30° 30° (y F A Bo20'7°20' B
Q: q = 150 kN/m 92 55° 1.5m 3m 2.7m
A: To find out the magnitude of q2.
Q: 4. From one side of a river 400m. wide, a level took a foresight to the other bank with a reading of…
A:
Q: SITUATION 1- Hissing Data Given: Traverse with omitted measurements. (a) Calculate the distance of…
A: LINE LENGTH (l) BEARING θ LATITUDE (L) DEPARTURE (D) AB 550 S 80.00° W 260° -95.51 -541.64…
Q: 60 N 0.75 m 1m 90 N
A:
Q: 0.20m Pp = 90 Pz = 12 %3D
A: Given:- fc=22Mpa fy=418MPa Pl=120kN Pd=90kN To find:- a) spacing of stirrups required by torsion b)…
Q: COURSE DISTANCE BEARING АВ 224.34m N 57°3' E S 89°32' E S 7°5' E BC 251.47m CD 301.79m DE 157.88m…
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Q: Missing data Length Bearing АВ 102.84 N 54d 47m E BC ? S 44d 46m E CD 77.25 S 54d 21m W
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Q: COURSE LENGTH (m) BEARING АВ 126.90 81°9' South of West BC 90.20 N 18°51' W CD 110.80 57°33' North…
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Q: SITUATION 1- Hissing Data Given: Traverse with omitted measurements. (a) Calculate the distance of…
A: A closed traverse is given in tabulated form and it has been asked to determine missing data. Please…
Q: Station Sight (+) Stadia Sight (-) Stadia Elev. ВМА 510,38 6,920 6,240 5,360 5,470 4,210 4,330 TP1…
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Q: √3=9 P₂=9 P3=9 27 = ? Pi=30 K^ D T √₁-3 D=450mm Ref Axis Z2-5 3 m K 545 за 23=4.5 7 12:2.5 1 2D₂=300…
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Q: Draw the longitudinal section ( profile) of the leveling process if vou know that the elevation of…
A: Assumptions:(1)- Since No Specific angle/Bearing is Given , so we are Considering The Offset as a…
Q: 20 kN y 30 -x- 45° 8 kN 4 kN Problem 2/80
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Q: At A , PA = Ka 9 : 0:219 x40 %3D 8.76 KN/m At B, Ka 9 + Ka Y Z 8.76+ 0.219 x 19 x 20 %3D 91.98 KN/m…
A: this is the active case in cohesion soilfirst of all you have to determine,active earth pressure at…
Q: An instrument was set up over TBM 1, and backsights onto point A before sighting onto B. The…
A: To Calculate :- The angle measured at point C (BCD)
Q: What is the azimuth from the north of line OC? 30 60° 55° C 45° B O a. 125 deg O b. 235 deg O c. S…
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Q: vertical angle of +14o 55' is read to a target 1.50m above point B. The measured inclined distance…
A:
Q: Find: Max. V =? Max. M= 7
A: x=83=2.67 from A ∑MA=0 15×8×4 + 20×2.67 - RB×8 = 0RB = 66.675 kN∑V=0RA+RB=15×8 + 20RA = 73.325 kN
Q: Convert N 25 deg 25 min W to an equivalent azimuth from North O 154 deg 35 min O 334 deg 35 min O 25…
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Q: A, B and Care stations on a straight level line of bearing 110° 16' 48". The distance AB is 314.12 m…
A: Bearing is defined as the angel used in surveying for the instrument or the object. Surveying is a…
Q: Given: g1 = - 2% g2 = + 3% %3D BVC Station = 16+50 %3D BVC Elevation = 112.00' L = 400.00' %3D What…
A: Given: The grade g1 is -2%. The grade g2 is +3%. The BVC station is 16+50. The BVC elevation is…
Q: COURSE DISTANCE BEARING N 57°3' E S 89°32' E S 7°5' E АВ 224.34m BC 251.47m CD 301.79m DA 498.96m N…
A:
Q: A vertical angle of +25°15' is read to a target 1.75m above point B. The measured inclined distance…
A:
Q: 0.0025
A: Area of the channel :A=120.25×2×0.25×tan60=0.108253m2Perimeter :P=0.25cos60=0.5m
Q: N87°30'E 589.06 m B 3 Unknown Bearing 654.06 m Unknown. Bearing 686.66 m .---- -- S 42°45'E 656.98…
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Q: Q3/ from the travers below find; A; the interior angles (F, G, H,J, K) B; bearing direction of ( FG,…
A: Given data Bearings To find Angles Directions
Q: 12KN/m of horizontal length d 5 4 9. a.
