributed Forces:
Centroids and Center of Gravity
1
MEM
MEM202 Engineering Mechanics - Statics
Centroid – An Introduction
r
F1
r r r
R = F1 + F2
r r
r
C = M 1 + M 2 = F1 x1 + F2 x2
r
F2
3
R
Simplify
x1
x2
x
r
r
Critirion for determining x : Moment due to Fi = Moment due to R
∑M = ∑F x
i
M
∑
x=
R
i
i i
Fx
∑
=
∑F
i i
= Rx = (∑ F )x
Sum of moment of forces
=
Sum of forces
2
MEM
MEM202 Engineering Mechanics - Statics
Centroid – An Introduction
R = ∑ (dR ) = ∫ f ( x )dx
L
0
C = ∑ (dM ) = ∑ ( x ⋅ dR ) = ∫ xf ( x )dx
L
0
C = ∫ xf ( x )dx = d ⋅ ∫ f ( x )dx
L
0
L
0
xf ( x )dx
Moment of f ( x ) about O
∫
⇒d =
=
∫ f (x )dx Total amount (area) of f (x )
L
0
L
0
= controid of w( x )
C = d ⋅ R = d ⋅ ∫ f ( x )dx
L
0
3
MEM
MEM202 Engineering Mechanics - Statics
5.2 Center of Gravity and Center of Mass
Center of Gravity
z
z
V
r
r
r
dW
V
y
z
Χ
V = ∫ dV
dV
r
rG r zG
W
xG
W = ∫ dW
x
V
O
O
y
x
r r
r
dM = r × dW
r
r
r
r
M = ∫ dM = ∫ r × dW
V
V
r
r r
r
rG × W = ∫ r × dW
(
)
V
dWx Wx
NOTE :
=
dW W
(
dWy
dW
=
x
yG
r r r
M = rG × W
)
Wy
W
y
dWz Wz
=
dW W
4
MEM
MEM202 Engineering Mechanics - Statics
5.2 Center of Gravity and Center of Mass
Center of Gravity
r
r
r
r r
rG × W = ( yGWz − zGWy ) i + (zGWx − xGWz ) j + (xGWy − yGWx ) k
r
r
r
r
r
r × dW = ∫ ( ydWz − zdWy ) i + (zdWx − xdWz ) j + (xdW y − ydWx ) k
∫(
V
)
V
[
(
)
]
yGWz − zGWy = ∫ ( ydWz − zdWy )
V
r
r r
r
rG × W = ∫ r × dW ⇒ zGWx − xGWz = ∫ (zdWx − xdWz )
V
V
xGWy − yGWx = ∫ (xdW y − ydWx )
V
W
⎛W ⎞
y ⎜ z ⎟dW = z ∫ ydW , etc.
W V
⎝W ⎠
W
W
yGWz − zGWy = ∫ ( ydWz − zdWy ) = z ∫ ydW − y ∫ zdW , etc
V
W V
W V
Recall
1
xG = ∫ xdW
W V
dWz Wz
=
⇒ ∫ ydWz = ∫
V
V
dW W
1
yG =
W
∫
V
ydW
1
zG = ∫ zdW
W V
r
1 r
rG = ∫ r dW
W V
5
MEM
MEM202 Engineering Mechanics - Statics
5.2 Center of Gravity and Center of Mass
Center of Mass
z
z
V
r
r
m
V
y
r
rG
m = ∫ dm
x
z
Χ
V = ∫ dV
dm
xG
V
O
O
y
x
xG
y
zG
yG
x
xdm ∫ xdm
∫
=
=
m
∫ dm
V
V
V
yG
∫
=
V
ydm
m
ydm
∫
=
∫ dm
V
V
zG
zdm ∫ zdm
∫
=
=
m
∫ dm
V
V
V
6
MEM
MEM202 Engineering Mechanics - Statics
5.3 Centroids of Volumes, Areas, and Lines
z
z
Volume
V
dA
z
y
x
y
yc
zc
V
V
V
V
V
V
V
V
dL
z
y
y
x
y
x
∫ dV = ∫ xdV V
= ∫ ydV ∫ dV = ∫ ydV V
= ∫ zdV ∫ dV = ∫ zdV V
V
C
C
z
xc = ∫ xdV
Line
L
A
dV
C
x
z
Area
xc = ∫ xdA
zc
A
A
A
y
x
∫ dA = ∫ xdA A
= ∫ ydA ∫ dA = ∫ ydA A
