250 



REPORT — 1865. 



through the centre, whatever may be the in- 

 clination. Thus 



where 1= the length of the side. 



And for a square hollow girder of uniform 

 thickness, T i n* 7 *\ 



where l x = the length of the interior side. 



(7) For a parallelogram CEFG about any axis AB passing through 

 the centre, 



I = T \K{1 2 sin 2 (0, + 6) + e 2 sin 2 flj, Fig. 9. 



where Z=CG=EF; e=CE = GF; E= area of 

 parallelogram; 0=angle GCE, 1 =angle B, the 

 angle which the side CE or CE produced makes 

 with the axis of moments. 



When 0=90°, or the surface becomes a rec- 

 tangle, 



(8) For angle-iron AMRFE about any axis 

 NO, 



where Z=AM=MR, Z 1= =EF=FL, x, x l = the 

 distance of the centres of the squares MC and 

 FC respectively from NO. Fig. 10. 



(9) If A and B be put for the moments of 

 inertia of a surface about its two principal axes, 

 perpendicular to each other respectively, then 

 the moment of inertia of the surface about any 

 axis passing through the centre will be 



I o =Acos s 1 + Bsin 2 1 , 



where t is the inclination of the oblique axis 



to the principal axis corresponding to the mo-0 c ST 



ment of inertia A. 



(10) For a plate KB in the form of a quadrant, about the axis NO parallel 

 to the radius CB, 



I= TB { r\^ + 4e^-r l \r^ + 4e') } +§<r*-r 1 s ). Fig. 11. 



where r is put for the external radius of the plate, r t 

 for its internal radius, and e is put for the distance of 

 the axis NO from the radius CB. 



When the thickness of the plate is small as com- 

 pared with its distance e from the axis, then 



I=lK(r 2 +4e 2 )+f<r 3 -r 1 3 ). 



Section IV. 



To find tlie Moment of a Complex Girder, such, as an iron sTiip, composed 



of a series of parts. 



18. Let a , a l9 a n represent the sectional areas of the portions com- 





