Chapter 5, Problem 5.1PFS

To determine

(a)

The critical buckling load using Euler equation for $L=4ft,d=1.5in$.

Answer to Problem 5.1PFS

$30.87kips$

Explanation of Solution

Given:

$\begin{array}{l}L=4ft\\ d=1.5in\end{array}$

Calculation:

The area of a given solid round bar is

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

Substitute the d value 1.5inches.

$\begin{array}{c}=\frac{\pi {\left(1.5\right)}^2}{4}\\ =1.77i{n}^2\end{array}$

Moment of inertia of solid round bar:

$\begin{array}{c}I=\frac{\pi d^4}{64}\\ putdas1.5in\\ =\frac{\pi {\left(1.5\right)}^4}{64}\\ =0.25i{n}^4\end{array}$

Calculating radius of gyration:

$r=\sqrt{\frac{I}{A}}$

Substitute the values of I and A in the formula:

$\begin{array}{c}r=\sqrt{\frac{I}{A}}\\ =\sqrt{\frac{0.25}{1.77}}\\ =0.376in\end{array}$

Calculate the slenderness ratio for L= 4ft:

$\begin{array}{c}slendernessratio=\frac{L}{r}\\ =\frac{4\times 12in}{0.375in}\\ =128<200\end{array}$

Determine critical buckling stress:

$\begin{array}{l}{F}_c=\frac{{\pi }^2\times E}{{\left(\frac{L}{r}\right)}^2}\\ where\\ E=\text{elastic modulus}\end{array}$

Substitute the values in formula:

$\begin{array}{c}{F}_c=\frac{{\pi }^2\times E}{{\left(\frac{L}{r}\right)}^2}\\ =\frac{{\pi }^2\times 29000ksi}{{\left(128\right)}^2}\\ =17.47ksi<36ksi\end{array}$

The buckling stress is less than the limit of 36 ksi. The column is in elastic range and safe.

Calculation of critical buckling load:

$P_{cr}=F_c\times A$

Put in the values

$\begin{array}{c}{P}_{cr}=F_c\times A\\ =17.47\times 1.767\\ =30.87kips\end{array}$

Conclusion:

Therefore, the critical buckling load is $30.87kips$.

To determine

(b)

The critical buckling load using Euler equation for $L=2ft9in,d=1.5in$.

Answer to Problem 5.1PFS

The buckling stress is not less than the limit of $36ksi$. The column is in inelastic range. Therefore, Euler equation is not applicable.

Explanation of Solution

Given:

$\begin{array}{l}L=2ft9in\\ d=1.5in\end{array}$

Calculation:

Calculate the slenderness ratio for L= 2ft 9in:

$\begin{array}{c}slendernessratio=\frac{L}{r}\\ =\frac{2ft+\frac{9in}{12}ft\times 12in}{0.375in}\\ =88<200\end{array}$

Determine critical buckling stress:

$\begin{array}{l}{F}_c=\frac{{\pi }^2\times E}{{\left(\frac{L}{r}\right)}^2}\\ where\\ E=\text{elastic modulus}\end{array}$

Substitute the values in formula:

$\begin{array}{c}{F}_c=\frac{{\pi }^2\times E}{{\left(\frac{L}{r}\right)}^2}\\ =\frac{{\pi }^2\times 29000ksi}{{\left(88\right)}^2}\\ =39.96ksi<36ksi\end{array}$

Conclusion:

The buckling stress is not less than the limit of $36ksi$. The column is in inelastic range. Therefore, Euler equation is not applicable.

To determine

(c)

To find: critical buckling load using Euler equation for $L=7ft6in,d=1.5in$.

Answer to Problem 5.1PFS

Euler equation is not applicable as the slenderness ratio exceeds 200.

Explanation of Solution

Given:

$\begin{array}{l}L=7ft6in\\ d=1.5in\end{array}$

Calculation:

Calculate the slenderness ratio for L= 7ft 6in:

$\begin{array}{c}slendernessratio=\frac{L}{r}\\ =\frac{7ft+\frac{6in}{12}ft\times 12in}{0.375in}\\ =240>200\end{array}$

Conclusion:

Therefore, Euler’s equation is not applicable as the slenderness ratio exceeds 200.