Assignment strength of materials
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7/25/2019 Assignment strength of materials
1/8
1). Find the centroid area of some geometry shapes below
a).
Find of the composite geometry shape
2
162 100 6200 ; ( , ) (50, 1)A m m m x y= = =
2
2!" 200 #600 ; ( , ) (0, 16)A m m m x y= = =
2
$"0 220 1600 ; ( , ) ( 10,!")A m m m x y= = =
2
! " " 6! ; ( , ) (!,!)A m m m x y= = =2 2 2 2 2
1 2 $ !6200 #600 1600 6! $$$$6A A A A A m m m m m= + + = + + =
1 1 2 2 $ $ ! !6200 50 #600 0 1600 10 6! ! 161"
!.01$$$$6 !16
i iA x A x A x A x A x
x mA A
+ + + = = = = =
1 1 2 2 $ $ ! ! 6200 1 #600 16 1600 !" 6! ! 250!! .52$$$$6 !16
i iA y A y A y A y A y
y mA A
+ + + = = = = =
92810020
80
62
8
40
x
y
m
m
andx y
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b).
Find of the parabole
!!
; ,H
y ax when y H x L aL
= = = =
!!
!
H y
y x x LL H
= =
!
51 5 !! !
1! !0 0 0!0
! ! !
555
Hy
H L HH
L L LH HLdA dxdy y dy y
H HH
= = = = =
!! $212 2
000 0 2$02
15 5 52
! ! 12"12
5 5
yy H LH LHH
Hx dyxdxdyxdA L LH
x y dy LHL HLA HL H
H
= = = = = =
!
5
!#1 0 5 !!
0 0 !5 50! !
5 5 5
! ! #! #
5 5
Hy
H LH
H
Ly dy
ydxdyydA HHy y dy HHL HLA
H H
= = = = = =
%
x
dA
y
H
L
!y ax=
andx y
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c).
Find of the semi&circle
2 2 2 2 2x y R x R y+ = =
2 2
2 2
0 0 02 2
R R y R
dA dxdy R y dy
= =
sin cosLet y R dy R d = =
0 0When y = =
2y R
= =
2 2 2 2 22 2 2
0 0 0
1 cos 22 (1 sin ) cos 2 cos 2
2dA R R d R d R d
+
= = = 2
22
0
sin2( )
2 2
RR
= + =
2 2
$2 2 $0 0
2 2 20
2 ! 1 2 !( ) ( )
2 $ $
2
R R y
RxdxdyxdA R Rx R y dy R
RA R R
= = = = = ?
x
y
xx
dy
2 2 2x y R+ =
2 22dA R y dy=
andx y
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2 2
2 22 2
2 2
0 2
2 2 0
2 1
2
2
R R y
R R yR y
R y
xdxdyxdAx x dy
RA R
= = =
2 2 2 2
2 0
1
( ) 0
R
x R y R y dyR = =
2 2
$
2 2 2 20 0 22 2 20
0
2 ! ! 1( )
$
2
R R yR
RydxdyydAy y R y dy R y
RA R R
= = = =
$
2 2 $22
! !( )
$ $
Ry R R R
R
= =
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2).
1! 2 ! 1! 2 11 $,61! 2 ! 1! $
i iA x
x cmA
+ = = = =
+
1! 2 1 ! 1! # 1# 6,$$1! 2 ! 1! $
i iA yy cmA
+ = = = = +
1 2
2$ $2
, ,
1! 2 1# ! 1! 1#1! 2 ( 1) ! 1! #
12 $ 12 $x x A x A
I I I = + = + + +
!6$56 211",6$
xI cm= =
1 2
2$ $2
y, y,
1! 2 11 ! 1! 111! 2 ( ) ! 1! 2
12 $ 12 $y A A
I I I = + = + + +
!2##6 ##",6
$Y
I cm= =
1 2 1 1 2 2, , 1 2
(0 ) (0 )XY XY A XY A x y x y
I I I A d d A d d= + = + + +
!11 1# 11 1# 22!01! 2 ( ) ( 1) ! 1! 2 (# ) !6,6$ $ $ $ $
XYI cm
= + = =
C
4 cm 10 cm
2 cm
14 cm
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'y carrying ot an inspection on the ohr circle, we get some info in it inclding*
!
min 625,$$!I cm=!
ma+ 2!#2,006I cm=
12 5$,1$
p = o
126,565
p = o
22 126,"p =
o
26$,!$5
p = o
12 p
22
p
!!6,6XYI cm=
!!6,6YX
I cm=
minI
ma+I
!rodct of inertia ( )cm
!oments of inertia ( )cmXI
YI
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-he ne+t step after getting sch these ales we can also draw a principle a+is
throghot the geometry/s center of graity.
$).
a).
$ $
!0,2 2 0,1$$$$12 12
x
bhI m
= = =
$ $
!0,2 2 0,001$$$12 12
x
b hI m= = =
C
126,565
p = o
26$,!$5
p = o
X
Y
x
y
X
Y
rinciple a+is
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0xyI =
b).
2 2cos sin 2 sin cosX x y xy
I I I I = +
2 2 !0.1$$$$ cos 11 0.001$$$ sin 11 0 0,12"52X
I m= + =o o
2 2sin cos 2 sin cosY x y xyI I I I = + +
2 2 !0,1$$$$ sin 11 0, 001$$$cos 11 0 0, 0061$"Y
I m= + + =o o
2 2sin cos sin cos (cos sin )XY x y xyI I I I = +
!sin11 cos11 (0,1$$$$ 0,001$$$) 0 0,02!2XY
I m= + =o o
c).
1
2 2 !0,1$$$$ 2 0,2 $ $,$$$$x x y
I I Ad m= + = + =
1
2 2 !0,001$$$ 2 0,2 2,5 2,501$$y y x
I I Ad m= + = + =
1 1
!0 2 0,2 2,5 $ $x y xy x y
I I Ad d m= + = + =
!).
$ $2 2 2 !6 12 6 12 16 1#2#6
12 12x X y y
bhI I Ad Ad m
= + = + = + =
$ $2 2 !6 12 6 12 0 216
12 12
y Y x x
b hI I Ad Ad m
= + = + = + =
0xy XY x y
I I Ad d= + =
2 2 2 2cos sin 2 (cos sin )x x y xy
I I I I = +
2 2 !1#2#6 cos 20 216 sin 20 0 106!,06!x
I m= + =o o
