Chapter 8.2, Problem 14P

8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ_m ≤ σ_all. For the selected design, determine (a) the actual value of σ_m in the beam, (b) the maximum value of the principal stress σ_max at the junction of a flange and the web.

8.14 Loading of Prob. 5.78 and selected S460 × 81.4 shape.

Fig. P5.78

To determine

The actual value of ${\sigma }_m$ in the beam.

Answer to Problem 14P

The actual value of ${\sigma }_m$ in the beam is $\underset{\_}{119.9\text{MPa}}$.

Explanation of Solution

Given information:

Refer to problem 5.78 in chapter 5 in the textbook.

The allowable normal stress of the beam is ${\sigma }_{\text{all}}=160\text{MPa}$.

Calculation:

Design of beam:

Show the free-body diagram of the beam as in Figure 1.

Determine the vertical reaction at point D by taking moment at point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -60\left(2.5\right)-40\left(5\right)+D_y\left(10\right)=0\\ -150-200+10D_y=0\\ D_y=35\text{kN}\end{array}$

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-60-40+D_y=0\\ A_y-100+35=0\\ A_y=65\text{kN}\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=65\text{kN}$

$\begin{array}{c}{V}_B-V_A=0\\ V_B^{Left}=V_A\\ =65\text{kN}\end{array}$

$\begin{array}{c}{V}_B^{Right}=65-60\\ =5\text{kN}\end{array}$

$\begin{array}{c}{V}_C-V_B=0\\ V_C^{Left}=0+V_B\\ =0+5\\ =5\text{kN}\end{array}$

$\begin{array}{c}{V}_C^{Right}=5-40\\ =-35\text{kN}\end{array}$

$\begin{array}{c}{V}_D-V_C=0\\ V_D=V_C\\ =-35\text{kN}\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	65
B (Left)	65
B (Right)	5
C (Left)	5
C (Right)	–35
D	–35

Plot the shear force diagram as in Figure 2.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_B-M_A=65\times 2.5\\ M_B=162.5+M_A\\ =162.5+0\\ =162.5\text{kN-m}\end{array}$

$\begin{array}{c}{M}_C-M_B=5\times 2.5\\ M_C=12.5+M_B\\ =12.5+162.5\\ =175\text{kN-m}\end{array}$

$M_D=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	162.5
C	175
D	0

Plot the bending moment diagram as in Figure 3.

Refer to the Figure 3;

The maximum bending moment in the beam is $\left|M_{\max }\right|=175\text{kN-m}$.

Write the section a property for a $S460\times 81.4$ rolled steel section as table 2.

Dimension	Unit($\text{mm}$)
d	457
$b_f$	152
$t_f$	17.6
$t_w$	$11.7$
$I$	$333\times {10}^6{\text{mm}}^{\text{4}}$
$S$	$1460\times {10}^3{\text{mm}}^{\text{3}}$

Here, d is depth of the section, $b_f$ is breadth of the flange, $t_f$ is thickness of the flange, $I_x$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=\frac{1}{2}d$ (1)

Substitute $457\text{mm}$ for d in Equation (1).

$\begin{array}{c}c=\frac{1}{2}\times 457\\ =228.5\text{mm}\end{array}$

Find the maximum value of normal stress $\left({\sigma }_m\right)$ in the beam using the relation:

${\sigma }_m=\frac{\left|M\right|}{S}$ (2)

Here, $\left|M\right|$ is maximum bending moment and $S$ is the section modulus.

Substitute $175\text{kN}\cdot \text{m}$ for $\left|M\right|$ and $1460\times {10}^3{\text{mm}}^{\text{3}}$ for $S$ in Equation (2).

$\begin{array}{c}{\sigma }_m=\frac{175\text{kN}\cdot \text{m}\left(\frac{{10}^3\text{N}\cdot \text{m}}{1\text{kN}\cdot \text{m}}\right)}{\begin{array}{l}1460\times {10}^3{\text{mm}}^{\text{3}}\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\hfill \end{array}}\\ =\frac{175\times {10}^3}{1.46\times {10}^{-3}}\\ =119,863,013.7\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =119.9\text{MPa}\end{array}$

Thus, the actual value of ${\sigma }_m$ in the beam is $\underset{\_}{119.9\text{MPa}}$.

To determine

The maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web

Answer to Problem 14P

The maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web is $\underset{\_}{110.9\text{MPa}}$.

Explanation of Solution

Calculation:

Find the value $y_b$ as follows:

$y_b=c-t_f$ (3)

Here, c is the centroid and $t_f$ is the thickness of flange.

