Chapter 5, Problem 5.6.5P

A cantilever beanie B is loaded by a uniform load q and a concentrated load P, as shown in the figure.

1. Select the most economical steel C shape from Table F-3(a) in Appendix F; use q = 20 lb/ft and P = 300 lb (assume allowable normal stress is cr_a= IS ksi).

Select the most economical steel S shape from Table F-2(a) in Appendix F; use q = 45 lb/ft and P = 2000 lb (assume allowable normal stress is tr_a= 20 ksi),

Select the most economical steel W shape from Table F-1(a) in Appendix F; use q = 45 lb/ft and P = 2000 lb (assume allowable normal stress is (T = 20 ksi). However, assume that the design requires that the W shape must be used in weak axis bending, i.e., it must bend about the 2-2 (or v) axis of the cross section.

Note: For parts (a), (b), and (c), revise your initial beam selection as needed to include the distributed weight of the beam in addition to uniform load q.

  

To determine

The most economical steel $C$shape.

Answer to Problem 5.6.5P

The most economical steel $C$shape is $C15\times \mathrm{33.9.}$

Explanation of Solution

Given information:

The uniform distributed load is $20\text{lb}/\text{ft}$, the load is $300\text{lb}$, maximum normal stress is $18\text{ksi}$.

The following figure shows the free body diagram:

Figure-(1)

Write the expression for the maximum moment of beam.

$M_{\max }=\frac{q{L}^2}{2}+P\left(6\right)$…… (I)

Here, the load is $P$, the uniform load on beam is $q$,and the length of the beam is $L$.

Write the expression for the section modulus.

$S=\frac{M_{\max }}{{\sigma }_{allow}}$…… (II)

Here, the maximum stress is ${\sigma }_{allow}$, and the maximum moment is $M_{\max }$,and the section modulus is $S$.

Write the expression for the maximum stress.

${\sigma }_{\max }=\frac{M_{\max }}{S}$……(III)

Here, the maximum stress is ${\sigma }_{\max }$.

Calculation:

Substitute $20\text{lb}/\text{ft}$for $q$, and $300\text{lb}$for $P$and $10\text{ft}$for $L$in Equation (I).

  $\begin{array}{c}{M}_{\max }=\frac{20\text{lb}/\text{ft}\left({10}^2{\text{ft}}^2\right)}{2}+\left(300\text{lb}\right)\left(6\text{ft}\right)\\ =\left(2800\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ \text{=33600}\text{lb}\cdot \text{in}\text{.}\end{array}$

Substitute $\text{33600}\text{lb}\cdot \text{in}\text{.}$for $M_{\max }$, and $18\text{ksi}$for ${\sigma }_{allow}$in Equation (II).

$\begin{array}{c}S=\frac{33600\text{lb}\cdot \text{in}}{18\text{ksi}}\\ =\frac{\left(33600\text{lb}\cdot \text{in}\right)}{\left(18\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)}\\ =\frac{\left(33600\text{lb}\cdot \text{in}\right)}{\left(18\text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)}\\ =1.87{\text{in}}^3\end{array}$

Refer to the table $\left(a\right)$of “Appendix F” to get the value of weight intensity $w=30\frac{\text{lb}}{\text{ft}}$, and the section modulus $S=2.05{\text{in}}^3$.

Substitute $20\text{lb}/\text{ft}$for $q$, and $300\text{lb}$for $P$and $10\text{ft}$for $L$, $30\frac{\text{lb}}{\text{ft}}$for $w$in Equation (I).

$\begin{array}{c}{M}_{\max }=\frac{\left(\left(20+30\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(300\text{lb}\times 6\text{ft}\right)\\ =\frac{\left(\left(20+30\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(1800\text{lb}\cdot \text{ft}\right)\\ =2150\text{lb}\cdot \text{ft}\end{array}$

Substitute $2150\text{lb}\cdot \text{ft}$for $M_{\max }$, $2.05\text{in}{\text{.}}^3$for $S$in Equation (III).

$\begin{array}{c}{\sigma }_{\max }=\frac{2150\text{lb}\cdot \text{ft}}{2.05\text{in}{\text{.}}^3}\\ =\frac{\left(2150\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{2.05\text{in}{\text{.}}^3}\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =\left(25797\text{psi}\right)\left(\frac{{10}^{-3}\text{ksi}}{1\text{psi}}\right)\\ =25.797\text{ksi}\end{array}$

Here, the maximum stress is greater than the calculated stress, so we neglect this section.

