Question

Question Design the cross section of a gravity dam as shown in the figure below. Assume the specific unit weight of the concrete is 24 kN/m² and the friction factor between the dam and the underlying layer is 0.65. Consider only water loads and weight of the dam. Find the normal stress and check against sliding and overturning, assuming full uplift. (8 mark) 10 m 8 44 m 1:10 3:10

Answer 1

lom 스 face slope = R 8m TH:10V Downstream slope 136m 3W; lov Pu use dw, from IP: triangle Abi Vwx36 44 E - 8 - nonm lom Z=36 3 ye x = hin from triangle die 212m X2 10 un 3 26 13.2 m no from triangle apa 36 hn =) 8=3.6m Forces Perte calculation fu he lever arm (km) (kN) (m) MR (km) M (kNm) w (fr) 24 7313-2 57313:35 10+13.2 +4.4 3 24.67 wa loxhux an IOS 60 13.24 10 / 1 92192.0 182 Wz (1324 x 13.2x 4n|x24 6969.6 dim X132-88 61332.58 PY (4 *3-6*36) * 22 1555.2 24.67 38 366.78 RA **9.814 36436 6 B36.6% 635688 12 76282.56 {fv = 27408.08 EfM= 6356188 EMR = 349204.6 EMO 576237

Taking Causes sc moment about toe. So, take reference about toe and cololate leverarm fine along negative resisting moment Force al moment along positive y aseis causes overturning {fra dinog kn EFHE and 6356.88 ICH EMRE 349204.6 kum Emoa 76282.56 INm Net moment AM = Eme-&Mo = 272922.oh lem ,2 . distance of resultant from the toe x = = 12: 75 m Ef .. factor of safety against overtoning log Ema {mo = 4.582 Fos against OT >1.S hence ok ii) fos against sliding l Efv 0.6521608 {FH 6356.88 2.1831 hence ok

eccentricity of the Resultant fore fone Ř from the centre 27.6 2 12.75 - 2 Ba base cuidth ( 1.05 m < P = 66 no tension] anywhere in the dom) developed normal streeses at toe Paces +6) ( ltre 11 21908 6x l-os 27.6 952.7 27.6 Kryon? at heel, Pn= EN Ibe B - 21408 1 681-05 27.6 :( 598.6 KM/m2 27.6 normal strees at toe = at heel a 952.7 Art/m2 59816 kr/m2 and the dom is is safe against overturning and Bloeding