A: CALCULATE The Diagram for shear force , bending moment, axial force ###
Q: From the following closed traverse. Compute the distance of the missing side * LINES AZIMUTH…
A:
Q: 59. The back staff reading on a B.M. of R.L. 500.000 m is 2.685 m. If foresight reading on a point…
A: given- Reduced level of BM=500 mBack staff reading BS=2.685 mforesight readig FS=1.345 m
Q: LINES BEARING DIST AB N 32 27' E 11 BC S 8 51' W 83 12 CD
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Q: Find the level of Poin A (Azimuth. OA.Oc) From the diagram below A verteal angle e = 47° 15m 4.5 B…
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Q: 200N 30 R, "y at point C
A:
Q: B) RA =, KN, Rg = KN/m %3D 45 KN/m 250 KN A 6.00 m Rg 1.50 m 4.00 m
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Q: LINE DISTANCE BEARING AB 370m N 15° E BC 650 S 86° E CD 475 S 19° W DA 589 N 78° W
A:
Q: VA = 0.6 m/s ав = ?
A:
Q: 13.35 kN 40° 1.22 m 1.83 m A \D 2,44 m Figure P4.137
A:
Q: 3. A turning point has an elevation of 65.15m and the foresight taken on that point is 0.30 m. If…
A:
Q: 0.8 m 643m 15 AN C LOM D VN 120
A: Given Data:UDL=6kN/mConcentrated Load = 15kNSpan =3m
Q: COURSE DISTANCE BEARING N 57°3' E S 89°32' E S 7°5' E АВ 224.34m BC 251.47m CD 301.79m DE 157.88m…
A: A traverse is said to be closed traverse, when a series of connected lines forms a closed circuit.…
Q: Station BS HI FS Elevation BM, 6.84 624.09 4.14 S, TP S3 TP2 BM2 6.90 5.72 5.33 4.80 3.21 4.54 3.78
A: Elevation refers to the height above the ground or another surface, or to a particular place or…
Q: 0.11 m%/s = 00 Coolm) I Vaxq81 m/<h %3D
A: the expression is given , the units should be kept in mind
Q: A dumpy level with an interval focusing telescope was set up on the left bank of a river and the rod…
A: given data: K=95.50mC=0.30m
Q: D00 psi. 23" 25" A=2.35 in? 10"
A:
Q: 4. The following data were obtained by a stadia measurement; vertical angle is 30°23', and observed…
A: Given :- Vertical angle = 30°23' Stadia interval factor at the measurement used = 95.50 m Observed…
Q: Converting azimuths to bearing. a. Azimuth line AB = 230° 30' b. Azimuth line BC = 112° 46' c.…
A:
Q: Compute for the coordinates of station O. Note that: A has a northing of 1620925.334 and easting of…
A: The coordinates of station O is calculated by computing the latitude and longitude from the given…
Q: Reading staff 3.100 with distance120m the Line of sight inclined (+3). * :Corrected reading is…
A:
Q: An instrument with K = 92.31 was used for determining the elevation point at D with a stadia…
A: Calculate the vertical elevation between the line of sight of the instrument and the stadia, V =…
Q: A W - 730 В 87° 54 45 77° 30 45 110° 37 45 83° 56 45
A:
Q: A; direction of azimuth (BC) (CD) (DA) (BA) B; compute clouser eror C; compute the correction angles…
A: We need to find the closure error, corrected angles and azimuth direction. For a closed traverse of…
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- Determine the axial force in members of the truss system in the attached photo: find: a) axial force on member marked with a in kN b) axial force on member marked with b in kNTHEORY OF STRUCTURES Use the substitute member to make a complete analysis of the complex truss shown.Use the substitute member to make a complete analysis of the complex truss shown. d 3m 25kN 4 m e1.5 m a. 2 at 4 m = 8 m %3D
- The truss shown in the figure is applied with force P= 5kN %3D 1.50 m 1.50 m 150 m 2 m 2 m What is the force in member DE in kN? A.Zero-Force Members are structural members that support No loading but aid in the stability of the truss. Tips on getting the zero-force members: 1. there is no external load or reaction at that joint 2. Insufficient structural members to satisfy or "balance" the system: Summation of forces along the horizontal and vertical 3. two of the members are collinear As shown in the figure, structural members of zero-force members can be noticed through observation. H G F A D E At a glance, it can be inferred that the member FG and FE are zero force members, because: First, there is no external load applied on the joint Secondly, taking the Joint F, if Summation of Forces along horizontal = 0 is executed, only the member FG has the horizontal component, or the "ability" to "balance" the system, and if Summation of Forces = 0 is executed, the member FE has the vertical component, or the "ability" to "balance" the system only. along vertical Aside from members FG and FE, through observation,…The truss ABC is constructed of three titanium (E Ti =17.4×10 3 ksi) members. A horizontal force of 1.8 kip and an unknown force P are applied to the truss at point B as shown. Member AB has a cross-sectional area of 0.1in 2 , member BC has a cross-sectional area of 0.0025ft 2 , and member AC has a cross-sectional area of 0.2 in2. b) Determine the magnitude of the force P that is required to displace the roller at point C to the right by .03 inches.
