Sta. BS HI FS IFS Elev. BM 1 2.22 56.70 m 0+00 2.7 2+00 3.1 0.8 4+00 ТР1 2.32 3.13 6+40 0.9 8+20 TP 2 1.2 2.77 2.16 11+10 15+00 18+30 20+50 BM 2 1.7 2.8 3.0 2.3 1.35 230 232

Q: 2k/ft 10 ft B + 5 ft 2k 5 ft SAL C

A: Draw the shear and moment diagram for the beam shown in problem 3.

Q: 80 kN MATHalino.com MATHalino.com 1 m E 1m F 1 m 0.75 m

A: First ∑MD=03AV+501=800.75AV= 3.33kN∑FH=0AH=80kN∑MA=03RD=502+800.75RD= 53.33kNAt joint…

Q: STA BS HI FS IFS ELEV BM-a 1.24 235.75 234.51m TP- 1 1.11 233.46 3.40 232.35 1+00 4.6 228.86 1+25…

A: At turning point we have both fore sight and back sight. Reading is more means , staff is in lower…

Q: KN 4m 上 4m 100 t Kat 200

A:

Q: Station Precipitation, mm Distance from Station x,y (km) (x.y) unknown 1. 5. 2. 10 2.4 3. 7.5 4 4.…

A: To find :- Probable value of precipitation at station x,y by using inverse distance method.

Q: 31. a) Deternine the elevation at each stations. b) Difference in elevation between the 1+20 and…

A: Given profile levelling HI = height of instrument FS = fore sight BS = back sight IS = intermediate…

Q: A lot has the following latitude and departure. Line ADJUSTED DEPARTURE(m) ADJUSTED LATITUDE(m) AB…

A: As per DMD method: Area=∑(DMD×Latitude)2…

Q: Situation 5. Determine Ix amd ly. 12" 6" 12" r=4" «9=J 6"

A:

Q: ADJUSTED CORR INT. ANGLE OBSERVED BEARINGS COMPUTED STA INT. ANGLE LINE LENGTH FORWARD BACK AB BC CD…

A: Given Closed Traverse with fore and back bearings.   To Determine. Interior Angles of the stations.…

Q: 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…

A:

Q: Station Chainage (m) BS IS FS Rise (+) Fall (-) RL (m) Remark 1 3.250 249.260 BM 2 20 1.755 0.750 СР…

A: Raise and fall. Method  Arithmetic check  Sum of BS - sum of FS = Sum of rise - sum of fall = last…

Q: Station Rainfall (in) 2.8 3.5 A B. 4.2 D 5.1 6.0 4.5

A:

Q: 1.283 BM-1 1.284 (а) 1.285 0.675 0.674 BM-2 0.677 (b) 0.674 0.677 0.678 Sum Mean INSTRUMENT SET UP…

A:

Q: ONS B.S. H.I. F.S. LF.S. ELEVATIONS -a 1.24 234.51m 1.11 3.40 4.6 2.8 3.6 2.4 3.21 2.46 1.2 3.2 2.8…

A: Given profile levelling HI = elevation +   BS elevation = HI - FS or IFS given elevation of BM a =…

Q: A STORm SEWER FLOWIR FULL HASs O.57. 4. A SLOPE OF D = /5 INCH FIND Opue m 2 0.013 CO PEN CHANVEL…

A: Slope = S = .5% = .5100 = .005Diameter = D = 15in.M = .013

Q: Follow directions and box your answers.  ANSWER: (a) 1.0     (b) 5/8, - 3/8     (c) 15/8…

A:

Q: STA BS HI FS IFS ELEV BMa 1.265 225.70 0+00 3.5 +15 2.8 TP1 3.478 0.95 0+20 2.1 +25 3.2 3.6 +30 +35…

A:

Q: mune void ratio ven peresity 0-347

A: Given data : Porosity (n)=0.347

Q: 200N/n A f O B 1.5m 1.5 1.5 m

A:

Q: Station BS HI FS Elevation BM1 1.56 564.3 m 1+00 0.70 1+100 2.90 1+200 3.50 BM2 0.70 1+300 4.90 ТР1…

A: Given:-

Q: 1.24 1.11 BM-a 222 ТР-1 1+00 3.40 4.6 1+25 1+50 2.8 3.6 1+75 2.4 ТР-2 3.21 2.46 2+00 1.2 2+25 3.2…

A: Profile levelling Given elevation of BM a station =  222 HI = elevation + BS elevation = HI - FS or…

Q: STA BS HI FS IFS ELEV BMa 0.95 125.50m 0 + 00 + 10 3.0 2.3 TP1 3.13 0.64 0 + 22 2.7

A: The elevation is generally the vertical distance from a datum that is the sea level, to a point on…

Q: 4. GNEH: watER 250 2m 44m SUめMERCED RECTAN GULAR SURPAOE OF WIDTH W = : (m

A:

Q: Complete the Differential Leveling Notes by providing data to the unknown quantities. STA BS HI FS…

A:

Q: POINT B.S I.S F.S HI DESIGN CUT+ FULL- ELEVATION 31.994 0.023 RL A 1.289 33.283 1.312 C 1.267 D…

A: There is A table which is Related to surveying.  It is Done by "Height of Instrument Method" Now…

Q: 4

A:

Q: Determine bar force member EF

A:

