Hi guys i have had help with maths stuff before, this is using scientific notation can anybody help with this type of question.?
A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 15 m. The mass of the lift when full of passengers is 700 kg.
The designer has decided to incorporate a safety factor of ten into the lift cable, which means the cable must be able to withstand ten times the load it will actually be exposed to in service before it fails.
The steel selected for the cable has a failure stress of 1200 MN m– 2.
a.Using this information, calculate the required diameter of the cable. Show all your working. Assume that the cable is a single piece of steel, with a circular cross-section. Ignore any effect of the mass of the cable in your calculation.
The downward force F on the cable is calculated by multiplying the total mass m (expressed in kg) by g, the acceleration due to gravity:
A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 9 m. The mass of the lift when full of passengers is 900kg.
The designer has decided to incorporate a safety factor of 10 into the lift cable. The steel selected for the cable has a failure stress of 1200 MN m^-2 Question: (a) Using this information, calculate the required diameter of the cable. Assume that the cable is a single piece of steel, with a circular cross-section. (Ignore any effect of the weight of the cable in your calculation.)
The downward force on the cable is calculated by multiplying the total mass by the acceleration due to gravity. F = m * g
Take the value of g to be 10ms^-2. Hint: The safety factor of 10 means that the cable will fail when it reaches 10 times its maximum load.
(b) By how much will the cable have extended owing to the weight of a full lift at the ground floor? The Youngs Modulus for steel is 210 GN m^-2
Answers: (a)Stress (S) = Force (F) / Area (A)
Force (F) = Mass (M) x Gravity (G)
(F) = 900 (M) x 10 (G)
So using this we can work out that the force is 9000N
900 (M) x 10 (G) = 9000 (F)
Right we know that the Stress is 1200 MN m‾2 but the designer has incorporated a safety factor of 10 into his design so we divide the 1200 by the factor of 10.
1200 / 10 = 120
So we now know the Stress and the Force so we can work out the area of the cable.
120 (S) = 9000 (F) / Area (A)
So 9000 / 120 = Area
Area = 75 So now that we know the area we can work out the diameter of the cable.
We take the area of the cable and divide it by pi. Then we take the square root of the answer we get and multiply it by 2 to find the diameter.
So We divide our area by pi: 75 / 3.14 = 23.88
Then we find the square root of our answer √23.88 = 4.89
Then we multiply our answer by 2 to give us the cable diameter: 4.89 x 2 = 9.78mm
(b) Youngs Modulus (E) = Stress (s)/ Strain (e)
210 = 120 / (e)
120 / 210 = 0.5714 (e)
Strain (e) = Extension (el) / Original length (l)
.5714 = (el) / 9
.5714*9 = (el)
.5714*9 = 5.1426
Then change this to metric as length stated is in meters so
5.1426 / 100 = .051426
or 5.1426cm
Here's something that might help....A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 9 m. The mass of the lift when full of
A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 9 m. The mass of the lift when full of passengers is 900kg.
The designer has decided to incorporate a safety factor of 10 into the lift cable. The steel selected for the cable has a failure stress of 1200 MN m^-2 Question: (a) Using this information, calculate the required diameter of the cable. Assume that the cable is a single piece of steel, with a circular cross-section. (Ignore any effect of the weight of the cable in your calculation.)
The downward force on the cable is calculated by multiplying the total mass by the acceleration due to gravity. F = m * g
Take the value of g to be 10ms^-2. Hint: The safety factor of 10 means that the cable will fail when it reaches 10 times its maximum load.
(b) By how much will the cable have extended owing to the weight of a full lift at the ground floor? The Youngs Modulus for steel is 210 GN m^-2
Answers: (a)Stress (S) = Force (F) / Area (A)
Force (F) = Mass (M) x Gravity (G)
(F) = 900 (M) x 10 (G)
So using this we can work out that the force is 9000N
900 (M) x 10 (G) = 9000 (F)
Right we know that the Stress is 1200 MN m‾2 but the designer has incorporated a safety factor of 10 into his design so we divide the 1200 by the factor of 10.
1200 / 10 = 120
So we now know the Stress and the Force so we can work out the area of the cable.
120 (S) = 9000 (F) / Area (A)
So 9000 / 120 = Area
Area = 75 So now that we know the area we can work out the diameter of the cable.
We take the area of the cable and divide it by pi. Then we take the square root of the answer we get and multiply it by 2 to find the diameter.
