Mechanics (The maths type) Question

[matty_doh]

Hi all, attempting to complete an assignment for work but I'm stumped on a question... Last ditch effort here, any engineers fancy giving me a hand !?

"A steel rod is subjected to a pull of force F, calculate the stress, total extension and strain

Force F = 1000KN
Steel Rod Length = 300mm
Steel Rod Diameter = 20mm
Modulus of Elasticity = 200GPa"

I know all the equations for working this out, but I'm struggling to apply it with the correct units..

Stress = F/A
Modulus of elasticity = (F*L)/(A*x)
Strain = x/L

Do I convert lengths into metres? Do I convert the modulus into pascals?

If anyone could help me out it would be greatly appreciated!

[Gulliver911]

1 Giga pascal is in 1 Newton per square meter, so all you have to do is convert the mm units in meters, and you are all good.

JL

[the_engineer]

Yep, convert everything to SI and it should be straightforward...

[LilPeteMordino]

Like above, Convert to Meters

Ive got mechanics stuff to be getting on with aswell thinking about it...

[mat205125]

the_engineer said:

Yep, convert everything to SI and it should be straightforward...

Doing calcs in cm or mm, and then trying to pick the bones out of the decimal places at the end is just a massive recipe for disaster.

[Syndrome]

Part of my job is strain testing optical fibers. We always quote the results in psi / Kpsi, although Nm is probably the correct way for an exam. convert all units to SI then do the maths.

[Matt_N]

matty_doh said:

Hi all, attempting to complete an assignment for work but I'm stumped on a question... Last ditch effort here, any engineers fancy giving me a hand !?

"A steel rod is subjected to a pull of force F, calculate the stress, total extension and strain

Force F = 1000KN
Steel Rod Length = 300mm
Steel Rod Diameter = 20mm
Modulus of Elasticity = 200GPa"

I know all the equations for working this out, but I'm struggling to apply it with the correct units..

Stress = F/A
Modulus of elasticity = (F*L)/(A*x)
Strain = x/L

Do I convert lengths into metres? Do I convert the modulus into pascals?

If anyone could help me out it would be greatly appreciated!

I always work in metres, its a bit annoying actually as one of our lecturers is Spanish and works in what ever he feels like, whilst we have always been told to work in meters.

I would work Force as 1E3, Youngs Modulus as 200E9.

Your force will be in Newton metres squared, extension in metres and strain has no unit.

[marshalla]

LilPeteMordino said:

Like above, Convert to Meters

Ive got mechanics stuff to be getting on with aswell thinking about it...

Although, since the size of a meter is pretty variable - measuring and counting devices being what they are, metREs might work better.

[matty_doh]

Right then, I'm stumped again - I'm no good at this..

Lets see if you can make sense of this one;



The diagram shows 2 tanks of liquid. A valve is opened and the liquid passes along to tank 2.

The flow rate is 5kg/s. Calculate;

- The velocity of the liquid flowing in the pipe
- The velocity of the surface of tank 1 as it falls
- The velocity of the surface of tank 2 as it rises

Tank 1 Sectional Area; 6m^2
Tank 2 Sectional Area; 4m^2
Pipe bore; 0.08m
Liquid Density; 920 kg/m^3


.... What on earth do I do for parts 2 and 3?

(No laughing at the back about my drawing!)

[TooLateForAName]

Is this homework due in tomorrow morning?

[matty_doh]

TooLateForAName said:

Is this homework due in tomorrow morning?

No. It's part of my BTEC coursework due in for late July

[TheEnd]

matty_doh said:

Right then, I'm stumped again - I'm no good at this..

Lets see if you can make sense of this one;



The diagram shows 2 tanks of liquid. A valve is opened and the liquid passes along to tank 2.

The flow rate is 5kg/s. Calculate;

- The velocity of the liquid flowing in the pipe
- The velocity of the surface of tank 1 as it falls
- The velocity of the surface of tank 2 as it rises

Tank 1 Sectional Area; 6m^2
Tank 2 Sectional Area; 4m^2
Pipe bore; 0.08m
Liquid Density; 920 kg/m^3


.... What on earth do I do for parts 2 and 3?

(No laughing at the back about my drawing!)

you know 5kg/s is moving, so you find out the volume of that 5kgs.

find out how deep that volume would be in a tank 6m^2, and it's that depth it moves in a second.
You won't even need to touch volume of cylinders, as you have the surface area already, just need the last depth dimension for the volume

[johnfm]

Matty

you can figure out 2 and 3 as you know the cross sectional area of each tank.

You know the mass flow rate and you kbnow the density of the liquid - so you can calc the volumetric flow rate.

So, if you know the rate of volume change in each tank, divide by the CSA to get the rate of level change.

