Shear stress calculation question

[tsukuba]

Homework Statement

I've been working on this question for a while but can't get anywhere.
I was able to calculate Force BD and Force CE
Now I am not sure how they calculated the area of section BD
the math isn't making much sense to me

Homework Equations

σ = P / A
A= π r²

The Attempt at a Solution

I tried doing A= pi x r² but I don't get the value of 160 x 10^-3 mm²

 

[SteamKing]

The link is not the pin, for which you calculated the area. The link is one of the 4 vertical pieces attaching the bar ABC to the web of the inverted T section under it. The dimensions of each link are 36 mm x 8 mm, according to the problem statement, and there are four links: two attached at point B and two attached at point C. Remember, the 16mm dia. hole for the pin can't be included in the net area of the link.

The solution is quite explicit about describing the area. See the line which states, "Net area of one link for tension = ..."

 

[tsukuba]

I am so confused with the math though. It taking the width of the link and multiplying it by the difference of the length and the diameter.
I would think it as 0.008 x 0.036 - 0.016

 

[SteamKing]

I am so confused with the math though. It taking the width of the link and multiplying it by the difference of the length and the diameter.
I would think it as 0.008 x 0.036 - 0.016

If you draw a sketch of the cross section of the link, and then draw a hole punched thru the link, it should be clear why you calculate the net area of the link in this manner.

It also helps to carry units. Your calculation is A = 0.008 m x 0.036 m - 0.016 m, which when simplified is A = 0.000288 m² - 0.016 m, which makes no sense. You can't subtract meters from square meters and obtain anything meaningful.

This is why parentheses are used in the solution: A = 0.008 * (0.036 - 0.016); you do the subtraction inside the parentheses before you multiply by the thickness of the link.

 

[tsukuba]

how come for the second part they just do 8mmx 36mm and don't subtract the 16mm

 

[SteamKing]

how come for the second part they just do 8mmx 36mm and don't subtract the 16mm

When the link is in tension, the pins are trying to pull out of the ends of each link, so you use the area of the link less the area of the pin in order to calculate the tensile stress.

When the link is in compression, the pins are bearing on the central portion of each link, so you use the total area of the link without subtracting the area of the pin.

 

[tsukuba]

Once again, Thank you very much!
you are of great help