Test bank For Mechanics of Materials 8th Edition by Russell C Hibbeler Link full download:Download here 2–1 An air-filled rubber ball has a diameter of in If the air pressure within it is increased until the ball’s diameter becomes in., determine the average normal strain in the rubber d0 = in d = in pd - pd e= 0= = 0.167 in./in Ans pd0 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 in., determine the average normal strain in the strip L0 = 15 in L = p(5 in.) L - L0 e = L0 5p - 15 = 15 = 0.0472 in.>in Ans 2–3 The rigid beam is supported by a pin at A and wires BD and CE If the load P on the beam causes the end C to be D displaced 10 mm downward, determine the normal strain E developed in wires CE and BD 4m P ¢LBD ¢LCE = ¢L BD = A B C (10) = 4.286 mm 2m 3m ¢LCE 10 Ans eCE = L = 4000 = 0.00250 mm>mm ¢LBD Ans eBD = L = 4000 = 0.00107 mm>mm 2m 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No 2–6 Nylon strips are fused to glass plates When moderately heated the nylon will become soft while the glass stays approximately rigid Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated y mm mm P mm mm mm mm g = tan -1 x Ans a10 b = 11.31° = 0.197 rad 2–7 If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown 18 in in 18 in Geometry: Referring to Fig2 a, the stretched length of the string is 2 L = 2L¿ = 18 + = 37.947 in Average Normal Strain: = e avg L - L0 L0 37.947 - 35.5 = 35.5 = 0.0689 in.>in Ans portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5 02 Solutions 46060 5/6/10 1:45 PM 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page 10 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page 11 02 Solutions 46060 5/6/10 1:45 PM Page 11 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page 12 11 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 02 Solutions 46060 5/6/10 1:45 PM Page 13 12 02 Solutions 46060 5/6/10 1:45 PM Page 14 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 13 ... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form