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Application Sheets:: Difference between revisions
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== Force Resolution: == | == Force Resolution: == | ||
'''Context''' Design of a tower crane | |||
'''Objective''' To assess the adequacy of the cable stay for the jib | |||
=== Concepts used in this application sheet === | |||
=== Concepts used in this application sheet | |||
* Force: applied load, moments, force resolution | * Force: applied load, moments, force resolution | ||
{| class="wikitable" style="float:right; | {| class="wikitable" style="float:right; | ||
|+ | |+ | ||
|[[File:Crane-stay-sc-model.png|center | |[[File:Crane-stay-sc-model.png|center|Jib Analysis Model|alt=|thumb|300x300px]] | ||
|- | |- | ||
|[[File:Crane-star-sch-pin.png|center|frame|Pinned Connection]] | |[[File:Crane-star-sch-pin.png|center|frame|Pinned Connection]] | ||
|- | |- | ||
|[[File:Crane-stay-sch-fbd.png|center | |[[File:Crane-stay-sch-fbd.png|center|Jib Free Body Diagram|alt=|thumb|300x300px]] | ||
|} | |} | ||
=== Engineering model === | |||
[[File:Crane-stay-sch.png|frameless|565x565px]] | |||
=== Structural Analysis === | |||
==== Analysis Model ==== | |||
* The self-weight of the jib is modelled as a uniformly distributed load along the length of the jib. | * The self-weight of the jib is modelled as a uniformly distributed load along the length of the jib. | ||
* As a simplifying assumption for this calculation, it is assumed that there is no moment continuity in the jib at the mast i.e. the end of the jib is pinned to the mast. | * As a simplifying assumption for this calculation, it is assumed that there is no moment continuity in the jib at the mast i.e. the end of the jib is pinned to the mast. | ||
| Line 30: | Line 29: | ||
* ''All forces and dimensions are given in kN and m respectively, unless otherwise stated'' | * ''All forces and dimensions are given in kN and m respectively, unless otherwise stated'' | ||
==== Calculate Tension Force in the Stay | ==== Calculate Tension Force in the Stay ==== | ||
Calculate the resultant load due to the self weight of the jib: <math>3.5\times50 = 175kN</math> at 25m. | Calculate the resultant load due to the self weight of the jib: <math>3.5\times50 = 175kN</math> at 25m. | ||
==== Take moments about A | ==== Take moments about A ==== | ||
<math>\sum{M}= (100\times45)+(175\times25)-(F_{y}\times30)=0</math> | <math>\sum{M}= (100\times45)+(175\times25)-(F_{y}\times30)=0</math> | ||
<math>F_y=296kN</math> | <math>F_y=296kN</math> | ||
==== Resolve forces to find the total cable force | ==== Resolve forces to find the total cable force ==== | ||
<math>\theta = \tan^{-1}(\frac{20}{30}) = 34^\circ </math> | <math>\theta = \tan^{-1}(\frac{20}{30}) = 34^\circ </math> | ||
<math>F_{t}=\frac{F_y}{\sin\theta}=\frac{296}{\sin45^\circ}=529kN</math> | <math>F_{t}=\frac{F_y}{\sin\theta}=\frac{296}{\sin45^\circ}=529kN</math> | ||
=== Assessment: === | |||
==== Criterion ==== | |||
Factor of Safety = σu/σw | |||
* σu is the ultimate tensile stress | |||
* σw is the working stress | |||
==== Data input ==== | |||
Force in stay Ft = 529 kN = 529*10^3 | |||
Diameter of the stay D = 25 mm | |||
Cable stay ultimate tensile stress σu = 345 N/mm2 | |||
Minimum required FoS = 3.0 | |||
==== Calculations ==== | |||
Area of stay A = πD2 /4 = π*25^2/4 = 491 mm2 | |||
Stress in stay σw = Ft/A = 529*10^3/491 = 1077 N/mm2 | |||
==== Apply the criterion ==== | |||
FoS = σu/σw = 345/1077 = 0.32 this is much less than the required FoS of 3.0 so the design is not acceptable | |||
Decision: cable size must be increased | |||
==== Validation ==== | |||
The force in the cable may be significantly less than the estimated value because of the assumption of the pin connection for the jib. Overestimating the force is a safe assumption. Crane collapses are relatively common and can result in deaths. Therefore, a higher than normal factor of safety is used to ensure the crane will not be overloaded. | |||
* | * | ||