Chapter 15, Problem 1QAP

To determine

The elastic modulus and tensile strength of the poly (methyl methacrylate) from the stress strain data given in Figure 15.3 at room temperature $\left[20°\text{C}\left(68°\text{F}\right)\right]$ and compare these values with the Table 15.1.

Answer to Problem 1QAP

The calculated elastic modulus of the poly (methyl methacrylate) is slightly greater than the maximum value of elastic modulus given in the Table 15.2 that is $3.33\text{}\text{GPa}\text{>}\text{3}\text{.24}\text{GPa}$.

The tensile strength (stress) found from the Figure 15.3 ( $52\text{MPa}$) is lies between the value of tensile strength that is given in the Table 15.1 ( $48.3\text{MPa}-72.4\text{MPa}$).

Explanation of Solution

Write the relation between elastic modulus, tensile strength and strain.

$E=\frac{{\sigma }_2-{\sigma }_1}{{\epsilon }_2-{\epsilon }_1}$ (I)

Here, $E$ is the elastic modulus, $\sigma$ is the tensile strength, $\epsilon$ is the strain and the suffixes $1,2$ indicates the initial and final stages.

Conclusion:

Refer to Figure 15.3. “The influence of temperature on the stress strain characteristics of poly (methyl methacrylate)”.

The stress and strain corresponding to the temperature curve of $20°\text{C}\left(68°\text{F}\right)$ is separately shown in Figure 1.

From Figure 1.

Let, the initial stress and strain corresponding to the temperature curve of $20°\text{C}\left(68°\text{F}\right)$ is

${\sigma }_1=0\text{MPa}$

${\epsilon }_1=0$

Let, the final stress and strain corresponding to the temperature curve of $20°\text{C}\left(68°\text{F}\right)$ is

${\sigma }_2=30\text{MPa}$

${\epsilon }_2=0.009$

Calculate the elastic modulus.

Substitute $30\text{MPa}$ for ${\sigma }_2$, $0.009$ for ${\epsilon }_2$ and $0$ for ${\sigma }_1$, ${\epsilon }_1$ in Equation (I).

$\begin{array}{c}E=\frac{30\text{MPa}-0\text{MPa}}{0.009-0}\\ =\frac{30}{0.009}\\ =3333.3333\text{MPa}\times \frac{{10}^{-3}\text{GPa}}{1\text{MPa}}\\ =3.333\text{GPa}\end{array}$

Thus, the elastic modulus of poly (methyl methacrylate) is $3.33\text{GPa}$.

Refer to Table 15.2, “Room-Temperature mechanical characteristics of some of the more common polymers”.

The elastic modulus corresponding to poly (methyl methacrylate) is $2.24\text{GPa}-3.24\text{GPa}$.

Hence, the calculated elastic modulus of the poly (methyl methacrylate) is slightly greater than the maximum value of elastic modulus given in the Table 15.2 that is $3.33\text{}\text{GPa}\text{>}\text{3}\text{.24}\text{GPa}$.

From Figure 1.

The tensile strength (stress) at the end of the curve corresponding to the temperature curve of $20°\text{C}\left(68°\text{F}\right)$ is $52\text{MPa}$.

Refer to Table 15.2, “Room-Temperature mechanical characteristics of some of the more common polymers”.

The tensile strength (stress) corresponding to poly (methyl methacrylate) is $48.3\text{MPa}-72.4\text{MPa}$.

Hence, the tensile strength (stress) found from the Figure 15.3 ( $52\text{MPa}$) is lies between the value of tensile strength that is given in the Table 15.1 ( $48.3\text{MPa}-72.4\text{MPa}$).