138 



STRENGTH OF MATERIALS 



If T Q = 0, that is to say, if the load is assumed to be concentrated 

 at a single point at the center of the plate, formula (84) becomes 



3P 



(85) 



P = 



If the load is uniformly distributed over the entire plate, then 

 r = r and P = 7rr*w, where w is the load per unit of area. In this 

 case formula (84) becomes . 



P = 



TT/i, 2 



(-!)- 



which agrees with the result of the preceding article. 



Problem 142. Show that the maximum concentrated load which can be borne 



by a circular plate is independent of the radius of the plate. 



118. Dangerous section of elliptical plate. Consider a homo- 

 geneous elliptical plate of semi-axes a and b and thickness //, and 

 c suppose that an axial cross is cut out 



of the plate, composed of two strips 

 Mi and CD, each of unit width, and 

 B intersecting in the center of the plate, 

 as shown in Fig. 98. 



Now suppose that a single concen- 

 trated load acts at the inters. 

 of the cross and is distributed to the 

 supports in such a way that the two 



beams AB and CD each deflect the same amount at the center. E 

 AB is of length 2 a, from Article 66, Problem 7:.. the deflection 



at the center of AB is D l = \ L From symmetry, the react 



4o JiiL 



at A and B are equal. Therefore, if each of these reactions is denoted 

 by RI, 2 RI = P, and, consequently, 



Similarly, if R z denotes the equal reactions at C and Z>, the deflec- 

 tion D 2 of CD at its center is 



