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MEM202 Engineering Mechanics - Statics MEM
Chapter 5 Distributed Forces:Centroids and Center of Gravity
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MEM202 Engineering Mechanics - Statics MEM
1Fr
2Fr
1x2x
21 FFRrrr
+=
R
x
221121 xFxFMMC +=+=rrr
Simplify
Centroid – An Introduction
RFx i
rr todue Moment todueMoment : gdetermininfor Critirion =
( )xFxRxFM iii ∑∑∑ ===
forces of Sumforces ofmoment of Sum
===∑∑∑
FxF
RM
x iii
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MEM202 Engineering Mechanics - Statics MEM
Centroid – An Introduction
( ) ( )∫∑ ==L
dxxfdRR0
( ) ( ) ( )∫∑∑ =⋅==L
dxxxfdRxdMC0
( ) ( )
( )( )
( )( )
( )xw
xfOxf
dxxf
dxxxfd
dxxfddxxxfC
L
L
LL
of controid
of (area)amount Totalabout ofMoment
0
0
00
=
==⇒
⋅==
∫∫
∫∫
( )∫⋅=⋅=L
dxxfdRdC0
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MEM202 Engineering Mechanics - Statics MEM
5.2 Center of Gravity and Center of MassCenter of Gravity
WdrMdrrr
×=
( )∫∫ ×==VV
WdrMdMrrrrx
y
y
z
xz
V
rr
dV
Wdr
O
WrM G
rrr×=x
y
Gy
z
GxGz
Χ
Grr
Wr
O∫∫
=
=
V
V
dWW
dVV
( )∫ ×=×VG WdrWr
rrrr
WW
dWdW
WW
dWdW
WW
dWdW zzyyxx === :NOTE
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MEM202 Engineering Mechanics - Statics MEM
5.2 Center of Gravity and Center of MassCenter of Gravity
( ) ( ) ( )kWyWxjWxWziWzWyWr xGyGzGxGyGzGG
rrrrr −+−+−=×
( ) ( ) ( ) ( )[ ]∫∫ −+−+−=×V xyzxyzV
kydWxdWjxdWzdWizdWydWWdrrrrrr
( )( )( )( )∫
∫∫
∫−=−
−=−
−=−
⇒×=×
V xyxGyG
V zxzGxG
V yzyGzG
VG
ydWxdWWyWx
xdWzdWWxWz
zdWydWWzWy
WdrWrrrrr
etc. , Recall ∫∫∫ =⎟⎠⎞
⎜⎝⎛=⇒=
Vz
Vz
V zzz ydW
WWdW
WWyydW
WW
dWdW
( ) etc ,∫∫∫ −=−=−V
y
Vz
V yzyGzG zdWWW
ydWWWzdWydWWzWy
∫∫∫ ===VGVGVG zdW
WzydW
WyxdW
Wx 111
∫=VG dWr
Wr rr 1
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MEM202 Engineering Mechanics - Statics MEM
5.2 Center of Gravity and Center of MassCenter of Mass
x
y
y
z
xz
V
rr
dm
O
x
y
Gy
z
GxGz
Χ
Grr
m
O∫∫
=
=
V
V
dmm
dVV
∫∫∫
∫∫∫
∫∫∫ ======
V
VVG
V
VVG
V
VVG dm
zdm
m
zdmz
dm
ydm
m
ydmy
dm
xdm
m
xdmx
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MEM202 Engineering Mechanics - Statics MEM
5.3 Centroids of Volumes, Areas, and Lines
x
y
y
z
dA
x
z
x
y
y
z
dV
x
z
x
y
y
z
dL
x
z
V AL
Volume Area Line
C CC
VzdVdVzdVz
VydVdVydVy
VxdVdVxdVx
VVVc
VVVc
VVVc
∫∫∫∫∫∫∫∫∫
==
==
==
AzdAdAzdAz
AydAdAydAy
AxdAdAxdAx
AAAc
AAAc
AAAc
∫∫∫∫∫∫∫∫∫
==
==
==
LzdLdLzdLz
LydLdLydLy
LxdLdLxdLx
LLLc
LLLc
LLLc
∫∫∫∫∫∫∫∫∫
==
==
==
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MEM202 Engineering Mechanics - Statics MEM
5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Rectangular Area
b
h
x
ydA
x
ybhdydxdAA
h b
A=⎟
⎠⎞⎜
⎝⎛== ∫ ∫∫ 0 `0
222111 2
0
2
0 0
bbhhbdyb
bhdyxdx
bhxdA
AAM
xhh b
A
yc ===⎟
⎠⎞⎜
⎝⎛=== ∫∫ ∫∫
22111 2
00 0
hbh
bhbydybh
dyydxbh
ydAAA
Myhh b
Ax
c ===⎟⎠⎞⎜
⎝⎛=== ∫∫ ∫∫
b
hy
dA
x
y
( )
2
1
1
0
h
bdyybh
ydAA
y
h
Ac
=
=
=
∫
∫ b
hdA
x
y
( )
2
1
1
0
b
hdxxbh
xdAA
x
b
Ac
=
=
=
∫
∫x
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MEM202 Engineering Mechanics - Statics MEM
5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Quarter Circle
- Double integral in rectangular coordinates
r
x
y
dA
x
y
dy
