Question

If the (square) beam in fig 12-6aassociated sample problem is of Douglasfir, what must be its thickness to keep the compressive stress on it to$\frac{\mathbf{1}}{\mathbf{6}}$ of its ultimate strength?

Short answer

The thickness of Douglas fir is, $\text{T}=0.031\text{\hspace{0.17em}m}$

Step-by-step solution

Understanding the given information

The force exerted on the beam,$\text{F}=7900\text{\hspace{0.17em}N}$(from sample problem of fig 12.6)

The Ultimate strength of Douglas fir, ${\text{S}}_{\text{u}}=50\times {10}^6\frac{\text{N}}{{\text{m}}^2}(\mathrm{From}\mathrm{table}12.1)$

Concept and formula used in the given question

You can use the concept of stress to find the thickness of Douglas fir. The formulas used are given below.

$\begin{array}{c}\mathbf{Stress}\mathbf{=}\mathbf{}\frac{\mathbf{F}}{\mathbf{A}}\mathbf{}\\ \mathbf{A}\mathbf{=}{\mathbf{T}}^{\mathbf{2}}\mathbf{}\end{array}$

Calculation for the thickness to keep the compressive stress on it to 16 of its ultimate strength

Rearranging the stress equation for area A, we get

$\mathrm{A}=\frac{\mathrm{F}}{\mathrm{Stress}}$

But we have given that the compressive stress is 1/6 of the ultimate strength that is,

$\mathrm{Stress}=\frac{1}{6}{\mathrm{S}}_{\mathrm{u}}$

So, the equation for the area will become,

$\begin{array}{c}\mathrm{A}=6\frac{\mathrm{F}}{{\mathrm{S}}_{\mathrm{u}}}\\ =6\times \frac{7900\mathrm{N}}{50\times {10}^6\mathrm{N}/{\mathrm{m}}^2}\\ =9.48\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Now, the shape of Douglas fir is square, so its area is

$\mathrm{A}={\mathrm{T}}^2$

Hence, the thickness T is,

$\begin{array}{c}\mathrm{T}=\sqrt{\mathrm{A}}\\ =\sqrt{9.48\times {10}^{-4}{\mathrm{m}}^2}\\ =0.031\text{\hspace{0.17em}m}\end{array}$