Chapter 8.2, Problem 9P

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.

8.9 Loading of Prob. 5.73 and selected W530 × 92 shape.

Fig. P5.73

To determine

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

Answer to Problem 9P

The actual value of ${\sigma }_m$ in the beam is $\underset{\_}{137.5\text{MPa}}$.

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $\text{W}530\times 92$.

Calculation:

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.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -70\left(3\right)-\left(5\times 11\right)\times \frac{11}{2}-70\left(8\right)+D_y\left(11\right)=0\\ -210-302.5-560+11D_y=0\\ D_y=97.5\text{kN}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-70-70-\left(5\times 11\right)+D_y=0\\ A_y-140-55+97.5=0\\ A_y=97.5\text{kN}\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=97.5\text{kN}$

$\begin{array}{c}{V}_B-V_A=-5\times 3\\ V_B^{Left}=-15+V_A\\ =-15+97.5\\ =82.5\text{kN}\end{array}$

$\begin{array}{c}{V}_B^{Right}=82.5-70\\ =12.5\text{kN}\end{array}$

$\begin{array}{c}{V}_C-V_B=-5\times 5-70\\ V_C^{Left}=-95+V_B\\ =-95+82.5\\ =-12.5\text{kN}\end{array}$

$\begin{array}{c}{V}_C^{Right}=-12.5-70\\ =-82.5\text{kN}\end{array}$

$\begin{array}{c}{V}_D-V_C=-5\times 3-70\\ V_D=-15-70+V_C\\ =-85-12.5\\ =-97.5\text{kN}\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$\begin{array}{l}\frac{12.5}{x}=\frac{12.5}{5-x}\\ 62.5-12.5x=12.5x\\ 25x=62.5\\ x=2.5\text{m}\end{array}$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_B-M_A=\left(82.5\times 3\right)+\frac{1}{2}\times \left(97.5-82.5\right)\times 3\\ M_B=247.5+22.5+M_A\\ =270+0\\ =270\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{\max }-M_B=\frac{1}{2}\times 12.5\times 2.5\\ M_{\max }=15.625+M_B\\ =15.625+270\\ =285.625\text{kN-m}\end{array}$

$\begin{array}{c}{M}_C-M_{\max }=-\frac{1}{2}\times 12.5\times 2.5\\ M_C=-15.625+M_{\max }\\ =-15.625+285.625\\ =270\text{kN-m}\end{array}$

$M_D=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

Refer to the Figure 3;

The maximum bending moment in the beam is $\left|M_{\max }\right|=285.625\text{kN-m}$.

Write the section a property for a $W530\times 92$ rolled steel section as table 2.

Dimension	Unit($\text{mm}$)
d	533 mm
$b_f$	209 mm
$t_f$	15.6 mm
$t_w$	$10.2\text{mm}$
I	$554\times {10}^6{\text{mm}}^{\text{4}}$
$S$	$2080\times {10}^3{\text{mm}}^{\text{3}}$

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 $533\text{mm}$ for d in Equation (1).

$\begin{array}{c}c=\frac{1}{2}\times 533\\ =266.5\text{mm}\end{array}$

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

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

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

Substitute $285.625\text{kN-m}$ for ${\left|M\right|}_{\max }$ and $2080\times {10}^3{\text{mm}}^{\text{3}}$ for $S$ in Equation (2).

$\begin{array}{c}{\sigma }_m=\frac{285.625\text{kN-m}\left(\frac{{10}^3\text{N-m}}{\text{kN-m}}\right)}{\begin{array}{l}2080\times {10}^3{\text{mm}}^{\text{3}}\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\hfill \end{array}}\\ =\frac{285.625\times {10}^3}{2.08\times {10}^{-3}}\\ =137,319,711.5\text{Pa}\left(\frac{1\text{Mpa}}{{10}^6\text{Pa}}\right)\end{array}$

${\sigma }_m=137.5\text{Pa}$

Thus, the actual value of ${\sigma }_m$ in the beam is $\underset{\_}{137.5\text{MPa}}$.

To determine

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

Answer to Problem 9P

The maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web is $\underset{\_}{129.5\text{MPa}}$.

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 $266.5\text{mm}$ for c and $\begin{array}{l}15.6\text{ mm}\hfill \end{array}$ for $t_f$ in Equation (3).

$\begin{array}{c}{y}_b=266.5-15.6\\ =250.9\text{mm}\end{array}$

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

$A_f=b_f{t}_f$ (4)

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

Substitute $209\text{mm}$ For $b_f$ and $15.6\text{mm}$ for $t_f$ in Equation (4).

