WEEK 7- Defleksi Beban Merata n PUNTIRAN

8/18/2019 WEEK 7- Defleksi Beban Merata n PUNTIRAN

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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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WEEK 7- Defleksi Beban Merata n PUNTIRAN

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