week 7- defleksi beban merata n puntiran
TRANSCRIPT
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Mekanika Kekuatan Material
Defleksi Batang Dengan Beban
Terdistribusi Merata
Dr. Arhami, S.T, M.T,
JTM – UNSYIAH
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w
Kurva Defleksi
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Example:
Misalkan suatu konstruksi batang ditumpu sederhana ABdengan panjang L mendapat beban terdistribusi merata
sebesar w per satuan panjang, seperti yang ditunjukkan
pada Gambar 1.a.
Tentukan: a). Persamaan kurva defleksi
b). Defleksi maksimum max
c). Sudut kemiringan A dan B Pada tumpuan
A dan B. (Kekakuan fleksural = EI)
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wL
DBB
R A
R B
wx
x Mx
X
R B
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Solution:
Dari geometri gambar dan penjumlahan momen bisadiketahui reaksi pada A:
0 A
M
0)()2
( L
R LwL B
)()
2
( L R L
wL B
B R L
L
wL)
2(
2
wL R B
0
y F
0 B A
RwL R
02
wLwL R A
2
wLwL R A
2
wL R A
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Misalkan suatu penampang X pada jarak x dari B. Kita bisacari momen bending pada penampang ini:
22
2wxwLx M X
0 X M
0)2()( X B M xwx x R
)2
()( xwx x R M B X
(1)
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Dengan demikian, momen bending pada satu titik X :
Dengan mengintegrasikan persamaan ini satu kali, maka akan
diperoleh:
X M dx
yd EI
2
2
22
2
2
2 wxwLx
dx
yd EI
1
32
64C
wxwLx
dx
dy EI
(2)
(3)
dimana C1 adalah konstanta integrasi pertama.
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Kondisi batas:
Kita tahu bahwa x = L / 2, maka dy/dx = 0. Substitusikan harga-harga kondisi batas ini ke persamaan
(3) di atas:
1
32
64C
wxwLx
dx
dy EI
1
32
26240 C
Lw LwL
1
33
48160 C
wLwL
24
3
1
wLC
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Substitusikan harga C1 ke persamaan (3):
1
32
64C
wxwLx
dx
dy EI
2464
332
wLwxwLxdxdy EI
Kemiringan maksimum akan terjadi pada A dan B. Jadi kemiringan
maksimum, substitusikan x = 0 ke persamaan (4).
2464
1 332 wLwxwLx
EI dx
dy x
(4)
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24641
332
wLwxwLx EI dx
dy x
246
0
4
01 332
wLwwL
EI B EI
wL
B 24
3
Tanda negatif artinya tangen A dengan sudut AB adalah negatif
atau berlawanan jarum jam.
Karena simetri maka:
EI
wL A
24
3
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• Dengan mengintegrasikan persamaan (3) di atas sekali
lagi, maka diperoleh persamaan defleksi batang:
21
43
2412C xC wxwLx y EI
2
343
242412
C xwLwxwLx
y EI
Kondisi batas:
Kita tahu bahwa pada x = 0 maka y = 0. Dengan mensubstitusikan
harga-harga ini ke persamaan (5), kita peroleh C2 = 0.
242412
343 xwLwxwLx y EI
(5)
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Persamaan defleksi pada sembarang bagian pada batang AB.
2424121
343
xwLwxwLx EI
y
Defleksi maksimum terdapat pada titik tengah batang. Dengan
mensubstitusikan harga x = L/2 ke persamaan (6) defleksimaksimal:
(6)
224224212
1 343
LwL Lw LwL
EI
y
4838496
1 444
)2/max(
wLwLwL
EI y L x
EI
wL y
384
5 4
max
Tanda negatif menunjukkan defleksi mempunyai arah ke bawah
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:
EI
wL y
384
5 4
max
EI
wL
384
5 4
L= 6 m = 6 x 103 mm
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Mekanika Kekuatan Material
PUNTIRAN
(Torsion)
Dr. Arhami, S.T, M.T,
JTM – UNSYIAH
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Objective:
Setelah mengikuti materi kuliah puntiran ini, mahasiswa
diharapkan;
Dapat memahami prinsip-prinsip puntiran.
Mampu memecahkan persoalan-persoalan elemen
mesin yang menerima beban puntir.
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Many machine parts are loaded in torsion, either totransmit power (like a driveshaft or an axle shaft in a
vehicle) or to support a dynamic load (like a coil spring
or a torsion bar).
Power transmission parts are typically circular solid
shafts or circular hollow shafts because these shapes are
easy to manufacture and balance, and because the
outermost material carries most of the stress.
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Torsi
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Poros yang mengalami torsi.
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Deformations of a circular bar in
pure torsion.
Since every cross section of the bar is identical, and
since every cross section is subjected to the same
internal torque T, we say that the bar is in pure torsion.
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Area a
r
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Shear Strains at the Outer Surface
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The magnitude of the shear strain at the outer surface ofthe bar, denoted max
where max is measured in radians, bb’ is the distance through
which point b moves, and ab is the length of the element
(equal to dx). With r denoting the radius of the bar, we can
express the distance bb’ as rd , where d also is measuredin radians. Thus, the preceding equation becomes
ab
bb'
max
dx
rd max
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In the special case of pure torsion, the rate of twist isequal to the total angle of twist divided by the length L,
that is, = /L. Therefore, for pure torsion only, we obtain
For interior elements with an interior cylinder of radius r
are also in pure shear with the corresponding shear
strains given by the equation
L
r
r max
maxmaxr 6
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Outer Surface Inner Surface
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Circular Tubes
in which r 1 and r 2 arethe inner and outer radii,
respectively, of the tube.
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The magnitudes of the shear stresses can be determined
from the strains by using the stress-strain relation for the
material of the bar. If the material is linearly elastic, we
can use Hooke’s law in shear
in which G is the shear modulus of elasticity and is the
shear strain in radians. Combining this equation with the
equations for the shear strains (Eq. 4 and 6), we get;
G
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in which max is the shear stress at the outer surface of the
bar (radius r), is the shear stress at an interior point (radius
r), and is the rate of twist. (In these equations, has
units of radians per unit of length.)
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The Torsion Formula
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To determine this resultant, we consider an element of
area dA located at radial distance from the axis of the
bar (Fig. 3-9). The shear force acting on this elementis equal to dA, where is the shear stress at radius r.
The resultant moment (equal to the torque T ) is the
summation over the entire cross-sectional area of all
such elemental moments:
IP is the polar moment of inertia of the circular cross section.
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For a circle of radius r and diameter d, the polar
moment of inertia is
An expression for the maximum shear stress can be
obtained by rearranging Torsion equation, as follows:
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Typical units used with the torsion formula are as
follows.
In SI, the torque T is usually expressed in newton meters
(Nm), the radius r in meters (m), the polar moment of
inertia IP in meters to the fourth power (m4
), and the shearstress in pascals (Pa).
If USCS units are used, T is often expressed in pound-feet
(lb-ft) or pound-inches (lb-in.), r in inches (in.), IP in inches
to the fourth power (in.4
), and in pounds per square inch(psi).
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Substituting r = d/2 and IP = d 4/32 into the torsion
formula, we get the following equation for the maximum
stress:
The shear stress at distance from the center of the bar is
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Reference
James M. Gere. 2004. Mechanics of Material. 6th Edition.
Thomson Learning, Inc.
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