Q: kN 20 kN 18 kN 4 kN/m A B C. 2m 6m 2m 2m 4m 2m E one: -0.60 0.40 1.00 0.50

A:

Q: STA BS HI FS IFS ELEV BM1 3.92 328.7 0+00 3.3 1+00 3.8 2+00 2.8 3+00 2.5 TP1 4.37 5.03 4+40 3.8 5+00…

A:

Q: Station B.S I.S F.S 1.95 Remarks В.М. 1 1 2. 0.65 3 2.60 1.45 4 -1.90 -1.60 0.55 6. 0.50 В.М. 2

A: Following problem can be done by the Height and Fall method  And  Height of instrument method…

Q: Ax 10 KN AY 230° 20 KN 2.5m 20 KN 20 KN I H IFY -√250/200 / 200 /M² 2.5m 2.5m 2.5m 20 KN VE U 3m

A:

Q: stream, the water flows 0.71m dee ace in the apron.

A: We know equation :2q2g=y1y2(y1+y2)q=Qw=18023.5=7.659m3s·mso,2(7.659)29.81=0.71y2(.71+y2)y2=3.7647m

Q: STA BS HI FS IFS ELEV ВМ1 3.92 328.7 0+00 3.3 1+00 3.8 2+00 2.8 3+00 2.5 TP1 4.37 5.03 4+40 3.8 5+00…

A:

Q: unknown quantities. STA BS HI FS IFS ELEV BM-a ТР. 1 1.24 234.51m 1.11 3.40 1+00 1+25 1+50 1+75 4.6…

A: Datas are calculated by the following expressions: Height of instrument is calculated as: HI =…

Q: STA BS HI FS IFS ELEV BМ — а 1.24 234.51 m ТР- 1 1.11 3.40 1 + 00 + 25 + 50 + 75 ТР - 2 4.6 2.7 3.6…

A: Height of instrument method HI = elevation + BS elevation = HI - FS

Q: Situation 5. Determine Ix amd ly. 12" 6" 12" r=4" 6"

A:

Q: Complete the Differential Leveling Notes by providing data to the unknown quantities. STA BS HI FS…

A: It is asked to fill the levelling notes from the given date in the table. HI = elevation + BS  HI is…

Q: STA sa 1.24 BM-a 234.51m TP- 1 1.11 3.40 1+ 00 4.6 1+25 2.8 3.6 1+50 1+75 2.4 TP-2 3.21 2.46 2+00…

A:

Q: NO. LOT BEARING DISTANCE S 36.00 W S 54.00 E 1-2 20.00 meters 2-3 (гoad) 12.57 meters 3-4 N 46.00 E…

A: To draw the lot bearing on scale of 1:100.

Q: find:- احسب القيم المفقودة-1 2- apply usual check Sta B.S I.S ES Fall Rise H.I R.L remark A 2.600…

A: Missing data is required to find and then usual check is to be applied.

Q: Solve for this prob. Formula: H.I. = Elevation of given B.M. + B.S. Elevation of T.P. = H.I. – F.S.…

A:

Q: SEATWORK NO. 1 3. If gage A reads -20 kPa, determine the elevations of points L, N, Q and D. 2 El.…

A: Consider the given data Pressure of gauge AP=-20 kpa=-20×0.14503=-2.90 psiSpecific gravity of…

Q: STA BS HI FS IFS ELEV BM-a 1.24 234.51m ТР-1 1.11 3.40 1+00 4.6 1+25 2.8 1+50 3.6 1+75 2.4 ТР-2 3.21…

A:

Q: 3130 KN 220 KN 20 20° 2.0 m E Av = 2.0 m Ah = Rp = D 3.5 m 3.5 m AB = DE = EC= BC = BE= CD= AE =

A: Using equilibrium equation

Q: Situation 5. Determine Ix amd ly 12" 6" 12" r=D4" r=6" 49

A:

Q: Rain gauge X was out of operation for a month during which there was a storm. The rain fall amounts…

A:

Q: FS (m) Eleva tion (m) Sta Bs (m) HI (m) Remar ks BM-1 2.00 | 001 ·0 1000 1P -1 1.41 1002.93 0 .48…

A:

Q: Station Chainage (m) BS IS FS Rise (+) Fall (-) RL (m) Remark 1 3.250 249.260 BM 2 20 1.755 0.750 CP…

A:

Q: 2- Using Meyerhof Eq 12.18 for Qp and Eq 12.42 & 12.43 Q, find Qult 0.5mø CONCRETE 4,=320 Y = 18.5…

A: Meyerhof method gives the resultant load at the bearing level is vertical and  the horizontal…

Q: STA BS HI FS IFS ELEV ВМ-а 1.24 1.11 234.51m 3.40 ТР-1 1+00 1+25 1+50 1+75 ТР-2 2+00 2+25 2+50 ТР-3…

A: To find :- elevation at each station using height of instrument station.

Unit Conversion

Standard units are also called SI units. SI units are used for measuring a physical quantity satisfying certain conditions. There are two types of units: fundamental units and derived units. Fundamental units are physical quantities that are independent, that is, they do not depend on other quantities like mass, length, or time. There are plenty of quantities in our lives which need to be measured and compared. Measuring is the comparison of a particular physical quantity with its reference value. A unit is a measure of a fixed physical quantity. A physical quantity is the product of a unit and the numerical value we need to compare with.