Chapter 10, Problem 128RP

(a)

To determine

Find the maximum allowable eccentricity e.

Answer to Problem 128RP

The maximum allowable eccentricity is $\underset{\_}{0.0987\text{in}\text{.}}$.

Explanation of Solution

Given information:

The effective length of the steel tube is $L_e=14\text{ft}$.

The magnitude of the vertical load is $P=55\text{kips}$.

The allowable yield strength in the column is ${\sigma }_Y=36\text{ksi}$.

The modulus of elasticity of the material is $E=29\times {10}^6\text{psi}$.

Calculation:

The outer dimension $\left(d_o\right)$ of the steel tube is $d_o=4\text{in}\text{.}$.

The inner dimension $\left(d_i\right)$ of the steel tube is;

$\begin{array}{c}{d}_i=4-2\left(\frac{3}{8}\right)\\ =3.25\text{in}\text{.}\end{array}$

Find the cross sectional area of the steel tube (A) using the equation.

$A=d_o^2-d_i^2$

Substitute 4 in. for $d_o$ and 3.25 in. for $d_i$.

$\begin{array}{c}A=4^2-{3.25}^2\\ =5.4375\text{in}{\text{.}}^2\end{array}$

Find the moment of inertia of the square cross section (I) using the equation.

$I=\frac{d_o^4}{12}-\frac{d_i^4}{12}$

Substitute 4 in. for $d_o$ and 3.25 in. for $d_i$.

$\begin{array}{c}I=\frac{4^4}{12}-\frac{{3.25}^4}{12}\\ =12.0361\text{in}{\text{.}}^4\end{array}$

Find the minimum radius of gyration (r) using the relation.

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

Substitute $12.0361\text{in}{\text{.}}^4$ for I and $5.4375\text{in}{\text{.}}^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{12.0361}{5.4375}}\\ =1.4878\text{in}\text{.}\end{array}$

Find the slenderness ratio $\frac{L}{r}$ using the equation.

$\frac{L}{r}=4.71\sqrt{\frac{E}{{\sigma }_Y}}$

Here, the modulus of elasticity of the material is E and the allowable yield stress is ${\sigma }_Y$

Substitute $29\times {10}^6\text{psi}$ for E and 36 ksi for ${\sigma }_Y$

$\begin{array}{c}\frac{L}{r}=4.71\sqrt{\frac{29\times {10}^6}{36\text{ksi}\times \frac{\text{1,000}\text{psi}}{\text{1}\text{ksi}}}}\\ =133.68\end{array}$

Find the ratio of effective length $\left(L_e\right)$ and the minimum radius of gyration $\left(r_y\right)$ as follows;

$\begin{array}{c}\frac{L_e}{r_y}=\frac{14\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}}{1.4878}\\ =112.92<\frac{L}{r}=133.68\end{array}$

Find the effective stress $\left({\sigma }_e\right)$ using the equation.

${\sigma }_e=\frac{{\pi }^{2}E}{{\left(L_e/r_y\right)}^2}$

Substitute $29\times {10}^6\text{psi}$ for E and 112.92 for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 29\times {10}^6\text{psi}\times \frac{\text{1}\text{ksi}}{\text{1,000}\text{psi}}}{{\left(112.92\right)}^2}\\ =22.447\text{ksi}\end{array}$

Find the critical stress $\left({\sigma }_{\text{cr}}\right)$ using the relation.

${\sigma }_{\text{cr}}=\left[{0.658}^{\frac{{\sigma }_Y}{{\sigma }_e}}\right]{\sigma }_Y$

Substitute 36 ksi for ${\sigma }_Y$ and 22.447 ksi for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=\left[{0.658}^{\frac{36}{22.447}}\right]\times 36\\ =18.398\text{ksi}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}=\frac{{\sigma }_{\text{cr}}}{1.67}$

Substitute 18.398 ksi for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{18.398}{1.67}\\ =11.017\text{ksi}\end{array}$

Find the moment acting in the column (M) using the relation.

$M=Pe$

Here, the vertical load is P and the eccentricity of the load is e.

Substitute 55 kips for P.

$\begin{array}{c}M=55\times e\\ =55e\end{array}$

The distance between the neutral axis and the outermost fibre is;

$c=\frac{d_0}{2}=\frac{4}{2}=2\text{in}\text{.}$.

Find the allowable stress in the column $\left({\sigma }_{all}\right)$ using the bending equation.

${\sigma }_{all}=\frac{P}{A}+\frac{Mc}{I}$

Substitute 11.017 ksi for ${\sigma }_{all}$, 55 kips for P, $5.4375\text{in}{\text{.}}^2$ for A, 55e for M, 2 in. for c, and $12.0361\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}11.017=\frac{55}{5.4375}+\frac{55e\times 2}{12.0361}\\ 9.13917e=11.017-10.11494\\ e=0.0987\text{in}\text{.}\end{array}$

Therefore, the maximum allowable eccentricity is $\underset{\_}{0.0987\text{in}\text{.}}$.

