# Question bout Shear force

#### iionas

I have just reg'd here and I have been lurking forever but its time to give in so I can get some help.

Ive been working through some examples in this book; and so far so good however Im on a question which does not make sense the way Im working it out seems to me to be correct however the answer directs me elsewhere.

The question is on hear force.

Determine the force required to punch 20 x 15 mm rectangular holes in am ild steel plate 8mm thick..

Now Ive done a similar question but with a circle instead of the rectangle..

I dont see the problem here..

Formulas which it prompts me to use is for Area = 20 x 15 x t (8) t = thickness..

Which is 2400 respectively. With this answer I multiply for the USS (ultimate shear strength of the material which is given mild steel = 360N .

And I get a figure of 864,000 or 864 KN..

The answer in the back of this book is 202kN. Im Stumped What Am I doing wrong?

Apologies if this is a bit to simple or It comes off as stupid. Its racking my brain...

#### Studiot

You are getting the wrong answer because you are not calculating the punching shear.

The area you are calculating is the surface area of the rectangle being punched out. This is not the area in shear.

The area in shear is the perimeter times the thickness.

The answer given by the book is correct.

Does this help?

Incidentally I strongly recommend taking more care with units, you are incorrectly stating stress in newtons. Attention to detail is all important in the technical world.

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#### iionas

20+20+15+15 x 8 = 70

70 * 8 = 560

560 * 360 = 201600 / 1000 = 201.6 = 202kN

I Thank you for the help. I may post more in here.

Im going to contribute as much as I can here. You've been more help than my tutor has.

Thanks again.

#### Studiot

You are welcome, glad I could help you see it for yourself, that's the best way, but don't be afraid to ask for more detail if needed. #### iionas

Im Going to post another question here...

This is something once again with shear...

I dont quite understand why this is.

Determine the required number of 10mm mild steel bolts to hold two over lapping strips of metal against a total shear force of 75.4kN if the allowable stress in shear is 120MPa

Im not quite certain how to go about this one.

However Im going to keep trying..

Can anyone provide some insight?

Thanks

#### Studiot

The bolts for two strips of steel are in single shear.

This means that there is only one cross section of the bolt under shear stress.

So total shear = shear stress on each bolt x number of bolts.

However you also have to check the bearing or tear out stress here.

This is failure of a metal strip, not a bolt. This stress is a bit like the punching shear in that it is (almost) perimeter times strip thickness.

A diagram helps.

#### iionas

Let me give it a shot; No diagram just working out of a book with next to no explanation.

#### iionas

Like I said this is a stab in the dark I dont want to come off as stupid...

so Im thinking

A = 2 x pie*10^2/4 = 157

then wed use f = F / A

so wed have 75400/157 = 480.25

then wed use what you suggested..

480.25 = 120 B

B = Bolts

this will equal 4 .. We multiply by 2 because of two strips.. So wed get 8..

This is the right answer in the back of the book however Im unsure that its correct...

Am I completely wrong?

#### Studiot

this will equal 4 .. We multiply by 2 because of two strips.. So wed get 8..
No this is not correct. If you draw a diagram you will see why. Just because there is no diagram in the book, does not stop you drawing one.

I really suggest you pay more attention to detail, including written English. It will help you enormously get things right. This is the biggest cause of wrong answers = lost marks.

I do indeed make it 8 bolts thus:-

$$N.\pi .\frac{{{{(10)}^2}}}{4}{.120.10^6}{.10^{ - 6}} = 75400$$

Where N is the number of bolts.

#### iionas

I will look over it again once Im home infront of my books. Thanks again for the help

#### nvn

Homework Helper
In case this helps, a multiplication dot in latex is {\cdot}, I think. For example,

a{\cdot}b gives $$a{\cdot}b$$, and
a\cdot b gives $$a\cdot b$$.​

#### Studiot

Thank you, nvn.

Not being a LaTex fiend I am using (learning) MathType. I have now located the multiplication point, which I was using for clarity and emphasis in this case, in MT.

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