900 



THE IRRIGATION AGE. 



slight yield of the dam by reason of its elacticity and this 

 yielding may cause a strain considerably greater than the 

 pressure that caused it. Thus supposing that the pressure 

 against the dam should be increased by a violent flood or 

 earthquake, the dam would then be forced against the abut- 

 ments and the elasticity of the curved walls would allow a 

 little yielding such that the face would be a little out of the 

 perpendicular. This condition would add to the strain 

 on the slab in question a small percentage of the superim- 

 posed dead weight. As this is an indeterminable factor, it is 

 necessary to have an unusually high factor of safety in the 

 face wall. 



The following notation and formula are used in comput- 

 ing the water pressure on the slab : 

 g=center of gravity of outer surface of slab exposed to 



water pressure. 

 d=vertical distance from g to water surface, measured in 



feet=184'5. 



A=area of slab exposed to pressure measured in sq. ft. 

 w=weight of a cubic foot of water in Ibs. 

 L=length between supports in inches=16S. 

 W=total pressure on slab. 



In this case 



A=14 sq. ft. (The slab is 20 ft. long center to center of but- 

 tresses, but only 14 ft. long in the clear.) 

 D=1S4.50. 

 w=62.50 (approximately). 



It is evident that 

 W : : A X w X D = 14 X 184.50 X 62.50 = 161,437 



The pressure moment in inch pounds = 

 W.L 161,437 X 168 



= 3,390,177 inch pounds. 



8 8 



Weight of face wall, central slab = 



8X2 



- X 184 X 20 X 150 = 2,760,000 (approximate). 



}'l weight of dirt above floor = 

 17.5 X 13 X HO X 80 



= 500,500 



Total weight = 3,260,500 Ibs. 



Our beam therefore must stand a pressure of 3,390,177 

 inch pounds and the foundation for this segment must be 

 strong enough to carry a load of 3,260,000 Ibs. 



The following are the notations, formulae and respective 

 dimensions used in the formulae to calculate the strength of 

 the face wall as assumed above. 



The coefficients have been calculated to correspond to the 

 material used, both with regard to concrete and steel. Con- 

 crete will be first class 1-3-5 mixture and steel of good quality. 

 See Webb and Gibson, "Reinforced Concrete," pp. 49 and 50. 

 Mo=Ultimate resistance moment in inch pounds. 

 b=breadth of slab transverse to pressure = 12". 

 d=depth from the face under compression to the center of 

 gravity of the steel reinforcement, in this case = 82" 

 as shown later. 

 C=compressive strength of concrete in Ibs. per sq. in. = 



2700 in our case. (Webb and Gibson.) 

 Ec=Initial modulus of elasticity of concrete = 3,000,000. 

 Es=Modulus of elasticity of steel = 30,000,000. 



EC 30,000,000 

 R=Ratio of Moduli - - = 10 



Es 3,000,000 



S=Tensile strength of steel = 55,000 Ibs. 

 p=Percentage of steel = 



7 C" R 7 X 2700 2 X 10 



.0121 

 12 S (CR+.667S)= 12X55,000 (2700 X 10 +.667X55,000) 



Webb and Gibson p. 62. 



This percentage of steel amounts to 96 X 12 X -0121 = 

 13.94 sq. in. of steel for each tabular segment 1 ft. thick. 

 This may be done by placing 6 round bars, l l /g," in diameter, 

 in layers 6" apart and spaced 4" from each other, beginning 

 4" from the face. This gives a value to "d" of 82". 

 K=Ratio of depth from compressive face to neutral axis to 



the total effective depth, "d." 



K=From table on page 61, Webb and Gibson, = .424. 

 Y=Distance from compressive face to center of gravity of 



compressive stresses. 



Y=.357 Kd = .357 X -424 X 82 = 12.4 

 (See Webb and Gibson p. 59.) 



Mo=p b d S (d Y) = .0121 X 12 X 82 X 55,000 X (82 

 12.4 = 48,016,750. 



