Chapter 6, Problem 6.1CP

Using appropriate measurements and data, explain how to determine the bending stress in the blade.

C6–1

To determine

The bending stress in the blade.

Answer to Problem 6.1CP

The bending stress in the blade is $\underset{\_}{\sigma =\frac{3}{4}\frac{w{\left(\pi r\right)}^2}{b{d}^2}}$.

Explanation of Solution

Given information:

• The steel saw blade passes over the drive wheel of the band saw.
• Use appropriate measurements and data.

Explanation:

The contact area of the cable is upper half portion of the drive wheel. The, the upper half portion of the wheel will undergo stress.

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

The force induced in the drive wheel will be uniformly distributed.

The circumferential distance of the circular section is $2\pi r$. Here, r is the radius of the circular section.

Convert the semi-circular section into beam section as in Figure 2.

Determine the tension in the cable:

Moment about point A:

Determine the tension in the cable at point B by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \left(w\times \pi r\times \frac{\pi r}{2}\right)-T_B\left(\pi r\right)=0\end{array}$ (1)

Along the vertical direction:

Determine the tension in the cable at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \left(w\times \pi r\right)-T_A-T_B=0\end{array}$ (2)

Show the calculation of values as follows:

Solve Equation (1).

$\begin{array}{l}{T}_B\left(\pi r\right)=w\times \pi r\times \frac{\pi r}{2}\\ T_B=\frac{w\pi r}{2}\end{array}$

Substitute $\frac{w\pi r}{2}$ for $T_B$ in Equation (2).

$\begin{array}{l}w\times \pi r-T_A-\frac{w\pi r}{2}=0\\ T_A=\frac{w\pi r}{2}\end{array}$

Maximum Bending moment:

The maximum bending moment will occur where the shear force changes sign

Consider a section at a distance x from left end of the beam.

Show the free body diagram of the section as in Figure 3.

Along the vertical direction:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -V+\left(w\times x-T_A\right)=0\end{array}$ (3)

Moment about the section:

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ M+T_A\left(x\right)-w\times x\times \frac{x}{2}=0\end{array}$ (4)

Substitute 0 for V and $\frac{w\pi r}{2}$ for $T_A$ in Equation (3).

$\begin{array}{l}-0+w\times x-\frac{w\pi r}{2}=0\\ wx=\frac{w\pi r}{2}\\ x=\frac{\pi r}{2}\end{array}$

Thus, the maximum bending moment will occur at a distance $\frac{\pi r}{2}$ from left end.

Substitute $\frac{w\pi r}{2}$ for $T_A$ and $\frac{\pi r}{2}$ for x in Equation (4).

$\begin{array}{l}{M}_{\max }+\frac{w\pi r}{2}\left(\frac{\pi r}{2}\right)-w\times \frac{\pi r}{2}\times \frac{\pi r}{4}=0\\ M_{\max }+\frac{w{\left(\pi r\right)}^2}{4}-\frac{w{\left(\pi r\right)}^2}{8}=0\\ M_{\max }=-\frac{w{\left(\pi r\right)}^2}{8}\end{array}$

Bending stress:

Calculate the bending stress in the blade using the flexure formula.

$\sigma =\frac{M_{\max }c}{I}$ (5)

Here, c is the distance between the centroid and the extreme fibre and I is moment of inertia of the band saw.

Consider the band saw is in rectangular cross section.

The value of c is $c=\frac{d}{2}$.

The moment of inertia of the band saw is $I=\frac{b{d}^3}{12}$.

Here, b is width of the section and d is depth of the section.

Substitute $\frac{w{\left(\pi r\right)}^2}{8}$ for $M_{\max }$, $\frac{d}{2}$ for c, and $\frac{b{d}^3}{12}$ for I in Equation (5).

$\begin{array}{c}\sigma =\frac{\frac{w{\left(\pi r\right)}^2}{8}\times \frac{d}{2}}{\frac{b{d}^3}{12}}\\ =\frac{w{\left(\pi r\right)}^2}{8}\times \frac{d}{2}\times \frac{12}{b{d}^3}\\ =\frac{3}{4}\frac{w{\left(\pi r\right)}^2}{b{d}^2}\end{array}$

Thus, the bending stress in the blade is $\underset{\_}{\sigma =\frac{3}{4}\frac{w{\left(\pi r\right)}^2}{b{d}^2}}$.