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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 (57) 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 {58) 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 (57} 30 ﬁnd 0g:
9, I_ a 63% pus
u
I sing Equation (59) 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 (59):
‘ o',+o'y=cx+o‘y I
8MPa+6MPa=fa,Iuy .v \ '
ox=2MPao*y // [1]
 Using equation [1] in equation {58}: /\ 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 (58):
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 (59):
o'x+o",=o:+cr, ay=16MPaa'x . ing equation [1] in Equation (58)": 03,—0 2 M = — 33.04 M‘Pa; 0.839 a; . 0316), 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'sfrarne is subjected to the components of plane stress 6', = 32 MPa, c'y = 16 MPa and
 x, = 24 MPa. if 39 = 20”. whatare 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
axia'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. 55 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. (57), derive Eq. (59). m1
~< m :93. .1111 ON:
:Jt stant by substituting the expression (9 + 90°) for 8 in equation (57). 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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 Spring '06
 Keil

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