Chapter 5.1, Problem 33P

A solid steel bar has a square cross section of side b and is supported as shown. Knowing that for steel ρ = 7860 kg/m³, determine the dimension b for which the maximum normal stress due to bending is (a) 10 MPa, (b) 50 MPa.

Fig. P5.33

To determine

the dimension b.

Answer to Problem 33P

The dimension b of the square cross section is $\underset{\_}{33.3\text{mm}}$.

Explanation of Solution

Given information:

The maximum normal stress due to bending is 10 MPa.

Determine the weight density $\left(\gamma \right)$ using the relation.

$\gamma =\rho g$

Here, the mass density of the bar is $\rho$ and the acceleration due to gravity is g.

Consider the acceleration due to gravity as $9.81{\text{m/s}}^{\text{2}}$.

Substitute $7860{\text{kg/m}}^3$ for $\rho$ and $9.81{\text{m/s}}^{\text{2}}$ for g.

$\begin{array}{c}\gamma =7860\times 9.81\\ =77,106.6{\text{N/m}}^3\end{array}$

Determine the dead load (W) of the solid steel bar using the relation.

$\begin{array}{l}\rho =\frac{W}{AL}\\ W=\gamma \times b\times b\times L\\ w=\frac{W}{L}=\gamma b^2\end{array}$

Here, the cross sectional area of the steel bar is A, the length of the beam is L, and the dimension of the bar is d.

Convert the mass density into weight density as follows;

Substitute $77,106.6{\text{N/m}}^3$ for $\gamma$.

$w=\frac{W}{L}=77,106.6b^2$

Determine the reactions of the beam.

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

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

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ \left(w\times 2.4\right)\times \frac{2.4}{2}-C_y\left(1.2\right)-\left(w\times 1.2\right)\times \frac{1.2}{2}=0\\ 2.88w-1.2C_y-0.72w=0\\ C_y=1.8w\end{array}$

Show the free-body diagram of the sections as in Figure 2.

Region AC (Section 1-1):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ wx\times \frac{x}{2}+M=0\\ M=-\frac{w{x}^2}{2}\end{array}$

Region CD (Section 2-2):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ wx\times \frac{x}{2}-1.8w\left(x-1.2\right)+M=0\\ \frac{w{x}^2}{2}-1.8wx+2.16w+M=0\\ M=-\frac{w{x}^2}{2}+1.8wx-2.16w\end{array}$

Region DB (Section 3-3):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ -M-w\left(3.6-x\right)\times \frac{\left(3.6-x\right)}{2}=0\\ -M-\frac{w}{2}{\left(3.6-x\right)}^2=0\\ M=-\frac{w}{2}{\left(3.6-x\right)}^2\end{array}$

Bending moment values:

Show the calculated bending moment values as in Table 1.

Location (x) m	Bending moment (M) N-m
A (0 m)	0
C (1-1) (1.2 m)	–0.72w
C (2-2) (1.2 m)	–0.72w
Mid-point (1.8 m)	–0.54w
D (2-2) (2.4 m)	–0.72w
D (3-3) (2.4 m)	–0.72w
B (3.6 m)	0

Plot the bending moment diagram as in Figure 3.

Refer to Figure 3;

The maximum bending moment is $\left|M_{\max }\right|=0.72w$

Determine the section modulus (S) of the square section using the equation.

$S=\frac{b^3}{6}$

Determine the maximum normal stress $\left({\sigma }_m\right)$ using the equation.

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

Substitute $0.72w$ for M and $\frac{b^3}{6}$ for S.

${\sigma }_m=\frac{0.72w}{\frac{b^3}{6}}$

Substitute 10 MPa for ${\sigma }_m$ and $77,106.6b^2$ for w.

$\begin{array}{c}10\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=\frac{0.72\times 77,106.6b^2}{\frac{b^3}{6}}\\ b=\frac{0.72\times 77,106.6\times 6}{10\times {10}^6}\\ =0.0333\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}\\ =33.3\text{mm}\end{array}$

Therefore, the dimension b of the square cross section is $\underset{\_}{33.3\text{mm}}$.