Refer to the table $\left(a\right)$of “Appendix F” to get the value of weight intensity $w=33.9\frac{\text{lb}}{\text{ft}}$, and the section modulus $S=3.09{\text{in}}^3$.

Substitute $20\text{lb}/\text{ft}$for $q$, and $300\text{lb}$for $P$and $10\text{ft}$for $L$, $33.9\frac{\text{lb}}{\text{ft}}$for $w$in Equation (I).

$\begin{array}{c}{M}_{\max }=\frac{\left(\left(20+33.9\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(300\text{lb}\times 6\text{ft}\right)\\ =\frac{\left(\left(20+33.9\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(1800\text{lb}\cdot \text{ft}\right)\\ =3595\text{lb}\cdot \text{ft}\end{array}$

Substitute $3595\text{lb}\cdot \text{ft}$for $M_{\max }$, $3.09\text{in}{\text{.}}^3$for $S$in Equation (III).

$\begin{array}{c}{\sigma }_{\max }=\frac{3595\text{lb}\cdot \text{ft}}{3.09\text{in}{\text{.}}^3}\\ =\frac{\left(3595\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{3.09\text{in}{\text{.}}^3}\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =\left(13961.16\text{psi}\right)\left(\frac{{10}^{-3}\text{ksi}}{1\text{psi}}\right)\\ =13.961\text{ksi}\end{array}$

Hence, from the table “Appendix F” we will use the value $C15\times 33.9$.

Conclusion:

The most economical steel $C$shape is $C15\times 33.9$.

To determine

The most economical steel $S$shape.

Answer to Problem 5.6.5P

The most economical steel $S$shape is $S8\times \mathrm{18.4.}$

Explanation of Solution

Given Information:

The uniform distributed load is $45\text{lb}/\text{ft}$, the load is $2000\text{lb}$, maximum normal stress is $20\text{ksi}$.

Write the expression for the maximum moment of beam.

$M_{\max }=\frac{q{L}^2}{2}+P\left(AC\right)$…… (IV)

Write the expression for the section modulus.

$S=\frac{M_{\max }}{{\sigma }_{allow}}$…… (V)

Write the expression for the maximum stress.

${\sigma }_{\max }=\frac{M_{\max }}{$ 20\text{lb}/\text{ft}$}$

S…… (VI)

Calculation:

Substitute for $q$, and $300\text{lb}$for $P$and $10\text{ft}$for $L$in Equation (IV).

$\begin{array}{c}{M}_{\max }=\frac{45\text{lb}/\text{ft}\left({10}^2{\text{ft}}^2\right)}{2}+\left(2000\text{lb}\right)\left(6\text{ft}\right)\\ =\left(14250\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ \text{=171000}\text{lb}\cdot \text{in}\text{.}\end{array}$

Substitute $171000\text{lb}\cdot \text{in}\text{.}$for $M_{\max }$, and $20\text{ksi}$for ${\sigma }_{allow}$in Equation (V).

$\begin{array}{c}S=\frac{171000\text{lb}\cdot \text{in}\text{.}}{\left(20\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)}\\ =\frac{171000\text{lb}\cdot \text{in}\text{.}}{\left(20000\text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)}\\ =8.55\text{in}{\text{.}}^3\end{array}$

Refer to the table $\left(a\right)$of “Appendix F” to get the value of weight intensity $w=17.2\frac{\text{lb}}{\text{ft}}$, and the section modulus $S=8.74{\text{in}}^3$.

Substitute $45\text{lb}/\text{ft}$for $q$, and $2000\text{lb}$for $P$and $10\text{ft}$for $L$, $17.2\frac{\text{lb}}{\text{ft}}$for $w$in Equation (I).

$\begin{array}{c}{M}_{\max }=\frac{\left(\left(45+17.2\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(2000\text{lb}\times 6\text{ft}\right)\\ ==\frac{\left(\left(45+17.2\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(12000\text{lb}\cdot \text{ft}\right)\\ =18220\text{lb}\cdot \text{ft}\end{array}$

Substitute $18220\text{lb}\cdot \text{ft}$for $M_{\max }$, $8.74\text{in}{\text{.}}^3$for $S$in Equation (VI).

$\begin{array}{c}{\sigma }_{\max }=\frac{\left(18220\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{8.74\text{in}{\text{.}}^3}\\ =\left(25016.01\text{lb}/{\text{in}}^2\right)\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =\left(25016.01\text{psi}\right)\left(\frac{{10}^{-3}\text{ksi}}{1\text{psi}}\right)\\ =25.016\text{ksi}\end{array}$

Here, the maximum stress is greater than the calculated stress, so we neglect this section.

