# Mse Sloution Manual

Topics: Young's modulus, Tensile strength, Elasticity Pages: 12 (2171 words) Published: April 19, 2012
CHAPTER 15

COMPOSITES

PROBLEM SOLUTIONS

Large-Particle Composites
15.1 The mechanical properties of cobalt may be improved by incorporating fine particles of tungsten carbide (WC). Given that the moduli of elasticity of these materials are, respectively, 200 GPa (30 × 106 psi) and 700 GPa (102 × 10 psi), plot modulus of elasticity versus the volume percent of WC in Co from 0 to 100 vol%, using both upper- and lower-bound expressions. Solution The elastic modulus versus volume percent of WC is shown below, on which is included both upper and lower bound curves; these curves were generated using Equations 15.1 and 15.2, respectively, as well as the moduli of elasticity for cobalt and WC given in the problem statement. 6

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

15.2 A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.70 and 0.30, respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow.

Specific Gravity Copper Tungsten 8.9 19.3

Modulus of Elasticity (GPa) 110 407

Solution There are two approaches that may be applied to solve this problem. The first is to estimate both the upper limits of elastic modulus [Ec(u)] and specific gravity () for the composite, using expressions of the form of c Equation 15.1, and then take their ratio. Using this approach

Ec( u) = ECu V + EWVW Cu
= (110 GPa)(0.30) + (407 GPa)(0.70)

= 318 GPa

And

 =  V +  VW c Cu Cu W

= (8.9)(0.30) + (19.3)(0.70) = 16.18

Therefore

Specific Stiffness =

Ec (u) 318 GPa  19.65 GPa  16.18 c

With the alternate approach, the specific stiffness is calculated, again employing a modification of Equation 15.1, but using the specific stiffness-volume fraction product for both metals, as follows:

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Specific Stiffness =

ECu
 Cu

V + Cu

EW

V  W W

=

110 GPa 407 GPa (0.30) + (0.70) = 18.47 GPa 8.9 19.3

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

15.3 (a) For a fiber-reinforced composite, the efficiency of reinforcement η dependent on fiber length l is according to



l  2x l

where x represents the length of the fiber at each end that does not contribute to the load transfer. Make a plot of η versus l to l =50 mm (2.0 in.) assuming that x =1.25 mm (0.05 in.). (b) What length is required for a 0.90 efficiency of reinforcement? Solution (a) The plot of reinforcement efficiency versus fiber length is given below.

(b) This portion of the problem asks for the length required for a 0.90 efficiency of reinforcement. Solving for l from the given expression

l =

2x 1 

Or, when x = 1.25 mm (0.05 in.) and = 0.90, then

Excerpts from this work may be reproduced by instructors for distribution on a...