Stressful Engineering coursework help!

[43021HST]

I'm studying level 3 engineering in college at the moment and I'm really stuck on this one question for a bit of coursework, the question goes like this:

A workshop guillotine is used to shear a sheet of metal 1.5m across and 1.6mm thick. If a shear force of 840kN is required to shear the metal, determine the ultimate shear stress of the metal.

So basically I need the formula for Ultimate Shear stress, I know the formula for working stress is force/area. I've tried googling it but that doesn't come up with anything useful.

By the way the ultimate stress is defined as the amount of stress the material is subjected to so that it breaks.

Does anyone on here know?

 

[rdeez]

It's been a while since I did this but isn't "ultimate shear stress" just the shear strength of the material? A Google for shear strength brings up the formula.

Sent from my C6903 using Tapatalk

 

[maniacmartin]

I think everything you need to find the answer is in the opening post. You have the force, and can work out the area

+1 for the witty thread title btw

 

[Nym]

Whats humerous is that any level 3 qualification can be classed as "Engineering".

How calculating shear stress in this circumstance is an engineer's job is beyond me, unless it leads into structural questions for shear?

Levels 6, 7 and 8 are Engineering qualification levels.

A level 5 qualification or equivalent experience is required for EngTech (Not an Engineer!).

This really does grind my gears sometimes...!

Nym
Mechatronic Engineer

 

[telstarbox]

Typical Nym post there, showing off but adding nothing useful.

 

[asylumxl]

Typical Nym post there, showing off but adding nothing useful.

And fails to acknowledge some people have to start somewhere. Personally I feel that it's better to start with the sciences and progress on to Bachelors etc.

 

[Nym]

Someone seems to be missing the point here that I'm ranting on about the course name compared with the content (yes I have looked at it). What is truely engineering is not reflected in any significant way by the course.

In addition to this, a Level 3 course entitled "Engineering" in addition to the level 2 course I am aware of, works wonders in de-valuing what is, in most of the world, a protected title for professionals. Not for the likes of a washing machine repairman or IT support agent(!)

I agree you need to start somewhere, but the courses required for this foundation already existed before these "Engineering" courses at lower levels throughout the NQA framework, one does think sometimes that the course designers either delibrately wished to de-value the title, or did not have the mind to think of a more appropriate name, or it is simply a matter of marketing; using a premium term to improve the perception of what is little more than a "Design and Technology" course.

PS: If this comes across as elitism, so be it, it is as much if not more of an elite profession as any other that holds a chartership or higher level professional registration status. See: Accountant, Surveyor, etc. and dare I say Doctor...?

 

[Clip]

<Snip>

so rather than spouting off about the course title and showing off about your qualifications(well done,shep) maybe give the OP a bit of assistance on his problem so that one day he can become as qualified as you are in the engineering feild.

Hopefully without your smug/Holier than thou attitude.

 

[rdeez]

Someone seems to be missing the point here that I'm ranting on about the course name compared with the content (yes I have looked at it).

With respect I think you're missing the point that the OP didn't ask for an opinion on their course and whether it is appropriately titled and you went off on that tangent without offering anything relevant to the thread, which is entirely why it provoked this kind of reaction. Your words are not likely to have offered them much encouragement in what they have chosen to do.

The last paragraph of the post quoted in particular provokes a reaction of "wow". Is engineering a more elite profession than medicine or accountancy? I'm sure I don't know, but I'm pretty sure it doesn't matter. Presumably the prestige comes (or should) from the knowledge that you're making useful contributions, not the titles.

 

[Nick W]

I'm studying level 3 engineering in college at the moment and I'm really stuck on this one question for a bit of coursework, the question goes like this:

A workshop guillotine is used to shear a sheet of metal 1.5m across and 1.6mm thick. If a shear force of 840kN is required to shear the metal, determine the ultimate shear stress of the metal.

So basically I need the formula for Ultimate Shear stress, I know the formula for working stress is force/area. I've tried googling it but that doesn't come up with anything useful.

By the way the ultimate stress is defined as the amount of stress the material is subjected to so that it breaks.

Does anyone on here know?

Assume the material shears cleanly and no deformation happens, i.e the sheet metal does nothing until suddenly shearing at 840kN.

 

[telstarbox]

So is the ultimate shear stress as follows:

840,000 N / (1.5 x 0.0016) m2 = 350,000,000 N/m2 or 350 MPa for short.

This is within the range of values for steel here: http://www.tribology-abc.com/calculators/uts.htm

 

[MichaelAMW]

So is the ultimate shear stress as follows:

840,000 N / (1.5 x 0.0016) m2 = 350,000,000 N/m2 or 350 MPa for short.

This is within the range of values for steel here: http://www.tribology-abc.com/calculators/uts.htm

And, perhaps to add something also useful: stress, force per unit area, is used as it is universal across the material, i.e. any lump of that particular stuff will have the same value for this. You then have to adjust for the area of the particular lump you're dealing with in order to find the force that will cause the shear in the particular case. From a practical point of view it's an exercise in how you relate what you look up in data tables with the reality of a particular case. See also extrinsic and instrinsic quantities.

 

[Nick W]

So is the ultimate shear stress as follows:

840,000 N / (1.5 x 0.0016) m2 = 350,000,000 N/m2 or 350 MPa for short.

This is within the range of values for steel here: http://www.tribology-abc.com/calculators/uts.htm

The hard part is whether that's an upper or lower bound.