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Chapter 16: Problem 7

(a) For a fiber-reinforced composite, the efficiency of reinforcement \(\eta\) is dependent on fiber length \(l\) according to $$\eta=\frac{l-2 x}{l}$$ where \(x\) represents the length of the fiber at each end that does not contribute to the load transfer. Make a plot of \(\eta\) versus \(l\) to \(l=50\) \(\mathrm{mm}(2.0 \text { in. })\) assuming that \(x=1.25 \mathrm{mm}\) \((0.05 \text { in. })\) (b) What length is required for a 0.90 efficiency of reinforcement?

Short Answer

Expert verified
Answer: Based on the given steps and calculations, the plot should display the relationship between efficiency of reinforcement and fiber length, with x being a constant 1.25 mm. The required fiber length for a 0.90 efficiency of reinforcement is 25 mm.

Step by step solution

01

Part(a): Plotting the efficiency of reinforcement as a function of fiber length

To plot the relation between efficiency of reinforcement (\(\eta\)) and fiber length (\(l\)) follow these steps: 1. Define the equation: \(\eta=\frac{l-2x}{l}\), where \(x = 1.25\) mm. 2. Create a list of fiber lengths \(l\) from 0 mm to 50 mm. 3. Calculate the efficiency of reinforcement \(\eta\) for each fiber length \(l\) using the given equation. 4. Create a plot with fiber lengths on the x-axis and efficiency of reinforcement on the y-axis. Note: The actual plotting will be done by the student using their preferred software or graphing calculator.
02

Part(b): Finding the required fiber length for 0.90 efficiency of reinforcement

To find the fiber length required for an efficiency of 0.90, follow these steps: 1. Set the efficiency of reinforcement equation equal to 0.90: \(\eta = 0.90 = \frac{l - 2x}{l}\), where \(x = 1.25\) mm. 2. Substitute \(x\) value into the equation and solve for \(l\): \(0.90 = \frac{l - 2(1.25)}{l}\) 3. Clear the fraction by multiplying both sides of the equation by \(l\): \(0.90l = l - 2.5\) 4. Rearrange the equation to solve for \(l\): \(0.10l = 2.5\) 5. Divide both sides by 0.10: \(l = \frac{2.5}{0.10} = 25\) mm Thus, the required fiber length for a 0.90 efficiency of reinforcement is 25 mm.

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Most popular questions from this chapter

Estimate the maximum and minimum thermal conductivity values for a cermet that contains 90 vol\% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and \(67 \mathrm{W} / \mathrm{m}\) -K for \(\mathrm{TiC}\) and \(\mathrm{Ni}\), respectively.

(a) What is a hybrid composite? (b) List two important advantages of hybrid composites over normal fiber composites.

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. $$\begin{array}{lcc}\hline & \begin{array}{c}\text {Specific} \\\\\text {Gravity}\end{array} & \begin{array}{c}\text {Modulus of} \\\\\text {Elasticity (GPa)}\end{array} \\\\\hline \text { Copper } & 8.9 & 110 \\\\\text { Tungsten } & 19.3 &407 \\\\\hline\end{array}$$.

(a) Cite several reasons why fiberglassreinforced composites are utilized extensively. (b) Cite several limitations of this type of composite.

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, \(200 \mathrm{GPa}\left(30 \times 10^{6} \mathrm{psi}\right)\) and \(700 \mathrm{GPa}\). \(\left(102 \times 10^{6} \mathrm{psi}\right),\) plot modulus of elasticity versus the volume percent of WC in Co from 0 to 100 vol\%, using both upper- and lowerbound expressions.
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