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Mechanics of Materials 10th Edition Hibbeler Solutions Manual

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Mechanics of Materials 10th Edition Hibbeler Solutions Manual Download at: https://goo.gl/4zaFnQ People also search: mechanics of materials 10th edition solutions pdf mechanics of materials hibbeler 10th edition pdf mechanics of materials 10th pdf mechanics of materials 10th edition hibbeler solutions pdf mechanics of materials 10th edition pdf free download mechanics of materials 10th edition ebook mechanics of materials 10th edition pdf download mechanics of materials 9th edition pdf free download
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  105     105   0  Mechanics of Materials 10th Edition Hibbeler Solutions Manual Full clear download (no formatting errors) at: http://testbanklive.com/download/mechanics-of-materials-10th-edition-hibbeler-solutions-manual/  2  –  1.  An air-filled rubber  ball has a diameter of 6 in. If the air  pressure within the ball is increased until the diameter  becomes 7 in., determine the average normal strain in the rubber  .  Solution   d  0 = 6 i n .  d = 7 i n .   p d -  p d  0  7 - 6P =  p d   = 6 = 0.167 i n . >  in. Ans.    106   106   Ans:  P = 0.167 i n . >  i n .  107     107    L   2  –  2.  A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip .  Solution    L 0 = 15 i n .   L =  p(5 i n . )    L -  L 0  5p - 15P = =   0  15 = 0.0472 i n . >  in. Ans.   Ans:  P = 0.0472 i n . >  i n .  108     108   2  –  3.  If the load P on the beam causes the end C to be displaced  10 mm downward, determine the normal strain in wires CE  D  E    and  BD .  4 m   P   Solution    A    B   C     L  BD   3 =    L  BD =    L C   E    7 3 ( 10 )  7 = 4.286 mm   3 m 2 m 2 m    L C   E    10 P CE =    L = 4000 = 0.00250 mm >  mm Ans.   P  BD =    L  BD    L =  4.286 4000 = 0.00107 mm >  mm Ans.   Ans:  P CE = 0.00250 mm >  mm, P  BD = 0.00107 mm >   mm
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