90 STRESSES IN A TRANSVERSE BENT 



Substituting the value of B given in (31) we have 



M b = H 1 d + C d C h (32a) 



Eliminating h and d by means of (41) and (42) we have finally 



Jf t = H 1 , (43) 



In like manner it can be shown that the moment at the base of the 

 windward column is 



<a) 



where w equals the wind load per foot of height. 



To find V we will take moments about the leeward column. The 

 moments M b and M bj at the bases of the columns respectively, resist 

 overturning and we have 



and since H 



Now if - is taken equal to y , we have after transposing 



(45) 



It will be seen that (45) is the same value of V and F 1 that we 

 would obtain if the bent were hinged at the point of centra-flexure. 



From (43) and (45) it will be seen that we can consider the col- 

 umns as hinged at the point of contra-flexure and solve the problem 

 as in Case I, taking into account the wind above the point of contra- 

 flexure only. The maximum shear in the column is shown in (d) 

 Fig. 52. 



The maximum positive moment occurs at the foot of the leeward 

 knee brace and is M k = H (d y ) ; the maximum negative moment 

 occurs at the base of the leeward column and is equal to M b = H y Q . 