- The force in BC of the truss shown in the figure below isDetermine the force on the EF and EG elements on the truss shown in the figure and consider P = 35kN. Final answer: EF force = 69.5kN T Force at EG = 250kN CZero-Force Members are structural members that support No loading but aid in the stability of the truss. Tips on getting the zero-force members: 1. there is no external load or reaction at that joint 2. Insufficient structural members to satisfy or "balance" the system: Summation of forces along the horizontal and vertical 3. two of the members are collinear As shown in the figure, structural members of zero-force members can be noticed through observation. Load P G Н, F A B D E At a glance, it can be inferred that the member FG and FE are zero force members, because: First, there is no external load applied on the joint Secondly, taking the Joint F, if Summation of Forces along horizontal = 0 is executed, only the member FG has the horizontal component, or the "ability" to "balance" the system, and if Summation of Forces = 0 is executed, the member FE has the vertical component, or the "ability" to "balance" the system only. along vertical Aside from members FG and FE, through…
- Zero-Force Members are structural members that support No loading but aid in the stability of the truss. Tips on getting the zero-force members: 1. there is no external load or reaction at that joint 2. Insufficient structural members to satisfy or "balance" the system: Summation of forces along the horizontal and vertical 3. two of the members are collinear. As shown in the figure, structural members of zero-force members can be noticed through observation. At a glance, it can be inferred that the member FG and FE are zero force members, because: First, there is no external load applied on the joint Secondly, taking the Joint F, if Summation of Forces along horizontal = 0 is executed, only the member FG has the horizontal component, or the "ability" to "balance" the system, and if Summation of Forces along vertical = 0 is executed, the member FE has the vertical component, or the "ability" to "balance" the system only. Aside from members FG and FE, through observation, determine the…1. For the roof truss shown in Figure 3, determine the force present in ele- ment GH and in each element located to the right of GH. Also, establish whether the elements are in tension or compression. (I suggest you start from node M applying the method of nodes until you get to element GH) -2.4 m--2.4 m-1.2 m-2.4 m--2.4 m--1.2 m- 1.5 kN 1.5 kN 1.5 kN 1.5 kN D VI 1.5 kN 1 kN В 2.5 m E Ук 1 kN 1m H F Figura 3. Cercha 3. 2.Determine the force in the elements ED, EH, and GH of the reinforcement (see Figure 4), and establish whether the elements are in tension or compres- sion. (Use the method of sections). Figura 4. Cercha 4. 50 kN 40 kN -2 m -2 m- -2 m- 1.5 m -30 kN 1.5 m B 40 kN 1.5 m1. For the roof truss shown in Figure 3, determine the force present in ele- ment GH and in each element located to the right of GH. Also, establish whether the elements are in tension or compression. (I suggest you start from node M applying the method of nodes until you get to element GH) -2.4 m--2.4 m-1.2 m-2.4 m--2.4 m--1.2 m- 1.5 kN 1.5 kN 1.5 kN 1.5 kN D VI 1.5 kN 1 kN В 2.5 m E K 1 kN 1m H M F