A square column carries a load P at the centroid of one of the quarters of the square. If a is the side of the main square, the combined bending stress will be A. $$\frac{{\text{P}}}{{{{\text{a}}^2}}}$$ B. $$\frac{{2{\text{P}}}}{{{{\text{a}}^2}}}$$ C. $$\frac{{3{\text{P}}}}{{{{\text{a}}^2}}}$$ D. $$\frac{{4{\text{P}}}}{{{{\text{a}}^2}}}$$

$$rac{{ ext{P}}}{{{{ ext{a}}^2}}}$$
$$rac{{2{ ext{P}}}}{{{{ ext{a}}^2}}}$$
$$rac{{3{ ext{P}}}}{{{{ ext{a}}^2}}}$$
$$rac{{4{ ext{P}}}}{{{{ ext{a}}^2}}}$$

The correct answer is $\frac{{3{\text{P}}}}{{{{\text{a}}^2}}}$.

The bending stress is calculated by the following formula:

$$\sigma = \frac{M}{I}$$

where:

  • $\sigma$ is the bending stress
  • $M$ is the bending moment
  • $I$ is the moment of inertia

The bending moment is calculated by the following formula:

$$M = P\cdot d$$

where:

  • $M$ is the bending moment
  • $P$ is the load
  • $d$ is the distance from the load to the centroid of the beam

The moment of inertia of a square beam is calculated by the following formula:

$$I = \frac{1}{36}{\text{a}^4}$$

where:

  • $I$ is the moment of inertia
  • $a$ is the side of the square

Substituting the expressions for $M$ and $I$ into the expression for $\sigma$, we get:

$$\sigma = \frac{{P\cdot d}}{{\frac{1}{36}{\text{a}^4}}}$$

$$\sigma = \frac{{36{\text{P}}}}{{{{\text{a}}^4}}}\cdot d$$

In this case, the load is applied at the centroid of one of the quarters of the square. The distance from the load to the centroid of the beam is $\frac{a}{2}$. Therefore, the bending stress is:

$$\sigma = \frac{{36{\text{P}}}}{{{{\text{a}}^4}}}\cdot \frac{a}{2}$$

$$\sigma = \frac{{3{\text{P}}}}{{{{\text{a}}^2}}}$$

Exit mobile version