14:4 univp:usity of Colorado studies 



A closing line imn' can be found which will make 2M=0 and 

 2M«=0, but unless the as-'umed pole C happens to be the true pole 

 D (which is not likely) the condition 2M?/ = cannot be satisfied 

 by the trial polygon. 



B/i = K7? / = the average of the trial polygon interscepts = 23.524 

 feet. The line nn' might therefore be the true closing line of the 

 trial polygon because since the vertical interscepts between nn' and 

 the broken line BIJK represent bending moments, their total sum 

 by construction is zero. Column 5 of Table II gives quantities cor- 

 responding to M.k; they are found by multiplying the interscepts of 

 column 3 by the lever arms of column 4. 



We must now find a line mra' , the true closing line of the trial 

 polygon, which will make not only 2M = but also 2M-^'=0. It 

 sometimes happens that nn' coincides with wm' but not usually. 



Draw the diagonal nK. in the parallelogram nn'li.^ dividing 

 it into the triangles nn'l^ and nKB. Scale the interscepts in 

 these triangles corresponding to the points of the rib. Thus we get 

 column of Table II. These interscepts could better be found by 

 the relations of similar triangles. For example 



, ?iB 23.524 X16.2U 



fg=ed=. x= — 



span 192 



Here //B and the span are constants for all interscepts and .t' is the 

 lever arm for each interscept in turn. One setting of a slide rule can 

 therefore give all the figures in column 6. 



Column 7 is found by multiplying the interscepts of column (> 

 by the lever arms of column 4 and gives quantities similar to mo- 

 ments M;/'. Find the sum of column 7. 



Divide the sum of column 5 by the sum of coluum 8. This 

 gives 31' = the distance of the line RT from the right springing point 

 v. In the line RT is found the centroidof the interscepts of the 

 trial polygon BIJKB. 



Divide the sum of column 7 by the sum of column 6. The 

 quotient, .v' ', gives the distance of the centroid of the interscepts of the 

 triangle 7iKB from the right springing point. The centroid of the 