$\begin{array}{c}{A}_f=209\times 15.6\\ =3,260.4{\text{mm}}^{\text{2}}\end{array}$

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

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

Substitute $266.5\text{mm}$ for c and $250.9\text{mm}$ for $y_b$ in Equation (5).

$\begin{array}{c}\overline{y}=\frac{1}{2}\left(266.5+250.9\right)\\ =258.7\text{mm}\end{array}$

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 $3,260.4{\text{mm}}^{\text{2}}$ for $A_f$ and $258.7\text{mm}$ for $\overline{y}$ in Equation (6).

$\begin{array}{c}\left(Q\right)=3,260.4\times 258.7\\ =843.47\times {10}^3{\text{mm}}^3\end{array}$

At mid span the value of $V=0$ and ${\tau }_b=0$.

Find the maximum value of principal stress ${\sigma }_{\max }$ 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 $137.5\text{MPa}$ for ${\sigma }_m$, $250.9\text{mm}$ for $y_b$, and $266.5\text{mm}$ for $c$ in Equation (7).

$\begin{array}{c}{\sigma }_b=\frac{250.9}{266.5}\times 137.5\\ =129.45\text{MPa}\end{array}$

At section B and C.

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

${\sigma }_m=\frac{M}{S}$ (8)

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

Substitute $270\text{kN-m}$ for $M$ and $2080\times {10}^3{\text{mm}}^{\text{3}}$ for $S$ in Equation (8).

$\begin{array}{c}{\sigma }_m=\frac{270\text{kN-m}\left(\frac{{10}^3\text{N-m}}{\text{kN-m}}\right)}{\begin{array}{l}2080\times {10}^3{\text{mm}}^{\text{3}}\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\hfill \end{array}}\\ =\frac{270\times {10}^3}{2.08\times {10}^{-3}}\\ =129,807,692.3\text{Pa}\left(\frac{1\text{Mpa}}{{10}^6\text{Pa}}\right)\end{array}$

${\sigma }_m=129.808\text{Pa}$

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

${\sigma }_b=\frac{y_b}{c}{\sigma }_m$ (9)

Substitute $129.808\text{MPa}$ for ${\sigma }_m$, $250.9\text{mm}$ for $y_b$, and $266.5\text{mm}$ for $c$ in Equation (9).

$\begin{array}{c}{\sigma }_b=\frac{250.9}{266.5}\times 129.808\\ =0.9414\times 129.808\\ =122.20\text{MPa}\end{array}$

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

${\tau }_b=\frac{VQ}{It}$ (10)

Substitute $A_f\overline{y}$ for Q in Equation (10).

${\tau }_b=\frac{V{A}_f\overline{y}}{I{t}_w}$ (11)

Substitute $82.5\text{kN}$ for V, $3,260.4{\text{mm}}^{\text{2}}$ for $A_f$, $258.7\text{mm}$ for $\overline{y}$, $554\times {10}^{-6}{\text{m}}^{\text{4}}$ for I, and $10.2\text{mm}$ for $t_w$ in Equation (11).

$\begin{array}{c}{\tau }_b=\frac{82.5\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 3,260.4{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^{\text{2}}}\right)\times 258.7\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{554\times {10}^{-6}\times 10.2\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{82.5\times {10}^3\times 3.26\times {10}^{-3}\times 0.2587}{554\times {10}^{-6}\times 0.0102}\\ =12,312,834.4\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =12.3143\text{MPa}\end{array}$

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

$R=\sqrt{{\left(\frac{{\sigma }_b}{2}\right)}^2+{\tau }_b^2}$ (12)

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

Substitute $122.20\text{MPa}$ for ${\sigma }_b$ and $12.3143\text{MPa}$ for ${\tau }_b$ in Equation (12).

$\begin{array}{c}R=\sqrt{{\left(\frac{122.20}{2}\right)}^2+{\left(12.3143\right)}^2}\\ =\sqrt{\left(3,733.2+151.641\right)}\\ =62.32\text{MPa}\end{array}$

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

${\sigma }_{\max }=\frac{{\sigma }_b}{2}+R$ (13)

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

Substitute $122.20\text{MPa}$ for ${\sigma }_b$ and $62.32\text{MPa}$ for R in Equation (13).

$\begin{array}{c}{\sigma }_{\max }=\frac{122.20\text{MPa}}{2}+62.32\\ =61.1+62.32\\ =123.4\text{MPa}\end{array}$

Based on results,

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

The maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web is $\underset{\_}{129.5\text{MPa}}$.