Any structural or civil engineers here?

[andy_con]

as title?

[Banthai]

well im an engineer and some people have said that my structure is very big

but sometimes im not as civil as i should be

all the best .. Karl

[andy_con]

strange some might say karl!

[T0mmm]

I'm not far from graduating from materials science and engineering a materials engineer is the only one who graduates without using a calculator :P

[danielbriggs]

I'm doing Mech Eng at uni...
So not much help on the Civil side then.

What was it you needing sorting Andy, as I could ask around if you want?

[andy_con]

im doing a NC in civil eingeering at college and need help with the following



A hollow steel tube 100mm external diameter and 80mm internal diameter and 3m long is subject to a tensile load of 400 kn. Calculate the stress in the material of the tube and also calculate the amount the tube stretches if youngs modulus is 200 000 n/mm 2 (squared)

A 150mm concrete test cube is crushed in a testing machine. A load of800kn produces a reduction in height if 0.4mm.
what is the stress in the cube (i know this one)
what is the resulting strain
what is the modulus of elasticity value of the concrete


im lost and need help!

[danielbriggs]

Ok,
Stress (sigma) = Force (F) / Area (A)
Area of tube = area of 100 - area of 80 diameters. Convert to meters, and you get: 2.8274x10^-3 m^2
Force = 400,000N (keep everything in SI units)
Stress (sigma) = 141.47 x 10^6 Pa = 142.47 MPa


Gotta dash out for a bit, so will sort the rest out later for you...

Dan

[tocket]

A 150mm concrete test cube is crushed in a testing machine. A load of800kn produces a reduction in height if 0.4mm.
what is the stress in the cube (i know this one)
what is the resulting strain
what is the modulus of elasticity value of the concrete

Well, I'm no civil engineer, but this one seems pretty easy...

Since you already know the stress, I'll skip to the strain and E modulus. The strain is simply the deformation; that is ΔL/L = -0.4/150 = -0.00267.

Once the stress and strain is known it is easy to calculate the E modulus; just divide the stress by the strain. Just be careful with the units.

What is NC in civil engineering by the way?

[andy_con]

Ok,
Stress (sigma) = Force (F) / Area (A)
Area of tube = area of 100 - area of 80 diameters. Convert to meters, and you get: 2.8274x10^-3 m^2
Force = 400,000N (keep everything in SI units)
Stress (sigma) = 141.47 x 10^6 Pa = 142.47 GPa


Gotta dash out for a bit, so will sort the rest out later for you...

Dan

easy....

youve lost me already. this is what ive got so far but its probably wrong



external tube = pie 0.1 squared = 0.31m

internal tube= pie 0.08 squared = 0.020m

external - internal = 0.31 - 0.02 = 0.29m

now add (or times) the 3 metre length = 0.29 x 3 = 0.87m


so if Stress = Force over (divided by) Area then

400 / 0.87 = 459.77 (not sure on the outcome unit?? )


can i just add im not looking for someone to do it for me just guidance

[danielbriggs]

easy....

youve lost me already. this is what ive got so far but its probably wrong



external tube = pie 0.1 squared = 0.31m

internal tube= pie 0.08 squared = 0.020m

external - internal = 0.31 - 0.02 = 0.29m

now add (or times) the 3 metre length = 0.29 x 3 = 0.87m


so if Stress = Force over (divided by) Area then

400 / 0.87 = 459.77 (not sure on the outcome unit?? )

Sorry Andy was in a rush before...

Area = Pi x Radius Squared
You're given diameters, so half the values and you get the radius. Then square that, and times by Pi. e.g. (0.1 / 2) = 0.05, squared = 0.0025, times pi = 0.007854 m^2
Same for the other dimension, (0.08 / 2) = 0.04, sqaured = 0.0016, times pi = 0.0050265 m^2
Subtract the smaller value from the larger value to get the effective area = 0.0028274 m^2

The 3m length is for the later part in the question. You don't need it for the stress. Stresses are based upon the 2D area.

Stress is given by the force applied divided by the area of the part.
We know the force applied (400kN = 400,000N) and we've just worked out the area (0.0028274 m^2)
So just plug the numbers in to get the stress:
Stress = 400,000 / 0.0028274 = 141470000 Pascals = 141.47 MPa (Mega = 10^6)

How's that?


Edit: And for the second part:

You know the pipe length = 3m
The Young's Modulus for the material = 200,000 N/mm^2 (1 N/mm^2 = 1MPa), so you've got 200,000 MPa = 200,000 x 10^6 Pa
Young's Modulus is found by, Stress divided by Strain
We have worked out stress, and we are given Young's modulus, so the only unknown part is Strain.

Strain is calculated by the change in length divided by the original starting length. = ∆L / L
So we can swap the strain value in the Young's modulus equation for the ∆L/L part.
To give:
Young's Modulus = Stress / (∆L / L)
Then re-arrange to give ∆L on it's own:
Change in length = (Stress / Young's Modulus) * Original Length
So plug in the numbers:
Change in length = (141.47 x 10^6 / 200,000 x 10^6) * 3 = 2.12205 x 10^-3 m
To convert into mm times by 1000 = 2.12mm extension

You've just got to watch out for the units, and the rest will pop into place.
Keep everything in S.I. units and you' can't go wrong.
Forces in Pascals: (Pa), Distances in Meters, m. etc...

Dan