AND DIAGRAMS OF FORCES. 203 



so that if f t g, h in the diagram of stress correspond to F, G, H in the 

 original figure, we have 



h=-p' 1 sinq<l> (28). 



Case of a Uniform Horizontal Beam. 



As an example of the application of the condition that the stresses must 

 be such as are consistent with an initial condition of no strain, let us take the 

 case of a uniform rectangular beam of indefinite length placed horizontally with 

 a load = h per unit of length placed on its upper surface, the weight of the 

 beam being k per unit of length. Let us suppose the beam to be supported 

 by vertical forces and couples in a vertical plane applied at the ends ; but 

 let us consider only the middle portion of the beam, where the conditions 

 applicable to the ends have no sensible effect. Let the horizontal distance x 

 be reckoned from the vertical plane where there is no shearing force, and let 

 the planes where there is no moment of bending be at distances from 

 the origin. Let y be reckoned from the lower edge of the beam, and let b 



be the depth of the beam. Then, if 17= , , is the shearing stress, the total 

 vertical shearing force through a vertical section at distance x is 



f03v-W -l^} 



\ fl iT I \ CM IT I 



and this must be equal and opposite to the weight of the beam and load 

 from to x, which is evidently (h + k)x. 



Hence, ^- = - (h + k) x<f> (y), where <(&)-<(0) = 1 ............... (29). 



From this we find the vertical stress 



The vertical stress is therefore a function of y only. It must vanish at 

 the lower side of the beam, where y = 0, and it must be A on the upper side 

 of the beam, where y = b. The shearing stress U must vanish at both sides 

 of the beam, or <j>'(y) = Q, when y = 0, and when y = b. 



262 