Piece de la piss, non?

Edited by johnfm on Sunday 13th June 21:48

[matty_doh]

Got it, cheers for the help fellas!

[sebhaque]

Sorry to steal your topic sir but coincidentally I also have a maths question I'm stuck on. I was away for the lessons (Bronchitis) so I haven't really been able to make much sense of these so far.

The question is in 5 parts. I'm sure once I understand the theory behind it I'll be able to prove them myself.

Anyway:

6. Solve the following trigonometric equatoiins in the range 0° <= x <= 360°:

a) 4cos x + 1 - =
b) 3.25 cosec x = 5.25
c) 5sin(squared) x + 3sin x = 4
d) 2sec(squared) θ + 5tan θ = 3
e) Henceforth solve the equation 5sin (θ-pi/6) = 8cos θ for values 0 <= θ <= 2pi

Any help would be very much appreciated. Sorry again for stealing the topic but it makes sense to use a recent, existing one rather than make my own.

[matty_doh]

sebhaque said:

Sorry to steal your topic sir....

Not a problem - I would try and help but I've never done that before!

Edited by matty_doh on Sunday 13th June 22:25

[james_gt3rs]

sebhaque said:

Sorry to steal your topic sir but coincidentally I also have a maths question I'm stuck on. I was away for the lessons (Bronchitis) so I haven't really been able to make much sense of these so far.

The question is in 5 parts. I'm sure once I understand the theory behind it I'll be able to prove them myself.

Anyway:

6. Solve the following trigonometric equatoiins in the range 0° <= x <= 360°:

a) 4cos x + 1 - =
b) 3.25 cosec x = 5.25
c) 5sin(squared) x + 3sin x = 4
d) 2sec(squared) θ + 5tan θ = 3
e) Henceforth solve the equation 5sin (θ-pi/6) = 8cos θ for values 0 <= θ <= 2pi

Any help would be very much appreciated. Sorry again for stealing the topic but it makes sense to use a recent, existing one rather than make my own.

Right...

a & b)
You simply rearrange to find x so cos x = -1/4...
With b you invert the equation at the end to make cosec into sin.

c) Think of it as a quadratic equation, so 5x^2 + 3x - 4 = 0

d) sec^2(x) = tan^2(x) +1 then do the same as c.

Not sure about e.

If you're not following this, then read this: http://www.mathsrevision.net/alevel/pure/

I can't really explain it with just a keyboard, HTH!

[Matt_N]

matty_doh said:

Right then, I'm stumped again - I'm no good at this..

Lets see if you can make sense of this one;



The diagram shows 2 tanks of liquid. A valve is opened and the liquid passes along to tank 2.

The flow rate is 5kg/s. Calculate;

- The velocity of the liquid flowing in the pipe
- The velocity of the surface of tank 1 as it falls
- The velocity of the surface of tank 2 as it rises

Tank 1 Sectional Area; 6m^2
Tank 2 Sectional Area; 4m^2
Pipe bore; 0.08m
Liquid Density; 920 kg/m^3


.... What on earth do I do for parts 2 and 3?

(No laughing at the back about my drawing!)

Thats a mass flow rate (m) (5kg/s), m = ñ(density).A(area).c(velocity) so re-arrange to find c, so c = m/ñA

[thinfourth2]

matty_doh said:

Right then, I'm stumped again - I'm no good at this..

Lets see if you can make sense of this one;



The diagram shows 2 tanks of liquid. A valve is opened and the liquid passes along to tank 2.

The flow rate is 5kg/s. Calculate;

- The velocity of the liquid flowing in the pipe
- The velocity of the surface of tank 1 as it falls
- The velocity of the surface of tank 2 as it rises

Tank 1 Sectional Area; 6m^2
Tank 2 Sectional Area; 4m^2
Pipe bore; 0.08m
Liquid Density; 920 kg/m^3


.... What on earth do I do for parts 2 and 3?

(No laughing at the back about my drawing!)

Is that how hard degree stuff is these days?

[Matt_N]

thinfourth2 said:

matty_doh said:

Right then, I'm stumped again - I'm no good at this..

Lets see if you can make sense of this one;



The diagram shows 2 tanks of liquid. A valve is opened and the liquid passes along to tank 2.

The flow rate is 5kg/s. Calculate;

- The velocity of the liquid flowing in the pipe
- The velocity of the surface of tank 1 as it falls
- The velocity of the surface of tank 2 as it rises

Tank 1 Sectional Area; 6m^2
Tank 2 Sectional Area; 4m^2
Pipe bore; 0.08m
Liquid Density; 920 kg/m^3


.... What on earth do I do for parts 2 and 3?

(No laughing at the back about my drawing!)

Is that how hard degree stuff is these days?

BTEC, as it says!