FBAB= S 81° 50' E FBBC= S 53° 05' E BBAF N 82° 15' w BBBC= N 52° 45' w FBCD= S 78° 40' w BBCD= N 78° 40' E FBDA= N 9° 25' W BBDA= S 9° 30' E

Q: d) 30 kN/m 60 30 kN/m kN AB C DE 2 m 5 m 5 m 2 m

A:

Q: W1 b. C 1.75 m d 5.00 m W4 kN/m W2 5 kN/m P 15 kN 1.25 m b 2.00 m Given: a

A:

Q: 40 kN 15 kN/m 5 kN 5 kNm A В 1m 1.5m 1.5m 1m 2m 1m 1.2 m

A:

Q: 3 kN A В 4 6 kN E D F 4 m 4 m 4 m 4 kN 4 kN

A: To find force in member BC FC and FE

Q: H.W 20 kN 12 kN/m Sm 10 AN Sm m

A: Moment distribution method: The moment distribution method is presented by Professor Hardy Cross.…

Q: 2.5m 1.5 m WST E C B A 2.0m H 1.0 m D G 3.5 m 15m 2.0 m 50 kN WST

A: Stress: The resisting force acting upon the object or a system per given unit area of the body is…

Q: 6 kN/m 6 kN/m A B 1.5 m 3 m + 1.5 m

A: Determine support reaction

Q: Line Forward Bearing Backward Bearing N 38° 10' E N 540 13' E S 490 17' E S 17° 20' W S 21° 54' W N…

A: The bearing of AF can be picked from the table.

Q: y at a distance 3m From 20 KN lom' OF R.1 21 R2

A: To find deflection at 3m from R1

Q: 20⁰ 30, A 6kN 2kN 0.4m 0.6m

A: Given:- To find:- Force in AB

Q: 9 m 60 KN 4.5 m A 7.5 m 4.5 15E

A: To find the force in member CD -

Q: Course Bearing Length (m) АВ S 60° E 50 ВС N 30° E 100 CD 200 DA 151

A: Bearing of line AB and BC and Length of 4 lines are given these four lines join to make a…

Q: Assume Cd = 0.60 * 1 ft 60° 0.5 ft weir plate

A:

Q: 40 kN 5 kN 15 kN/m 5 kNm A В 1m 1.5m 1.5m 1m 2m 1m 1.2 m +

A:

Q: Determine: RA KN, RB KN, V Diagram (2 decimal places), Vmax (KN 2 dec), Area V (2 dec), M Diagram (2…

A:

Q: 2 kN 8 kNm 12 kN.m B 2

A:

Q: 2 kN/m C 1.5 kN 5 kN/m 6 m 1m TEE 4 m 0.8 m

A: (a) ∑Fy=0; VA=0∑Fx=0; HA=2×4=8 KN∑MA=0; MA=2×4×2=16 KN-m Member Axial force diagram Shear force…

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

A:

Q: a KN/m b KN 2.4 m A 21 31 |41 31 51 b KN a KN/m 3.6 m F| 21 31 41 31 51 G H em em d m

A:

Q: Find the discharge of rectangular orifice 3.0 m wide and 2 m deep fitted to a water tank. The water…

A: To find the discharge (Q) -

Q: 6 kN 5 kN 3KN30° 13 0.2 m 4 B A 2 m 4 m 2 m

A: Let,RA=vertical reaction at support ARB=vertical reaction at support Bsince support B is…

Q: kN 20 kN 18 kN 4 kN/m A B C 2m 6m 2m 2m 4m one: 0.40 .0.60 -0.33 . 0.67

A:

Q: B 0.90 kN hinge 1.75 kN R = 1.50 m 1 m 480 0.5 m

A:

Q: A simple beam 16 ft long carries the loading shown inFigure 1.21a. Find the reactions.

A: To determine the force in members we have to calculate the support reactions by taking moments at…

Q: 2.8 m. 43¹ 5.6 KN 3.2 m 6.0 KN R BC.CO 3.2 m

A:

Q: 2 m 18 kN 2 m B 2 m A -36m-| Fig. P6.79

A:

Q: Pipe L (m) Diameter (mm) f 1052 600 0.025 890 600 0.025 3 890 650 0.025 4 1055 650 0.025 2 25 m-

A:

Q: 3 kN C 2.5 m 1 kN/m ||||||||| A 6 m B 1 kN 2 m D

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: 6. A block of wood (sg = 0.78) floats in water with 600 cm³ above the water surface. What is the…

A:

Q: 10 kPa B 0.4 m |+0.5 m++0.5 m++0.5 m+0.5 m- Soil A 1 m Exit Datum 0.75 m 0.5 m FIGURE P6.2

A:

Q: 1.0 m 3.0 m C 3.0 m B 1.50 m

A:

Q: 30 KN/m 50 KN/m D A 1.5 m C 1.5 m 2 m Rol RA

A: Free Body Diagram of the given problem: Calculate reactions at A and D:

Q: 3P (kN/m) C B 3m 60° 5m P (kN/m)

A:

Q: Wi A 1. a

A: Note: As per our policy we are not allowed to give handwritten solutions. So, we apologies for this…

Q: 1.5 m ".. 0-3 m Figure (4) 1.2 m 0.9m 03m 2.2m

A:

Q: 5 kN rn 15 kN.m 10 KN.m 20 kN m

A:

Q: A 5.0 m 5 KN/m 5.0 m B 5KN 2.0 m

A:

Q: posite open channel of varying roughness ng roughness coefficient (ne) The vertical and slantıng…

A: According to given data

Q: PLL 3. 25 kN 4m 2m WLL = 14 kN/m WDL = 18 kN/m A В

A:

Q: 3 kN 2.5 kN 1.5 kN 30° 5 13 A 2 m 4 m 2 m B.

A: to find resultant and moment about B

Q: 200 kN 150 kN D 6m F A C Sm 5m

A: A loaded truss is given and it has been asked to determine roller support reactions which has been…

Q: Water flows in a rectangular concrete channel that is 12m wide at a depth of 2.5m. The channel slope…

A: To find Velocity in a rectangular channel.

Q: A) RA=. KN, R = KN 50 KN 20 KN/m 75 KN 15 KN/m 1m 1m A 3.00 m 1m 1m

A: Given  Simply supported beam with  Point load  Udl  UVL

Q: 20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of…

A: B = 6m  Side slope 2:1 = n:1 ---> n = 2  V = 0.6 m/s

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: 3 m 00 D A 2 m 2 m 2 m 5 kN 5 kN 5 kN

A: to find forces in member

Q: An earth canal in good condition having a bottom width of 4 meters an slide slopes of 2 horizontal…

A:

Q: er surface. If the block is placed the surface of oil. 24. Det. Th 126 kN 236 kN 338 kN 414 kN

A:

Q: a height of 10 m. The H nds is 1m. The coeff to two decimal places)

A:

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.