Revised Calculations

From the analysis done to work out how much length was needed for an object to move 4m maximum from one point to another it was calculated that the beam would have to be a minimum standard length of 5.5m

Revising the calculations done at an earlier date a delfection of 12.29mm was established with the same CSA.

Buckling calculations for the legs, foot extender and crane shoulder corrected

recently went over some of my old calculations for the legs and realised i had used the wrong Euler formula for the crane shoulder legs and foot extender. Here are my corrected calculations.

P.S these calculations do not imply that any changes should be made to the crane, material or design wise. I simply wanted to correct my mistakes.

Beam Length Analysis

The beam length is determined on a few conditions and they are:

• The position of the struts at the base of the leg
• The dimensions of the object to be lifted

1/

The struts position is fixed in relation to the beam so therefore

length of leg at full extension = 1.145m

angle leg sits at = 30 degrees

s.t.f:

1.145 x sin 30 = 0.5725m

however the position of the leg in relation to the object is not that simple. Each leg is 120 degrees from each other and two of the legs sit 30 degrees from the plane which runs parallel with the leg itself.

s.t.f

0.5725 x sin 30 = 0.28625m = 286.25mm

2/

The dimensions of the material is not as big a problem as it would seem, because if we for example take the material to be of a cuboid nature (like in the load dimensions calculated) then if the worst case scenario set is that the material is 4.16m long then we can just work at a different angle by position the crane so we work on a plane with gives more room for the object to moved. However if we come upon a scenario where the material is the same distance no matter what way you take it, for example circular.

The volume of the worst case scenario = 0.3744m^3

s.t.f

pi x r^2 x l = 0.3744m^3

in the previous examples l = 450mm so it will be used in these calculations.

s.t.f

r = 0.515m

So from these two calculated results we can say that if the load has to be moved 4m then the length of the beam must be:

4m + 0.515m + (2 x 0.28625m) = 5.0875m

The beam has also got to have an extra 0.290m added to this length since it must sit in the shoulders. So therefore the overall minimum length needed is 5.3775m

With this it can be deduced that we need to use the standard length of 5.5m for the beam so that in the worst case scenario picking up the most common material.

approximate prices for crane materials

Aluminium alloys
According to London Market Exchange figures on http://www.metalprices.com/ the averegae price of an alluminium alloy is 1 us dollar/ pound. Using this info and my estimates on the mass of the legs iv calculated that the approximte cost of the legs.

mass of both legs = 27.4 kg = 60.28 lb

cost = $60-$61 = £40

Carbon steel
The LME doesn't give a specific price for carbon steel (at least i couldn't find one) however its most recent price on steels in general was £350 per tonne. however http://www.meps.co.uk/ has a specific prices for carbon steels that are used to build beams but the prices are 4 months old. As meps gives a specified price i will use this to calculate the price of the boom.

mass of ub178 boom = 94.77 kg
cost of carbon steel = $691 per tonne
cost of boom = $65.5 = £42.55

The prices used are bulk purchase prices, individually the boom and the legs may cost more. Also these prices dont take into consideration the cost of manufacturing the boom and the legs.

This is the calculated deflection and characteristics of the beam when its length is 5m long. As you can see form the previous set it is heavier and has a larger deflection. So for the benefit of the design it would be beneficial to make the beam 4.7m long.

Revised Beam Analysis

This is the revised calculations of the beam as requested at a length of 4.7m with the specified CSA. I will get the calculations for 5m and the legs hopefully tomorrow.

Calculations for pins using the smallest and largest pins

Using the info mark published on hex bolts i decided to calculate the shear stress that would be experienced M12/ smallest bolt (largest stress) and the M36/ biggest bolt (smallest stress). I also calculated the stress on the leg pins by using the dimensions in mark's cad drawings. I found that the largest stress obtained was 92mpa when using the M12 bolts, which is not a problem as carbon steel generally has a shear stregnth of 42000psi = 289.5mpa.

