beam - 4

[logistic_guy]

Determine the minimum dimension \(\displaystyle a\) to the nearest \(\displaystyle \text{mm}\) of the beam’s cross section to safely support the load. The wood has an allowable normal stress of \(\displaystyle \sigma_{\text{allow}} = 10 \ \text{MPa}\) and an allowable shear stress of \(\displaystyle \tau_{\text{allow}} = 1 \ \text{MPa}\).

 

[khansaheb]

Determine the minimum dimension \(\displaystyle a\) to the nearest \(\displaystyle \text{mm}\) of the beam’s cross section to safely support the load. The wood has an allowable normal stress of \(\displaystyle \sigma_{\text{allow}} = 10 \ \text{MPa}\) and an allowable shear stress of \(\displaystyle \tau_{\text{allow}} = 1 \ \text{MPa}\).

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[logistic_guy]

Let us design the dimensions of the beam.



From accumulated knowledge, we know that the bending stress is given by:

\(\displaystyle \sigma = \frac{Mc}{I}\)

then

\(\displaystyle \sigma_{\text{allow}} = \frac{M_{\text{max}} \ c}{I} = \frac{M_{\text{max}} \ a}{\frac{1}{12}a(2a)^3} = \frac{M_{\text{max}}}{\frac{8}{12}a^3} = \frac{3M_{\text{max}}}{2a^3}\)


We have two unknowns, so we will have to find a way to get \(\displaystyle M_{\text{max}}\).

 

[logistic_guy]

We can get \(\displaystyle M_{\text{max}}\) by calculating the torques around \(\displaystyle x = 2 \ \text{m}\).

\(\displaystyle M_{\text{max}} = 6000(2) + 6000(1) = 18000 \ \text{N} \cdot \text{m} = 18 \ \text{kN} \cdot \text{m}\)

 

[logistic_guy]

\(\displaystyle \sigma_{\text{allow}} = \frac{3M_{\text{max}}}{2a^3}\)

Plug in numbers.

\(\displaystyle 10 \times 10^6 = \frac{3 \times 18000}{2a^3}\)

This gives:

\(\displaystyle a = 0.13925 \ \text{m} = \textcolor{green}{139.25 \ \text{mm}}\)

\(\displaystyle \large\textcolor{purple}{\text{Is this a safe dimension?}}\)

 

[logistic_guy]

From accumulated knowledge, we know that the shear stress is given by:

\(\displaystyle \tau = \frac{VQ}{It} = \frac{VA\overline{y}}{It} = \frac{Va^2\frac{a}{2}}{\frac{1}{12}a(2a)^3a} = \frac{3V}{4a^2}\)

In this formula, I want to get \(\displaystyle \tau\) to compare it with \(\displaystyle \tau_{\text{allow}}\), but I have two unknowns, so we will have to figure out a way to get \(\displaystyle V \rightarrow\) shear force.

 

[logistic_guy]

If we place \(\displaystyle V\) at \(\displaystyle x = 2 \ \text{m}\), it becomes \(\displaystyle V_{\text{max}}\). If we calculate the vertical forces, we get:

\(\displaystyle V_{\text{max}} - 6000 - 6000 = 0\)

Or

\(\displaystyle V_{\text{max}} = 12000 \ \text{N} = 12 \ \text{kN}\)

 

[logistic_guy]

\(\displaystyle \tau = \frac{3V_{\text{max}}}{4a^2} = \frac{3 \times 12000}{4(0.13925)^2} = 464143 \ \text{Pa} = 0.464143 \ \text{MPa}\)

\(\displaystyle a = 0.13925 \ \text{m} = \textcolor{green}{139.25 \ \text{mm}}\)

\(\displaystyle \large\textcolor{purple}{\text{Is this a safe dimension?}}\)

Since \(\displaystyle \tau = 0.464143 \ \text{MPa}\) \(\displaystyle \ \large\boldsymbol{<} \ \) \(\displaystyle \tau_{\text{allow}}= 1 \ \text{MPa}\)

Yes, the dimension \(\displaystyle a = \textcolor{blue}{139.25 \ \text{mm}}\) can safely support the load.

Determine the minimum dimension \(\displaystyle a\) to the nearest \(\displaystyle \text{mm}\)

\(\displaystyle a = \textcolor{red}{139 \ \text{mm}}\)