MANALO, IVY L.CE Competency Appraisal II BSCE 5-2 ENGR. RICO ASUNCION

HYDRAULICS

1. A 300 mm dia. pipe is equipped a 100 mm dia. orifice plate as shown. By calibration, the discharge coefficient was determined to be 0.63. What is the discharge in the pipe if he differential pressure registered during the flow equals 82.9 kPa?a. b. 0.048c. 0.064d. 0.068e. 0.044f.

Solution:Q = CA h = h = h = 8.45 mQ = 0.63 (0.100)2 Q = 0.064 m3/s

2. A cylindrical water storage 12 m in dia. and 8 m high, is filled through a 100 mm pipe carrying a velocity of 40 m/s. How long it will take to fill the tank?a. b. 48.025 minc. 39.065 mind. 25.125 mine. 52.065 min

Solution:Q = AVQ = (0.10)2 (40)Q = 0.314 m3/s t = = t = 2881.5 sec t = 48.025 min

3. Determine the discharge in L/sec of a trapezoidal the weir having a head of 0.30 m with sides inclined 14.04 with the vertical and a length of crest of 2 m. a. b. 629 L/secc. 719 L/secd. 611 L/sece. 721 L/sec

Solution:tan 14.04 = 0.25tan = The weir is a Cipolletti weirQ = 1.859 LH3/2Q = 1.859 (2) (0.3) 3/2Q = 0.611 m3/sQ = 611 L/sec

4. A sharp crested, suppressed weir is under a head of 0.67 m. The weir length is 3 m and its height is 1.22 m.a. b. 0.6522c. 0.5623d. 0.3783e. 0.7282

Solution:Cw = 0.611 + Cw = 0.611 + Cw = 0.6522

5. A channel of rectangular cross section conveys water at a rate of 9 m3/s with a velocity of 1.2 m/s. Compute the hydraulic mean depth.a. b. 1.936c. 1.047d. 0.863e. 0.968

Solution:

b = 2dA = bdA = 2d2Q = AV9 = 2d2 (1.2)d = 1.936

P = 1.936 (2) + 3.872P = 7.744 mA = 1.936 (3.872)A = 7.496 m2

R = R = R = 0.968 (hydraulic mean depth)

GEOTECHNICAL ENGINEERING I

1. A saturated clay layer has a thickness of 10 m with a water content of 51% and a sp.gr. of 2.72. Compute the total stress at the bottom.a. b. 168.7 kPac. 176.8 kPad. 156.7 kPae. 187.7 kPa

Solution:

Saturated unit wt. of the clay:S = 100 = e = 1.39

sat = sat = sat = 16.87 kN/m3

Total Stress at the bottom: = 16.87 (10) = 168.7 kPa

2. A soil sample has a mass ofof 1830 g taken from the field having a volume of 1x10-3 m3. It has a sp.gr. of 2.60 and a water content of 10%. Determine the void ratio.a. b. 0.26c. 0.43d. 0.56e. 0.34

Solution:Dry Unit wt.moist = moist = moist = 1830 kg/m3dry = dry = dry = 1664 kg/m3

Void Ratiodry = 1664= e = 0.56

3. From a given formation of soil, the water table is located 1.0 m below the ground surface. Determine the effective stress at a depth of 4.2 m.a. b. 54.29 kPac. 49.76 kPad. 63.83 kPae. 64.83 kPa

Solution:

Total stress = 20.4 (4.2) = 85.68 kPa

Pore pressureUA = 9.81 (3.2)UA = 31.39 kPa

Effective Stress = Total stress Pore pressureEffective Stress = 85.68 kPa - 31.39 kPaEffective Stress = 54.29 kPa

4. A vertical retaining wall 6 m high with a cohesionless horizontal backfill having a unit weight of 15.5 kN/m3. The backfill has an angle of internal friction of 30. Determine the active static force on the wall if it is frictionless.a. b. 85 kNc. 95 kNd. 83 kNe. 93 kN

Solution:Pa = Pa = Pa = 93 kN

5. A soil sample was determined in the laboratory to have a liquid limit of 41% and a plastic limit of 21.1%. If the water content is 30%, what is the characteristic of soil?a. b. Liquidc. Plasticd. Densee. Brittle solid

Solution:Liquidity IndexLI = LI = LI = 0.447

Note:LI < 0brittle solid0 < LI < 1plasticLI > 1liquidSince LI < 1, it is plastic

GEOTECHNICAL ENGINEERING II

1. A retaining wall 6m high is supporting a horizontal backfill having a dry unit weight of 1600 kg/m3. The cohesionless soil has an angle of friction of 32 and a void ratio of 0.68. Compute the Rankine active force on the wall.a. b. 76.86 kNc. 86.76 kNd. 97.68 kNe. 98.67 kN

Solution: = = 15.70 kN/m3

Pa = Pa = Pa = 86.76 kN

2. A cantilever sheet pile is 8.2 m long with a depth of embedment of 32 m. Angle of friction of the soil supported by the sheet pile is 34 and has a unit weight of 1.91 g/cc. There is water table below the base of the sheet pile. Compute the passive force acting on the sheet pile. = 9.81 kN/m3

a. 338.70 kN/mb. 233.60 kN/mc. 133.30 kN/md. 178.30 kN/m

Solution: = = 18.74 kN/m3 Pp = Pp = Pp = 338.70 kN/m3. A square footing has a dimension of 1.2 m x 1.2 m and has its bottom 1 m below the ground surface. Find the ultimate bearing capacity of the soil if the ground water table is at the bottom of the footing.From table: Nc = 35; Nq = 22; Ny = 19

a. 1392.17 kPab. 1749.48 kPac. 1102.64 kPad. 1201.14 kPa

Solution:qu = 1.3CNc + qNq + 0.40BNyqu = 1846 (1)qu =1846 kg/m2 = = 1965 1000 = 965 kg/m3

qu = 1.3CNc + qNq + 0.40BNyqu = 1.3 (1605) (35) + 1846 (22) + 0.40(965)(1.2)(19)qu =122440.3qu = qu =1201.14 kPa

4. A group of friction piles in deep clay is shown on the figure. The total load on the piles reduced by the weight of soil displayed by the foundation is 1700 kN. Thickness of silt is 1.5 m and that of clay is 16 m. Compute the compression index of the clay layer.

a. 0.288b. 0.328c. 0.238d. 0.388

Solution:L.l. = 0.80 = LL = 41.965

Cc = 0.009 (LL - 10)Cc = 0.009 (41.965 - 10)Cc = 0.288

5. A group of friction piles in deep clay is shown on the figure. The total load on the piles reduced by the weight of soil displayed by the foundation is 1700 kN. Thickness of silt is 1.5 m and that of clay is 16 m. Compute the approximate total settlement of the pile foundation.

a. 126.21 mmb. 156 mmc. 138 mmd. 148.34 mm

Solution: = = 30.36 kPaSp = Sp = Sp = 0.138Sp = 138 mm