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Mathematical Physing xy,, —— Newsy, , et ., at enerRY stored in the stretched Wire, y, , elastic potenti ¢, , , , , , , , , , Une Mm, e2exld oe are given that in .Om. Awd |, e 2 ay ¥, Sa oxtorim and Y= 2.0% 10 Nims th j, Je 2.0mm = eh 1 2.0x10 ., Longitudinal strain = 7°" 2.0, be Volume of wite = AL = AOX 10™ X2.0= 8.0% 19, , , , stored in the stretched wi, , , , potential energy:, , , , , , , , , , , , , , , , , , , , , , , , , , , , ‘ Elastic 1 ., Ua 5X stress X strain X volumg BG ge 7 VOLES jagt j, 1 Strain = I 0.001 t, =5% (Young's Modulus x strain) L 05m 7802, tg modulus of rubber,, * strain x ., Younk§ . 14 %10°N 7.2, “lo, 7 a Volt o - stress _ 7.4 %10"N fm?, 55 X2.0x 10" x (L0™)? x g.9y 10-6 . Y strain 0.002, U=0.82 Y =3.7%10°N /m?, . 5 se di * or 4: The rectangular cr, :, Example 2. A steel wire of 0.1. em radius is bent in the ample 4 F : a ‘085-8ection, crear are of rains 50 em. Cateualte() the bending yiol® git? Mee the same Toad’ where (Gell he ei, (ii) The maximum stress. Given Y for steel =2x10}2 dynesiemt re) the longer side is vertical. ( the Te side is, . . jc + He}, yr Rohillchane 2o9q 1, The depression at the loaded (or free) end of a ang, . , ., YT , , east ; ntilew, Sol. (i) The bending moment is R? where I is the Beometrieg) etangular cross-section under oe Wis er of, moment of inertia of the cross section of the wire and R is the radi b= “ > We Mg, the circular arc. NS of . Yod', 4 4 i h, b breadth and d is thick: ;, . ‘ _ mr _ 3.14x (0.1) shore Lis length, z 1s thickness of the cantilever rod, For a wire of radius r, 7 = Ts 4 and 24 be the sides (ratio = 1 : 2) of the cross-section. ir rod. Leta, R=50cm | (i) When smaller side is vertical, b=2a,d=«a, 3 3, . YI_ 2x10! x3.14x (0.1)! _ 5 . 3, - AWE _ LAW, «bending moment = z- 0x4 = 3,14 x 10° dyne-cm “ 1 YQo)a® 2 Yat, (ii) In bending of wire, the longitudinal stress on the cross section of | (ii) When the longer side is vertical, b=a, d= 2a, 7, a filament at a distance Z from the neutral surface is 4 -Its value will be bys awe’ _1(awi, : | a 3 Bl vad, maximum on the surface of wire i.e. for Z =r. | Ya(2a)* $\ Yo, =. Visi Yr_ 2x10! x0.1 | om Sy 2A, .M eb 1 9 2 UIs a.", faximum stress = > = > = 4.0 10 dyne /em’ | 8) WS 1, Example 3. Calculate Y for rubber if a rubber tube 0.5 m long, | 8, 289 =4:1, , whose external and internal diameters are 0.01 m and 0.004 m, , , , Scanned with CamScanner

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ogsion 6 at the fro,, press free ong, , , , Example 5. Young’s modulus and modul di, ua ice :, Gre 2.0x10!!N/m? and 0.8x10''N/m? reg, near isin, 1, PE gos DY Fa beam, 'Pectivel. Y fo, 90" jg 8 4 he, values of Poisson’s ratio and bulk modulus foy a 4 ley) , Mle af is b= Wet ™ “ther ,, Sol. From the relation Y = 2n(1+6), we have eek Ate he ft BYT nd ig, Y 2.0x 101 d, L is the length, ee a , the 10a . » MBH Of the, an axosx1o! 25-1=0.95 aM inertia OFS section, eam ang ) th, y a, Now from the relation, Y = 3K(1-20), we have hott ° if eross-section, T= art = 2lx(n 7 Peimetriy, Ka Y= 29x10 N/m? Fog 4 ty, 3(-20) 3 (1-20.25) = 283x101 7 a ye ME 5:5 X 20008 96. ye 4 0785, W=5. 6., jon 5.39x 108 dyne, , , , times that of rigidity modulus. What will be its Poisson'gt® isang © jeopem¥ =9x 10" dynes fem?, Oti 0 ¥, Sol. Given Y = 2.4n Rohitkhang % “ § = 6.39% 10 dynes), From the relation Y = 2n(1+6), we have ae 8xQx 10" dynes Peery, 5=0.549 em=5,49 ~, ot imple 9. For the same cross-sectional, a square cross-section ig Stiffer 4 {rens show that ji, yet ion of the same length and material, i One of che, a o=0.2 ery oad the eae the ratio 3:5, pat Hat fore, Example 7. One end of a wire 1.0 mm in radius and 59 op, , pol es sia a ty ee end of a beam whose other aa, length is twisted through 45° relative to the other end, Find in ype ho Ww is, angle of shear on its surface. [Rohilkhang Mi $= -—, Sol. Let L be the length of the wire, r the radius and 6 the twist i ze, the free end. Then, the angle of shear at a radial distance x from the ath stere WV is the load, L the length of the beam, I the geometrical, of the wire is inertia of its cross-section, and Y is the Young's modulus of the face, eet? he beam. rial, L For a rectangular beam,, At the surface of the wire, x=r, so that ba, ou T= y+ b= breadth & d=depth, L Ifthe beam has square section, b=d, Given that ; r= 1.0mm = 1.0x 107m, L = 50cm = 0.5 m, and 9 = 45° oof, g = (1.0% 108m) x45° _ 4 ogy gg 43, Om + Depression for this beam, 5; = we = an ~(), Example 8. A light metallic rod of length 60 cm and of radius 1 cm v1 bo yo?, is clamped at one end. It is loaded at the free end with 5.5 kg. Calculate y’ (, , uming Young’s modulus, , [Rohilkhand 2006 If the beam has circular cross-section of radius r, then, , the depression of the free end of rod assi, Y =9x101 dynes/em? and g = 980 cm/sec”., , Scanned with CamScanner