= ∫ zdA ∫ dA = ∫ zdA A
A
yc
x
A
A
A
A
A
xc = ∫ xdL
∫ dL = ∫ xdL L
= ∫ ydL ∫ dL = ∫ ydL L
= ∫ zdL ∫ dL = ∫ zdL L
L
yc
zc
L
L
L
L
L
L
L
L
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MEM
MEM202 Engineering Mechanics - Statics
5.3 Centroids of Volumes, Areas, and Lines
Example: Centroid of A Rectangular Area
y
b
⎛
A = ∫ dA = ∫ ⎜ ∫ dx ⎞⎟dy = bh
0 ⎝ 0`
A
⎠
h
b
b2h b
1
1 h⎛ b
1 h b2
⎞
xc =
dy =
= ∫ xdA =
=
⎜ ∫0 xdx ⎟dy =
∫
∫
A
0
0
⎠
A
A
bh ⎝
bh 2
2bh 2
My
x
dA
h
y
x
bh 2 h
Mx 1
1 h⎛ b
1 h
⎞
bydy =
yc =
= ∫ ydA =
=
⎜ ydx ⎟dy =
⎠
A
A A
bh ∫0 ⎝ ∫0
bh ∫0
2bh 2
y
y
1
ydA
∫
A
A
1 h
=
y (bdy )
∫
0
bh
h
=
2
yc =
b
dA
h
y
x
x
1
xdA
∫
A
A
1 b
=
x (hdx )
∫
0
bh
b
=
2
xc =
b
dA
h
x
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MEM202 Engineering Mechanics - Statics
MEM
5.3 Centroids of Volumes, Areas, and Lines
Example: Centroid of A Quarter Circle
- Double integral in rectangular coordinates
⎛ r2 − x2
⎞
M x = ∫ ydA = ∫ ⎜ ∫
ydy ⎟dx
A
0
⎝ 0
⎠
r
x2 + y2 = r2
y
x = r2 − y2
y = r2 − x2
A = ∫ dA =
r
dy
A
dA
y
x
r2 − x2 ⎞
r
⎛
2
2
2
r⎜⎡ y ⎤
r r −x
⎡ r2 x x3 ⎤ r3
⎟
dx = ∫
dx = ⎢
= ∫ ⎜⎢ ⎥
− ⎥ =
⎟
0
0
2
2
2
6 ⎦0 3
⎜ ⎣ ⎦0
⎟
⎣
⎝
⎠
dx
M
yc = x =
A
x
π r2
4
∫
A
ydA
A
r3 3
4r
=
=
π r 2 4 3π
M y ∫AxdA 4r
=
=
xc =
A
A
3π
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MEM
MEM202 Engineering Mechanics - Statics
5.3 Centroids of Volumes, Areas, and Lines
Example: Centroid of A Quarter Circle
- Single integral using a horizontal strip
x +y =r
2
y
2
2
2 32⎤
⎡
r
(
)
−
r
y
r3
2
2
M x = ∫ ydA = ∫ y r − y dy = ⎢−
⎥ =
A
0
3
⎢⎣
⎥⎦0 3
r
2
x = r2 − y2
y = r2 − x2
dy
M
yc = x =
A
r
∫
A
ydA
A
⎛ r2 − y2
x
dM y = dA = ⎜
⎜
2
2
⎝
dA
y
x = r2 − y2
r3 3
4r
=
=
π r 2 4 3π
x
M y = ∫ dM y = ∫
A
r
0
(
)
2
2
⎞ 2
r
y
−
2
⎟ r − y dy =
dy
⎟
2
⎠
r
⎡ r2 y y3 ⎤ r3
r −y
dy = ⎢
− ⎥ =
2
6 ⎦0 3
⎣ 2
2
2
M y 4r
=
xc =
A 3π
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MEM202 Engineering Mechanics - Statics
MEM
5.3 Centroids of Volumes, Areas, and Lines
Example: Centroid of A Quarter Circle
- Double integral using polar coordinates
M x = ∫ ydA = ∫
A
r π 2
∫ (ρ sinθ )(ρdθdρ )
0 0
π 2
r
r
π 2
= ∫ ρ 2 ⎛⎜ ∫ sin θdθ ⎞⎟dρ = ∫ ρ 2 [− cosθ ]0 dρ
0
0
⎠
⎝ 0
y
r
⎡ ρ 3 ⎤ r3
2
= ∫ ρ dρ = ⎢ ⎥ =
0
⎣ 3 ⎦0 3
r
dy
y
dθ
dρ
dA
ρ
Mx
r3 3
4r
=
=
yc =
A π r 2 4 3π
θ
x
x
x = ρ cosθ
M y 4r
=
xc =
A 3π