( source: http://www.tpub.com/content/construction/14250/css/14250_14.htm)

Full calculations for legs (corrected)

references : http://www.matweb.com
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm
ps. i have already estimated the masses of the legs (see older blog)

Recently discovered issues with the IPE160 beam and alternative solutions

For the given types of load exerted on the standard IPE160 beam the deflection, BMmax and SFmax were obtained to indicate worst case conditions that the given material (steel grade) would need to withstand:



The next stage was to calculate the maxBM and maxSF for the given structural steel. If these values are found to be higher than the calculated maxBM and maxSF under load then the beam will not fail. The following formulas were used:

M = Z *("fy") where Z = I/y (for the moment)
Where “fy” = steel grade

V= "dw"*"tw"*("fy"/(3)^(1/2)) (for the shear force)

Formulas and following information courtesy of:
The Corus interactive “blue book” to BS 5950-1
BS 4: part1:1993
BS 5950-12000 part 1
Elements of Strength of Materials, International Student Edition, by S Timoshenko

To calculate the maxBM:

M = Z ("fy")
Note:
Z = I/y and y= D/2

“fy”= common steel grade S250= 250 N/mm^2
( i.e for the given 5mm thickness the required design strength is 250N/mm^2)



I=8.34E-06 m^4 or 8.34E+06 mm^4

D/2 =160mm/2= 80mm

Therefore Z =I/y = (8.34E+06 mm^4/80mm)
Z =104250mm^3

M=Z* “fy”

M=104250mm^3 * 250 N/mm^2
M=26.06E+06 Nmm
M=26.06 kNm

In order to obtain the maxBM a factor of safety rm=1.05 and rf=1.1 provided in the Corus interactive “blue book” to BS 5950-1 need to be included:

Md =M/rm*rf
Md =26.06/1.05*1.1
Md =22.5 kNm

This is the maxBM that this steel grade could withstand.

For max shear force that the beam would need to withstand:
V ="dw"*"tw"*("fy"/(3)^(1/2))
V= 5* 145.2*(250/(3)^(1/2))
V=104789N or 104.789kN

However another factor mentioned in the Corus interactive ” blue book” is to have a safety capacity related to Buckling (some tables are given ):



Therefore a major issue with this beam is the maxBM allowable due to consideration of safety factors and buckling.
Further research also found IPE160 to be a castellated beam type which further impacts on strength but also increases the general cost of the material.

To obtain a more suitable beam a different standard will be used and a steel grade S355:



The next stage was to calculate the maxBM and maxSF for the given structural steel. If these values are found to be higher than the calculated maxBM and maxSF under load then the beam will not fail.

I=1.356E-05 m^4 or 1.356E+07 mm^4

D/2 =177.8mm/2= 88.9mm
Therefore Z =I/y = (1.356E+07 mm^4/88.9mm)
Z=152530.9336mm^3

M=Z* “fy”

M=152530.9336mm^3 * 355 N/mm^2
M=54.148E+06 Nmm
M=54.148 kNm

In order to obtain the maxBM a factor of safety rm=1.05 and rf=1.1 provided in the “blue book” to BS 5950-1 need to be included:

Md =M/rm*rf
Md =54.148/1.05*1.1
Md =46.881 kNm

This is the maxBM that this steel grade could withstand.

For max shear force that the beam would need to withstand:
V ="dw"*"tw"*("fy"/(3)^(1/2))
V= 4.8* 146.8*(355/(3)^(1/2))
V=144422.553N or 144.423kN

However another factor mentioned in the Corus interactive "blue book” is to have a safety capacity related to Buckling (the following value is from table “4.1.3. bending. Advance UKB. Moment capacity and buckling resistance moment for Advance 355”):

Md =17kNm



This beam clearly passes the basic required criteria and so long as these loading conditions are maintained it should not fail for the duration of the service for the disaster relief (if a 4.7m beam were to be used the buckling resistance moment would be higher than 17kNm).

The following are the lists of tables used from the BS standards :

references for calculations

The following scanned images are from 2 selected books which have been used to obtain the equations used to calculate the stresses and deflections for the boom.

Courtesy of:
Mechanics of engineering materials, second edition, by PP Benham, RJ Crawford & CG Armstrong;
Elements of Strength of Materials, International Student Edition, by S Timoshenko;

In order to obtain the following equations in the calculations:



A formula sheet from "Elements of Strength of Materials, International Student Edition, by S Timoshenko" where by the principle of superposition two factors (deflection due to a uniformly distributed load and deflection due to a point load) were combined providing the final deflection value.


The theory of superposition is explained in the following paragraphs along with the diagrams provide a visual image of how the beam will behave when subjected to varying loads (this is an important situation as it's behaviours in the following example are very similar to the behaviours we could expect our crane to experience):





The following scans are proofs of the formula obtained for the varying deflection loads and can be used as a template for our mathematical equations (for double checking values or if an error was made in the final value where could it have occurred by following the template).

For a point load our calculations would need to follow the following format:




For a uniformly distributed load the calculations would need to follow the following format:




For the particular type of I beam our crane will be using, due to the shape we have selected, it was required to use the following formula for is second moment of area.


