Situation 01. In a survey within the catchment area of a lake, Thiessen polygons were established bysetting up a transit at Point Z, which is the inner portion of a four-sided lake MNOP. The readings areas follow:BEARINGN 40°30' WN 38° ES 70° E| S 60° 15' W | 695.10kmRAINFALL, ftDISTANCE420.35kmLINEZM0.211ZN530.15km0.26zo480.75km0.239ZPIf the rainfall from the area of bounded by ZMP is the Arithmetic Mean of the other rainfalls, computethe average precipitation (in cm) by Thiessen Polygon Method.Situation 02. Given the following data gathered on December 4-23, 2011, compute for the requiredinflow (in cu.ft/hr) in order to have an increase in storage of 14cm.Outflow8.23 cmsSeepage Loss68 mmPrecipitationEvaporation12 cm2.45 inArea21.21 sq.mi.Situation 03. Given the following rainfall data, draw the hyetograph, and determine at what time isthe highest intensity of the rainfall.TimeRainfall, mm7:4027:40-7:48437:48-7:56677:56-8:04878:04-8:12448:12-8:20218:20-8:28138:28-8:366.

Answer (1) Let us first draw the figure of a four sided lake MNOP using the given data We are given…

Situation 01. In a survey within the catchment area of a lake, Thiessen polygons were established by setting up a transit at Point Z, which is the inner portion of a four-sided lake MNOP. The readings are as follow: LINE ZM ZN ZO Z.P 0.211 Outflow Seepage Loss Precipitation Evaporation Area RAINFALL, ft 8.23 cms 68 mm 0.26 If the rainfall from the area of bounded by ZMP is the Arithmetic Mean of the other rainfalls, compute the average precipitation (in cm) by Thiessen Polygon Method. 12 cm 2.45 in 21.21 sq.mi. 0.239 Situation 02. Given the following data gathered on December 4-23, 2011, compute for the required. inflow (in cu.ft/hr) in order to have an increase in storage of 14cm. BEARING N 40°30' W N 38° E $70° E S 60° 15' W DISTANCE 420.35km 530.15km 480.75km 695.10km Situation 03. Given the following rainfall data, draw the hyetograph, and determine at what time is the highest intensity of the rainfall.

Situation 01. In a survey within the catchment area of a lake, Thiessen polygons were established by setting up a transit at Point Z, which is the inner portion of a four-sided lake MNOP. The readings are as follow: LINE ZM ZN ZO Z.P 0.211 Outflow Seepage Loss Precipitation Evaporation Area RAINFALL, ft 8.23 cms 68 mm 0.26 If the rainfall from the area of bounded by ZMP is the Arithmetic Mean of the other rainfalls, compute the average precipitation (in cm) by Thiessen Polygon Method. 12 cm 2.45 in 21.21 sq.mi. 0.239 Situation 02. Given the following data gathered on December 4-23, 2011, compute for the required. inflow (in cu.ft/hr) in order to have an increase in storage of 14cm. BEARING N 40°30' W N 38° E $70° E S 60° 15' W DISTANCE 420.35km 530.15km 480.75km 695.10km Situation 03. Given the following rainfall data, draw the hyetograph, and determine at what time is the highest intensity of the rainfall.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Line MN, 250 m long, was paced five Himer by the surveyor and We recorded as follows." TRIAL NO. DISTANCE 121 2 122 3 120 4 123 121 line OP was paced four Hones dgain with the following resuilts'. Ano ther DISTANCE TRIAL NO. 1 321 S 13 3 310 4 312
C. Table: Determination of Stadia Interval Factor (K) D. TRIAL 1 2 3 4 5 6 7 8 DISTANCE STADIA HAIR READINGS UPPER LOWER 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 Where: SKETCH: 1.319 m 1.335 1.358 1.367 1.415 1.435 1.460 1.445 1.169 m 1.135 1.108 1.065 1.035 1.035 1.010 0.935 S = Stadia Intercept K = Stadia Interval Factor STADIA STADIA INTERCEPT INTERVAL FACTOR (K) COMPUTATIONS: 1. Computing Stadia Intercept (S) and Stadia Interval Factor (K). a.) Stadia Intercept = Upper Rod Reading - Lower Rod Reading b.) Stadia Interval Factor (K) = Distance / Stadia Intercept (S) AVERAGE K = (Show computations on the computed stadia intercept and stadia interval factor.) (Draw the resulting figure based on the data given data. Refer to the illustration above.)
Q.3 A level circuit is shown in figure. The survey, needed for local engineering project, commenced at BM 35, the elevation of new benchmarks 36, 37, 38 and 39 were determined, and then the level survey was looped back to BM 35, the point of commencement. Fined the adjusted new BMs. ВМ 35 Elevation(m) 190.000(fixed) 35 36 37 38 39 35 188.242 186.299 188.251 186.397 190.033 IM 36 09 km DM 35 0.8 km OBM 37 08 km O6 km DM 39 BM 30 0.7 km
Q.3 A level circuit is shown in figure. The survey, needed for local engineering project, commenced at BM 35, the elevation of new benchmarks 36, 37, 38 and 39 were determined, and then the level survey was looped back to BM 35, the point of commencement. Fined the adjusted new BMs. ВМ 35 Elevation(m) 185.000(fixed) 35 36 37 38 39 35 184.242 187.299 186.251 188.397 184.933 BM 36 0.9 km. вм 35 0.8 km OBM 37 0.8 km 0.6 km BM 39 BM 38 0.7 km
22.) Line MN, 251m long, was paced five times by the surveyor and recorded as follows: Trial No. Distance 1 121 2 122 3 120 4 124 5 122 Another line OP was paced four times again with the following results: Trial No. Distance 1 122 2 314 3 311 4 322 Determine the actual length of line OP.
1. Complete the differential level notes shown below and perform the customary arithmetic check Level Notes from BM1 to BM4: Station BS HI FS ELEVATION BM1 1.256 127.133 m ТР1 1.116 1.886 ТР2 1.228 1.527 BM2 1.189 2.246 BM3 1.070 2.017 ТРЗ 1.831 2.656 BM4 2.765
ineed solutions ANSWER Vs = 120453.0446 N
1) Complete the missing data and determine the difference in elevation of BM1 and BM2. Show complete solution. STA BS HI FS ELEV (m) BM1 1.097 504.233 TP1 1.661 1.998 ТР2 1.012 2.015 ТРЗ 1.905 2.396 BM2 2.761
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