y = ρ sin θ
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MEM
MEM202 Engineering Mechanics - Statics
5.4 Centroids of Composite Bodies
M y = A1 xc1 + A2 xc2
y
A1
AC = A1 + A2
A2
xc =
× C2
yc2
× C1
x
xc1
xc2
A1
10 in
10 in
A2
10 in
xc1 = 6.67 in
xc2 = 15.0 in
yc1 = 3.33 in
yc1 = 5.0 in
Ac
=
A1 xc1 + A2 xc2
A1 + A2
∑Ax
∑A
=
i ci
i
M x = A1 yc1 + A2 yc2
yc1
Example
My
yc =
M x A1 yc1 + A2 yc2
=
=
Ac
A1 + A2
∑Ay
∑A
i
i
Part
Ai (in2)
xci (in)
My (in3)
yci (in)
Mx (in3)
1
50.0
6.67
333.3
3.33
166.7
2
100.0
15.0
1,500.0
5.0
500.0
Σ
150.0
xc =
∑M
∑A
y
=
1,833.3
= 12.22 in
150
1,833.3
yc =
ci
666.7
∑M
∑A
x
=
666.7
= 4.44 in
150
12
MEM
MEM202 Engineering Mechanics - Statics
5.4 Centroids of Composite Bodies
=
+
+
+ −
Table 5-1, Page 212
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MEM202 Engineering Mechanics - Statics
MEM
5.6 Distributed Loads on Beams
w = f (x )
Find a concentrated force R that is
equivalent to the distributed load w.
O
dR = wdx
O
x
dx
R = ∫ dR = ∫ wdx = ∫ f ( x )dx
M O = ∫ dM O = ∫ xdR = ∫ xwdx = ∫ x f ( x )dx
R
O
M O = Rxc
xc
M O ∫ xf ( x )dx
xc =
=
R
∫ f (x )dx
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MEM202 Engineering Mechanics - Statics
MEM
5.6 Distributed Loads on Beams
1
1
b1h1 = (6)(100) = 300 N
2
2
2
2
xc1 = b1 = (6) = 4 m
3
3
Example
F1 = A1 =
w = 100 N/m
O
6m
300 N
15 m
1,500 N
O
4m
13.5 m
24 m
2,250 N
9m
450 N
F2 = A2 = b2h2 = (15)(100) = 1,500 N
1
1
xc2 = 6 + b2 = 6 + (15) = 13.5 m
2
2
1
1
F3 = A3 = b3h3 = (9 )(100) = 450 N
2
2
1
1
xc3 = 21 + b3 = 21 + (9 ) = 24 m
3
3
R = F1 + F2 + F3 = 2,250 N
M O = F1 xc1 + F2 xc2 + F3 xc3
= (300)(4 ) + (1,500)(13.5) + (450)(24 )
= 32,250 N - m
O
14.3 m
xc = M O R = 14.3 m
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MEM202 Engineering Mechanics - Statics
MEM
5.7 Forces on Submerged Surfaces
dR = pdA = dV ps
R = ∫ dR = ∫ pdA = ∫ dV ps = V ps
Rd x = ∫ xdR = ∫ xpdA = ∫ xdV ps = xc psV ps
Rd y = ∫ ydR = ∫ ypdA = ∫ ydV ps = yc psV ps
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MEM202 Engineering Mechanics - Statics
MEM
5.7 Forces on Submerged Surfaces
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MEM202 Engineering Mechanics - Statics
MEM
5.7 Forces on Submerged Surfaces
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