Chapter 1.5, Problem 41P

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 bar AB has a cross-sectional area of 225 mm², determine (a) the factor of safety for bar AB, (b) the cross-sectional area of bar AC if it is to have the same factor of safety as bar AB.

Fig. P1.40 and P1.41

To determine

The factor of safety for bar AB.

Answer to Problem 41P

The factor of safety for bar AB is $\underset{\_}{3.97}$.

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 factor of safety for bar AB using the relation:

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

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

Modify Equation (5).

$F.S=\frac{{\sigma }_U{A}_{AB}}{F_{AB}}$ (6)

Substitute $300\times {10}^6\text{Pa}$ for ${\sigma }_U$, $17\text{kN}$ for $F_{AB}$ and $225{\text{mm}}^{\text{2}}$ for $A_{AB}$ in Equation (6).

$\begin{array}{c}F.S=\frac{300\times {10}^6\times 225{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^{\text{2}}}\right)}{17\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)}\\ =\frac{300\times {10}^6\times 225\times {10}^{-6}}{17\times {10}^3}\\ =\frac{67,500}{17\times {10}^3}\\ =3.97\end{array}$

Thus, the factor of safety for bar AB is $\underset{\_}{3.97}$.

To determine

The cross sectional area of bar AC.

Answer to Problem 41P

The cross sectional area of bar AC is $\underset{\_}{265{\text{mm}}^{\text{2}}}$.

Explanation of Solution

Calculation:

Show the expression factor of safety for bar AC using the relation:

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

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

Modify Equation (7).

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

Substitute $300\times {10}^6\text{Pa}$ for ${\sigma }_U$, $20\text{kN}$ for $F_{AB}$ and $3.97$ for $\text{FS}$ in Equation (8).

$\begin{array}{c}{A}_{AC}=\frac{20\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 3.97}{300\times {10}^6}\\ =\frac{79,400}{300\times {10}^6}\\ =264.67\times {10}^{-6}{\text{m}}^{\text{2}}{\left(\frac{{10}^6\text{m}}{1\text{mm}}\right)}^2\\ =265{\text{mm}}^{\text{2}}\end{array}$

Thus, the cross sectional area of bar AC is $\underset{\_}{265{\text{mm}}^{\text{2}}}$.