WEEK 7 - Defleksi Beban Merata N PUNTIRAN [PDF]

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Mekanika Kekuatan Material Defleksi Batang Dengan Beban Terdistribusi Merata



Dr. Arhami, S.T, M.T,



JTM – UNSYIAH



w



Kurva Defleksi



Example: 



Misalkan suatu konstruksi batang ditumpu sederhana AB dengan 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)



DBB wL



RA



RB



X



wx Mx



x



RB



Solution: Dari geometri gambar dan penjumlahan momen bisa diketahui reaksi pada A:



MA



0



L wL ( ) RB ( L) 0 2 L wL ( ) RB ( L) 2 wL L ( ) RB L 2 wL RB 2



Fy



RA



RA RA RA



0



wL RB



0



wL wL 0 2 wL wL 2 wL 2







Misalkan suatu penampang X pada jarak x dari B. Kita bisa cari momen bending pada penampang ini:



MX



0



x RB ( x) wx ( ) M X 0 2 x M X RB ( x) wx ( ) 2 wLx wx 2 MX 2 2



(1)











Dengan demikian, momen bending pada satu titik X:



d2y EI 2 dx



MX



d2y EI 2 dx



wLx 2



wx 2 2



(2)



Dengan mengintegrasikan persamaan ini satu kali, maka akan diperoleh:



dy EI dx



wLx 2 4



wx 3 6



C1



dimana C1 adalah konstanta integrasi pertama.



(3)



Kondisi batas: 







Kita tahu bahwa x = L / 2, maka dy/dx = 0. Substitusikan harga-harga kondisi batas ini ke persamaan (3) di atas:



dy EI dx 0



0



2



3



wLx wx C1 4 6 2 wL L w L 4 2 6 2 wL3 16



wL3 48



C1



3



C1



C1



wL3 24







Substitusikan harga C1 ke persamaan (3):



dy EI dx dy EI dx dy dx



2



wLx 4 2 wLx 4 x



3



wx C1 6 3 3 wx wL 6 24 2 3 1 wLx wx EI 4 6



3



wL 24



(4)



Kemiringan maksimum akan terjadi pada A dan B. Jadi kemiringan maksimum, substitusikan x = 0 ke persamaan (4).



dy dx



x



B



1 wLx 2 EI 4 1 wL 0 EI 4



wx 3 6 2



w0 6



wL3 24 3



wL3 24



B



wL3 24EI



Tanda negatif artinya tangen A dengan sudut AB adalah negatif atau berlawanan jarum jam. 3



Karena simetri maka: A



wL 24EI



• Dengan mengintegrasikan persamaan (3) di atas sekali lagi, maka diperoleh persamaan defleksi batang:



wLx 3 wx 4 EI y 12 24 3 4 wLx wx EI y 12 24



C1 x C2 3



wL x C2 24



(5)



Kondisi batas: Kita tahu bahwa pada x = 0 maka y = 0. Dengan mensubstitusikan harga-harga ini ke persamaan (5), kita peroleh C2 = 0.



EI y



wLx 12



3



4



wx 24



3



wL x 24







Persamaan defleksi pada sembarang bagian pada batang AB.



y



1 wLx 3 EI 12



wx 4 24



wL3 x 24



(6)



Defleksi maksimum terdapat pada titik tengah batang. Dengan mensubstitusikan harga x = L/2 ke persamaan (6) defleksi maksimal:



y



1 wL L EI 12 2



ymax( x



L / 2)



3



w L 24 2



1 wL4 EI 96



4



3



wL 24



wL4 wL4 384 48



L 2 4



ymax



5wL 384EI



Tanda negatif menunjukkan defleksi mempunyai arah ke bawah



:



L= 6 m = 6 x 103 mm



ymax



5wL4 384EI



5wL4 384EI



Mekanika Kekuatan Material PUNTIRAN (Torsion)



Dr. Arhami, S.T, M.T,



JTM – UNSYIAH



Objective: Setelah mengikuti materi kuliah puntiran ini, mahasiswa diharapkan;  Dapat memahami prinsip-prinsip puntiran.  Mampu memecahkan persoalan-persoalan elemen mesin yang menerima beban puntir.







Torsion refers to the twisting of a straight bar when it is loaded by moments (or torques) that tend to produce rotation about the longitudinal axis of the bar.







Many machine parts are loaded in torsion, either to transmit 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.



Torsi



Poros yang mengalami torsi.



Regangan dan Tegangan geser akibat torsi.



=0



=0



Regangan dan Tegangan geser bernilai maksimum pada daerah terluar poros, dan bernilai 0 pada sumbu poros.



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.



Area a



r



Shear Strains at the Outer Surface







The magnitude of the shear strain at the outer surface of the bar, denoted max '



max







bb ab



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 measured in radians. Thus, the preceding equation becomes max



rd dx







We will denote d /dx by the symbol and refer to it as the rate of twist, or the angle of twist per unit length:



d dx 



we can now write the equation for the shear strain at the outer surface



max



rd dx



r 4







In the special case of pure torsion, the rate of twist is equal to the total angle of twist divided by the length L, that is, = /L. Therefore, for pure torsion only, we obtain max







r L



r



For interior elements with an interior cylinder of radius r are also in pure shear with the corresponding shear strains given by the equation max



r



max



6



Outer Surface



Inner Surface



Circular Tubes



in which r1 and r2 are the inner and outer radii, respectively, of the tube.







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



G 



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;







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.)



The Torsion Formula



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 element is 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.







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:



 







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 shear stress 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).







Substituting r = d/2 and IP = d4/32 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



Reference 



James M. Gere. 2004. Mechanics of Material. 6th Edition. Thomson Learning, Inc.