STA BS HI FS ELEVATION BM-A ТР 1 TP 2 1.73 346.50 m 2.50 1.85 2.76 1.05 TP 3 3.15 1.20 BM-B 1.93 0.74 TP 4 1.67 0.98 TP 5 2.12 1.84 BM-C 1.77

Q: n = 18 kN/m $i'= 40° c;' = 0 0,7m Qsi(net) B=0,6m . 0,5m L=1,2m Y2 = 16 kN/m³ $2'= 30°

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Q: PROBLEM 01: The horizontal wye fitting splits into two equal flow rates. At section 1, Q1 = 0.11…

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Q: Ve waler ) = los Square PiPe with dimen Sin 25o m e =o. 25m m, L-120m hL=Sm C Ve waler ) = 10°m?/sec

A: Step 1: e =0.25mm γ=10^-5 m2/s hL = 5 m per 120m length

Q: 0.6 m - 0.6 m → 0.6 m -- 0.6 m 900 N/m 50° B Cl 2

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Q: 20 kN 40 kN/m Al B. 150 LN-m -S m -3m

A: Free body diagram of the given problem:

Q: A plug sits in a hole in the side of a tank. z = 2 m, ro-ring = 0.2 m, rplug = 0.25 m. O-ring 2 m…

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Q: 5m C KN m 2 KN lm 2m 2m JKN 3. B

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Q: Hiwi- clasify the structures below Lor stabilit. minagy. and deter Inte Trnk

A: As per authoring guidelines, if the question number is not specified, the first question or three…

Q: VII. Data Collection and Analysis Staff Reading (m) Rise Fall RL Remarks Station BS IS FS (m) (m)…

A: To complete the table by rise and fall method

Q: Hope your help hydraulic CEE 340

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Q: Q3: Answer ONE Section only. b. For the open channel find the discharge if channel slope (1/200).…

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Q: DIFFERENTIAL LEVELING NOTES Elevation Sta BS (m) HI (m) FS (m) Remarks (m) BM-1 2.00 1000 ТР-1 1.41…

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Q: A regime channel carries discharge of 25 m³/s. If the dmm-0.25 mm. Find the slope of bed.

A: given- discharge Q=25 m3/smean size dmm=0.25 mm

Q: A flow of 280 lit/s of water is carried in a 90° – V-notch flume with n-0.011. Find the critical…

A: Q=280L/s= 0.280m3/sec Theta=90o

Q: DIFFERENTIAL LEVELING NOTES Elevation Sta BS (m) HI (m) FS (m) Remarks (m) BM-1 2.00 1000 ТР-1 1.41…

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Q: Open Channel: ( Activity 3 J- Find the reauired diameter for circular pipes such that two Of them…

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Q: 2 kN 45° B A P = 4 kN D 2 m 2 m 2 m

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Q: 2.) Determine RA RB 5K 3K/fr 64 业业土 4' 2 材

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Q: Ra and Rb 10ok 200 kN 500KN.m 200 KN.M Ra Ro

A: as per our policy we should answer only one question at a time. so, i am answering only first…

Q: 4.0 m A=?- 1.6 m P=? 3.6 m T=? -2.0 m

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Q: AP=120 kPa 2m Sand V ground water table Sand 3m e=0.50 G=2.70 Clay 2.5m Sm 0.60 LI40%

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Q: Given E1 = 2*10" Kn.m² Suttlementate 2cm to the right Find all reachions 100kn Using moment…

A: Solution: Moment Distribution Method: Fixed end moment: Span DB: MBD=100×1=100 kN Span BC & BA:…

Q: 3 m b ▬▬ 4.5 m PD=12 kN PL=24 KN с 3 m WD=7.3 kN/m W₁ = 14.6 kN/m d

A: Given :-  Steel Beam  W16×19 of 1992 PD=12 KN PL=24 KN WD=7.3KN/m WL=KN/m L1=3m L2=4.3 L3. =3m To…

Q: 4. Find : O 9 :? Tmax 1. 25 P © e = ? Cmax a 0.80P member AC, of P- 200KN AAc:6000m? iF Stress

A: An object is strained when it is under tension. The term compression refers to something being…

Q: Staff Reading (m) Rise Fall RL Remarks Station BS IS FS (m) (m) (m) 100.0 Bench Mark 1.47m 0-53m…

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Q: ind Elő under each concentrated load. -200 -400 3.00 4.00

