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190_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 190_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.6.qxd 184 9/25/08 11:40 AM Page 184 CHAPTER 2 Axially Loaded Members 45 degrees (c) on on u y face x face xcos( 2 xcos( ) 5.45 ksi xsin ( ) cos ( ) 5.45 ksi Problem 2.6-10 A plastic bar of diameter d compressed in a testing device by a force P shown in the figure. ; ; 32 mm is 190 N applied as xsin( p 2 u+ )2 5.45 ksi )cos( ) 5.45 ksi P = 190 N 100 mm (a) Determine the normal and shear stresses acting on all 0°, faces of stress elements oriented at (1) an angle 22.5°, and (3) an angle 45°. In each (2) an angle case, show the stresses on a sketch of a properly oriented element. What are max and max? (b) Find max and max in the plastic bar if a re-centering spring of stiffness k is inserted into the testing device, as shown in the figure. The spring stiffness is 1/6 of the axial stiffness of the plastic bar. 300 mm 200 mm Re-centering spring (Part (b) only) Plastic bar u d = 32 mm k Solution NUMERICAL DATA d 32 mm A P 190 N A a 22.50 degrees 100 mm b (2) p2 d 4 804.25 mm2 300 mm on x face )2 807 kPa xcos( ; xsin( ) cos( ) 334 kPa ; (a) Statics - FIND COMPRESSIVE FORCE F & STRESSES IN on PLASTIC BAR F P( a + b) a F A sx x F 0.945 MPa max (1) x x 472 kPa 0 degrees xsin( ) cos( ) 334.1 kPa 945 kPa 45 degrees (3) 945 kPa max sx 2 on 945 kPa x face )2 472 kPa xcos( 472 kPa x u+ )2 138.39 kPa from (1), (2) & (3) below max u xcos( 760 N or y face ; ; xsin( ) cos( ) 472 kPa ; p 2 ...
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