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Unformatted text preview: Mg ‘14 ’ Sow‘no VS 09—17. Determine the equivalent state of stress on an
element at the same point which represents (a) the principal
stress, and (b) the maximum inplane shear stress and the
associated average normal stress. Also, for each case,
determine the corresponding orientation of the element
with respect to the element shown. Sketch the results on
each element. Normal and Shear Stress: 0x = 125 MPa 0, = —75 MPa 7;, = —50 MPa In  Plane Principal Stresses: 03,0 0' —a' 2
y x y
012: :l: ( ) +¢xy2 2 2 125 + —75 125  —75 2
= __g i (A) + HO):
2 2
= 25:]:V12500
01 = 137 MPa 0'2 = ——86.8 MPa Ans. Orientation of Principal Plane: Txy —50
20 = —— = ——— = —o.5
tan P (a,r — a,)/2 (125—(—75))/2 0,, = ~13.28° and 76.72° Substitute 6 = —13.28° into (n+0), (Ix—(7y a" = 2 + —2—cos 20 + TX), sin 20
125 + (—75) 125 — (—75) ,
= ————2—— + ————3—cos(—26.57°)+(—50) sm(—26.57°)
= 137 MPa = 01
Thus, '
(11p)1 = —13.3° and (tap)2 = 76.7° Ans. 125 — (—75)/(—50)
The element that represents the state of principal stress is shown in Fig.a. Maximum In  Plane Shear Stress: a" — 0' 2 ’ ' “ 2
T = < y) + 7x 2 = (M) + 252 = 112MPa ADS
ma 2 y 2
Illp ne Orientation of the Plane of Maximum In  Plane Shear Stress: TU 50 93 = 31.7° and 122° 9—17. Continued By inspection, 1 has to act in the same sense shown in Fig. b to maintain
$35k“
equilibrium. Average Normal Stress: ax + (Ty _ 125 + (—75)
2 — 2 = 25 MPa Ans. O'avg = . The element that represents the state of maximum in  plane shear stress is shown in
Fig. c. 993M?“ 9—30. The cantilevered rectangular bar is subjected to the
‘force of 5 kip. Determine the principal stress at points A
and B. = le—(sxé) = 54 in“ A = (6)(3) = 18 in2 A = 2.25(1.5)(3) = 10.125 in3 QB = 2(2)(3) = 121113 Point A:
P sz 4 45(15) .
= — + = — + = .472 k
”A A I 18 54 1 5‘
QA 3(10.125)
Z .
= = — = . k
TA It 54(3) 0 1875 $1
ox = 1.472 ksi a, = 0 TU = 0.1875 ksi
a“: + a". o" — cry 2
. + . — 2
= 1472 0 i (1472 0) + 0.18752
2 2
0'1 = 1.50 ksi
0'2 = —0.0235 ksi
Point B:
_ P sz _ 4 45(1) _ .
0'3 — A I — 18 54 — 0.6111ksx
_ ,zQB__3(12)_ .
TB — It — 54(3) — 0.2222 k51
0x = —0.6111 ksi a", = 0 TX, = 0.2222 ksi
0' + 0' a — a' 2
X Y X y
01,2 = —2 :1: (————2 ) + rxyz = —0.611 + 0 :1: (—0.6111 — 0 2
+ . 2
2 2 > 0222 01 = 0.0723 ksi 0'2 = —0.683 ksii ‘— ﬂ'N75 4'5! —{ / #74 6‘? .——. ~— uua r»: H a. m a. _. E5 Ans. 9—38. A paper tube is formed by rolling a paper strip in
a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is
at 30° from the vertical, when the tube is subjected to an .
axial force of 10 N. The paper is 1 mm thick and the tube has
an outer diameter of 30 mm. 10N ION 5 10 = ——.—— = 109.76 kP
A §(0.032 — 0.0282) a
0,, = 109.76 kPa a, = 0 7,, = 0 e = 30° O‘x—O'y 7,3,: = ~—2— sin 20 + 7,}, cos 20 _106.76 — 0
2 I I #9776 if“ sin 60° + 0 = ~47.5 kPa Ans. 9—39. Solve Prob. 938 for the normal stress acting
perpendicular to the seam. ION 10N P 10
= — = ————— = 109.76 kP
‘7 A g (0.032 — 0.0282) a (n+0), axa'y .
0,, = ,——— + ~——cosZ(9 + TxySII'lza 2 2
109.76 + 0 + 109.76 — 0 2 2 cos (60°) + 0 =_ 82.3 kPa Ans. ...
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 Fall '10
 Dr.Pollock

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