help w/ hwk!

[zakfab]

This is a homework question, it doen't have to be handed in, so there is no rush to answer, but the question has totally confused me!

 

A steel bolt has a diameter of 6.35mm and a single start thread of pitch 1mm. The bolt is used to fasten together parts of a machine.When the parts are just in contact, without load on the bolt, the distance betwee the bolt head and the nut is 254mm. Estimate the tension in te bolt if the nut is tightened by 1/6 of a turn. State your assumptions. Would you expect your result to be larger or smaller than the actual load? Why?

 

 

 

The answer is 4.3KN, larger.

 

 

Any pointers/or a method would be greatly appreciated!

[YT2095]

unless there`s something missing about steel strength in the equasion, the 6.35mm dia is to mislead you.

 

1/6`th of a turn would pull the nut tighter by 1/6`th of a mm

spread that tension over the elasticity of the 25.4cm long bolt....

that`s as far I get

 

but from a practical standpoint, as a single bolt, it seems a bit inadequate as a single fastener, perhaps part of many would be the case?

 

[zakfab]

I thought I should use the diameter to work out the cross sectional area and so something with that.

 

Tension is just force right?

 

if I use strain = change in length/original length

 

strain = 1/1524

 

Then if I knew Young's Modulus or something I could work out stress, then work it out with the cross sectional area (3.167*10^-5) * the stress.

 

Alas, no Young's Modulus, so then I'm stuck.

 

I think I will just presume the question was supposed to include Young's Modulus for steel for the moment. E = 207 * 10^9 N/m^2

 

It works that way.

 

If anyone can do it with just the original information please tell me how!

[YT2095]

according to my book of data here, the young modulus` would be:

 

mild steel: 200

high tensile steel: 200 also

 

E= young modulus

 

so: E/GN-2

 

[Kedas]

It states that it is steel so the young modulus is given. (look it up)

[zakfab]

So you are supposed to know the modulus if it says the material?

 

I will have to remember that!

[Kedas]

You shouldn't know it but you should know that you can look that up.

 

later at a job they wont tell you 'that is the modulus' but they will say 'it is made of that'.

[YT2095]

sorry then, didn`t look at all yer post (numbers scare me) so use 207 instead of the 200 I said

 

after all this book is from 1971

[Kedas]

Numbers don't scare me, well at least not all of them, that number that represents my age is scary.

[YT2095]

Shhhhh......!

 

say nothing!

[zakfab]

Another question, which means I'm stuck again, lol.

 

A 60mm diameter sphere, with modulus of elasticity E, has parallel flats on it 20mm each side of the central axis. If a compressive load F is applied perpendicular to the flats, demonstrate the decrease in length along the loading axis is given by:

2F            0.73rad                  d(theta)
--------      (integral sign)        ----------
r*pi*E         0                        cos (theta)

 

Where (theta) is the angle between the central axis and any point along the surface of the sphere, and r the radius of the sphere.

 

looks like this:

____

(____)

 

I hope you get that :S

[YT2095]

yeah we get it, it`s like top and tailing an orange and trying to squash together the flat bits.

 

well how compressable is it for the force applied, that will be your answer

the 20mm and 60mm part, you can factor into a finite number 1`st...

you never mentioned the material to be used tho? as I`m sure bread dough and titanium would give different results

 

[Kedas]

'E' is given as a variable so material isn't needed.

So the solution must be valid for all materials.

(there is an assumtion that it is homogeneous)

[Kedas]

Solved it.

 

Due to the fact that it is a sphere we can't just use the regular formula but we can use the regular formula on dx(along the loading axis) and then integrate it over the whole part.

 

first the integration start and end:

Start is 0 because we start in the middle of the sphere going to the side and then we just mutiply it by 2 because the parts are identical. (hence the 2 in the formula)

The end point:

r=30mm cut to 20 meaning 2/3 or cos(pi/2-end)=2/3 --> end = 0.7297rad

(you can also go from -0.73 to 0.73 and forget about the 2)

 

The bigger the surface the less it will change. So the denominator is:

R.R.pi or better r.r.pi.cos^2(theta)

because R=r.cos(theta)

With R = radius of one slice.

 

The numerator is how thick it is that is dx. (along the axis)

Now dx in function of d(theta)

dx = r.cos(theta). d(theta)

 

put numerator and denominator together and you get:

d(theta) / (r.pi.cos(theta))

 

Multiply with F/E and put the constants before the integral sign and it's done

[YT2095]

Holy crap Kedas!

 

do you take Vitamins or something!?

respect to the max dude!

[Kedas]

YT2095 said in post #15 :

do you take Vitamins or something!?

 

LOL, No I'm supposed to be able to solve things like that according to my degree but haven't done it in years so I actually enjoyed it.

 

I 'love(d)' this kind of integral applications.

I also liked the 'power' of differential equations a lot

You can solve problems with that like:

your beer is 25°C you put it in a fridge of 3°C and after 15min. it's 20°C how long do you have to wait until it is 10°C

[zakfab]

I swear we have never done this in class.

 

Thanks Kedas I'm gonna have to totally analyse your answer

[zakfab]

Kedas: Could you elaborate on how you got the r cos (theta) parts, and how you know to multiply by F/E?

 

Thanks