To determine

the dimension b.

Answer to Problem 33P

The dimension b of the square cross section is $\underset{\_}{6.66\text{mm}}$.

Explanation of Solution

Given information:

The maximum normal stress due to bending is 50 MPa.

Determine the weight density $\left(\gamma \right)$ using the relation.

$\gamma =\rho g$

Here, the mass density of the bar is $\rho$ and the acceleration due to gravity is g.

Consider the acceleration due to gravity as $9.81{\text{m/s}}^{\text{2}}$.

Substitute $7860{\text{kg/m}}^3$ for $\rho$ and $9.81{\text{m/s}}^{\text{2}}$ for g.

$\begin{array}{c}\gamma =7860\times 9.81\\ =77,106.6{\text{N/m}}^3\end{array}$

Determine the dead load (W) of the solid steel bar using the relation.

$\begin{array}{l}\rho =\frac{W}{AL}\\ W=\gamma \times b\times b\times L\\ w=\frac{W}{L}=\gamma b^2\end{array}$

Here, the cross sectional area of the steel bar is A, the length of the beam is L, and the dimension of the bar is d.

Convert the mass density into weight density as follows;

Substitute $77,106.6{\text{N/m}}^3$ for $\gamma$.

$w=\frac{W}{L}=77,106.6b^2$

Determine the reactions of the beam.

Show the free-body diagram of the beam as in Figure 4.

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

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ \left(w\times 2.4\right)\times \frac{2.4}{2}-C_y\left(1.2\right)-\left(w\times 1.2\right)\times \frac{1.2}{2}=0\\ 2.88w-1.2C_y-0.72w=0\\ C_y=1.8w\end{array}$

Show the free-body diagram of the sections as in Figure 5.

Region AC (Section 1-1):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ wx\times \frac{x}{2}+M=0\\ M=-\frac{w{x}^2}{2}\end{array}$

Region CD (Section 2-2):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ wx\times \frac{x}{2}-1.8w\left(x-1.2\right)+M=0\\ \frac{w{x}^2}{2}-1.8wx+2.16w+M=0\\ M=-\frac{w{x}^2}{2}+1.8wx-2.16w\end{array}$

Region DB (Section 3-3):

Determine the bending moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M=0}\\ -M-w\left(3.6-x\right)\times \frac{\left(3.6-x\right)}{2}=0\\ -M-\frac{w}{2}{\left(3.6-x\right)}^2=0\\ M=-\frac{w}{2}{\left(3.6-x\right)}^2\end{array}$

Bending moment values:

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) N-m
A (0 m)	0
C (1-1) (1.2 m)	–0.72w
C (2-2) (1.2 m)	–0.72w
Mid-point (1.8 m)	–0.54w
D (2-2) (2.4 m)	–0.72w
D (3-3) (2.4 m)	–0.72w
B (3.6 m)	0

Plot the bending moment diagram as in Figure 6.

Refer to the Figure 6;

The maximum bending moment is $\left|M_{\max }\right|=0.72w$

Determine the section modulus (S) of the square section using the equation.

$S=\frac{b^3}{6}$

Determine the maximum normal stress $\left({\sigma }_m\right)$ using the equation.

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

Substitute $0.72w$ for M and $\frac{b^3}{6}$ for S.

${\sigma }_m=\frac{0.72w}{\frac{b^3}{6}}$

Substitute 50 MPa for ${\sigma }_m$ and $77,106.6b^2$ for w.

$\begin{array}{c}50\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=\frac{0.72\times 77,106.6b^2}{\frac{b^3}{6}}\\ b=\frac{0.72\times 77,106.6\times 6}{50\times {10}^6}\\ =6.66\times {10}^{-3}\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}\\ =6.66\text{mm}\end{array}$

Therefore, the dimension b of the square cross section is $\underset{\_}{6.66\text{mm}}$.