Chapter 1.5, Problem 40P

Members AB and BC of the truss shown are made of the same alloy. It is known that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be achieved for both bars, determine the required cross-sectional area of (a) bar AB, (b) bar AC.

Fig. P1.40 and P1.41

To determine

The required cross sectional area of member AB.

Answer to Problem 40P

The required cross sectional area of AB is $\underset{\_}{181.3{\text{mm}}^2}$.

Explanation of Solution

Given information:

The ultimate load $P_U$ is $120\text{kN}$.

The factor of safety F.S is $3.2$.

The area (a) of square cross section is $20\text{mm}$.

Calculation:

Refer to Figure P1.40 in the text book.

Find the length of member $AB$ using the relation:

$\begin{array}{c}{l}_{AB}=\sqrt{{0.75}^2+{0.4}^2}\\ =\sqrt{0.5625+0.16}\\ =0.85\text{m}\end{array}$

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

Here, $A_x$ is the horizontal component of reaction at point A and $A_y$ is the vertical component of reaction.

Refer to Figure 1.

Calculate the horizontal reaction A by using equilibrium Equation as follows:

$\sum M_C=0$

$\begin{array}{l}1.4A_x-\left(0.75\times 28\right)=0\\ 1.4A_x-21=0\\ 1.4A_x=21\\ A_x=15\text{kN}\end{array}$

Calculate the vertical reaction $\left(A_y\right)$ using the relation:

$\sum F_y=0$

$\begin{array}{l}{A}_y-28=0\\ A_y=28\text{kN}\end{array}$

Sketch the free body diagram of joint A as shown in Figure 2.

Refer to Figure P1.40 in the text book.

Refer to Figure 2.

$\sum F_x=0$

$\begin{array}{l}\frac{0.75}{0.85}{F}_{AB}-A_x=0\\ 0.8823F_{AB}-A_x=0\end{array}$ (1)

Substitute $15\text{kN}$ for $A_x$ in Equation (1).

$\begin{array}{l}0.8823F_{AB}-\left(15\text{kN}\right)=0\\ 0.8823F_{AB}=15\\ F_{AB}=\frac{15}{0.8823}\\ F_{AB}=17\text{kN}\end{array}$

Refer to Figure 2.

$\sum F_y=0$

$A_y-F_{AC}-\frac{0.4}{0.85}{F}_{AB}=0$ (2)

Substitute $28\text{kN}$ for $A_y$ and $17\text{kN}$ for $F_{AB}$ in Equation (2).

$\begin{array}{l}28-F_{AC}-\frac{0.4}{0.85}\left(17\right)=0\\ 28-F_{AC}-8=0\\ -F_{AC}+20=0\\ F_{AC}=20\text{kN}\end{array}$

Find the area of test bar (A) using the relation:

$A=a^2$ (3)

Substitute $20\text{mm}$ for a in Equation (3).

$\begin{array}{c}A={20}^2\\ =400{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =400\times {10}^{-6}{\text{m}}^2\end{array}$

Find the ultimate load for the material using the formula:

${\sigma }_U=\frac{P_U}{A}$ (4)

Here, $P_U$ is ultimate load.

Substitute $120\text{kN}$ for $P_U$ and $400\times {10}^{-6}{\text{m}}^2$ for A in Equation (4).

$\begin{array}{c}{\sigma }_U=\frac{120\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)}{400\times {10}^{-6}}\\ =\frac{120\times {10}^3}{400\times {10}^{-6}}\\ =300\times {10}^6\text{Pa}\end{array}$

Determine the area of member $\left(A_{AB}\right)$ using the relation as follows:

Show the expression of factor of safety as follows:

$F.S=\frac{P_U}{F_{AB}}$ (5)

Here, $F_{AB}$ is force in member AB.

Modify Equation (5).

$\begin{array}{l}F.S=\frac{{\sigma }_U{A}_{AB}}{F_{AB}}\\ A_{AB}=\frac{\text{F}\text{.S}\left(F_{AB}\right)}{{\sigma }_U}\end{array}$ (6)

Substitute 3.2 for F.S, $300\times {10}^6\text{Pa}$ for ${\sigma }_U$, and $17\text{kN}$ for $F_{AB}$ in Equation (6).

$\begin{array}{c}{A}_{AB}=\frac{3.2\left(17\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)}{300\times {10}^6}\\ =\frac{54,400}{300\times {10}^6}\\ =181.333\times {10}^{-6}{\text{m}}^{\text{2}}{\left(\frac{{10}^6\text{m}}{1\text{mm}}\right)}^2\\ =181.3{\text{mm}}^{\text{2}}\end{array}$

Thus, the required cross sectional area of AB is $\underset{\_}{181.3{\text{mm}}^2}$.

To determine

The required cross sectional area of AC.

Answer to Problem 40P

The required cross sectional area of AC is $\underset{\_}{213{\text{mm}}^2}$.

Explanation of Solution

Determine the area of member $\left(A_{AC}\right)$ using the relation as follows:

Show the expression of factor of safety as follows:

$F.S=\frac{P_U}{F_{AC}}$ (7)

Modify Equation (7).

$\begin{array}{l}F.S=\frac{{\sigma }_U{A}_{AC}}{F_{AC}}\\ A_{AC}=\frac{\text{F}\text{.S}\left(F_{AC}\right)}{{\sigma }_U}\end{array}$ (8)

Substitute 3.2 for F.S, $300\times {10}^6\text{Pa}$ for ${\sigma }_U$, and $20\text{kN}$ for $F_{AC}$ in Equation (8).

$\begin{array}{c}{A}_{AC}=\frac{3.2\left(20\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)}{300\times {10}^6}\\ =\frac{3.2\times 20\times {10}^3}{300\times {10}^6}\\ =213.33\times {10}^{-6}{\text{m}}^2{\left(\frac{{10}^6\text{m}}{1\text{mm}}\right)}^2\\ =213.3{\text{mm}}^{\text{2}}\end{array}$

Thus, the required cross sectional area of AC is $\underset{\_}{213{\text{mm}}^2}$.