(b)

To determine

Find the maximum allowable eccentricity e.

Answer to Problem 128RP

The maximum allowable eccentricity is $\underset{\_}{0.787\text{in}\text{.}}$.

Explanation of Solution

Given information:

The effective length of the steel tube is $L_e=14\text{ft}$.

The magnitude of the vertical load is $P=35\text{kips}$.

The allowable yield strength in the column is ${\sigma }_Y=36\text{ksi}$.

The modulus of elasticity of the material is $E=29\times {10}^6\text{psi}$.

Calculation:

The outer dimension $\left(d_o\right)$ of the steel tube is $d_o=4\text{in}\text{.}$.

The inner dimension $\left(d_i\right)$ of the steel tube is;

$\begin{array}{c}{d}_i=4-2\left(\frac{3}{8}\right)\\ =3.25\text{in}\text{.}\end{array}$

Find the cross sectional area of the steel tube (A) using the equation.

$A=d_o^2-d_i^2$

Substitute 4 in. for $d_o$ and 3.25 in. for $d_i$.

$\begin{array}{c}A=4^2-{3.25}^2\\ =5.4375\text{in}{\text{.}}^2\end{array}$

Find the moment of inertia of the square cross section (I) using the equation.

$I=\frac{d_o^4}{12}-\frac{d_i^4}{12}$

Substitute 4 in. for $d_o$ and 3.25 in. for $d_i$.

$\begin{array}{c}I=\frac{4^4}{12}-\frac{{3.25}^4}{12}\\ =12.0361\text{in}{\text{.}}^4\end{array}$

Find the minimum radius of gyration (r) using the relation.

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

Substitute $12.0361\text{in}{\text{.}}^4$ for I and $5.4375\text{in}{\text{.}}^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{12.0361}{5.4375}}\\ =1.4878\text{in}\text{.}\end{array}$

Find the slenderness ratio $\frac{L}{r}$ using the equation.

$\frac{L}{r}=4.71\sqrt{\frac{E}{{\sigma }_Y}}$

Substitute $29\times {10}^6\text{psi}$ for E and 36 ksi for ${\sigma }_Y$

$\begin{array}{c}\frac{L}{r}=4.71\sqrt{\frac{29\times {10}^6}{36\text{ksi}\times \frac{\text{1,000}\text{psi}}{\text{1}\text{ksi}}}}\\ =133.68\end{array}$

Find the ratio of effective length $\left(L_e\right)$ and the minimum radius of gyration $\left(r_y\right)$ as follows;

$\begin{array}{c}\frac{L_e}{r_y}=\frac{14\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}}{1.4878}\\ =112.92<\frac{L}{r}=133.68\end{array}$

Find the effective stress $\left({\sigma }_e\right)$ using the equation.

${\sigma }_e=\frac{{\pi }^{2}E}{{\left(L_e/r_y\right)}^2}$

Substitute $29\times {10}^6\text{psi}$ for E and 112.92 for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 29\times {10}^6\text{psi}\times \frac{\text{1}\text{ksi}}{\text{1,000}\text{psi}}}{{\left(112.92\right)}^2}\\ =22.447\text{ksi}\end{array}$

Find the critical stress $\left({\sigma }_{\text{cr}}\right)$ using the relation.

${\sigma }_{\text{cr}}=\left[{0.658}^{\frac{{\sigma }_Y}{{\sigma }_e}}\right]{\sigma }_Y$

Substitute 36 ksi for ${\sigma }_Y$ and 22.447 ksi for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=\left[{0.658}^{\frac{36}{22.447}}\right]\times 36\\ =18.398\text{ksi}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}=\frac{{\sigma }_{\text{cr}}}{1.67}$

Substitute 18.398 ksi for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{18.398}{1.67}\\ =11.017\text{ksi}\end{array}$

Find the moment acting in the column (M) using the relation.

$M=Pe$

Substitute 35 kips for P.

$\begin{array}{c}M=35\times e\\ =35e\end{array}$

The distance between the neutral axis and the outermost fibre is;

$c=\frac{d_0}{2}=\frac{4}{2}=2\text{in}\text{.}$

Find the allowable stress in the column $\left({\sigma }_{all}\right)$ using the bending equation.

${\sigma }_{all}=\frac{P}{A}+\frac{Mc}{I}$

Substitute 11.017 ksi for ${\sigma }_{all}$, 35 kips for P, $5.4375\text{in}{\text{.}}^2$ for A, 35e for M, 2 in. for c, and $12.0361\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}11.017=\frac{35}{5.4375}+\frac{35e\times 2}{12.0361}\\ 5.81584e=11.017-6.43678\\ e=0.787\text{in}\text{.}\end{array}$

Therefore, the maximum allowable eccentricity is $\underset{\_}{0.787\text{in}\text{.}}$.