So We divide our area by pi: 75 / 3.14 = 23.88
Then we find the square root of our answer √23.88 = 4.89
Then we multiply our answer by 2 to give us the cable diameter: 4.89 x 2 = 9.78mm
(b) Youngs Modulus (E) = Stress (s)/ Strain (e)
210 = 120 / (e)
120 / 210 = 0.5714 (e)
Strain (e) = Extension (el) / Original length (l)
.5714 = (el) / 9
.5714*9 = (el)
.5714*9 = 5.1426
Then change this to metric as length stated is in meters so
5.1426 / 100 = .051426
or 5.1426cm
Here's something that might help....A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 9 m. The mass of the lift when full of
Dunno why that posted twice...here's another example I lifted...
A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 12 m. The mass of the lift when full of passengers is 1200 kg. The designer has decided to incorporate a safety factor of 8 into the lift cable: i.e., the cable must be able to withstand 8 times the load it will actually be exposed to in service, before it fails. The steel selected for the cable has a failure stress of 1000 MN m– 2. Using this information, calculate the required diameter of the cable. Show all your working. Assume that the cable is a single piece of steel, with a circular cross-section. (Ignore any effect of the weight of the cable in your calculation.) The downward force F on the cable is calculated by multiplying the total mass m (expressed in kg) by g, the acceleration due to gravity: F = m x g Take the value of g to be 10 ms–2. (In reality there is extra force needed to accelerate the lift upwards, but as this is relatively small, there is no need to consider it here.) (Hint: the safety factor means that the cable will fail – the stress will reach its failure stress – when it is loaded to 8 times the intended design load. Use this to calculate the cross-sectional area of the cable, and so its diameter. I found the diameter to be around 11 mm.)
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Paul Paul Best Answer - Chosen by Asker 1 Calculate the force that the cable will need to support We are told that F = m * g and that m is 1200kg and g is 10msˉ² Therefore F = 1200 * 10 = 12,000 N 2 Allow for the safety factor The designer wants the cable to support eight times the expected load so multiply the actual load by eight. So the cable must be able to support 12,000 * 8 = 96,000 N 3 Calculate Steel area The steel will support 1000 MN per square metre. 96,000N is 0.096 MN so we would need 0.000096 square metres of steel to support it 4 Calculate Cable diameter The cable is round so the cross section area is given by Area = πr² So 0.000096 = π r² That means r² = 0.000 031 and r = 0.005 568m The diameter is twice the radius so that is 0.011136m That's in metres in milimetres it would be 11.136 mm
Dunno why that posted twice...here's another example I lifted...A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 12 m. The m
If you read those answers it looks like one contradicts the other,in essence. Intuitively I would say the second method (ie multiplying to cover the stress rather than dividing) is correct. Interesting problem.
If you read those answers it looks like one contradicts the other,in essence.Intuitively I would say the second method (ie multiplying to cover the stress rather than dividing) is correct.Interesting problem.
personally I would use a titanium/vanadium mixture giving you a yield strength permanent deformation of 0.2% of the original dimension which would also resist corrosion and give a final diameter of 9.1872 mm. Did you know I can also weld milk bottle tops together?
personally I would use a titanium/vanadium mixture giving you a yield strength permanent deformation of 0.2% of the original dimension which would also resist corrosion and give a final diameter of 9.1872 mm. Did you know I can also weld milk bottle
The 15m length is irrelevant seeing as self weight is supposed to be neglected. This is a pretty basic calculation i think for anyone who has done a level physics.
The 15m length is irrelevant seeing as self weight is supposed to be neglected. This is a pretty basic calculation i think for anyone who has done a level physics.
anybody help me with displaying the data in proper scientific notation?
example = as 1 mm is 10^-3 m, 1mm(squared) is 10^-6 m(squared)
dear guys of betfair chit chat forum,anybody help me with displaying the data in proper scientific notation?example = as 1 mm is 10^-3 m, 1mm(squared) is 10^-6 m(squared)
what you do is get a pair of chebs, fondle them. preferably stick your n0b between them and give them a massage.
after spaffing your load go to sleep. if you wake up tomorrow morning and feel that the problem persists, seek a forum that is relevant to your question, rather than asking a general betting forum.
what you do is get a pair of chebs, fondle them. preferably stick your n0b between them and give them a massage.after spaffing your load go to sleep. if you wake up tomorrow morning and feel that the problem persists, seek a forum that is relevant to
Right we know that the Stress is 1200 MN m‾2 but the designer has incorporated a safety factor of 10 into his design so we divide the 1200 by the factor of 10.
1200 / 10 = 120
120? 120 what, that's what im asking.
MN is mega newtons
but stress is 1200 mega newtons to power of -2
Right we know that the Stress is 1200 MN m‾2 but the designer has incorporated a safety factor of 10 into his design so we divide the 1200 by the factor of 10.1200 / 10 = 120120? 120 what, that's what im asking.MN is mega newtonsbut stress is 1200