However in order to find the maximum bending moment and the maximum shear force the following diagrams were used along with the knowledge that maximum bending moment will occur when the 1000kg load is placed in the middle of the beam and that the max shear force when the load is placed at its ends:



the diagrams and calculations allowed through the theory of superposition to combine:
for max shear:
Pa/l (for point load where a=l)+ wl/2 (uniform load)
therefore obtaining the equation: wl/2 + P

for max BM:
(Pab)/l (for point load where a=l/2 and b=l/2) + (wl^2)/8
Therefore obtaining the equation: (wl^2)/8 +Pl/4


The final thing which would need to be calculated is the max BM and SF that the particular beam we have selected can handle due to its dimensions and type of structural steel component. ( if these values are higher than the ones obtained for the calculations that our beam will create with its load, it will mean the beam is safe to use under these conditions as it will not fail under them):





the following equations will be used:
M = Z ("fy") where Z = I/y (for the moment)
V= "dw""tw"("fy"/(3)^(1/2))

"fy"= specific design strength of the steel selected from a list of steel grades (e.g. are given on previous blogs)

Another factor that could be calculated in its stead includes the working stress at a distance point y. It would need to be found using the yield point or ultimate tensile stress for that particular grade of steel and divide it by a safety factor e.g. 2 (as given in the example). This will ensure that the steel will remain in the safe elastic region and avoid becoming permanently deformed.

calculations for IPE 160 beam and legs

To find the stresses on each leg, I had to go over the calculations for the IPE160 beam using the dimensions sandy published. This also gave me a chance to check sandy's answers (they are all correct). The calculations show that the maximum amount of stress experienced by any part of the legs is 4.818mpa. The lowest compressive yield strength for an aluminium alloy that I have come across is 105mpa, therefore this comfirms that the vast majority aluminium alloys would be able to be used on the legs.Now its just a question of finding the most cost effective one.

Calculations for the glide rail

The following are the formulas used in the calculations for the designed boom of the crane:


Equations courtesy of:

http://www.efunda.com/math/areas/SquareIBeam.cfm

Mechanics of engineering materials, second edition, by PP Benham, RJ Crawford & CG Armstrong

Elements of Strength of Materials, International Student Edition, by S Timoshenko

Legend
E – Young’s modulus of the material (Pa)
I – second moment of area (m^4)
w – Uniformly distributed load (N/m)
P –point load (N)
L –total length of glide rail (m)

Using the properties of superposition it is possible to obtain the combined loading effect on a structure by adding the individual effects caused by the different loading conditions.


Courtesy of:
Mechanics of engineering materials, second edition, by PP Benham, RJ Crawford & CG Armstrong

Elements of Strength of Materials, International Student Edition, by S Timoshenko

The ORIGINAL I BEAM possesses the following values:



This particular glide rail beam possesses several disadvantages, particularly the total mass. This would make it difficult to be carried by 4 people over 100m of rough ground. The deflection however complies with the BS 5950-12000 part 1:



The maximum deflection for the given length would be 5.1m/180=0.02833m.

Several improvements would need to be carried out on the beam particularly choosing a beam with a low manufacturing cost. Currently choosing this type of beam would have large cost implications due to its custom made quality.

Alternative options were therefore researched (courtesy of http://www.structural-drafting-net-expert.com/steel-selections-IPE.html):





Using the values of the original I beam structure as a template three standards were chosen IPE140, IPE120 and IPE160.
These were also selected as the beam width fell within an acceptable range 50-220mm to be able to use the selected travel trolleys for the hoists.

The following results were obtained:



From these calculations it can be deduced that not only has an effect been created on the mass but also on the deflection value, falling within the acceptable range.

According to the standard, the new max deflection value= 4.7m/180=0.02611.

The length was selected due to the dimensions of the loads that will be dealt with.

It is likely that beam IPE140 will not be used in the final design as the deflection is too close to the upper range of the specified BS 5950-12000 part 1 standard.


As we can see the mass has decreased to a weight which could be carried by 2 people however the I beam dimensions caused the deflection to be greater than the accepted standard for structural steel. Therefore beam IPE120 will not be used in the final design.




It is clear from these results and from the δmax equation that the higher the second moment of area the lower the deflection but the greater the area and the overal mass of the beam, an appropriate balace always needs to be found.

My calculations for current boom (corrected)

in my final page i forgot to intergrate the step function. however the value of the deflection is unaffected due to the fact that i copied these calculations from my original rough notes, and in those i did intergrate the step function correctly.

System Calculations

These are the calculations that have been worked out for the design we chose.