Chapter 8, Problem 65RP

(a) Knowing that σ_all = 24 ksi and σ_all = 14.5 ksi, select the most economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for τ_m, τ_m, and the principal stress σ_max at the junction of a flange and the web of the selected beam.

Fig. P8.65

To determine

Select the most economical wide flange shape section.

Answer to Problem 65RP

The most economical wide flange shape section is $\underset{\_}{\text{W}14\times 22}$.

Explanation of Solution

Given information:

The allowable bending stress ${\sigma }_{\text{all}}$ is $24\text{ksi}$.

The shear stress ${\tau }_{\text{all}}$ is $14.5\text{ksi}$.

Calculation:

Sketch the free body diagram of beam as shown in figure 1.

Here, $R_A$ is reaction at point A and $R_D$ is reaction at point D.

Calculate the reaction for the given structure:

Taking moment about D,

$\sum M_D=0$

$\begin{array}{l}-15R_A+\left(10.5\times 9\times 2\right)+\left(3\times 12.5\right)=0\\ -15R_A+\left(189\right)+\left(37.5\right)=0\\ -15R_A+226.5=0\\ R_A=15.1\text{kips}\uparrow \end{array}$

Summation of vertical force zero.

$\sum F_y=0$

$\begin{array}{l}-15.1+18=-12.5+R_D\\ R_D=15.4\text{kips}\end{array}$

Calculate the shear force diagram as follows:

Shear force at A is $S{F}_A=0$.

Shear force at B,

$\begin{array}{c}S{F}_B=15.1-2\times 9\\ =15.1-18\\ =-2.9\text{kips}\end{array}$

Shear force at C,

$\begin{array}{c}S{F}_{@c}=15.1-2\times 9-12.5\\ =15.1-18-12.5\\ =-15.4\text{kips}\end{array}$

Shear force at D,

$\begin{array}{c}S{F}_D=15.1-\left(2\times 9\right)-12.5+15.4\\ =15.1-18-12.5+15.4\\ =0\end{array}$

Maximum shear force occurs at the point where $SF=0$.

Consider a section $x-x$ at a distance x from A.

$\begin{array}{l}S{F}_{@x-x}=15.1-\left(2\times x\right)=0\\ 2x=15.1\\ x=7.55\text{ft}\end{array}$

Calculate the bending moment as follows:

Bending moment at point A,

$B{M}_A=0$

Bending moment at point B,

$\begin{array}{c}B{M}_B=15.1\times 9-\left(2\times 9\times \frac{9}{2}\right)\\ =135.9-81\\ =54.9\text{kips}\cdot \text{ft}\end{array}$

Bending moment at point C,

$\begin{array}{c}B{M}_C=15.1\times 12-\left(2\times 9\left(\frac{9}{2}+3\right)\right)\\ =181.2-135\\ =46.2\text{kips}\cdot \text{ft}\end{array}$

Bending moment at point D,

$\begin{array}{c}B{M}_D=15.1\times 15-2\times 9\left(\frac{9}{2}+6\right)-12.5\times 3\\ =226.5-189-37.5\\ =0\end{array}$

Sketch the shear force and bending moment diagram as shown in Figure 2.

Refer to Figure 2.

The maximum bending moment is $57.003\text{kips}\cdot \text{ft}$.

Find the value of $S_{\min }$ using the relation:

$S_{\min }=\frac{{\left|M\right|}_{\max }}{{\sigma }_{\text{all}}}$ (1)

Here, ${\left|M\right|}_{\max }$ is maximum moment.

Substitute $57.003\text{kips}\cdot \text{ft}$ for ${\left|M\right|}_{\max }$ and $24\text{ksi}$ for ${\sigma }_{\text{all}}$ in Equation (1).

$\begin{array}{c}{S}_{\min }=\frac{57.003\times 12}{24}\\ =28.502{\text{in}}^{\text{3}}\end{array}$

Write the section properties shape as shown in Table 2.

Shape	$S\left({\text{in}}^{\text{3}}\right)$
$W16\times 26$	38.4
$W14\times 22$	29.0
$W12\times 26$	33.4
$W10\times 30$	32.4
$W8\times 35$	31.2

Select the section $W14\times 22$ wide flange section.

The most economical wide flange shape section is $\underset{\_}{W14\times 22}$.

Write the section properties wide flange section as shown in Table 3.

Shape	W14x22
Area, A	$6.49{\text{in}}^{\text{2}}$
Depth, d	$13.74\text{in}$
Web thickness, $t_w$	0.230 in.
Width $b_f$	5.000 in.
Thickness $t_f$	0.335 in.

Find the area of web $\left(A_{\text{web}}\right)$ using the relation:

$A_{\text{web}}=d{t}_w$ (2)

Here, d is the depth of the section and $t_w$ is thickness of web.

Substitute $13.7\text{in}\text{.}$ for d and $0.230\text{in}\text{.}$ for $t_w$ in Equation (2).

$\begin{array}{c}{A}_{\text{web}}=13.7\times 0.230\\ =3.151{\text{in}}^{\text{2}}\end{array}$

To determine

The value to be expected for ${\sigma }_m,{\tau }_m$ and the principal stress ${\sigma }_{\max }$ at the junction of a flange and the web of the selected beam.

Answer to Problem 65RP

The normal stress ${\sigma }_m$ at point is $\underset{\_}{23.6\text{ksi}}$

The shear stress ${\tau }_m$ at point is $\underset{\_}{4.89\text{ksi}}$

The maximum principal stress is $\underset{\_}{22.4\text{ksi}}$

Explanation of Solution

Calculation:

Point E:

Find the normal stress at point E using the relation:

${\sigma }_m=\frac{M_E}{S}$ (3)

Here, $M_E$ moment at point E and S is the section property.

