CHAP, in.] SUPPLEMENT TO CHAP. VH. Ill 



products obtained by multiplying the mass of every element of a given crota- 

 section by the square of its distance from that axis. 



If a parallelogram stand on end, and then its support be suddenly pulled 

 away from under it, it will fall over backwards. But to knock it over thus 

 requires force. The force which in this case overturns it is that due. to 

 inertia. At every point of the surface there is, then, a force acting, depend- 

 ing upon the mass of this point. But not alone upon the mass. A force 

 at the top acts evidently with more effect to turn the body over than one at 

 the bottom, which merely tends to make it slide. The moment of each ele- 

 ment of the area is, then, a measure of the force which at each point causes 

 rotation, and the sum of these moments is, then, the measure of the over- 

 turning action of the whole force of inertia upon the surface. The moment 

 of this latter force, or the sum of the moments of the moments, is, then, the 

 moment of inertia of the cross- section. Each element of the surface must 

 then be multiplied by the square of its lever arm, and the sum of all the 

 results thus obtained taken. In other words, the moment of each element 

 is itself considered as a force, and then its moment again taken. The sum 

 is denoted by I. For any given dimensions and axis it is a perfectly defi- 

 nite quantity, and may thus often replace expressions containing unknown 

 quantities. 



The principles of the calculus just developed will enable us to deter- 

 mine it in some cases, at least, very readily. Its value for various forms of 

 cross-section, in terms of the given dimensions, is given in every text-book 

 upon the strength of materials. 



Let us suppose a rectangular cross-section of breadth b and-height h, and 

 take the bottom as axis. The area of any elementary strip is, then, b d x. 

 If its distance from the bottom is x, we have for its moment b x d JK, and for 

 its moment of inertia, then, b x* d x. Integrating this expression, we have 



b X s 

 bx^dx- -y + O. 



This integral is to be taken between the limits * = and x = h. For * = 0, 



b h 3 



b x'' d x = 0, and hence O = 0. For x = h, then, we have -. If the axis 



3 



had been taken through the centre of gravity, we should have the above 



integral between the limits + - and For+ we have + O. For 

 2 2 2 24 



-, + O. Subtracting one from the other (Art. 6), we have 



for the moment of inertia. For a triangle of height h and base b, we have 

 for axis through centre of gravity, from Art. 7, for the area of the very 



small strip at distance x, -bdx x d x. Multiplying this by a?, we have 



3 h 



for its moment of inertia -bx*dx -x 9 d x. The integral of this ia 

 3 n 



2 4- 



For x = h, this becomes 

 o 243 



