regangan bidang analisa struktur lanjutan
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Kuliah ANALISA STRUKTUR LANJUTAN Tegangan Bidang (Plane Stress)y zy z yz zx yx xy x Z Y X xy = yx xz = xz yz = zy This state of stress in called triaxial
xz
General three dimensional state of stress. In plane stress, two faces of the cubic element are free of stress. It is conventional to assume the stress-free faces to be the front and the back faces of the cubic element. There for z = zx = zy = 0 y yz = xy xy
.yx
x
An examples of plane stress we can see as follows F1 F2
P P MT
EQUATIONS FOR THE TRANSFORMATION
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
OF PLANE Stress
Y dA.Cos
xy x
X
x xy
dA
yx dA.Sin y
Equilibrium equation :
X = 0 Y = 0
x . dA - x (dA . Cos ) Cos - xy (dA . Cos ) . Sin - y (dA . Sin ) Sin - xy (dA . Sin ) Cos = 0
xy . dA + x (dA . Cos ) . Sin - xy (dA . Cos ) . Cos - y (dA . Sin ) Cos + xy (dA . Sin ) Sin = 0
x = x . Cos2 + y . Sin2 + 2 xy . Sin . Cos xy = - (x - y) (Sin . Cos ) + xy (Cos2 - Sin2 )Using the trigonometri relations : Cos2 = 1 + Cos 2 1 Cos 2 : Sin2 = 2 2 Sin 2 = 2 Sin . Cos Cos 2 = Cos2 - Sin2 we have : x =
x + y + x - y . Cos 2 + xy . Sin 22 2
+ xy = - x 2 y . Sin 2 + xy . Cos 2
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Prinsipal Stresses
x =
x + y x - y . Cos 2 + xy . Sin 2 + 2 2 x + y x - y . Cos 2 + . Sin 2 ) = 0 xy + 2 2
dx d = d d
- (x - y) . Sin 2 + 2 xy . Cos 2 = 0
It can be solved for as 2 xy tg 2p = - x y the angle p call be called principal planes. we can also interpretive with this figure
xy 2 p x - y 2 2 xy x - y
tg 2 p =
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
MAXIMUM SHEAR STRESS
xy = -
x - y . Sin 2 + . Cos 2 xy 2
d xy d = d d
- x - y . Sin 2 + xy . Cos 2) = 0 2
- (x - y) Cos 2 - 2 xy . Sin 2 = 0 tg 2s = - x - y 2 xy s the angle which define the maximum shear stress.
R 2 s
-
x - y 2
xy
tg 2s = -
x - y 2 xy
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Mohrs circle for Plane stress Analysis
x - x + y = x - y . Cos 2 + xy . Sin 2 2 2 xy = x - y . Sin 2 + . Cos 2 xy22 2
x - x + y2
+ (xy - 0)2 =
x - y2
+ (xy)2
(x - ave)2 + (xy - 0)2 = R2 (x - a)2 + (y - b)2 = r2
(y, xy)
ave
(x, - xy)
a = ave b=0 r=R=
x - y2
2
+ (xy)2
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
xy
xy
+How to draw the mohr circle ?
-
y
x
x xy y
Step : 1. Write down the two stress coordinates X,Y to be plotted. X = (x, xy) ; Y = (y, - xy) 2. Determine the center C of the circle from C = ave =
x + y2
Where C is always on the horizontal axis
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
3. Sketch the circle using points X, Y, and C, as shows in figure bellow.
X
S X 2p
2s P2 C D
P1
Y (y, - xy)
4. Determine the radius R of the circle by observing from circle that2 2 R = CD + XD
Tg 2 s =
CD XC
Tg 2 p = XD CD
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Contoh soal :
9.000
N Cm2
24.000
N x Cm2
xy9.000
24.000 N Cm2
N Cm2
yx
max maxCara I :
?
max :tg 2p =
18.000 2 . 9000 x - y = 24.000 - 0 = 24.000 = 0,75 p = 18,4
2 yx
2p = 36,80
max =p
x + y + x - y . Cos 2 + xy . Sin 22 2 18,40 24.000 + 0 24.000 + 0 . Cos 36,8 + 9.000 . Sin 36.8 + 2 2N
max =
= 27.000 /cm2
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
max :tg 2 s = -
x - y - 24.000 - 0 = = - 1,33 2 . 9.000 2xy
2 s = 53.10
max = -
x - y . Sin 2 s + . Cos 2 s xy2
0 0 = - 24.000 - 0 . Sin 53,1 + 9.000 . Cos 53,1 2
= 15.000 /cm2 Cara II : Lingkaran Mohr
N
(N/cm2)
O
S 2 s 2 p C D
max.