Refer to the table $\left(a\right)$of “Appendix F” to get the value of weight intensity $w=18.4\frac{\text{lb}}{\text{ft}}$, and the section modulus $S=14.4{\text{in}}^3$.

Substitute $45\text{lb}/\text{ft}$for $q$, and $2000\text{lb}$for $P$and $10\text{ft}$for $L$, $18.4\frac{\text{lb}}{\text{ft}}$for $w$in Equation (I).

Substitute $18340\text{lb}\cdot \text{ft}$for $M_{\max }$, $14.4\text{in}{\text{.}}^3$for $S$in Equation (VI).

$\begin{array}{c}{\sigma }_{\max }=\left(15283.33\text{lb}/{\text{in}}^2\right)\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =\left(15283.33\text{psi}\right)\left(\frac{{10}^{-3}\text{ksi}}{1\text{psi}}\right)\\ =15.283\text{ksi}\\ =12.64\text{ksi}\end{array}$

Hence, from the table “Appendix F” we will use the value $S8\times 18.4$.

Conclusion:

The maximum value of load $P$is $S8\times 18.4$.

To determine

The most economical steel $W$shape.

Answer to Problem 5.6.5P

The most economical steel $W$shape is $W8\times 35$.

Explanation of Solution

Given Information:

The uniform distributed load is $45\text{lb}/\text{ft}$, the load is $2000\text{lb}$, maximum normal stress is $20\text{ksi}$.

Write the expression for the maximum moment of beam.

$M_{\max }=\frac{q{L}^2}{2}+P\left(AC\right)$…… (VII)

Write the expression for the section modulus.

$S=\frac{M_{\max }}{{\sigma }_{allow}}$…… (VIII)

Write the expression for the maximum stress.

${\sigma }_{\max }=\frac{M_{\max }}{S}$…… (IX)

Calculation:

Substitute $20\text{lb}/\text{ft}$for $q$, and $300\text{lb}$for $P$and $10\text{ft}$for $L$in Equation (V)

$\begin{array}{c}{M}_{\max }=\frac{45\text{lb}/\text{ft}\left({10}^2{\text{ft}}^2\right)}{2}+\left(2000\text{lb}\right)\left(6\text{ft}\right)\\ =\left(14250\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ \text{=171000}\text{lb}\cdot \text{in}\text{.}\end{array}$

Substitute $171000\text{lb}\cdot \text{in}\text{.}$for $M_{\max }$, and $20\text{ksi}$for ${\sigma }_{allow}$in Equation (VI)

$\begin{array}{c}S=\frac{171000\text{lb}\cdot \text{in}\text{.}}{\left(20\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)}\\ =\frac{171000\text{lb}\cdot \text{in}\text{.}}{\left(20000\text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)}\\ =8.55\text{in}{\text{.}}^3\end{array}$

Refer to the table $\left(a\right)$of “Appendix F” to get the value of weight intensity $w=35\frac{\text{lb}}{\text{ft}}$, and the section modulus $S=10.6{\text{in}}^3$.

Substitute $45\text{lb}/\text{ft}$for $q$, and $2000\text{lb}$for $P$and $10\text{ft}$for $L$, $35\frac{\text{lb}}{\text{ft}}$for $w$in Equation (I).

$\begin{array}{c}{M}_{\max }=\frac{\left(\left(45+35\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(2000\text{lb}\times 6\text{ft}\right)\\ =\frac{\left(\left(45+35\right)\frac{\text{lb}}{\text{ft}}\right)\times {10}^2{\text{ft}}^2}{2}+\left(12000\text{lb}\cdot \text{ft}\right)\\ =10000\text{lb}\cdot \text{ft}\end{array}$

Substitute $18220\text{lb}\cdot \text{ft}$for $M_{\max }$, $8.74\text{in}{\text{.}}^3$for $S$in Equation (VI).

$\begin{array}{c}{\sigma }_{\max }=\frac{\left(10000\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{10.6\text{in}{\text{.}}^3}\\ =\frac{\left(10000\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{10.6\text{in}{\text{.}}^3}\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =11320.75\text{psi}\left(\frac{{10}^{-3}\text{ksi}}{1\text{psi}}\right)\\ =11.320\text{ksi}\end{array}$

Hence, from the table “Appendix F” we will use the value $W8\times 35$.

Conclusion:

The maximum value of load $P$is $W8\times 35$.