RETAINING WALLS 



341 



BATTERED BACK 



For a wall with a battered back, Fig. 2, the line of maximum 

 pressure is the one bisecting the angle ibt formed by the vertical 

 bt and the slope of repose. The prism of maximum pressure 

 is one whose cross-section is cbs. The point of application e of 

 the force P is such that be = %bc; X is perpendicular to be, and 

 its magnitude is determined as follows: Calculate the weight 

 of the prism of maximum pressure for a unit length and lay it 

 off to any convenient scale on a vertical line drawn through e. 

 Let this weight be represented t u k 



by el. Through I draw Ix, 

 parallel to bs and through e 

 draw ex perpendicular to be; ex 

 gives the magnitude and posi- 



] / 



I / 

 I / 



FIG. 2 



FIG. 3 



tion of A'. Then, as before, on a line at right angles to ex, lay 

 off xp=fX=Y\ ep then determines the position and magni- 

 tude of P, and R may now be found as in the case of a wall 

 with a vertical back. 



SURCHARGED WALL 



With a surcharged wall, Fig. 3, the line of maximum pressure 

 is determined as before, and the maximum pressure is considered 

 as being caused by the earth lying between the broken line bet 



