Chapter 8.2, Problem 10P

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.

8.10 Loading of Prob. 5.74 and selected W250 × 28.4 shape.

Fig. P5.74

To determine

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

Answer to Problem 10P

The actual value of ${\sigma }_m$ in the beam is $\underset{\_}{155.8\text{MPa}}$.

Explanation of Solution

Given information:

Refer to problem 5.74 in chapter 5 in the textbook.

The rolled steel section is $\text{W}250\times 28.4$.

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ \left(50\times 3.2\right)\times 2.4-B_y\left(3.2\right)=0\\ 384-3.2B_y=0\\ B_y=120\text{kN}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ B_y-\left(50\times 3.2\right)+D_y=0\\ B_y-160+120=0\\ B_y=40\text{kN}\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=0$

$\begin{array}{c}{V}_B-V_A=-50\times 0.8\\ V_B^{Left}=-40+V_A\\ =-40+0\\ =-40\text{kN}\end{array}$

$\begin{array}{c}{V}_B^{Right}=-40+120\\ =80\text{kN}\end{array}$

$\begin{array}{c}{V}_C-V_B=-50\times 2.4+120\\ V_C=-120+120+V_B\\ =0-40\\ =-40\text{kN}\end{array}$

$\begin{array}{c}{V}_D-V_C=0\\ V_D=V_C\\ =-40\text{kN}\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	0
B (Left)	–40
B (Right)	80
C	–40
D	–40

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.

$\begin{array}{l}\frac{80}{x}=\frac{40}{2.4-x}\\ 192-80x=40x\\ 120x=192\\ x=1.6\text{m}\end{array}$

The maximum bending moment occurs at a distance 2.4 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_B-M_A=-\frac{1}{2}\times 40\times 0.8\\ M_B=-16+M_A\\ =-16+0\\ =-16\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{\max }-M_B=\frac{1}{2}\times 80\times 1.6\\ M_{\max }=64+M_B\\ =64-16\\ =48\text{kN-m}\end{array}$

$\begin{array}{c}{M}_C-M_{\max }=-\frac{1}{2}\times 40\times 0.8\\ M_C=-16+M_{\max }\\ =-16+48\\ =32\text{kN-m}\end{array}$

$M_D=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	–16
Max BM	48
C	32
D	0

Plot the bending moment diagram as in Figure 3.

Refer to the Figure 3;

The maximum bending moment in the beam is $\left|M_{\max }\right|=48\text{kN-m}$.

$\left|M_{\max }\right|=48\text{kN-m}$ at section E, that lies $1.6\text{m}$ to the right of B.

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

Dimension	Unit($\text{mm}$)
d	259 mm
$b_f$	102 mm
$t_f$	10.0 mm
$t_w$	$6.35\text{mm}$
I	$40.1\times {10}^6{\text{mm}}^{\text{4}}$
$S$	$308\times {10}^3{\text{mm}}^{\text{3}}$

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 $259\text{mm}$ for d in Equation (1).

$\begin{array}{c}c=\frac{1}{2}\times 259\\ =129.5\text{mm}\end{array}$

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 $48\text{kN-m}$ for ${\left|M\right|}_{\max }$ and $308\times {10}^3{\text{mm}}^{\text{3}}$ for $S$ in Equation (2).

$\begin{array}{c}{\sigma }_m=\frac{48\text{kN-m}\left(\frac{{10}^3\text{N-m}}{\text{kN-m}}\right)}{\begin{array}{l}308\times {10}^3{\text{mm}}^{\text{3}}\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\hfill \end{array}}\\ =\frac{48\times {10}^3}{308\times {10}^{-04}}\\ =155,844,155.8\text{Pa}\left(\frac{1\text{Mpa}}{{10}^6\text{Pa}}\right)\end{array}$

${\sigma }_m=155.8\text{MPa}$

Thus, the actual value of ${\sigma }_m$ in the beam is $\underset{\_}{155.8\text{MPa}}$.

To determine

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

Answer to Problem 10P

The maximum value of principal stress ${\sigma }_{\max }$ at the junction of a flange and the web is $\underset{\_}{143.8\text{MPa}}$.

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 $129.5\text{mm}$ for c and $10\text{mm}$ for $t_f$ in Equation (3).

