Question

Machine Desgin Problem

For the wire of circular cross section below. find the deflection of point B in the direction of force using superposition. Assume a =4, b= 10" e = 10, F = 100 lbf, and dia. D= 0.75": E = 30 Mpsi: v = 0.3. Ignore the effect of fillet radii for this problem. I for circular cross-sections is given by 1 = D. Assume G = 2(2+)

The answer to the problem is:

Delta 1=0.458 e-2

Delta 2=0.0715

Delta 3=0.044

Delta 4=0.75 e-4

Delta 5=0.214

Delta 6=0.034

Total Deflection=0.368 in

Please show all work and steps

Answer 1

Answer:

As given in the solution below, free body diagrams are taken by breaking the body in sections and internal balancing of forces and moments also has been shown. With the help of superposition theorm, in each section, a variable x is taken and used as variable of integration. Different elements as divided and these are in different loading conditions as shown below:

Step by step solution is given below

Answer:- Gwen valus a=4" 10" f = 100 lbf D= .75" b 11 to V=.3 I= XD4 64 G= 2(1+0) stub making free body diagrams B (1) X MB - fxxg. VBC= McD = FX F g Tep=fa cins a NCD=F x Mp4, - FЬ О MDG=fa (۱)

Element bending only Be is in bending so from supeposition thom, strain energy can be given by JUBC ㅗ Fa3 mo F(a) dx EI DF ЗЕТ 0 8 = Fas ЗЕІ as we know that Iz ddy tu I = I x (AS) ².01552 int 64 2 A = Id² 44156 in ² 4 = Now 8, = Fa3 3 EI 100x(4) 3 3X30X "01552 X 106 - « 458 X 10 2inch Auswer. 61 Element CB is in Bending as well as tortion. ou T dI Xl a fi a fi as as tortion is constart where le length for tortion in Member Ср fi=F, T= fa lab l= length of member So

Now by putting the values = (fa) ха хь OUCH OF Tortion GJ Fa2 Ь GJ ut us take it as 83 = Tortion) &z= due to bending Nao G= E и мо 2 Clv) Зо с? - 11:33 2 (143) Т- 8xI — XX •o) SS2 103) оч go 82 - 100 х42х10 from eq 1) (3gx •o3)оч x 4 5. — o Qo, 9 ) и 1 Arsue O3 = О-4" due to Bending in co 6) ICP f (a)nda Bend Of S2 Pb3 ЗГІ

6, a FЬ3 100 x 103 3x30x.01552 3ti d = .071S Answer Membu Da és loaded and in anially bending in two planes L for amal = f of Х force du Jfi dri AE where oupal OF Janial (н Fc AE TOOX10 76.48 106 44156430 = .75X10 inch Answer So Coupe fc as a tref hyby a) a da Bending moment in each plane Du are constant along the legth MDU2=fb, MDG i=fa b)(b) dxt/ OF bending Jopa Feaz OF 1 86 CI Fc be + EI ter

&s= fcb2 100 x 10X102 30X.01552x104 os = > 214 inch Answer 2 fo- fca2 100x Joxu? EI 30x: 015528 106 do *034 inch Anown Now adding all the equation ( + @ +0 +0 +0 I- I DY , J = 2I, A - Z d² and pulting 64 G= E we will &(1+y) get (SB)* (Sp)= (110$,b9) +48 c(0*36%)48{1+)a ºb +3cd2] 3 KEd4 103) +49x1064*163) +48 (1+.3]x4°x10 (82) Putting the valuess 3 x 9 30251" [ 6[44710 [ 16 + 3X10X102) 447[17024 + 55680 +9989+ 16-87) 1368 368598.5 (8) F 106 Answer. (OB/f = 1368 inch