HW 6 - _L Threcampld'rients o'f plain sfiess a point p of...

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Unformatted text preview: __L_ Threcampld'rients o'f plain sfiess a point p. of a material are ‘3,.=20MPa 9}: 0333:3930. I33: .45" whataretne stresses 6,, a; and“, at point p? - Using Equatlon (5-7) to fmd 6;: a“; = 6', Ir?" + a.“ g 0" (cos'20)+ rxy(sin 26?) a; 23331? + o +1233333 — 3 («393° )+ 0 ANS 9.5mm Using Equation (5. 9) to find 6,: a; =.- 0‘ :0? — a" 2 0’ (335243)— 3363123) ¢;=2ma+o__ I ,7“'°(cos90°)—o 2 2 INS ' I .2 Using Equation {5-8) to find If: 7' = -. 'a'. 2 533323)” ”(60523) -—-—---———(s'i'n90°)+ 0 Th'eI comma ' if bfplanastressataipointpofamaterial are ax= 0 fiOand1w=25 I33. 33- 45“; whatarefhest'essesaxlmy and ILatpointp? mum: Using Equafi'on (5-7} 30 find 0g: 9-, -I_- a 63% pus u I sing Equation (5-9) to find cry: .2- a".-(¢os 26) + Ty (sin 26') {363909)+ (2513:)(33 93°) Hw L — .4— I W The Imponems of plane stress at T §nguTION: Adding equations (5—7) and (5-9): ‘ o',+o'y=cx+o‘y I -8MPa+6MPa=fa,-I-uy .v \ -' ox=-2MPa-o*y // [1] | Using equation [1] in equation {5-8}: /\ I l 4? = _ “x “a? (sin 25% 73(c0526) \ 4 a" = —- —16Afl’a = {Mikel} 4o°)+ 7,,(cos4o°) t point p of a material referred to the x’y’z' coordinate system are o',‘ = —8 MPa, at y = 6 MPa and ‘l.' x, = -16 MP8. if e = 20°. what are the stresses 5,, cry. and In, at point p? 2 - 16 Mpa = (1 MPa :4 5,) (sin 40°) + 1,, (cos 40°) 1x, = - 21.73 MP8 - 0.839 a, [2] Using equations [1] and [2] in Equation (5-8): a" + cry a" - 0' y 2 2 y (c052€)— raisin 26) -2MPa—a' +0 —2MPa——a' —a'. . swaJ 2 I) y _£_m;Ls(cos4o°)- [.. 21.73MPa - 0.339crylsm 40°) . - 7.736 MPa = 1.305 c,- , .15 g, = - .9-- MP3 g = . MPa 1,, = -16.75 MP8 4"- _‘ _ l‘ I N: Ming Equations (5-?) and (5-9): o'x+o",=o:+cr, ay=16MPa-a'x .- ing equation [1] in Equation (5-8)": 03,—0- 2 M = — 33.04 M‘Pa; 0.839 a; . 03-1-6), axvay 0y: 2 ._ . 32MPa = 2 1;), =- 2 y (sin 260+ 73(o03219) — 241m: = {W]Gin 4o°)+ rxy (cos 40°) - 24 MPa 2 (a, + 3 Mpa) (sin 40°) + 1,, (cos 40°) or +(16MPa—a’,) _ a" -(16MPa- 0‘1) pontip of'car's-frarne is subjected to the components of plane stress 6', = 32 MPa, c'y = -16 MPa and - x, = -24 MPa. if 39 = 20”. what-are the stresses 5,, cry. and 1,, at point p? _4 ng equations [11 and [2] in Equation (so): (cos 29)- :1} (sin 26) 2 (cos4o")+[—33.04M19a40.839chr 5111409) 32 MPa = 8 MP8 + 0.735 axl- 6.128 MPa - 24.45 MP8 +0539 0', 5 9M. 9;, J1 -2-§...a Mpg 5, = - .97 MP3 I 247 l ' i E’. Sfiflflhmm -* mint p of the supporting structure of the antenna shown in Problem 5—2 32 is subjected to plane stress The .Jonents o, = -20 MPa. and 1,9,: 30 MPa. The allowable normal stress of the material (in tension and in ac haression) Is 80 MPa. Based on thislcnterion what' ts the allowable range of values of the stress component e LLELthtz ' 2 ax-i-a'y a, a", or — +22 | a 02+2acr +02 I | o12=°'—2*+7”: ##y =%-2°’:Pai alt—tn 40" +100+900 SOMPa = ff—IOMPafiI‘CZ—EHOJI +1000. "i t note (if the equation are: AiS g1=21ME§ g2=- El gateway ‘t t element in Fig. 5-5 is a cube with d‘irn'ension b. By summing moments about the x and y axes. show that = 1,, and 1,2 = Ta if the material is in eqitilibrium. (In fact, these results hold even if the material is not in at iilibrium.) :31 mm: *t t :ouple 1:2, exerts a moment of: M2,. = 12, A (b) couple 1,: exerts a moment of: Mw=tyzA(b) :si :e the sum of moments on- the element must be zero: r“ 5 inf—1y: 2W ittfiihg an equal to e + 90° in Eq. (5-7), derive Eq. (5-9). m1 ~< m :93. .1111 ON: :Jt stant by substituting the expression (9 + 90°) for 8 in equation (5-7). gr; = a, :0, +-—-—~——— or, 2 0’ (c0526)+ ti ”(sin 2t?) :3; _ ‘7’ :‘r’ + a ‘7 (cosz(a+9;o°)+ r (sin2(€+ 90°) 0—" i: ”1:”? + a "’0' (eos(2t9+1$0°)+ rxy(sin(2t9+180°) .o iecail that sin (e + 180”)“ — - sin (a) and cos (a + 130°)— — - cos (9) So the second and third terms of the above - on beoome negative when a: 6 + 90". which changes equation (5-?) to: m = ft" “ (mwi- thine) 261 ...
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