dx
22
22
222
xry
yrx
ryx
−=
−=
=+
36222
3
0
32
0
22
00
2
0 0
22
22
rxxrdxxrdxy
dxydyydAM
rrr
xr
r xr
Ax
=⎥⎦
⎤⎢⎣
⎡−=
−=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛⎥⎦
⎤⎢⎣
⎡=
⎟⎠
⎞⎜⎝
⎛==
∫∫
∫ ∫∫−
−
4 2rdAA
A
π== ∫
ππ 34
4 3
2
3 rr
rA
ydA
AMy Ax
c ==== ∫
π34r
A
xdA
AM
x Ayc === ∫
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MEM202 Engineering Mechanics - Statics MEM
5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Quarter Circle
- Single integral using a horizontal strip
( )33
3
0
2322
0
22 ryrdyyryydAMr
r
Ax =⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−=−== ∫∫
ππ 34
4 3
2
3 rr
rA
ydA
AMy Ax
c ==== ∫
r
22 yrx −=
y
dA
x
y
dy
22
22
222
xry
yrx
ryx
−=
−=
=+
( )
3622
222
3
0
32
0
22
2222
22
ryyrdyyrdMM
dyyrdyyryrdAxdM
rr
A yy
y
=⎥⎦
⎤⎢⎣
⎡−=
−==
−=−⎟
⎟⎠
⎞⎜⎜⎝
⎛ −==
∫∫
π34r
AM
x yc ==
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MEM202 Engineering Mechanics - Statics MEM
5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Quarter Circle
- Double integral using polar coordinates
( )( )
[ ]
33
cossin
sin
3
0
3
0
2
0
20
2
0
2
0
2
0
2
0
rd
ddd
ddydAM
rr
rr
r
Ax
=⎥⎦
⎤⎢⎣
⎡==
−=⎟⎠⎞⎜
⎝⎛=
==
∫
∫∫ ∫
∫ ∫∫
ρρρ
ρθρρθθρ
ρθρθρ
ππ
π
y
dA
x
y
dy
θ
θd
ρ
ρd
x
ππ 34
4 3
2
3 rr
rA
My xc ===
π34r
AM
x yc ==
θρθρ sincos == yx
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MEM202 Engineering Mechanics - Statics MEM
5.4 Centroids of Composite Bodiesy
x
2cx
2cy
1cx
1cy
×× 1C
2C2A1A
∑∑=
++
==
+=
+=
i
cicc
c
yc
C
ccy
AxA
AAxAxA
AM
x
AAA
xAxAM
i
21
21
21
21
21
21
∑∑=
++
==
+=
i
cicc
c
xc
ccx
AyA
AAyAyA
AMy
yAyAM
i
21
21
21
21
21
ExamplePart Ai (in2) xci (in) My (in3) yci (in) Mx (in3)
1 50.0 6.67 333.3 3.33 166.72 100.0 15.0 1,500.0 5.0 500.0Σ 150.0 1,833.3 666.7
in 10
in 10 in 10
2A1A
in 44.4150
7.666in 22.12150
3.833,1======
∑∑
∑∑
AM
yA
Mx x
cy
cin 0.5
in 0.15
in 33.3
in 67.6
1
2
1
1
=
=
=
=
c
c
c
c
y
x
y
x
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MEM202 Engineering Mechanics - Statics MEM
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
( )xfw =
O
Find a concentrated force R that is equivalent to the distributed load w.
( )
( )dxxfxdxxwxdRdMM
dxxfwdxdRR
OO ∫∫∫∫∫∫∫
====
===
x dx
wdxdR =
O
cx
R
O
( )( )∫
∫==dxxf
dxxxfR
Mx Oc
cO RxM =
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MEM202 Engineering Mechanics - Statics MEM
5.6 Distributed Loads on Beams
O
O
O
m 6 m 51 m 9
N/m 100=w
N 300 N 500,1 N 450
m 42m 3.51
m 4
( )( )
( ) m 4632
32
N 300100621
21
1
1111
1===
====
bx
hbAF
c
( )( )
( ) m 5.1315216
216
N 500,110015
2
2222
2=+=+=
====
bx
hbAF
c
( )( )
( ) m 2493121
3121
N 450100921
21
3
3333
3=+=+=
====
bx
hbAF
c
Example
m 4.31
N ,2502
N 250,2321 =++= FFFR
( )( ) ( )( ) ( )( )m-N 250,32
244505.13500,14300321 321
=++=
++= cccO xFxFxFM
m 3.14== RMx Oc
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MEM202 Engineering Mechanics - Statics MEM
5.7 Forces on Submerged Surfaces
psdVpdAdR ==
psps VdVpdAdRR ==== ∫∫∫
pscpsx VxxdVxpdAxdRRdps
==== ∫∫∫
pscpsy VyydVypdAydRRdps
==== ∫∫∫
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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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