Shear stress calc -confused

[Bert2]

Hi,

I have been asked to review a document at work on a lifting beam design;
After reviewing calculations for the rigging all is fine.
However it then looks at the shear stress in different areas.

The specific component in question is the plate at the end of the beam which (sling end has the shackle through).

Now the way I would calculate the shear stress is as follows.

Taking the highest load in this case XTe.

Find the surface area of the face of which the pin / bolt will apply the shear force
(circumference of hole, halved multiplied by the depth of plate)

Then divide the load over area to give shear stress eg; N/mm².

However this is my question; the way they have calculated the surface area is not clear to me; (this is the calc for the length for the area, the depth is given)

(230/2-(72/2 x cos (40xPi/180)) - see doc attached for drawing of plate.
they get a length of 87.42mm -
-if i calculate that i get 79mm ????

x depth (25mm) = 5532mm^2 (which is 2(two plates)x87.42x25)
=118.9N/mm^2

if i calculate that i get 79mm ????

I understand my method of assuming half the circumference as the contact area is a bit lackadaisical and possible not the best way.

But can anyone shine a light on why they have calculated it this way ?

My calculation; is attached along with the plate drawing

Many thanks - hope i have made this clear....

C.

 

[rb1957]

the doc doesn't match your post.

the post has "shear area" as pi*d/2*t = bearing area

the shear tear-out area is going to be close to (230-115)*t, the net area of the link.

 

[Bert2]

Sorry that wasnt clear;

i meant bearing area / the seat of the bolt / pin within the hole.

really all im asking is how they come to the figure of 87 for the length of the area (bearing)

 

[Bert2]

this will make it more clear why i dont understand.

Length 'A' is the question.

 

[kingnero]

(230/2-(72/2 x cos (40xPi/180))

Where does the "72" comes from?

When I do the math, I get 64.9 mm, see attached file.

 

[rb1957]

this looks like a shear tear-out area calc, i'd've thought "72" should be "115" ... i'm guessing tat 72 is the bolt diameter, and that there is a bushing in the hole.

you're calc'ing the width of the lug, parallel with the CL, at a point where the hole radius is at 40deg (clear as mud) ... sqrt(115^2-(115/2*sin(40))^2)-115/2*cos(40)

 

[Bert2]

@ kingnero thats my point i cant see where they get the 72 from..?

@ rb1957 that would make sense then as you can see there is a bush in but this is not shown on the 2D drawing.

ill look through the calc again but it yours does make sense of it now.

cheers.!

 

[Bert2]

@ kingnero just looked at you calc thats the same way i went also.

but looks like rb1957 has got it.