Application Sheets:: Difference between revisions

Application Sheets: (view source)

Revision as of 18:10, 25 December 2020

261 bytes added , 18:10, 25 December 2020

no edit summary

Liam

Bureaucrats, Interface administrators, Administrators

165

edits

@@ Line 1: / Line 1: @@
-== Force Resolution: ==
+== Crane Tension Cable: ==
 '''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 shee ===
 * Force: applied load, moments, force resolution
@@ Line 45: / Line 45: @@
 ==== Criterion ====
-Factor of Safety = σu/σw
+Factor of Safety <math>=\frac{\sigma_u}{\sigma_w}</math>
-* σu is the ultimate tensile stress
+* <math>\sigma_u
-* σw is the working stress
+</math> is the ultimate tensile stress
+* <math>\sigma_w
+</math> is the working stress
 ==== Data input ====
-Force in stay Ft = 529 kN = 529*10^3
+Force in stay: <math>F_t=529kN=5.29\times10^5N
+</math>
-Diameter of the stay D = 25 mm
+Diameter of the stay: <math>D=25mm
+</math>
-Cable stay ultimate tensile stress σu = 345 N/mm2
+Cable stay ultimate tensile stress: <math>\sigma_u=345N/mm^2
+</math>
-Minimum required FoS = 3.0
+Minimum required: <math>FoS = 3.0
+</math>
 ==== Calculations ====
-Area of stay A = πD2 /4 = π*25^2/4 = 491 mm2
+Area of stay: <math>A=\frac{\pi{}D^2}{4}=\frac{\pi\times25^2}{4}=491mm^2
+</math>
-Stress in stay σw = Ft/A = 529*10^3/491 = 1077 N/mm2
+Stress in stay: <math>\sigma_w=\frac{F_t}{A}=\frac{5.29\times10^5}{491}=1077N/mm^2
+</math>
 ==== 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
+<math>FoS = \frac{\sigma_u}{\sigma_w}=\frac{345}{1077}=0.32
+</math> <math>\ll 3.0
+</math>
+This is much less than the required FoS of 3.0 so the design is not acceptable
 Decision: cable size must be increased