Substitute $57.003\text{kips}\cdot \text{ft}$ for $M_E$ and $29{\text{in}}^3$ for S in Equation (3).

$\begin{array}{c}{\sigma }_m=\frac{57.003\text{kips}\cdot \text{ft}\left(\frac{12\text{kips}\cdot \text{in}}{1\text{kips}\cdot \text{ft}}\right)}{29.0}\\ =\frac{57.003\times 12}{29.0}\\ =23.587\text{ksi}\\ \simeq \text{23}\text{.6}\text{ksi}\end{array}$

Thus, the normal stress ${\sigma }_m$ at point is $\underset{\_}{23.6\text{ksi}}$

Find the value of C using the relation:

$c=\frac{d}{2}$ (4)

Here, d is the depth of section.

Substitute $13.7\text{in}$ for d in Equation (4).

$\begin{array}{c}c=\frac{13.7}{2}\\ =6.85\text{in}\text{.}\end{array}$

Find the value of $y_b$ using the relation:

$y_b=c-t_f$ (5)

Here, $t_f$ is the thickness of flange.

Substitute $6.85\text{in}\text{.}$ for c and $0.335\text{in}\text{.}$ for $t_f$ in Equation (5).

$\begin{array}{c}{y}_b=6.85-0.335\\ =6.515\text{in}\text{.}\end{array}$

Find the normal stress ${\sigma }_b$ at point E using the relation:

${\sigma }_b=\frac{y_b}{c}{\sigma }_{\text{m}}$ (6)

Here, ${\sigma }_{\text{m}}$ is the normal stress,

Substitute $6.515\text{in}\text{.}$ for $y_b$, $6.85\text{in}\text{.}$ for c, and $23.587\text{ksi}$ for ${\sigma }_{\text{m}}$ in Equation (6).

$\begin{array}{c}{\sigma }_b=\frac{6.515}{6.85}\times \text{23}\text{.587}\\ \text{=0}\text{.9510}\times \text{23}\text{.587}\\ \text{=22}\text{.43}\text{ksi}\end{array}$

Find the shear stress ${\tau }_m$ at point E using the relation:

The point E is located at top. Since Q is zero.

Thus, the shear stress at point ${\tau }_m$ is zero.

Point C:

Find the normal stress at point E using the relation:

${\sigma }_m=\frac{M_c}{S}$ (7)

Here, $M_c$ moment at point C and S is the section property.

Substitute $46.2\text{kips}\cdot \text{ft}$ for $M_C$ and $29{\text{in}}^3$ for S in Equation (7).

$\begin{array}{c}{\sigma }_m=\frac{46.2\text{kips}\cdot \text{ft}\text{kips}\cdot \text{ft}\left(\frac{12\text{kips}\cdot \text{in}}{1\text{kips}\cdot \text{ft}}\right)}{29.0}\\ =\frac{46.2\times 12}{29.0}\\ =19.1172\text{ksi}\end{array}$

Find the shear stress point C using the relation:

${\tau }_m=\frac{\left|V\right|}{A_{\text{web}}}$ (8)

Here, V is shear force and $A_{\text{web}}$ is the area of web.

Substitute $15.4\text{kips}$ for V and $3.151{\text{in}}^{\text{2}}$ for $A_{\text{web}}$ in Equation (8).

$\begin{array}{c}{\tau }_m=\frac{15.4}{3.151}\\ =4.8877\text{ksi}\end{array}$

Thus, the shear stress ${\tau }_m$ at point is $\underset{\_}{4.89\text{ksi}}$

Find the ${\sigma }_b$ point c using the relation:

${\sigma }_b=\frac{y_b}{c}{\sigma }_{\text{m}}$ (9)

Substitute $6.515\text{in}\text{.}$ for $y_b$, $6.85\text{in}\text{.}$ for c, and $19.1172\text{ksi}$ for ${\sigma }_{\text{m}}$ in Equation (9).

$\begin{array}{c}{\sigma }_b=\frac{6.515}{6.85}\times 19.1172\\ \text{=0}\text{.9510}\times \text{19}\text{.1172}\\ \text{=18}\text{.1823}\text{ksi}\end{array}$

Find the R using the relation:

$R=\sqrt{{\left(\frac{{\sigma }_b}{2}\right)}^2+{\tau }_m}$ (10)

Here, ${\sigma }_b$ is the normal stress at point c and ${\tau }_m$ is shearing stress at point C.

Substitute $\text{18}\text{.1823}\text{ksi}$ for ${\sigma }_b$ and $4.8877\text{ksi}$ for ${\tau }_m$ in Equation (10).

$\begin{array}{c}R=\sqrt{{\left(\frac{\text{18}\text{.1823}}{2}\right)}^2+4.8877}\\ =\sqrt{82.64+23.88}\\ =10.3216\text{ksi}\end{array}$

Find the maximum principal stress using the relation:

${\sigma }_{\max }=\frac{{\sigma }_b}{2}+R$ (11)

Substitute $\text{18}\text{.1823}\text{ksi}$ for ${\sigma }_b$ for $10.3216\text{ksi}$ for R in Equation (11).

$\begin{array}{c}{\sigma }_{\max }=\frac{\text{18}\text{.1823}}{2}+10.3216\\ =9.090615+10.3216\\ =19.41\text{ksi}\end{array}$

Compare the results,

Select the maximum value of stress for Point B is controls.

Thus, the maximum principal stress is $\underset{\_}{22.4\text{ksi}}$.