A: Draw free body diagram showing reactions at left end and right end supports next calculate reactions…

Q: Use 3ME Method 2 14KN/ 25 Im 2M 714 kN/m Im

A: Given data:- Below diagram is given

Q: Water is flowing in trapezoidal channel at a depth of 1.20 m with a bottom width of 7.00 m and has a…

A: Froude number: Froude number is defined as the ratio of inertia force to the gravitational force.…

Q: Ex find the critical load with Bifurcation approach and Energy approach (spring stiffness k). Fp=…

A: In an incompressible fluid flow, the density of small vol. element does not changes as pressure…

Q: Given the following differential level notes: STA BS HI FS ELEV BM-1 1.73 346.50 TP-1 2.50 185 TP-2…

A: Given data in question  Back sight Fore sight Elevation of BM1 To find out  Elevations of all points…

Q: Opo 3 m A В C D -2 m 2 m- -2 m 5 kN 5 kN 5 kN

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Q: Given the following differential notes below in the table: STA BS HI FS ELEV BM-A 1.73 346.50 ТР-1…

A: given - leveling table  to find out - point of highest elevation                    - difference in…

Q: 30 kN 60 kN 6 kN/m d. d /3 . R 36 kN R= 27.9 kN R 80.1 kN -3 m -3 m--5 m- -4 m-

A: To determine: Shear force diagram and Bending moment diagram of the given loading diagram.

Q: Q3. Estimate the precipitation depth at station X with coordinates (0,0) using four quadrant meth…

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Q: I"- 450 kN 1.5 O kN Val-1350 kN 0 kN 0.75 m y =16.4 kN/m • -25", e - 35 kPa Gs - 2.65, e - 0.71 GWT…

A: It has been asked to determine ultimate bearing capacity for above fig using hansen eq.

Q: 180 lb/ft 120 Ib/ft D B |C 5t 5t- 10 ft Compute for the Ely value at the midspan using Moment Area…

A: L = 20ft

Q: Problem #1 270 kN 180 kN El = constant E-200 GPa B 1-420x10° mm* Required: 6max en Og 3m 4.5m 3m

A: The Procedure for solving this problem is  Draw the Moment diagram of the above beam This can be…

Q: ras 8.2 in. Time (hr) Rainfall (in/hr) 0-1 0.85 1-3 1.2 3-5 1.6 5-7 2.9 7-10 2.4 10-12 1.8

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Q: Complete the differential level notes shown below, and determine the d of TP2 and BM11 STA BM 10…

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Q: problem 8-9 Water slopc of 0.0017. Usc n 0.014. velocity of 1.2 m/s. Find the normal depth of flow…

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Q: 6kN 2 kN\m 3 kN D 6kNm A B A CA 6m

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Q: Example Q=1000m/d, D.T= 2hr D,=0.06 mm Pp=1200 kg/m2, µ=1.027x103 N.s/m2 -V/70%, find L,W,h if L-4W…

A: Q=1000m3/dDT=2hrdp=0.06mmρ=1200Kg/m3μ=1.027×10-3Ns/m2Vscour=Vh0.7scoir velocity for primary…

Q: G2: The five fres and a Couple of fig comstitute Copliner force System. Determine the resultant…

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Q: Determine the reactions at the supports for the beam shown in Figure .

A: Consider the figure,

Q: 4 m 50 KN Fixed B A 777777 1 100 KN D C FA El, constant 2m 2m Roller

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Q: Three pipes are connected in parallel LA = 1800 m, DA = 0.45 m, fA = 0.012. LB = 600 m, DB = 0.15 m,…

A: The given figure is shown below:

Q: Critical path, Total float, Free float, Independent float A - 4 B - 3 C - 4 D A 5 E B 7 F C 4 G D 5…

A: Find critical path, total float, free float and independent float

Q: Q5: A vertical arc of 500 meters in length connects a line that rises with slope +1.35 and meets…

A: PVCSTATION=PVISTA-L2PVCsta=1800-5002=15+50PVCELV=PVIELV-g1L2PVCelv=886.1-0.0135×5002PVCelv=882.725ft

Q: 25 ... sat:= 18 -KN /m C-- .... -5 K=2.6 *lo. m/sec. 2.25m Datum

A: Saturated unit weight,  = 18 kN/m3 Hydraulic conductivity, k     = 2.6 x 10-5 m/sec…