0
$\begingroup$

In this notes , i dont understand how the Q=(Ay) of the is considered ...

I think the both Q at part C and part B should be = (60)(15)(mm^2) ... i have redrawn the structure , i am not sure whether it is correct or not... From the definition of shear flow , A is the cross section area of segment that is connected to the beam at juncture where the shear flow is calculated .

shear flow formula question sketch

$\endgroup$
1
  • $\begingroup$ No question here to answer. What's the question? $\endgroup$
    – wwarriner
    Nov 14 '16 at 17:19
4
$\begingroup$

For starters, I'm a little confused by your units; so I'll explain my assumptions.

The dimensions you give are completely different from the dimensions in the picture. It looks like you're taking inches and multiplying by 10 to get mm (e.g. your "60 mm x 15 mm" is 6" and 1.5" in the drawing; these units are not equivalent). Maybe you just prefer metric and don't really care about the units or the answer; but still. In North America it's good practice to be able to work in metric or imperial units. Regardless; for your reference and future information 1" is 25.4 mm; and other conversion factors are usually found in the front or back of your textbook (or online).

Anyways, that's not really relevant to your question, I just wanted to make sure you aren't making that mistake when it matters.

The two cross sectional areas they have make sense. Both the nails are supporting different loadings. In the area you propose to use, you would be trying to account for both of the different loadings on the nails at the same time. In their case, they split the problem into the cases where each nail is expected to experience the shear force.

Taking B for example; the nail is in one side; but nail C isn't doing the same thing as nail B. Instead, the shear is being resisted by the wood at B', so they take the area of 7.5"x1.5" knowing that the wood should be able to resist the shear as well. The same idea is used for C and C'.

If they used your section, both Qb and Qc would be the same. Just looking at the structure of the wooden box, that doesn't seem reasonable since the two nails are acting 90 degrees from each other while the applied loading is only in one direction.

If this isn't clear let me know where it gets confusing.

$\endgroup$
10
  • $\begingroup$ Can you label the exact location of B' and C' in the diagram ? the location of them are not clear enough in the original diagram... Why you said that I am trying to account for both of the different loadings on the nails at the same time? $\endgroup$ Nov 10 '16 at 17:11
  • $\begingroup$ They are at mirrored locations to the nail locations. i.imgur.com/w0rrfBW.png $\endgroup$
    – JMac
    Nov 10 '16 at 17:38
  • $\begingroup$ Also, for the loadings, if we took your area as the area where it was acting, it doesn't really make sense. We need the "symmetric" portion to actually solve the problem since we don't know anything about the loading on the nails. We instead assume that the force will be equally acting on nail B and the imaginary nail at B' (which we aren't worried about failing). We then do the same for C and C'. My best advice is to think about the loading and what the nails are doing. One nail is vertical, the other is horizontal; all the loading is horizontal... $\endgroup$
    – JMac
    Nov 10 '16 at 17:48
  • 1
    $\begingroup$ If we used your area to determine the shear flow that would say that the loading on both the vertical and horizontal nails are the same. Considering we only have a force acting vertically; that doesn't really make sense that both of those nails would have the same shear flow. It does make sense that the symmetrical sections would have the same shear flow; which is why they are used. $\endgroup$
    – JMac
    Nov 10 '16 at 17:50
  • 1
    $\begingroup$ We can ignore its contribution because we're assuming it's holding together. It has nothing to do with the other nail (and considering the answer the two nails will rarely be in the same spot anyways). $\endgroup$
    – JMac
    Nov 11 '16 at 18:29

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .