hwsoln14

# hwsoln14 - 1000 0 1000-0 Tx'x — 2 2 c0560°-500)sin60° =...

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Unformatted text preview: _ 1000+0 + 1000-0 Tx'x' — 2 2 c0560° +(-500)sin60° = 500 + (500)(.500) - (500)(.866) = 317 psi _ 1000+0 _ 1000-0 0 - 1y.y. - 2 2 c0560 (-500)sin60° = 500-250+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+é-6x105) + 3x105;6xlof cosze+0 a = tan‘1 f% = 26.565° s = 63.435° 6 = 180-3 = 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° = 590-293 ' 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 lb-ft. 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? ' _ gzgg-107.92 = 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.

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hwsoln14 - 1000 0 1000-0 Tx'x — 2 2 c0560°-500)sin60° =...

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