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[[File:Station-truss-sch-truss-cut.png|thumb|539x539px|Truss with cut for further analysis]] | [[File:Station-truss-sch-truss-cut.png|thumb|539x539px|Truss with cut for further analysis]] | ||
Calculate the force in the chords at the centre of the truss, use a cut just to the left of the central post. | Calculate the force in the chords at the centre of the truss, use a cut just to the left of the central post. | ||
Take moment equilibrium about joint G: | |||
<math>-F_{DE}\times5-(250\times30)+(25\times25)+50\times(20+15+10+5)=0</math> | |||
<math>F_{DE}=-875kN</math> (i.e. compressive) | |||
=== Assessment === | |||
==== Criterion ==== | |||
Factor of Safety <math>=\frac{\sigma_u}{\sigma_w}</math> | |||
*<math>\sigma_u | |||
</math> is the ultimate compressive stress | |||
*<math>\sigma_w | |||
</math> is the working stress | |||
==== Data Input ==== | |||
Minimum required FoS: <math>2.0</math> (given) | |||
Ultimate compressive stress <math>\sigma_u=240N/mm^2 | |||
</math> (this value is an older steel classification that is no longer used. However it is appropriate here because the truss was built in the late 1800s) | |||
Area of section, <math>A_s = 3870mm^2</math> | |||
==== End diagonal AB assessment: ==== | |||
Axial force <math>F_{AB}=356kN=3.56\times10^5N</math> (from analysis) | |||
<math>\sigma_w=\frac{F}{A}=\frac{356000}{3870}=92.0N/mm^2</math> | |||
<math>FoS=\frac{240}{92.0}=2.6</math> | |||
Decision: The chosen steel section is suitable for the compressive forces it has to withstand and meets the minimum FOS criterion. | |||
==== Top chord DE assessment ==== | |||
Axial force <math>F_{AB}=875kN=8.75\times10^5N</math> (from analysis) | |||
<math>\sigma_w=\frac{F}{A}=\frac{875000}{3870}=226.1N/mm^2</math> | |||
<math>FoS=\frac{240}{226.1}=1.06</math> | |||
Decision: The FOS is less than 2 and hence a bigger section should be chosen to ensure that the structure is safe. | |||
The next steps can either be: | |||
* Choosing an arbitrary section and performing the same calculations as above or | |||
* In this case it is possible to rearrange the equations to solve for the minimum required section area | |||
==== Adjustment calulations for top chord DE ==== | |||
Substitute the equations for working and ultimate stress into the factor of safety equation and setting the required FoS to 2: | |||
<math>FoS=\frac{\sigma_u}{\sigma_w}=\frac{240}{\frac{F}{A_s}}=\frac{240}{\frac{875000}{A_s}}=2</math> | |||
Solve for <math>A_S</math>: | |||
<math>\frac{875000}{A_s}=\frac{240}{2}</math> | |||
<math>A_s=\frac{875000}{120}=7292mm^2</math> | |||
Therefore, any section with an area greater than <math>7292mm^2</math> will be acceptable. | |||