$\begin{array}{c}{y}_b=129.5-10\\ =119.5\text{mm}\end{array}$

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 $102\text{mm}$ For $b_f$ and $10\text{mm}$ for $t_f$ in Equation (4).

$\begin{array}{c}{A}_f=102\times 10\\ =1,020{\text{mm}}^{\text{2}}\end{array}$

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

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

Substitute $129.5\text{mm}$ for c and $119.5\text{mm}$ for $y_b$ in Equation (5).

$\begin{array}{c}\overline{y}=\frac{1}{2}\left(129.5+119.5\right)\\ =\frac{249}{2}\\ \text{=124}\text{.5}\text{mm}\end{array}$

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 $1,020{\text{mm}}^{\text{2}}$ for $A_f$ and $124.5\text{mm}$ for $\overline{y}$ in Equation (6).

$\begin{array}{c}\left(Q\right)=1,020\times 124.5\\ =126.99\times {10}^3{\text{mm}}^3\left(\frac{{10}^{-9}{\text{m}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\right)\\ =126.99\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

At midspan the value of $V=0$ and $M=M_{\max }$.

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 $155.8\text{MPa}$ for ${\sigma }_m$, $119.5\text{mm}$ for $y_b$, and $129.5\text{mm}$ for $c$ in Equation (7).

$\begin{array}{c}{\sigma }_b=\frac{119.5}{129.5}\times 155.8\\ =0.922\times 155.8\\ =143.76\text{MPa}\end{array}$

Find the shear stress ${\tau }_b$ as follows:

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

At midspan the value of $V=0$.

Therefore, the shear stress ${\tau }_b$ is zero.

At section C,

The shear force at point C is $40\text{kN}$.

The bending moment about C is $32\text{kN}\cdot \text{m}$.

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

${\sigma }_b=\frac{M{y}_b}{I}$ (8)

Substitute $32\text{kN}\cdot \text{m}$ for $M$, $119.5\text{mm}$ for $y_b$, and $40.1\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I$ in Equation (8).

$\begin{array}{c}{\sigma }_b=\frac{32\text{kN}\cdot \text{m}\left(\frac{{10}^3\text{N}\cdot \text{m}}{1\text{kN}\cdot \text{m}}\right)}{40.1\times {10}^{-6}}\times 119.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =\frac{32\times {10}^3}{40.1\times {10}^{-6}}\times 0.1195\\ =95,361,596.01\\ =95.36\text{MPa}\end{array}$

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

${\tau }_b=\frac{VQ}{It}$ (9)

Substitute $40\text{kN}$ for V, $126.99\times {10}^{-6}{\text{m}}^{\text{3}}$ for $Q$, $40.1\times {10}^6{\text{mm}}^{\text{4}}$ for I, and $6.35\text{mm}$ for $t_w$ in Equation (9).

$\begin{array}{c}{\tau }_b=\frac{40\text{kN}\frac{{10}^3\text{N}}{1\text{kN}}\times \left(126.99\times {10}^{-6}\right)}{40.1\times {10}^6{\text{mm}}^{\text{4}}\left(\frac{{10}^{\text{-12}}{\text{m}}^{\text{4}}}{1{\text{mm}}^4}\right)\times 6.35\text{mm}\left(\frac{1{\text{m}}^3}{{10}^3\text{mm}}\right)}\\ =\frac{40\times {10}^3\times 126.99\times {10}^{-6}}{4.01\times {10}^{-05}\left(0.00635\right)}\\ =\frac{5.0796}{2.54635\left({10}^{-7}\right)}\\ =19.95\text{MPa}\end{array}$

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

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

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

Substitute $95.36\text{MPa}$ for ${\sigma }_b$ and $19.95\text{MPa}$ for ${\tau }_b$ in Equation (10).

$\begin{array}{c}R=\sqrt{{\left(\frac{95.36}{2}\right)}^2+{\left(19.95\right)}^2}\\ =\sqrt{\left(2,273.38+398.00\right)}\\ =51.68\text{MPa}\end{array}$

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

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

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

Substitute $95.36\text{MPa}$ for ${\sigma }_b$ and $51.68\text{MPa}$ for R in Equation (12).

$\begin{array}{c}{\sigma }_{\max }=\frac{95.36}{2}+51.68\\ =47.68+51.68\\ =99.6\text{MPa}\end{array}$

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 $\underset{\_}{143.8\text{MPa}}$.