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Unformatted text preview: _ 1000+0 + 10000 Tx'x' — 2 2 c0560° +(500)sin60°
= 500 + (500)(.500)  (500)(.866) = 317 psi _ 1000+0 _ 10000 0 
1y.y.  2 2 c0560
(500)sin60°
= 500250+434 =
‘ tx.y. = 0 $000 sin60° + (500)c0560° ll (500)(.866)  (500)(.500)  3 si "7.3. [7.2] A plate of thickness 25 mm (Fig. P13) is loaded
:uniformly by forces F1 and F2 having values of 2250 N and
f 9000 N, respectively. What is the normal stress normal to the
diagonals? What are the shear stresses for a pair of axes rotated 30° counterclockwise from the horizontal and vertical
directions? a) Consider a diagonal as shown. ‘1 .3
0411! ifllht 1
F1 2250
rxx = K; = TTEEETTTET = 3 x 105 Pa
F2 9000
ryy = K; =  Tfazgjffgy = 6 x 105 Pa
Tnn = 3x105+é6x105) + 3x105;6xlof cosze+0
a = tan‘1 f% = 26.565°
s = 63.435° 6 = 1803 = 116.565“ .. 5 5
b) Tx.y. = §5}£L§§¥;9— sin60° + 0
1x.y. = 3.897 x 105 Pa ~_————————_——_ 7.4. [7.2] The stresses at a point'in a wooden beam are showu in Fig. P.7.4. What are the stresses normal and parallel
to the grain? 500 psi 150 psi 5,000 psi ' From the ﬁgure, we can treat the grain angle as a
rotation of the axes of 600 ﬁ'om the x—axis, as
shown below. ' ‘ $.29 = 36.0° ; 216° 6 (Tnn)l _ 2 ’ 2 107.9 sin 36° Now TXX yy 12 = 625 p51 Max. normal stress = 625 psi . 18° + 45° = 63° = 590293
' Tns 2 = 183.6 psi 1 l8.0° ; 1085 293+590 + gg§;§gg + T‘ = (293+590) Max. shear = 183.6 psi CO 7.14. [7.4] A thin cylindrical tank (Fig. P.7.14) is subject to
an internal gage pressure of 10 psi and is at the same time
subject to a twisting torque T of 200 lbft. If the outside diam
eter D = 1 ft and the thickness of the cylinder is .1 in., what is
the maximum normal stress at A away from the ends? What
is the maximum shear stress away from the ends? (Hint:
Since the compression stress coming directly from the inter
nal pressure is so very small compared to the other stresses,
we neglect it, thus making the stress at a point on the cylinder
wall away from the ends a plane stress problem.) Consider element A shown in the diagram. Note
From the stress transformation equations, we get: _ _
'r = 0 'r  'r  0
21' 2X 23’ 5500 4500 a . a 1'”. = —2——+ coleO +15031n120 Compute Txx
.7 , . =—5—5£g— 4500 005120” —1505in120°
y y 2 We can see that x’ is parallel to the grain and y ’ ‘—
is normal to the grain. So, It 1mm, = 3,745 psi [parallel = 10 11.8 2
rcweck! (rxx)(w)(11.9)(.1) = “4
"1,755 + 3745 = 5,000 + 5.00 ok .
xxx — 293 p51
Compute 1X9
Since we have plane stress: ' , tar?
' _ gzgg107.92 =
tan 26  293 _ 590 .727 TH
6:9: ,.
the  ‘ 11.9 _ ‘
(rx9)(n)(11.9)(.1)(—§—) — (zoo)<12) 1 = 107.9 psi 0 _ . . S 1x6 258 P51 From sign convention 1x8 = 107.9
+ 258 (T2) «Compute r 2ryyl1)(.l) = (10)(1)(11.8‘) ryy = 590 p51 sin126°107.9c95126° ...
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This note was uploaded on 07/15/2008 for the course ME 311 taught by Professor Pitteressi during the Spring '08 term at Binghamton.
 Spring '08
 PITTERESSI

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