In other words, our slab will withstand a pressure of 

 48,016,750 inch pounds, and the maximum normal load that 

 can come on it is 3,390,177 inch pounds. 



The above calculations show that when the reservoir is 

 full, the face wall has a factor of safety against normal water 

 pressure alone of about 14, considering the face wall as a 

 simple beam supported at both ends, but it is also a cantilever. 

 Owing to the reversal of pressure when the dam is empty 

 the wall must be reinforced on both sides and hence with 

 regard to the buttresses it is both simple beam and cantilever, 

 which doubles the factor of safety, making it 28, and this is 

 disregarding entirely the fact that the face wall is backed up 

 with earth which, in turn, is held in place by all auxiliary 

 walls. It is evident, I think, that so far as the important 

 feature of water pressure is concerned, the face wall is more 

 than amply strong. 



It has been suggested that the dirt filling of the cells 

 would act as a liquid in exerting pressure, because this filling 

 will be packed with water. I think this extreme view is 

 unwarranted ; nevertheless, it is easy to compare the relative 

 strains of external and internal pressure by assuming such 

 a condition. We have taken 110 Ibs. per sq. ft. as the weight 

 of a cubic foot of water-packed earth and rock. The weight 

 of a cubic foot of water is commonly taken at 62'5 Ibs. If, 

 then, the earth filling were a liquid, it would exert a pressure 

 greater than that of water in proportion to the relative 

 weights or 



110 



= 1.76 



62.5 



The earth column is, however, divided by the sill-floor so 

 that we have 105 feet of earth as against 184.5 ft, of water. 

 The factor safety, then, will vary inversely as the relative 



105 X HO 

 height and weights - - = 1.003. 



62.5X1S4.5 



In other words, the face wall as designed is equally 

 strong to resist internal or external pressure, even if the 

 earthen filling is regarded as a liquid. It is obvious, however, 

 that the horizontal reinforcing of the buttress must be fast- 

 ened by tie bars to those of the face wall and be strong 

 enough to .prevent parting, so it is planned to hook the hori- 

 zontal reinforcing bars of the buttress over the horizontal 

 bars of the logitudinal walls. 



The next test in order is to ascertain what crushing strain 

 this wall will stand. 



Kidder" gives the crushing strength of unreinforced con- 

 crete (1 part cement, 2 parts sand, 6 parts broken rock), at 

 440,000 Ibs. per square foot. As we have in the bottom slab 

 20' X 8' = 160 square feet, the resisting strength is 70,400,- 

 000 Ibs., while, as previously computed, the weight above this 

 slab is only 3,260,500, and so we have a factor of safety of 

 20 when disregarding the reinforcement. 



The face wall is then abundantly strong from all stand- 

 points. 



We will now consider the buttresses. These are sub- 

 jected to two main strains. There is shear strain and pres- 

 sure exerted by the overturning moment and the direct water 

 pressure. As in this design the resultant of the graphic com- 

 putation of the overturning moment strikes the base line 1/3 

 the distance from the toe, it is sometimes assumed that all 

 of the resultant pressure is applied along the imaginary 

 line of the resultant and is met by the resistance of the butt- 

 ress wall equidistant on both sides of this line. This means 

 that the pressure will gradually increase from face to rear 

 and the extreme rear will bear twice the average pressure. 

 As the buttress is 165 feet long and 3 feet thick at the bot- 

 tom, we have 165 X 3 = 495 sq. ft. of material in compres- 

 sion and each square foot will stand 440,000 Ibs. (not count- 

 ing the reinforcement), or a total of 217,800,000 Ibs. which 

 acts through a moment arm of 110 ft., the product being 

 23,958,000,000. The water pressure on the panel supported 

 by this buttress is 21,332,720 Ibs. (as shown above), which 

 acts through an arm of moment of 61.6 feet, the product 

 being 1,314,067,272, the factor of safety by this method of 

 computation being 17, on the average of 8J4 for the rear- 

 most portion. This will be much increased by the reinforce- 

 ment and by the supplementary resistance of the earth filling 



