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hwsoln15 - XX T1/1 T X.Y K g ‘5—/l 2‘11 3.5 x 106 ‘...

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Unformatted text preview: XX T1/1! T X.Y K g ‘5— //l 2‘11? 3.5 x 106 ‘ ta” °‘ = [14 {-721x106 2 u = 18.43° T +1 _ 5 oc = —-¥¥""2 = L—i—U—M + 27 "10 = 3.5x10‘E' T -r r = [(_§§§_¥¥)2+(Txy)2]1/2 = 1050x101 I [(%>2 + (3.5)211/2x106 fl = 11.07x106 Pa .. \ Tx'x' = CC + n cos(90-a) y = 3.5x106 + 11.07x106 cos(90°- 18.43“) 'T . . = 0C - r cos(90°-18.43°) -yy = 0 Pa Tx'y' = -r sin(90°-18.43°) = -10.50 x 106 Pa 11 = CC + T' = 3.5x10‘ 1* .‘!.1.O7x10‘5 = 14.57 x 106 Pa 12 = 0C - r = 3.5x106 - 11.07x10s = -7.57 x 105 Pa 7.22. [7.5] A thin-walled tank in Fig. P122 has air inside at a pressure of 100 psi gage. A torque of 50,000 ft-lb is applied 3‘ each end as shown. Using Mohr’s circle, what is the maxi- . mum stress at point A? The wall thickness is .01 ft. Figure P.7.22. (TXX)(2V)(1.995)(12)(.01)(12) =‘ (100) 3%)“: x 144 Txx = 9925 psi (rxy)(21)(l.995)(12)(.01)(12) (1.995)(12) = (50,000)(12) Txy = 1388.5 err [<— l’—4 >(Iyy)(1){-01)(12)(2) = (100)(1)(3.98)(12) = 19,900 si Tyy P r = 1338.52+K19,900-9925Y2.T= 5177 psi. oc = % (9925 + 19,900) = 14,9125 psi x = 14,9125 + 5177 = 20,0fiopsi —————-— 7.23. [7.5] Given the following stresses at a point, a = 15 . 94° 7“ = —3000 psi = 4000 psi rxy = 1000 psi 7}? Hence 5 = 46.60 - 15.94° = 30.7? ROTATES @/1 = 15. Ben-i. what angle must we rotate from the x axis to get an axis with a normal stress of - 2000 psi and a positive shear stress? Use Mahr‘s circle. qag' tl-ho gpoint C is the desired point since it has a normal stress of -2000 psi and a positive shear stress (basic convention). We must find 8 . - First compute r : r = [10002 + (4000 Z 3000)2]1/2 = 3640 psi Find GC: 60 = {r24[zooo + %(4000 - 3000)] }Vi , [36402 - 25002)”2 = 2646 ' tan(u+5) : ‘fiL—_ (2000)+ 5(4000 - 3000) _ 2646 _ - m - 1.058 n+3 = 116.60 _ 1000 _ tanu - 3500 - .2857 ...
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