Substitute $228.5\text{mm}$ for c and $17.6\text{mm}$ for $t_f$ in Equation (3).

$\begin{array}{c}{y}_b=228.5-17.6\\ =210.9\text{mm}\end{array}$

Find the area of flange $\left(A_f\right)$ of section using the relation:

$A_f=\frac{1}{2}{b}_f{t}_f$ (4)

Here, $b_f$ is the width of the flange and $t_f$ is the thickness of the flange.

Substitute $152\text{mm}$ For $b_f$ and $17.6\text{mm}$ for $t_f$ in Equation (4).

$\begin{array}{c}{A}_f=152\times 17.6\\ =2,675.2{\text{mm}}^{\text{2}}\end{array}$

Find the centroid of flange $\overline{y}$ using the relation:

$\overline{y}=\frac{1}{2}\left(c+y_b\right)$ (5)

Substitute $228.5\text{mm}$ for c and $210.9\text{mm}$ for $y_b$ in Equation (5).

$\begin{array}{c}\overline{y}=\frac{1}{2}\left(228.5+210.9\right)\\ =\frac{439.4}{2}\\ \text{=219}\text{.7}\text{mm}\end{array}$

Find the first moment about neutral axis $\left(Q\right)$ as follows:

$\left(Q\right)=A_f\overline{y}$ (6)

Here, $A_f$ is the area of flange and $\overline{y}$ is the centroid.

Substitute $2,675.2{\text{mm}}^{\text{2}}$ for $A_f$ and $\text{219}\text{.7}\text{mm}$ for $\overline{y}$ in Equation (6).

$\begin{array}{c}\left(Q\right)=2,675.2\times 219.7\\ =587.74\times {10}^3{\text{mm}}^{\text{3}}\end{array}$

At section C,

Find the value of ${\sigma }_b$ as follows:

${\sigma }_b=\frac{y_b}{c}{\sigma }_m$ (7)

Here, actual value of normal stress $y_b$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $119.9\text{MPa}$ for ${\sigma }_m$, $210.9\text{mm}$ for $y_b$, and $228.5\text{mm}$ for $c$ in Equation (7).

$\begin{array}{c}{\sigma }_b=\frac{210.9}{228.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\times 119.9\\ =110,664.814\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =110.630\text{MPa}\end{array}$

Find the shear stress at $\left({\tau }_b\right)$ as follows:

${\tau }_b=\frac{VQ}{I{t}_w}$ (8)

Modify Equation (8).

${\tau }_b=\frac{V{A}_f\overline{y}}{I{t}_w}$

Substitute $35\text{kN}$ for V, $2,675.2{\text{mm}}^{\text{2}}$ for $A_f$, $333\times {10}^6{\text{mm}}^{\text{4}}$ for $I$, $219.7\text{mm}$ for $\overline{y}$, and $11.7\text{mm}$ for $t_w$ in Equation (8).

$\begin{array}{c}{\tau }_b=\frac{35\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 2,675.2{\text{mm}}^{\text{2}}\left(\frac{{10}^{-6}{\text{m}}^{\text{2}}}{{\text{mm}}^{\text{2}}}\right)\times 219.7\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{11.7\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times 333\times {10}^6{\text{mm}}^{\text{4}}\left(\frac{1{\text{m}}^{\text{4}}}{1{\text{mm}}^{\text{4}}}\right)}\\ =\frac{35\times {10}^3\times 2,675.2\times {10}^{-6}\times 219.7\times {10}^{-3}}{11.7\times {10}^{-3}\times 333\times {10}^{-6}}\\ =\frac{20.570}{3.896\times {10}^{-06}}\\ =5.2799\text{MPa}\end{array}$

Find the maximum shearing stress (R) using the relation:

$R=\sqrt{{\left(\frac{{\sigma }_b}{2}\right)}^2+{\tau }_b^2}$ (9)

Here, ${\sigma }_b$ is normal bearing stress and ${\tau }_{\text{m}}$ is the shearing stress.

Substitute $110.631\text{MPa}$ for ${\sigma }_b$ and $5.2799\text{MPa}$ for ${\tau }_b$ in Equation (9).

$\begin{array}{c}R=\sqrt{{\left(\frac{110.631\text{MPa}}{2}\right)}^2+{\left(5.2799\right)}^2}\\ =\sqrt{\left(3,059.80+27.877\right)}\\ =55.566\text{MPa}\end{array}$

Determine the maximum value of the principle stress using the relation:

${\sigma }_{\max }=\frac{{\sigma }_b}{2}+R$ (10)

Here, R is the maximum shearing stress and ${\sigma }_b$ is normal bearing stress.

