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Unformatted text preview: Structural Failure and Assessment A structural analysis is conducted to determine if the structure is ﬁt for
its purpose, is. does it meet is design speciﬁcation. For aircraft
structures one very strong requirement is that the structure has the
lowest possible weight. In addition. there are three aspects of the structural response under load
that must be satisﬁed: '
1. That the deﬂections are not excessive. 2. That the stresses are below the design limit. 3. That there is no buckling of members or part of the structure
that affects the integrity of the structure as a whole. In this section of the course we will be dealing with: Transtormation of stresses
«Determination of Principal stresses
~Mohr’s Circle Two Dimensional Stress Sate In general the state of stress at a point is This is 0&9" Simpliﬁed to the plane Stress characterized by six independent normal State
and shear components which act on the
face of an element of material located at a 1“ sufﬁx = direction of the normal to the piane “at”:
2"6 sufﬁce = the direction in which it acts ' ' Stress Transformation Stress Transformation Arbitrary Rotated Axis System Consider the case‘ where the axes are rotated by some arbitrary angle. 0,
then, consider the triangle of material OAB. For uquililurium
u! rllus‘d have "2:: = “xx Rotated Axis System To simplify the algebra, we will consider the rotated axis system for
normal and shear stress states separately. Rotated Axis System: Normal Stresses
Resolving the forces aids sine _. , I
n :1 Gzlsrtdl 62 2 till K
x b \
K
\
Ozztds 4—6:)
I
K
B curds cos 6' D 0' tdz cos (9
I I e ‘ 55
‘/ assrdz sin :9
GSStclz ﬁgsolvmq the forces normal to the inclined plane 02.51%? = ontds cos 6‘ + oﬂtdz sin :9 since d3 = CH 0058 dz = d1 sin 6 (ii$2152? = ogztdl cos2 6’ + gird! sin2 6 dz=dlsin9 A 62's Ml ozrztdl x /
x x ’ \
ozztds 4—313)
I
curds cos 6’ “ B , ’ Iﬁ‘A
o'sstdz sin 6 1" Usstdz Besalving the forces garaliel to the inclined plane 03.x.IdI = — O'xtdSSiI‘l 9 + aﬂtdz cost? since dszdicoso dz=disine d5: dlcose
child! = — axtdf sin 6’ cos 9 + amid! cos 9 sin 8 crst sm 6 A a _ In“ (TlMl ‘ is
x b
“a
\
numb: H \
. U“,
B l“! cry 30520036
, ’ ‘é‘A ‘
‘1 655.1ch Rotated Axis System: Shear Stresses
Resolving the forces blostds cos 6 Rotated Axis System: Shear Stresses
Resolving the forces Resolving the forces normal to the inclined plane
J;,:.rdl = amid; sin 6 + azsrdz cos 6 since ds = cl] €050 d7. = dl sin 8
03:.th = amid! sin 6 cos 6 + oﬂt'dl cos 6 sin 6 Resolving the forces parallel to the inclined plane d3=d13050 a__._w.tdl = axtds 0056  sztdz sin6 dzidlsine since ds = dice58 dz: dlsinO 0,516?! = 02921120032 6 — 03rd! sin2 6 Rotated Axis System: Normal &Shear Stresses Resolving the forces normal and parallel to the plane AB gives: (Sm1d]: outdlcosz 6 + osstcllsin2 6 + outdlsinecose + ontdlcosesinﬁ
51.5.tdl = — ouldl sin 0 case + osstdl cos 95in 8 + cmtdl cos2 6 — oﬁtdl sin2 0 tdl is a common factor so that:
on. = 6”. cos2 B + 6_, sin2 6 + or,A sin 29 a V = ( on + 6”]sin 29 + on 00526 i
’ 2
If the ortho onal die onal lane is considered then: or“. = a“, cos2 9 + on sin 1 G r on sin 29 ow =%{ on + :3“) sin 26 + a” c0329 Rotated Axis System: Normal &Shear
Stresses ﬂ 3 . 2 .
02.2. — on cos 6 + 0'SS sm 9 + on 5111 26 03.3. = é—(w on + 0'55)sin 29 + <5its c0528 The stress rotation can be written in matrix form as a tensor transformation cose sine csZZ eras cost) —sinG —sin_6 cost} 0' 0 51116 0059 82 SS Principal Stresses There will be some anglee , where the shear stress 6», is zero. This is given by 20'” l aﬂkin 20 + 6,5 c0526 80 that Ian 29 = (—H—3
5:7. "Us: 0= F6 72: l
2 Shear stress Direct stress Substituting these gives the principal stresses as '—'_'—2—~'
! ( ) +46; l'__'—“—‘— 9
~ girls”. «so? +45; Three Dimensional Stress Field It 12 13 0.“ 5. . ml I1 Uh! 0' .x'x' x ) o o 5.. = :111 m2 rn3 om ow 6 r112 :12 54*. Viv. y z x'z’ 5,. 5.... 11' n2 113 on on 0 m3 213 Where l,m,n are the direction cosines of the rotated stress ﬁeld, For the three
dimensional system there are three principal stresses acting normal to the
principal planes where the shear stresses 0 6 Ur“. ,_..‘ x a are zero. Stress Invariants it can be shown that some combinations of the stresses are the same
irrespective of the orientation of the axis system used to compute them. These
are called the stress invariants. For a general stress system there are three stress invariants.
—i First Stress ll =61 +6” +0”, =6“ +orw +0” =const      SD invariant 51:61 +6“ =0“ + U_ Eons: . . . .. .. 2D Lil1e sum of the diagonal terms on the stress tensor is constant. Second Stress 1F2 2 Gig.” 4‘ O'Hara + Omar
invariant = (Tqu + awe: + agou  0'; — of,  of, = cons: 12:53“ zone“ —o’.: =const Third Stress f“ z altoNam
invariant = Juan": + Ema03:91:  (Gugl + G's0'2". + cacti») = Cons: 1J : o . _ .. Mohr‘s Circle In this lecture we will see that equations for plane stress transformation have a
graphical solution that is often convenient to use and easy to remember. This is achieved through the use of Mohr's circle developed by the German
engineer Otto Mohr. O'lei How to Draw Mohr’s Circle Assume values of 5,1,6” and 6,3 are known 1') Locate the points ems” and 53:66 on the circle and hence draw the circle Shear stress ‘ Normal stress ii} Locate the centre of the circle on the horizontal axis for normal stress —Iv Normal stress How to Draw Mohr’s Circle Shear Stress Normal stress 1O How to Draw Mohr’s Circle iii) To ﬁnd normal direct stress 6.. and tangential shear stress a. on a plane
angle 0 anticlockwise: From the known stress draw a radius at an angle 2 9
clockwise from the radial line of known stress Shear Stress Normal stress Mohr’s Circle: Principal Stresses The principal stresses are the two values where the circle crosses the
horizontal axis Shear Stress Max shear stress Principal stress Principal stress
\ Normal stress Min shear stress ‘l'l Example —— Mohr‘s Circle a'___ =+16(}!\’:";1':'w'.2
0'3 =—120N!mm2 a” =+112..'\.rmm2 l 160men12. 1 13Wn‘lr1'12 ‘20Nimm2, 113Nimm2 ’12 ...
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 Winter '09
 S.PinhoandM.Aliabadi

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