X (24.000, 9.000)
(N/cm2) max.
Y (0,9.000)
xX= (24.000,
xy9.000)
yY= (0,
- xy-9.000)
Analisa Struktur Lanjutan hal.
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Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
R = CD2 + XD2 = 122 + 92 = 15.000
OC = ave = 24.000 + 0 2 = 12.000
max. = 15 N 2Cm
max. = OC + R= 12 + 15 = 27 tg 2s = CD = XD 12 9
2s = 53.10 tg 2p = XD = CD 2p = 36.80 9 12
Analisa Struktur Lanjutan hal. 10
Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
PLANE STRAIN (Regangan Bidang) Y 2 x Y y 1 X x . dx y . dy dy X 3
Y
1 dx
2
xy
3 X
x = x dx x = x . dx d = 1 + 2 + 3
y = y dy y = y . d y
d = x . dx . Cos + y . dy . Sin + xy dy d d d x = dL = x . dx . Cos + y . y . Sin + xy . y . Cos d d d dx = Cos d dy = Sin d x = x . Cos2 + y . Sin2 + xy . Sin . Cos Using trigonometric relation
x =
x + y x - y . Cos + xy . Sin 2 + 2 2 2
Analisa Struktur Lanjutan hal. 11
Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
1 =
x . dx . Sin d
;
2 =
y . dy . Cos d
;
1 =
xy dy . Sin d
1 = x . Cos . Sin ; 2 = y . Sin . Cos ; 3 = xy . Sin2
= - 1 + 2 - 3 + 900 + 900 = - (x y) . Sin ( + 900 ) . Cos ( + 900 ) - xy . Sin2 ( + 900 ) Using the Trigonometric : Sin ( + 900 ) = Cos Cos ( + 900 ) = - Sin + 900 = (x y) . Cos . Sin - xy . Cos2 xy = - + 900 = - 2 (x y) Sin . Cos + xy . (Cos2 - Sin2) Using the Trigonometric identities : Cos2 = Sin2 = 1 + Cos2 2 1 - Cos2 2 Sin2 2 d X = - (x y) Sin . Cos - xy . Sin2)
Sin . Cos =
xy = (x y) Sin 2 + xy . Cos 2
Analisa Struktur Lanjutan hal. 12
Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Prinsipal Angle p : xy tg 2p = x y
The angle assosiated with the maximum shen stain is tg 2s = x y xy
Morhs circle for plane strain 1. X = (x, xy ) ; Y = (y, 2 x + y 2 xy ) 2 - xy 2
2. C = ave =
3. Gambar titik X, Y dan C
4. R = CD2 + XD2xy/2
(y, xy ) 2
S (ave, max ) 2
D C 2 p
X (y, -
xy ) 2
xy xy 2 XD = x - y = x - y 2 p = CD 2
Analisa Struktur Lanjutan hal. 13
Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
x = 400 ; y = 300 ; xy = 600
xX= (400,
yY= (300,
xy 2 600 2 xy 2 600 2
2
Y (300, 300)
2 p
X (400, - 300) 350 x - y 400 + 300 = 2 2
C = ave = = 350
Analisa Struktur Lanjutan hal. 14
Prof. Dr.Ing. Johannes Tarigan Teknik Sipil USU
Kesimpulan : Plane Stress :
1,2 = x + y 2
2
x - y2
2
+ (xy)
2
max =
x - y2
2 + (xy) = R
max = -
x - y2
2
2 + (xy)
Plane Strain :
+ 1,2 = x 2 y max = 2
2
x - y2 + xy 2
2
+2
xy 2
x - y2
Analisa Struktur Lanjutan hal. 15
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