Substitute $110.631\text{MPa}$ for ${\sigma }_b$ and $55.566\text{MPa}$ for R in Equation (10).

$\begin{array}{c}{\sigma }_{\max }=\frac{110.631\text{MPa}}{2}+55.566\\ =55.3155+55.566\\ =110.9\text{Mpa}\end{array}$

At section B,

Find the maximum value of normal stress $\left({\sigma }_m\right)$ in the beam using the relation:

${\sigma }_m=\frac{M}{S}$ (11)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $162.5\text{kN-m}$ for $M$ and $1,460\times {10}^3{\text{mm}}^{\text{3}}$ for $S$ in Equation (11).

$\begin{array}{c}{\sigma }_m=\frac{162.5\text{kN-m}\left(\frac{{10}^3\text{N-m}}{\text{kN-m}}\right)}{\begin{array}{l}1,460\times {10}^3{\text{mm}}^{\text{3}}\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\hfill \end{array}}\\ =\frac{162.5\times {10}^3}{1,460\times {10}^{-6}}\\ =111,301,369.9\text{Pa}\left(\frac{1\text{Mpa}}{{10}^6\text{Pa}}\right)\end{array}$

${\sigma }_m=111.301\text{MPa}$

Find the value of ${\sigma }_b$ as follows:

${\sigma }_b=\frac{y_b}{c}{\sigma }_m$ (12)

Substitute $111.301\text{MPa}$ for ${\sigma }_m$, $210.9\text{mm}$ for $y_b$, and $228.5\text{mm}$ for $c$ in Equation (12).

$\begin{array}{c}{\sigma }_b=\frac{210.9}{228.5}\times 111.301\\ =0.922\times 111.301\\ =102.728\text{MPa}\end{array}$

Find the shear stress at b $\left({\tau }_b\right)$ as follows:

${\tau }_b=\frac{VQ}{It}$ (13)

Substitute $A_f\overline{y}$ for Q in Equation (13).

${\tau }_b=\frac{V{A}_f\overline{y}}{I{t}_w}$ (14)

Refer to figure 3.

Substitute $65\text{kN}$ for V, $2,675.2\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_f$, $219.7\text{mm}$ for $\overline{y}$, $336\times {10}^{-6}{\text{m}}^{\text{4}}$ for I, and $11.7\text{mm}$ for $t_w$ in Equation (14).

$\begin{array}{c}{\tau }_b=\frac{65\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 2,675.2\times {10}^{-6}{\text{m}}^{\text{2}}\times 219.7\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{333\times {10}^{-6}\times 11.7\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{65\times {10}^3\times 2,675.2\times {10}^{-6}\times 219.7\times {10}^{-3}}{333\times {10}^{-6}\times 11.7\times {10}^{-3}}\\ =\frac{38.203}{3.896\times {10}^{-6}}\\ =98,047,630.9\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =9.80\text{MPa}\end{array}$

Find the maximum shearing stress (R) using the relation:

$R=\sqrt{{\left(\frac{{\sigma }_b}{2}\right)}^2+{\tau }_b^2}$ (15)

Here, ${\sigma }_b$ is normal bearing stress and ${\tau }_{\text{m}}$ is the shearing stress.

Substitute $102.728\text{MPa}$ for ${\sigma }_b$ and $9.80\text{MPa}$ for ${\tau }_b$ in Equation (15).

$\begin{array}{c}R=\sqrt{{\left(\frac{102.728\text{MPa}}{2}\right)}^2+{\left(9.80\right)}^2}\\ =\sqrt{\left(2,638.26+96.04\right)}\\ =52.290\text{MPa}\end{array}$

Determine the maximum value of the principle stress using the relation:

${\sigma }_{\max }=\frac{{\sigma }_b}{2}+R$ (16)

Here, R is the maximum shearing stress and ${\sigma }_b$ is normal bearing stress.

Substitute $102.728\text{MPa}$ for ${\sigma }_b$ and $52.290\text{MPa}$ for R in Equation (16).

$\begin{array}{c}{\sigma }_{\max }=\frac{102.728}{2}+52.290\\ =51.364+52.290\\ =103.7\text{MPa}\end{array}$

Based on results,

Select the maximum value of principal stress ${\sigma }_{\max }$.

Thus, the maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web is $\underset{\_}{110.9\text{MPa}}$.