1840. J 



THE CIVIL ENGINEER AND ARCHITECTS JOURNAL. 



81 



bottom was porous, k should be increased to about 3X1S-4 — 5 = 

 55.2 — 5=.'iO feet. This shows the importance of securing the bed of 

 the dam from water b)' dredging, or otherwise clearing away all porous 

 materials. 



Problem VIII. 



To find the strength of a dam (form fig. 4) sufficient to resist the 

 pressure of a given depth of water. 



Fig. 4. 



By using the same notation as before, putting/ for EK, and k for 



d, k 125r" c 



DE, we get »hdy,{--\-k)-\-skf'X.-=——- X- for the equation of 



2t Z id o 



equilibrium from which s 6 (l--\-2 s bdk-\-s/k- ■■ 



12 5e^ 



This equation 



, , „ , 125c3 k-'f . ,2hd, 125c3 bd"^ „ 

 gives us cZ-+2fZA=-:rT -• and k"-\ -^k=-x — -, j- . From 



these we find a=v}^+ ^''~{K ~k (1), and 

 36s 



i^^^^JlljJ'l^J'J:.. (2). From these values for d and * 

 3»/ f f- b 

 we can fiud one when the other is given. 



Example \. — Having given *=4 feet, /= 10 feet, i = 21 feet, and 

 c=n feet, to find the value oi di 



/125c3 {i—f)k^ 7014125 Vt'Q , 



Bj^equatio,, (i) ,=V^ +L_Z_ _,=^__+___4 = 



V 108-3+8-4 - 4 = V 1 IIJ'^ - 4 = 10-S - 4 = G-S, or 7 feet nearly, the 



value required. 



Example 2.— Suppose *=10 feet, f~\7 feet, i=33 feet, and c=30 



feet, what is the width {d) equal to? 



„ , T25X3U' ItiXlU ,,^ , 1250U , IGOO , 

 Here d=./ ■ -L ^r 10=V- — -l - 1(J = 



3X33X90 



33 



33 



33 



V 



141U0 

 33 



-10=v/427-3-10=20-7-10=10'7 feet, the width re- 



quired. 



Example 3.— To find tlie value of k wheud=6 feet, the otlier 

 dimensions remaining the same as in Example 1. 



Frpm equation (2) *=V!!!f!_i^+^_*-i= 

 38/ / +/•-' / 



V 



614125 756 , 15876 126 



jyy--^=V227-5-75-6+158-8-12-G = 



2700 10 

 V310-7— 12-6 =17'6— 12-6 =5 feet, the value sought. 



Example 4. — To find the value of k when (^=10 feet, the other 

 dimensions remaining the same as in Example 2. 



Here t=:/\/ ^^^^^O' 33X10^ 33nU'^ 33xlO_^ 

 3X90X17 17 ^ 172 17 "^ 



V 



12500—3300 . 108900 330 



==V&4 1-2+37 G-8— 19-4 = 



17 ' 289 17 



V918— 19-4 =i3Q-3— 194 *10-9 feet, the value sought. 



Probiem IX. 



To find the strength of a coffer-dam fig. 5, sufficient to resist the 

 pressure of a given depth of water. 



Fig. 5. 



Here, by putting i-' for F E, and/' for K F, we have 



(I' + <^ ) X sk'f + I X s.i6 =i^' X g. 



for the equation of equilibrium by disregarding the vertical pressure 

 of the water above K G, and thence s&'^/' + 28 dk'f -\-d-sb — 



; from this equation we get 



3 



b 36s 6 



125 c' d- b 

 and /i" + 2^*" = -^ — -:; — -77-; these equations give 



A/ 386 6 ■'' 6^ 6 



„„ I 1, /l25c' bd' . ,, J 



(!)• 



(■2). 



Example 1. — When k' is equal 4 feet, what is the value of d, tiie 

 other dimensions being the same as those in example 1, problem 8. 



From equation (1) rf= ^ / X ^.Y^^V _ fXU) 10^x1 

 V 3x90x21 21 ^ 21^ 



C4 _ /g 14, 125 IGO 1600 _ 40 



t "y 5G70 ■ 21 "^ Hr 21 



lOxj 

 21 



= y/ 108-3 — 7-G + 3-G — 1-9 — v'i04-3 - 1-9 



= 10-2 — 1-9 = 8-3 feet. 



Example 2. — Using the same dimensions as in example 2, problem 

 8, what is the value of d ? 



d = / 125x30' _ 10^ X 17 10- X 17' 

 a/ 3x33x'JO 33 "*" 33- 



10x17 

 33 



V 



1-2,500 — 1700 28,900 

 + 



• 5-2 = 



33 ' 1080 



V327-3 + 2G-5 — 5-2 — V353^ — 5-2 = 18-9 — 5-2 = 13-7 feet. 

 Example 3. — Using the same dimensions as those in example 3, 

 problem 8, what is the value of i? 



From equation (2) h := 



V125: 

 3xi 



Xl7' 



90x10 



21x6' 

 ItT 



+ 6' 



= t/ 227-5 — 75-6 + 86 — 6 = ^ 187-9 — 6 = 13-7 — 6 = 7-7 feet. 

 Example 4. — Using the same dimensions as those for example 4, 

 problem 8, what is the value of k' ? 



,^ /I25x30^_10;;x^3 

 V 3XWX17 --^[7—^ 



*' = 



M 



