# strain 205 conflict with 206 kinda,,

/reviewer/mechanics-and-strength-of-materials/solution-to-problem-205-axial-deformation

please see this,,, sir/madam,,, why is it in problem 206

/reviewer/mechanics-and-strength-of-materials/solution-to-problem-206-axial-deformation

unit mass of 7850kg/m3 is not used to solve for the deformation due to weight directly to the formula,Deformation= (unit mass)(gravity)(L^2)/(2)(E) or solve for the force which is equal to the weight = (unit mass)(g)(AxL) which is equal to (7850kg/m3)(9.81m/s^2)(300x10^-3m^2)(150m)[my understanding] which the answer unit is (kg-m)/s^2 that is also equal to newton

rather

weight=force

P= 7850(1/1000)3(9.81)[300(150)(1000)] which i do not know where (1/1000)^3 came from and also the multiplier 1000 and where has 1x10^-3 for the milli go?

uhm... if i where to use my solution for the deformation,,, what comes as answer was different from the answer provided in 206.

which i guess is quite wrong or my knowledge with the problem is lacking

please reply ,,,, i really want to know why.

thank you,

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## Re: strain 205 conflict with 206 kinda,,

The multiplier 1000 is from meter to millimeter and the (1/1000)^3 is from (1/m

^{3}) to (1/mm^{3}).$V = 300 ~ \text{mm}^2 (150\,000 ~ \text{mm}) = 45\,000\,000 ~ \text{mm}^3$

Unit weight

$\gamma = 7850 ~ \dfrac{\text{kg}}{\text{m}^3} \times 9.81 ~ \dfrac{\text{m}}{\text{s}^2} \times \left(\dfrac{1 ~ \text{m}}{1000 ~ \text{mm}} \right)^3$

$\gamma = 77.0085 \times 10^{-6} ~ \text{N/mm}^3$

Weight

$W = \gamma V = (77.0085 \times 10^{-6})(45\,000\,000)$

$W = 3465.3825 ~ \text{N}$

You may also use the formula derived in Problem 205:

$\delta = 0.00433 ~ \text{m} = 4.33 ~ \text{mm}$

## Re: strain 205 conflict with 206 kinda,,

thanks for the comment sir,,, i just got the wrong idea with mm^2,I thought it is equal to 10^-3 m^2 but i got my mind clear with it that mm^2 is (1/1000 m)^2,,, one of my classmates told me about it,,,we had an argument yesterday,,, hahhaha,, thanks to you and her ,sir,,, i won't approach my professor now,,, which I have in mind since i got wrong with the solution i had in my midterm exam,,, thanks again sir ,,I have escaped embarrassment hahahha,,, promise ill do it right in my final exam ,,, smile!

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