Chapter 10.1, Problem 27P

Each of the five struts shown consists of a solid steel rod. (a) Knowing that the strut of Fig. (1) is of a 20-mm diameter, determine the factor of safety with respect to buckling for the loading shown. (b) Determine the diameter of each of the other struts for which the factor of safety is the same as the factor of safety obtained in part a. Use E = 200 GPa.

Fig. P10.27

To determine

Find the factor of safety with respect to buckling.

Answer to Problem 27P

The factor of safety with respect to buckling is $\underset{\_}{2.55}$.

Explanation of Solution

The dimeter of the strut (1) is $d_1=20\text{mm}$.

The centric load in the strut (1) is $P_0=7.5\text{kN}$.

The modulus of elasticity of the column is $E=200\text{GPa}$.

Determine the moment of inertia of the strut (1) $\left(I_1\right)$ using the relation.

$I_1=\frac{\pi d_1^4}{64}$

Here, the diameter of the strut 1 is $d_1$.

Substitute 20 mm for $d_1$.

$\begin{array}{c}{I}_1=\frac{\pi \times {20}^4}{64}\\ =7,853.98{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^4\\ =7.85398\times {10}^{-9}{\text{m}}^4\end{array}$

Both the ends are pin connected.

The effective length of the column $\left(L_e\right)$ is equal to the length of the column (L).

$L=L_e=900\text{mm}$

Determine the critical load $\left(P_{\text{cr}}\right)$ using the equation.

$P_{cr}=\frac{{\pi }^{2}E{I}_1}{{\left(L_e\right)}^2}$

Here, the modulus of elasticity is E.

Substitute 200 GPa for E, $7.85398\times {10}^{-9}{\text{m}}^4$ for $I_1$, and 900 mm for $L_e$.

$\begin{array}{c}{P}_{\text{cr}}=\frac{{\pi }^2\times 200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\times 7.85398\times {10}^{-9}}{{\left(900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2}\\ =19,140\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =19.14\text{kN}\end{array}$

Determine the factor of safety (FOS) using the relation.

$FOS=\frac{P_{cr}}{P_0}$

Here, the allowable load in strut 1 is $P_0$.

Substitute 19.14 kN for $P_{\text{cr}}$ and 7.5 kN for $P_0$.

$\begin{array}{c}FOS=\frac{19.14}{7.5}\\ =2.55\end{array}$

Therefore, the factor of safety with respect to buckling is $\underset{\_}{2.55}$.

To determine

Find the diameter of the other struts for the condition that the factor of safety is same.

Answer to Problem 27P

The diameter of the strut (2) is $\underset{\_}{28.3\text{mm}}$.

The diameter of the strut (3) is $\underset{\_}{14.14\text{mm}}$.

The diameter of the strut (4) is $\underset{\_}{16.73\text{mm}}$.

The diameter of the strut (5) is $\underset{\_}{20\text{mm}}$.

Explanation of Solution

Determine the factor of safety (FOS) using the relation.

$\begin{array}{l}FOS=\frac{P_{cr}}{P_0}\\ FOS\infty P_{\text{cr}}\end{array}$

Therefore, the factor of safety is directly proportional to the critical load.

$\begin{array}{l}FOS\infty \frac{{\pi }^{2}E{I}_1}{{\left(L_e\right)}_1^2}\\ FOS\infty \frac{I_1}{{\left(L_e\right)}_1^2}\\ \frac{I_i}{{\left(L_e\right)}_i^2}=\frac{I_1}{{\left(L_e\right)}_1^2}\\ \frac{d_i^4}{{\left(L_e\right)}_i^2}=\frac{d_1^4}{{\left(L_e\right)}_1^2}\end{array}$ (1)

Here, the moment of inertia of ith strut is $I_i$¸the effective length of the ith strut is ${\left(L_e\right)}_i$, the moment of inertia of the first strut is $I_1$, the effective length of the first strut is ${\left(L_e\right)}_1$, the diameter of the ith strut is $d_i$, and the diameter of the first strut is $d_1$.

Strut (2);

Show the effective length of the strut (2) as in Figure 1.

The effective length of the strut (2) is twice the length of the strut (1).

${\left(L_e\right)}_2=2L$

Substitute 2 for i, 2L for ${\left(L_e\right)}_2$, 20 mm for $d_1$, and L for ${\left(L_e\right)}_1$ in Equation (1).

$\begin{array}{l}\frac{d_2^4}{{\left(2L\right)}^2}=\frac{{20}^4}{L^2}\\ d_2=28.3\text{mm}\end{array}$

Therefore, the diameter of the strut (2) is $\underset{\_}{28.3\text{mm}}$.

Strut (3);

Show the effective length of the strut (3) as in Figure 2.

The effective length of the strut (3) is half the length of the strut (1).

${\left(L_e\right)}_3=\frac{L}{2}$

Substitute 3 for i, $\frac{L}{2}$ for ${\left(L_e\right)}_3$, 20 mm for $d_1$, and L for ${\left(L_e\right)}_1$ in Equation (1).

$\begin{array}{l}\frac{d_3^4}{{\left(\frac{L}{2}\right)}^2}=\frac{{20}^4}{L^2}\\ d_3=14.14\text{mm}\end{array}$

Therefore, the diameter of the strut (3) is $\underset{\_}{14.14\text{mm}}$.

Strut (4);

Show the effective length of the strut (4) as in Figure 3.

The effective length of the strut (4) is 0.7 times the length of the strut (1).

${\left(L_e\right)}_4=0.7L$

Substitute 4 for i, $0.7L$ for ${\left(L_e\right)}_4$, 20 mm for $d_1$, and L for ${\left(L_e\right)}_1$ in Equation (1).

$\begin{array}{l}\frac{d_4^4}{{\left(0.7L\right)}^2}=\frac{{20}^4}{L^2}\\ d_4=16.73\text{mm}\end{array}$

Therefore, the diameter of the strut (4) is $\underset{\_}{16.73\text{mm}}$.

Strut (5);

Show the effective length of the strut (5) as in Figure 4.

The effective length of the strut (5) is equal to the length of the strut (1).

${\left(L_e\right)}_5=L$

Substitute 5 for i, L for ${\left(L_e\right)}_5$, 20 mm for $d_1$, and L for ${\left(L_e\right)}_1$ in Equation (1).

$\begin{array}{l}\frac{d_5^4}{L^2}=\frac{{20}^4}{L^2}\\ d_5=20\text{mm}\end{array}$

Therefore, the diameter of the strut (5) is $\underset{\_}{20\text{mm}}$.