Q.2 50 kN 50KN - 20m 20 m. a 41

Q: 400 N R 2.12 m -2.12 m Fig.D

A: Moment is 248.334 N.m acts in anticlockwise direction.

Q: p=50kN , W=90KN/m , h=7m , b=5 m

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Q: D 32 ft d. 64k

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Q: 4.) 30 kN/m C 3m 3m D 7m 9m

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Q: Determine the yof the shaded area. L= 2.9m L -y = x³

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Q: Example Draw S.F.D. and B.M.D. SOKN 50KN вщиния •30KN.M -3M 3m d پیسے 12m • W = 10 KN/m )

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Q: at OF 3m Fro R1

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Q: 2.8m- 3m 34 kN 18 kN/m 15 kN/m B D 21 AE 21 4m 3m 5m 3.6m

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Q: 3 kN C 5m 5 m B.

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Q: between B and D having stadia intercept reading of 2.42 m at D at a vertical angle of +6°30'. a)…

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Q: B) 35 kN A -3m B 48kN/m 9 m со 55 kN-m

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Q: 100KN AM 4m 6M

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Q: L- 60 kN %3D D- 15 kN/m -5m

A: Solution:- We know that, the deflection in cantilever due to point load at free end is; ∆=P×L33×EI…

Q: A)le 3 -m →le 3 30 kN 25 K 120 KN. Find the Resaltant and the angle Find the 9. moment a bout Point…

A: Since you have asked multiple question, we will solve the first question (Q1) for you. If you want…

Q: L K 60° 4 m 10 kN В D E F -3 m - 3 m+ 3 m - 3 m -- 3 m-|- 3 m -| 20 kN 30 kN 40KN Figure 2

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Q: P KN w kN/m B- 10m 5m 5m

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Q: LINE AB BC CD DE EF FORWARD BEARING N 44°38' E S 42°30' W N 22°35' W BACK BEARING N 42°30' W S…

A: Bearing of AB= N44°38' E         Azimuth of AB= 44°38' Bearing of BC= S 42°30'E      Azimuth of BC=…

Q: B.S-1.85 m. F.S-3.15m Elev. = 30 m %3D 8.

A: Given Elevation of ground=30m BS=1.85m FS=3.15m To find height of B from ground

Q: 40KN ↓ A 27 kN/m ↓ ↓ ↓ ↓ ↓ ↓ fim B C tih с + qm 3m 3m 60 KN-m D

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Q: 5.43 KN/m 12-8 KN 37.23 KN 30. 266 KN 37.23 KN 12-8°KN 个 Gm

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Q: ucture below. 12KN/m 25KN A C 4m- 5m 2m+2m

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Q: 10 kN 8 KNIM mm 3m

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Q: 4. Locate the centroid (7, 5) of the shaded area. 1 in, 3 in. - 3 in.- -3 in.- Prob. 9-64

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Q: Determine the rof the x of shaded area. L= 3.9m y -y =x L

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Q: 45 40 KN 4m

A: to find moment about point O

Q: Pi= 1200 KN P = 1200KN %3D +3m Find the location of the point of inflection from the left end of the…

A: A footing is a part of the foundation that needs support the foundation so that the structure does…

Q: 40 kN- 60 kN D- B -6 m

A: The free-body diagram of the top portion of the frame where hinges are placed: As all the members…

Q: 24RN /m 12AN/m 3m Sm

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Q: Q5// Locate the centriod (xe.ye) for the shaded area shown in figure 5. 1cm 1cm 1pm 2cm 2cm (Figure…

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Q: = 20kN/m -8 kN/m

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Q: 2 kulm 8KN 4m

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Q: -b- d, b = 250 mm d, = 200 mm d2 = 100 mm d,

A: Given,Net loss=10%Force applied at X=750 kNb=250mmd1=200mmd2=100mm

Q: C -3m- kN/m

A: Pin support - it has 2 reactions One is horizontal and vertical reactions Roller support - it have…

Q: An OPEN HaNatt is T BE DESIGNER TO lit/ce Q = 2000 compuTE TE DEPTH OF MOST 0.0008 EFFICIENTSECTION…

A: Discharge, manning’s constant, slope is given and it has been asked to determine the depth of most…

Q: Find the velocity of the block when v. -4 [m/s]. %3D Va BLOCK

A: When C point moves x distance upward then Block will move 2x upward because there will be…

Q: 4m 35 kN B N 48° M С 4m oooo +=+= 4m 4m 15 kN 5m 5m 5m D H -5m 15 KN 24° E G 15 kN 5m

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Q: The reduced level of a floor is 99.995 m the staff reading on the floor is 1.4 m. If the inverted…

A: Given Data : Reduce level (R.L)=99.995 mStaff reading =1.4 mInverted staff reading =1.795 m

Q: E (600)тm EA EA Р- 180 kN EA Oy A В C 0-ɛ diagram of the bars 300 тm 300 тm 300 тm

A: The displacement of the system is equal to the displacement of all the individual components in the…

Q: 8 kN B 5m D E

A: The virtual work method is also known as the method of virtual force and the unit-load method and…

Q: B - 2 m

A: At CFCA =FCA-1i+2j+2sin 60 k8FCB =-0.354FCAi+ 0.707FCABj_0.612FCBKFCD = FCDj∑Fx =0       -0.254…

Q: Solve Problem using Cramer's Rule 2х + Зу + 2z%3D -3 y + z = 0 -x + 2y - 3z = -1

A: Cramer's rule can be solve using Matrix form. Detail solution is given below

Q: oblem #3 Am -2 m B BkN 2 m- C KN -2m- Y KN

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Q: 4m 5m Soit- 4m

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Q: 24KN/m

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Q: 80 kN 120 kN F 3m G TH 200 kN 4 Panels @ 4m 16m %3D D.

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Q: A 10 m y D B 60 KN 40 KN 80 KN 30 mr 50m 50m -30m

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Q: 11m 16 KN 20 KN

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Q: 3 m

A: Draw free body diagram Consider the summation of forces in the vertical direction…

Q: Find the RL of the top of the flag post located over a building. The angle of elevation from a point…

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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…

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.