Chapter 1.5, Problem 56P

In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter is to be used at A. Assuming that all other specifications remain unchanged, determine the allowable load P if an overall factor of safety of 3.0 is desired.

1.55 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired.

Fig. P1.55

To determine

The allowable load P when an overall factor of safety of 3.0 is desired.

Answer to Problem 56P

The allowable load P when an overall factor of safety of 3.0 is desired is $\underset{\_}{3.97\text{kN}}$.

Explanation of Solution

Given information:

The diameter (d) of each pin B and D is $12\text{mm}$.

The diameter (d) of pin A is $10\text{mm}$.

The ultimate shearing stress $\left({\tau }_U\right)$ is $100\text{MPa}$.

The ultimate normal stress $\left({\sigma }_U\right)$ is $250\text{MPa}$.

Calculation:

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

Refer to figure 1.

Take a moment about B.

$\sum M_B=0$

$\begin{array}{l}0.20F_A-0.18P=0\\ -0.18P=-0.20F\\ P=1.111F_A\end{array}$

Take a moment about A.

$\sum M_A=0$

$\begin{array}{l}0.20F_{BD}-0.38P=0\\ -0.38P=-0.20F\\ P=0.526F_{BD}\end{array}$

Find the area of pin at A using the relation:

$A=\frac{\pi d^2}{4}$ (1)

Substitute $10\text{mm}$ for d in Equation (1).

$\begin{array}{c}A=\frac{\pi {\left(10\text{mm}\right)}^2}{4}\\ =\frac{314.15}{4}\\ =78.54{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =78.54\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the value of $F_A$ based on double shear pin A using the relation:

$F_A=\frac{2{\tau }_{U}A}{F.S.}$ (2)

Here, A is the double shear pin A.

Substitute $100\text{MPa}$ for $\left({\tau }_U\right)$ and $78.54\times {10}^{-6}{\text{m}}^{\text{2}}$ for A in Equation (2).

$\begin{array}{c}{F}_A=\frac{2\times \left(100\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)78.54\times {10}^{-6}}{3.0}\\ =\frac{15,708}{3}\\ =5,236\text{N}\end{array}$

Find the value of P using the relation:

$P=1.111F_A$

Substitute $5,236\text{N}$ for P.

$\begin{array}{c}P=1.111\left(5,236\text{N}\right)\\ =5,817\text{N}\end{array}$

Find the area of double shear pin at B and D using the relation:

$A=\frac{\pi d^2}{4}$ (3)

Substitute $12\text{mm}$ for d in Equation (3).

$\begin{array}{c}A=\frac{\pi {\left(12\right)}^2}{4}\\ =\frac{452.38}{4}\\ =113.09{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =113.09\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the force in member BD based on double shear in pins at B and D using the relation:

$F_{BD}=\frac{2{\tau }_{U}A}{F.S.}$ (4)

Substitute $100\text{MPa}$ for $\left({\tau }_U\right)$ and $113.10\times {10}^{-6}{\text{m}}^{\text{2}}$ for A in Equation (4).

$\begin{array}{c}{F}_{BD}=\frac{2\times \left(100\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)113.10\times {10}^{-6}{\text{m}}^{\text{2}}}{3.0}\\ =\frac{22,620}{3}\\ =7.540\times {10}^3\text{N}\end{array}$

Find the value of P using the relation:

$P=0.526F_{BD}$

Substitute $7.540\times {10}^3\text{N}$ for P.

$\begin{array}{c}P=0.526\left(7.54\times {10}^3\right)\\ =3.97\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =3.97\text{N}\end{array}$

Find the area based on compression in links BD for one link as follows:

$A=bd$ (5)

Here, d is the diameter of pin and b is the width of the section.

Substitute $20\text{mm}$ for b and $8\text{mm}$ for d in Equation (5).

$\begin{array}{c}A=20\times 8\\ =160{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =160\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the force in member BD of pin at B and D for one link using the relation:

$F_{BD}=\frac{2{\tau }_{U}A}{F.S.}$ (6)

Here, A is the area based on compression in links BD.

Substitute $250\text{MPa}$ for $\left({\tau }_U\right)$ and $160\times {10}^{-6}{\text{m}}^{\text{2}}$ for A in Equation (6).

$\begin{array}{c}{F}_{BD}=\frac{2\times \left(250\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)160\times {10}^{-6}{\text{m}}^{\text{2}}}{3.0}\\ =\frac{80,000}{3}\\ =26.7\times {10}^3\text{N}\end{array}$

Find the value of P using the relation:

$P=0.526F_{BD}$

Substitute $26.7\times {10}^3\text{N}$ for P.

$\begin{array}{c}P=0.526\left(26.7\times {10}^3\text{N}\right)\\ =14.04\times {10}^3\text{N}\end{array}$

Based on results,

Select the smaller value of P is $3.97\times {10}^3\text{N}$.

Thus, the allowable load P when an overall factor of safety of 3.0 is desired is $\underset{\_}{3.97\text{kN}}$.