Chapter 1, Problem 66RP

In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes.

Fig. P1.66

To determine

The largest load P that can be applied at A.

Answer to Problem 66RP

The largest load P that can be applied at A is $\underset{\_}{1.683\text{kN}}$.

Explanation of Solution

Given information:

The diameter of the pin at C is $6\text{mm}$.

The diameter of the pin at B and D is $10\text{mm}$.

The ultimate normal stress $\left({\tau }_U\right)$ is $150\text{MPa}$.

The ultimate normal stress $\left({\sigma }_U\right)$ is $400\text{MPa}$

The factor of safety is 3.0.

Calculation:

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

Refer to Figure 1.

Taking moment about C.

$\sum M_c=0$

$\begin{array}{l}0.280P-0.120F_{BD}=0\\ 0.280P=0.120F_{BD}\\ P=0.42857F_{BD}\end{array}$ (1)

Taking moment about B

$\sum M_B=0$

$\begin{array}{l}0.160P-0.120C=0\\ 0.160P=0.120C\\ P=0.75C\end{array}$ (2)

Find the area of tension on net section of link BD using the relation:

$A_{\text{net}}=d_c\left(t-d_{BD}\right)$ (3)

Here, t is the thickness of the section, $d_c$ is the diameter of pin C, and $d_{BD}$ is diameter of pin at B and D.

Substitute $6\text{mm}$ for $d_c$, $18\text{mm}$ for t, and $10\text{mm}$ for $d_{BD}$ in Equation (3).

$\begin{array}{c}{A}_{\text{net}}=6\left(18-10\right)\\ =48{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^6\text{mm}}\right)}^2\\ =4.8\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

Find the force in member BD on net section of link BD using the relation:

$F_{BD}=\sigma A_{\text{net}}$ (4)

Here, $A_{\text{net}}$ is the area of tension on net section of link BD.

Modify Equation (4).

$F_{BD}=\frac{{\sigma }_U}{F.S.}{A}_{\text{net}}$ (5)

Substitute $400\text{MPa}$ for ${\sigma }_U$, 3 for F.S., and $4.8\times {10}^{-5}{\text{m}}^{\text{2}}$ for $A_{\text{net}}$ in Equation (5).

$\begin{array}{c}{F}_{BD}=\frac{400\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}{3}\times \left(4.8\times {10}^{-5}\right)\\ =\frac{400\times {10}^6}{3}\times \left(4.8\times {10}^{-5}\right)\\ =6,400\text{N}\end{array}$

Find the area of shear in pins BD using the relation:

$A_{\text{pin}}=\frac{\pi d^2}{4}$ (6)

Substitute $10\text{mm}$ for $d$ in Equation (6).

$\begin{array}{c}{A}_{\text{pin}}=\frac{\pi {\left(10\right)}^2}{4}\\ =78.54{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\\ =7.854\times {10}^{-5}{\text{m}}^2\end{array}$

Find the force in member BD shear in pins BD using the relation:

$F_{BD}=\tau A_{\text{pin}}$ (7)

Here, $A_{\text{pin}}$ is the shear in pins at BD.

Modify Equation (7).

$F_{BD}=\frac{{\tau }_U}{F.S.}{A}_{\text{pin}}$ (8)

Substitute $150\text{MPa}$ for ${\tau }_U$, 3 for F.S., and $7.854\times {10}^{-5}{\text{m}}^{\text{2}}$ for $A_{\text{net}}$ in Equation (8).

$\begin{array}{c}{F}_{BD}=15\frac{400\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}{3}\times \left(4.8\times {10}^{-5}\right)\\ =\frac{150\times {10}^6}{3}\times \left(7.8539\times {10}^{-5}\right)\\ =3,926\text{N}\end{array}$

Select the smaller value of $F_{BD}$ is $3,926\text{N}$.

Find the value of P as follows:

Substitute $3,926\text{N}$ for P in Equation (1).

$\begin{array}{c}P=0.428\times 3,926\\ =1,682.5\\ =1,683\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =1.683\text{kN}\end{array}$

Find the shear in pin at C using the relation:

$A_{\text{pin}}=\frac{\pi d^2}{4}$ (9)

Substitute $6\text{mm}$ for $d$ in Equation (9).

$\begin{array}{c}{A}_{\text{pin}}=\frac{\pi {\left(6\right)}^2}{4}\\ =28.27{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\\ =2.8274\times {10}^{-5}{\text{m}}^2\end{array}$

Find the value of C shear in pins C using the relation:

$C=2\tau A_{\text{pin}}$ (10)

Here, $A_{\text{pin}}$ is the shear in pins at C.

Modify Equation (10).

$C=2\left(\frac{{\tau }_U}{F.S.}\right)A_{\text{pin}}$ (11)

Substitute $150\text{MPa}$ for ${\tau }_U$, 3 for F.S., and $2.8274\times {10}^{-5}{\text{m}}^{\text{2}}$ for $A_{\text{net}}$ in Equation (11).

$\begin{array}{c}C=2\left(\frac{150\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}{3}\right)\times \left(2.8274\times {10}^{-5}\right)\\ =2\left(\frac{150\times {10}^6}{3}\right)\times \left(7.8539\times {10}^{-5}\right)\\ =2,827\text{N}\end{array}$

Find the largest load P such that be applied at A.

Substitute $2,827\text{N}$ for C in Equation (2).

$\begin{array}{c}P=0.75\times 2,827\\ =2,120\text{N}\end{array}$

Select the smaller value of P.

Thus, the largest load P such that be applied at A is $\underset{\_}{1.683\text{kN}}$.