112 SUPPLEMENT TO CHAP. VH. [OHAP. HI. 



For a? = j-A, we have- & A'+O. 

 8 972 



27 1 



Subtracting one from the other (Art 6), we have & A 1 , or I A 1 for 



972 36 



the moment of inertia. The moment of inertia of the rectangle I = 



12 



may be written*^- = x Ax , or the moment of inertia of the half 

 ~ 24 32 



parallelogram is equal to its area, into the distance of its centre of gravity 

 multiplied by ds. of its height We see at once that when we consider, 

 then, the statical moments as themselves forces, the centre of action of these 

 moment forces does not coincide with the centre of gravity of the area. This 

 principle we have already noticed in Chap. VL, Art. 60. 



I A* /I \ a 1 



We can also put - = = ( - A ) . This value A is called the 



A 12 \2V 3 ) 2/3 



radius of gyration. It is evidently the distance from the axis to that point 

 at which, if the mass were concentrated or sum of all the forces were con- 

 sidered as acting, their moment of inertia would be that of the cross-sec- 



I 

 tion itself. The value of is, in general then, the square of the radius of 



A 



gyration. We have already shown in Chap. "VT. how to find it graphically 

 for various cross-sections. 



We are now ready to take up the case of a deflected beam, and to find 

 the differential equation of its curve of deflection. 



11. Change of hape of the Axi. In the Fig. given in the 

 Supplement to Chap. XIV., we have represented a beam deflected from its 

 original straight line by outer forces. Let the two sections A O, B D be 

 consecutive sections, parallel before flexure, and remaining plane after. Let 

 the length of the axis m a be , then n a = d *, and let d <j> be the very 

 small angle between the sections after flexure. 



If the deflection is small, s will be approximately equal to x, and d s to 

 d x. The elongation of any fibre at a distance v from the centre is, then, 

 d <f>. The unit force corresponding to this elongation is from (7) T = 



E 5* v. If d a is the cross-section of any fibre as d c, then the whole force 



d x 



of extension is 



Tl i n. n. n. ej\ 



~~das ' 



The moment of this force is, then, -= -. The integral of this be- 

 tween the limits + - and will give the entire moment of rupture. But 



2 2 



this is equal and opposite to the moment M of all the outer forces ; hence 



r* 



I t* dc 



J _* 



