Chapter 1.2, Problem 7P

Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section, and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E.

Fig. P1.7

To determine

The maximum value of average normal stress in the links connecting at point B and D.

Answer to Problem 7P

The maximum value of average normal stress in the links connecting at point B and D is $\underset{\_}{101.6\text{MPa}}$.

Explanation of Solution

Given information:

The size of rectangular cross section is $8\times 36\text{mm}$.

The diameter (d) of the each pin is $16\text{mm}$.

Calculation:

Sketch the free body diagram of link ABC as shown in Figure 1.

Here, $F_{BD}$ and $F_{CE}$ are force applied by the link BD and the link CE.

Refer to Figure 1.

Apply the moment equilibrium condition at the point C.

$\sum M_C=0$

$\begin{array}{l}\left(0.4\right)F_{BD}-\left(0.25+0.4\right)\left(20\right)=0\\ 0.4F_{BD}-13=0\\ 0.4F_{BD}=13\\ F_{BD}=\frac{13}{0.4}\end{array}$

$\begin{array}{l}{F}_{BD}=32.5\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ F_{BD}=32.5\times {10}^3\text{N}\left(\text{Tension}\right)\end{array}$

Refer to Figure 1.

Apply the moment equilibrium condition at the point B.

$\sum M_B=0$

$\begin{array}{l}\left(-0.4\right)F_{CE}-\left(0.25\right)\left(20\right)=0\\ -0.4F_{CE}-5=0\\ -0.4F_{CE}=5\\ F_{CE}=-12.5\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ F_{CE}=-12.5\times {10}^3\text{N}\left(\text{compression}\right)\end{array}$

Calculate the net area of one link for tension as follows:

$A_{\text{net}}=b\left(h-d_{\text{Pin}}\right)$ (1)

Here, b is the width of the rectangular cross section, h is the depth of the rectangular cross section, and $d_{\text{Pin}}$ is the diameter of each pin.

Substitute $8\text{mm}$ for b, $36\text{mm}$ for h, and $16\text{mm}$ for $d_{\text{Pin}}$ in Equation (1).

$\begin{array}{c}{A}_{\text{net}}=8\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\left(36\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)-16\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =0.008\left(0.036-0.016\right)\\ =0.008\times \left(0.02\right)\\ =160\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the area of network for two parallel links as follows:

$\begin{array}{c}{A}_{\text{net}}=2\left(160\times {10}^{-6}{\text{m}}^{\text{2}}\right)\\ =320\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the average normal stress $\left({\sigma }_{BD}\right)$ in the links connecting at point B and D.

${\sigma }_{BD}=\frac{F_{BD}}{A_{\text{net}}}$ (2)

Here, $F_{BD}$ is force in link BD and $A_{\text{net}}$ is area of network of one link in tension.

Substitute $32.5\times {10}^3\text{N}$ for $F_{BD}$ and $320\times {10}^{-6}{\text{m}}^{\text{2}}$ in Equation (2).

$\begin{array}{c}{\sigma }_{BD}=\frac{32.5\times {10}^3\text{N}}{320\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =101,562,500\frac{\text{N}}{{\text{m}}^{\text{2}}}\times \left(\frac{1\text{MPa}}{{10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ =101.563\text{MPa}\\ \simeq \text{101}\text{.6}\text{Mpa}\end{array}$

Thus, the maximum value of average normal stress in the links connecting at point B and D is $\underset{\_}{101.6\text{MPa}}$.

To determine

The maximum value of average normal stress in the links connecting at point C and E.

Answer to Problem 7P

The maximum value of average normal stress in the links connecting at point C and E is $\underset{\_}{-21.7\text{MPa}}$.

Explanation of Solution

Calculation:

Calculate the net area of one link for tension as follows:

$A_{\text{net}}=bh$ (3)

Substitute $8\text{mm}$ for b and $36\text{mm}$ for h in Equation (3).

$\begin{array}{c}{A}_{\text{net}}=8\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\left(36\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =0.008\times 0.036\\ =288\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the area of network for two parallel links as follows:

$\begin{array}{c}{A}_{\text{net}}=2\left(288\times {10}^{-6}{\text{m}}^{\text{2}}\right)\\ =576\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the average normal stress $\left({\sigma }_{CE}\right)$ in the links connecting at point B and D.

$\left({\sigma }_{CE}\right)=\frac{\left(F_{CE}\right)}{A_{\text{net}}}$ (4)

Here, $F_{CE}$ is force in link CE and $A_{\text{net}}$ is area of network of one link in compression.

Substitute $-12.5\times {10}^3\text{N}$ for $F_{CE}$ and $576\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{net}}$ in Equation (4).

$\begin{array}{c}{\sigma }_{CE}=\frac{-12.5\times {10}^3\text{N}}{576\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =-21,701,388.8\frac{\text{N}}{{\text{m}}^{\text{2}}}\times \left(\frac{1\text{MPa}}{{10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ =-21.701\text{MPa}\end{array}$

Thus, the maximum value of average normal stress in the links connecting at point C and E is $\underset{\_}{-21.7\text{MPa}}$.