MURPHY : MAXIMUM LOAD ON A I.INTEI.. • 33 



Substituting for q a \-alue 144 in (2); that is assuming the brick- 

 work to weigh 144 lbs per cub. ft. and solving for H^ we have 



"'=;r to- 



From the second member of (2) we see that the resisting moment 

 of the beam varies as H'-^ or 



Q=cH2 (4), 



c being a constant and Q the resisting moment. Equation (4) 

 represents a parabola whose axis is horizontal and is the curve OBD 

 in the figure. 



Any horizontal line, as LN, between the vertical and OB is 

 proportional to the moment of the weight of the wall of height H. 

 Any line as LK is proportional to the resisting moment of the wall 

 of height H and the difference of the lines LN and LK is propor- 

 tionally to the load carried by the lintel when wall is at height H. 

 This weight carried by the lintel is zero at B since the wall whose 

 height is Hg is self supporting. 



From a property of the parabola the intercept between the curve 

 and a chord is a maximum at the middle of the chord, hence 



Hn=iH3 (5). 



That is, the height of the wall which produces the maximum stress 

 in the lintel equals one half the height of the self-supporting wall. 



r rom the parabola we have— — g^-^^^— - :=:J. 



AB ^s 



CE H 



From similar triangles — " =^, hence 



AB Hj5 



J)E=^ CE (6). 



That is, when the height of the wall is such as to produce a maximum 



load on the lintel only one-half the weight of this wall is carried by 



the lintel. 



From (2), (5) and (6) we find the maximum load carried by the lintel 



T, 18 t S3 ^ , . ^ 



P=: — lbs., nearly (7). 



K. 



