02-04ChapGere.0007

# 02-04ChapGere.0007 - u 5 22.5 8 1 90 8 5 112.5 8 s u 5 s x...

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128 CHAPTER 2 Axially Loaded Members Problem 2.6-10 A plastic bar of diameter d 5 30 mm is compressed in a testing device by a force P 5 170 N applied as shown in the figure. Determine the normal and shear stresses acting on all faces of stress elements oriented at (a) an angle u 5 0°, (b) an angle u 5 22.5°, and (c) an angle u 5 45°. In each case, show the stresses on a sketch of a properly oriented element. Solution 2.6-10 Plastic bar in compression Plastic bar 100 mm P = 170 N d = 30 mm 300 mm u F REE - BODY DIAGRAM F 5 Compressive force in plastic bar F 5 4 P 5 4(170 N) 5 680 N P LASTIC BAR ( ROTATED TO THE HORIZONTAL ) (a) u 5 0 8 (b) u 5 22.5 8 Use Eqs. (2-29a) and (2-29b) s u 5 s x cos 2 u 5 ( 2 962.0 kPa)(cos 22.5 8 ) 2 52 821 kPa 52 962.0 kPa (Compression) s x 52 F A 52 680 N p 4 (30 mm) 2 t u 52 s x sin u cos u 52 ( 2 962.0 kPa)(sin 22.5 8 )(cos 22.5 8 ) 5 340 kPa
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Unformatted text preview: u 5 22.5 8 1 90 8 5 112.5 8 s u 5 s x cos 2 u 5 ( 2 962.0 kPa)(cos 112.5 8 ) 2 5 2 141 kPa t u 5 2 s x sin u cos u 5 2 ( 2 962.0 kPa)(sin 112.5 8 )(cos 112.5 8 ) 5 2 340 kPa N OTE : All stresses have units of kPa. (c) u 5 45 8 s u 5 s x cos 2 u 5 ( 2 962.0 kPa)(cos 45 8 ) 2 5 2 481 kPa t u 5 2 s x sin u cos u 5 2 ( 2 962.0 kPa)(sin 45 8 )(cos 45 8 ) 5 481 kPa N OTE : All stresses have units of kPa. Plastic bar 100 mm P = 170 N d = 30 mm 300 mm u 100 mm P = 170 N 300 mm F d = 30 mm F F x y u x y 962 kPa 962 kPa y x 821 141 141 340 821 340 u = 22.5 ° y x 481 481 481 481 481 481 u = 45 ° A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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