Mr. B. C. Laws on Strength of the Thin-plate Beam. 173 



subject to a uniformly distributed load w per unit length 

 due to a head of water h. 



Take the axis -of X at the mid-surface of the plate, i. e. in 

 a plane distant t/2 from the faces of the supporting frames, 

 the origin midway between the supports, and the axis of 

 Y downwards. 



The diagrammatic representation of the beam is shown in 

 fig. 2. 



Fig. 2. 



Distribution of forces acting on beam subject to load w 

 per square inch. 



The solution is worked out on lines similar to those indi- 

 cated in the previous paper, and we are led to the fundamental 

 equation : — 



B.I.g=M.+.p.y-!££=*3. 



where P denotes the pull along the beam. 

 PutP = ro s E.L 

 The solution to this equation is : — 





(a) 



giving the bending moment at any point of the beam, where 

 e is the base of the natural or Naperian logarithms. 



The deflexion y at any point of the beam is given by : — 



m m i v w - a { ( «r + «.-~> -(«?*+«."") j . «(«'-«*) 



• 'H m \ e m.a_ e -m.a J 2 



...(b) 



The greatest bending moment occurs at d and is obtained 

 by putting x = a in equation (a) ; and the maximum deflexion 

 ^is obtained by putting # = in equation (7>). 



