(a) For a fiber-reinforced composite, the efficiency of reinforcement \(\eta\) depends 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 in.), assuming that \(x=1.25 \mathrm{~mm}(0.05 \mathrm{in} .)\). (b) What length is required for a \(0.90\) efficiency of reinforcement?

Short Answer

Expert verified
Question: Given the efficiency of reinforcement formula $\eta = \frac{l - 2x}{l}$, the fiber length up to 50 mm, and the distance $x = 1.25$ mm that does not contribute to load transfer, create a plot of efficiency against fiber length and determine the required length for an efficiency of 0.9. Answer: A plot should be created for the efficiency of reinforcement ($\eta$) against fiber length ($l$) using the formula $\eta = \frac{l - 2(1.25)}{l}$, with the range of fiber length being 0 to 50 mm. To achieve an efficiency of 0.9, the required fiber length is 25 mm.

Step by step solution

01

Insert the given values into the formula

First, we need to plug the given values of \(x\) into the formula for \(\eta\). We have \(x = 1.25\) mm and we will plot the graph for \(l\) up to 50 mm. $$ \eta=\frac{l-2(1.25)}{l} $$
02

Create the plot of \(\eta\) versus \(l\)

Using the given range for \(l\) i.e., from \(0\) to \(50\) mm, create a plot for \(\eta\) versus \(l\) based on the equation: $$ \eta=\frac{l-2.5}{l} $$
03

Find the length for 0.90 efficiency

To find the length required for an efficiency of 0.90 (part b), insert the desired efficiency into the equation and solve for \(l\): $$ 0.90 = \frac{l-2.5}{l} \\ 0.90l = l - 2.5 \\ 0.10l = 2.5 \\ l = \frac{2.5}{0.1} = 25\, mm $$ So, for an efficiency of 0.9, the required fiber length is 25 mm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reinforcement Efficiency Formula
Understanding the reinforcement efficiency formula is critical when working with fiber-reinforced composites. This formula is fundamental for optimizing materials to ensure they are strong enough for their intended use. The formula is given by:
\[\begin{equation} \[\eta=\frac{l-2 x}{l}\]\end{equation}\]
Here, \(\eta\) represents the efficiency of reinforcement, \(l\) is the length of the fiber, and \(x\) is the length at each end of the fiber that does not contribute to load transfer. This could be due to factors such as inadequate bonding or fiber end conditions.
To put this into practical terms, let's say a certain fiber has a negligible load contribution region of 1.25 mm at each end. If we want to calculate the efficiency of this fiber when it is 50 mm long, we plug the values into the formula to get: \[\begin{equation} \eta = \frac{50 - 2(1.25)}{50}\end{equation}\]By computing this, we find out the efficiency of the reinforcement for a given fiber length.
Plotting Efficiency vs. Fiber Length
The relationship between fiber length and reinforcement efficiency can be visualized by plotting \(\eta\) versus \(l\). Plotting this relationship helps us understand how increasing the fiber length affects the efficiency of load transfer in a composite material.
To create such a plot, we evaluate the efficiency formula using a range of fiber lengths. For instance, with an \(x\) value of 1.25 mm, the plot will show the efficiency values from \(l = 0\) mm up to \(l = 50\) mm. This graphic representation allows us to observe the trend that, in general, longer fibers will have higher reinforcement efficiency, up until a certain point where increases in length have diminishing returns due to practical limits such as fiber waviness or other composite structure characteristics.

Visualization and Interpretation

Visualizing the data makes it easier to understand the significance of each fiber length. In the plotted graph, the efficiency curve rises sharply with an increase in fiber length at the beginning and then levels off as the length continues to increase. This leveling off indicates that after a certain fiber length, there is no substantial gain in reinforcement efficiency.
Determining Fiber Length for Specific Efficiency
To determine the fiber length required for a specific reinforcement efficiency, we rearrange the reinforcement efficiency formula to solve for \(l\). For instance, if an efficiency of 0.90 is needed, we set up the formula with such efficiency in mind:
\[\begin{equation} 0.90 = \frac{l - 2 x}{l}\end{equation}\]
After inserting the non-contributing length, \(x\), and rearranging for \(l\), we find that a longer fiber length increases the efficiency, up to a certain limit. In the case of our exercise, if \(x = 1.25\) mm, the equation to determine the required fiber length would be, after simplification through algebraic steps:\[\begin{equation} l = \frac{2.5}{0.1} = 25\, mm\end{equation}\]
This calculation is vital for engineers and designers who wish to achieve a certain performance level with their composite materials. Knowing the necessary fiber length for the desired efficiency enables them to design more effectively and predict the behavior of the composite under different load conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Cite one similarity and two differences between precipitation hardening and dispersion strengthening.

A continuous and aligned fibrous reinforced composite having a cross-sectional area of \(970 \mathrm{~mm}^{2}\) (1.5 in. \(\left.^{2}\right)\) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 215 MPa (31,300 psi) and \(5.38\) MPa (780 psi), respectively, the force sustained by the fiber phase is \(76,800 \mathrm{~N}\left(17,265 \mathrm{lb}_{\mathrm{f}}\right)\), and the total longitudinal composite strain is \(1.56 \times 10^{-3}\), determine the following: (a) The force sustained by the matrix phase (b) The modulus of elasticity of the composite material in the longitudinal direction (c) The moduli of elasticity for fiber and matrix phases

(a) What is the distinction between cement and concrete? (b) Cite three important limitations that restrict the use of concrete as a structural material. (c) Briefly explain three techniques that are used to strengthen concrete by reinforcement.

(a) Briefly describe sandwich panels. (b) What is the prime reason for fabricating these structural composites? (c) What are the functions of the faces and the core?

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 the modulus of elasticity versus the volume percent of WC in Co from 0 to 100 vol \(\%\), using both upperand lower- bound expressions.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free