Chapter 2, Problem 134RP

The aluminum test specimen shown is subjected to two equal and opposite centric axial forces of magnitude P. (a) Knowing that E = 70 GPa and σ_all = 200 MPa, determine the maximum allowable value of P and the corresponding total elongation of the specimen, (b) Solve part a, assuming that the specimen has been replaced by an aluminum bar of the same length and a uniform 60 × 15-mm rectangular cross section.

Fig. P2.134

To determine

Find the maximum allowable load value P and corresponding total elongation length of the specimen.

Answer to Problem 134RP

The maximum allowable load value P is $\underset{\_}{92.3\text{kN}}$.

The total elongation length of the specimen is $\underset{\_}{0.791\text{mm}\text{.}}$

Explanation of Solution

Given information:

The modulus of elasticity (E) is $70\text{GPa}$.

The allowable stress $\left({\sigma }_{\text{all}}\right)$ is $200\text{MPa}$.

The width (D) of the specimen is $75\text{mm}$.

The width (d) of the fillet is $60\text{mm}$

The radius of the fillet is $6\text{mm}$

The uniform rectangular cross section is $\left(60\times 15\right)\text{mm}$.

Calculation:

Determine the area (A) of the cross section as follows:

$A=bt$ (1)

Here, b is the width of the specimen and t is thickness of specimen.

Substitute $60\text{mm}$ for b and $15\text{mm}$ for t in Equation (1).

$\begin{array}{c}A=60\times 15\\ =900{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =900\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Test specimen:

Calculate the ratio of $\left(\frac{D}{d}\right)$ as follows:

Substitute $75\text{mm}$ for D and $60\text{mm}$ for d.

$\begin{array}{c}\frac{D}{d}=\frac{75}{60}\\ =1.25\end{array}$

Calculate the ratio of $\frac{r}{d}$ as follows:

Substitute $6\text{mm}$ for r and $60\text{mm}$ for d.

$\begin{array}{c}\frac{r}{d}=\frac{6}{60}\\ =0.10\end{array}$

Refer to Figure 2.52b (flat bars with fillets), “Stress concentration factors for flat bars under axial loading” in the textbook.

Get the stress concentration factor (K) using the ratio of $\frac{D}{d}$ and $\frac{r}{d}$.

Get the value of K is 1.95 for the corresponding value of $\frac{D}{d}=1.25$ and $\frac{r}{d}=0.10$

Calculate the maximum allowable load using the expression as follows:

$\begin{array}{l}{\sigma }_{\max }=K\frac{P}{A}\\ P_{\text{all}}=\frac{A{\sigma }_{\max }}{K}\end{array}$ (2)

Here, K is stress concentration factor and ${\sigma }_{\max }$ is maximum stress.

Substitute $900\times {10}^{-6}{\text{m}}^{\text{2}}$ for A, $200\text{MPa}$ for ${\sigma }_{\max }$, and 1.95 for K in Equation (2).

$\begin{array}{c}{P}_{\text{all}}=\frac{900\times {10}^{-6}\times 200\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}{1.95}\\ =\frac{900\times {10}^{-6}\times 200\times {10}^6}{1.95}\\ =92.308\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =92.3\text{kN}\end{array}$

Thus, the maximum allowable value P is $\underset{\_}{92.3\text{kN}}$.

Determine the wide area $\left(A_{\text{w}}\right)$ of the cross section as follows:

$A_{\text{w}}=bt$ (3)

Substitute $75\text{mm}$ for b and $15\text{mm}$ for t in Equation (3).

$\begin{array}{c}{A}_{\text{w}}=75\times 15\\ =1125{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =1.125\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Determine the total elongation of the specimen using the relation:

$\begin{array}{c}\delta =\sum \frac{P_i{L}_i}{A_i{E}_i}\\ =\frac{P}{E}\sum \frac{L_i}{E_i}\end{array}$ (4)

Substitute $92.308\times {10}^3\text{N}$ for P, $70\text{GPa}$ for E, $\left[\frac{0.150}{1.125\times {10}^{-3}}+\frac{0.300}{900\times {10}^{-6}}+\frac{0.150}{1.125\times {10}^{-3}}\right]$ for $\frac{L_i}{E_i}$ in Equation (4).

$\begin{array}{c}\delta =\frac{92.308\times {10}^3}{70\text{GPa}\left(\frac{{10}^6\text{Pa}}{1\text{GPa}}\right)}\sum \left[\frac{0.150}{1.125\times {10}^{-3}}+\frac{0.300}{900\times {10}^{-6}}+\frac{0.150}{1.125\times {10}^{-3}}\right]\\ =\frac{92.308\times {10}^3}{70\times {10}^9}\times \left[\frac{0.150}{1.125\times {10}^{-3}}+\frac{0.300}{900\times {10}^{-6}}+\frac{0.150}{1.125\times {10}^{-3}}\right]\\ =7.91\times {10}^{-4}\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.791\text{mm}\end{array}$

Thus, the total elongation of the specimen is $\underset{\_}{0.791\text{mm}\text{.}}$

To determine

Find the maximum allowable load value P of an aluminium bar and total elongation length of the aluminum bar.

Answer to Problem 134RP

The maximum allowable load value P is $\underset{\_}{180.0\text{kN}}$.

The total elongation length of the aluminum bar is $\underset{\_}{\text{1}\text{.714}\text{mm}\text{.}}$

Explanation of Solution

Calculation:

Calculate the maximum allowable load of uniform bar as follows:

$P=A{\sigma }_{\text{All}}$ (5)

Substitute $900\times {10}^{-6}{\text{m}}^{\text{2}}$ for A and $200\text{MPa}$ for ${\sigma }_{\max }$ in Equation (5).

$\begin{array}{c}P=900\times {10}^{-6}\times 200\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ =900\times {10}^{-6}\times 200\times {10}^6\\ =180\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =180\text{kN}\end{array}$

Thus, the maximum allowable value P is $\underset{\_}{180.0\text{kN}}$.

Calculate the total elongation of uniform bar as follows:

$\delta =\frac{PL}{AE}$ (6)

Substitute $180\text{kN}$ for P, $600\text{mm}$ for L, $900\times {10}^{-6}{\text{m}}^{\text{2}}$ for A and $70\text{GPa}$ for $E$ in Equation (6).

$\begin{array}{c}\delta =\frac{180\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 600\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{900\times {10}^{-6}\times 70\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}}\\ =\frac{180\times {10}^3\times 0.6}{900\times {10}^{-6}\times 70\times {10}^9}\\ =1.714\times {10}^{-6}\text{m}\left(\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)\\ =1.714\text{mm}\end{array}$

Thus, the total elongation of the uniform bar is $\underset{\_}{\text{1}\text{.714}\text{mm}\text{.}}$