maximum load

[logistic_guy]

here is the question

If the allowable bearing stress for the material under the supports at \(\displaystyle A\) and \(\displaystyle B\) is \(\displaystyle (\sigma_b)_{allow} = 1.5\) MPa, determine the maximum load \(\displaystyle P\) that can be applied to the beam. The bearing plates \(\displaystyle A'\) and \(\displaystyle B'\) have square cross sections of \(\displaystyle 150 \) mm \(\displaystyle \times \ 150\) mm and \(\displaystyle 250 \) mm \(\displaystyle \times \ 250\) mm, respectively.



my attemb
to solve this question, i need first to convert units to \(\displaystyle SI\)
i don't understand how \(\displaystyle 150\) mm \(\displaystyle \times \ 150\) mm \(\displaystyle = 0.0225\) m\(\displaystyle ^2\)

 

[topsquark]

here is the question

If the allowable bearing stress for the material under the supports at \(\displaystyle A\) and \(\displaystyle B\) is \(\displaystyle (\sigma_b)_{allow} = 1.5\) MPa, determine the maximum load \(\displaystyle P\) that can be applied to the beam. The bearing plates \(\displaystyle A'\) and \(\displaystyle B'\) have square cross sections of \(\displaystyle 150 \) mm \(\displaystyle \times \ 150\) mm and \(\displaystyle 250 \) mm \(\displaystyle \times \ 250\) mm, respectively.

View attachment 38990

my attemb
to solve this question, i need first to convert units to \(\displaystyle SI\)
i don't understand how \(\displaystyle 150\) mm \(\displaystyle \times \ 150\) mm \(\displaystyle = 0.0225\) m\(\displaystyle ^2\)

Basic unit conversions. Yet another review topic.

Convert 150 mm to m. Square that number.

-Dan

 

[logistic_guy]

Basic unit conversions. Yet another review topic.

Convert 150 mm to m. Square that number.

-Dan

that's simple?

\(\displaystyle 150 \times mm \times \frac{m}{1000 \ mm} = 0.150 \ m\)

\(\displaystyle 0.150^2 \ m^2 = 0.0255 \ m^2\)

it's confusing me when i work with two numbers in the same time
i can't believe i do it easily in this way