30, 75.50 kN 35 kN 225 kN.m Fixed 15 kN.m 10 kN Fosm + 1m 0.75m

Q: n = 18 kN/m $i'= 40° c;' = 0 0,7m Qsi(net) B=0,6m . 0,5m L=1,2m Y2 = 16 kN/m³ $2'= 30°

A:

Q: K L 20 kN- 4 m F G H E 40 kN – 4 m B C D A 4 m 5 m - 4 m

A:

Q: 3. 5P (kN) P (kN/m) 2.7P (kNm) 50° 3m u. 2m 3m 3m

A:

Q: 80 kN 60 kN D 5 m C 5 m В E5 m - 5 m

A:

Q: Determine the T2, If y=18m. * A 10 m B 60 KN 40 KN 80 KN 30 m -50m- 50m -30m O 164.746 KN O 181.707…

A:

Q: L = 2.7 m, P = 49 kN, and w = 26 kN/m. B VA =| mm.

A:

Q: 30 KN B 4 m 5m C 2m 워 D

A:

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

A:

Q: Al C 2.5 m 20 KN/m 7.5 m B

A:

Q: 50 kN 60 kN/m 140 kN-m B. C| 3 m 1 m 1 m

A:

Q: D at 2m 20 kN/m 4m 3m look

A:

Q: 2.5 m B 20 kN 40 kN 40 kN - 3 m - 3 m-

A: in the truss their is not any bending moment in the truss their is only axial and transverse force.…

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

A:

Q: 1.8 m 1.8 m 1.38 kN 0.80 kN B 25° 15m.- 2.1 m

A: lets assume the reaction at A is VA and HA and at B is VB and HB now free body diagram is

Q: Draw the shear and moment curves and sketch the deflected shape.

A: The Shear Force and Bending Moment Diagram Calculation is Shown in Steps Below one by one.

Q: 3 kN C 5m 5 m B.

A:

Q: 20 kN 20 kN F -20 kN 3 m -40 kN E 3 m D 3 m -40 kN- C 3 m B -6 m- m-

A: To find the force in member EC ,DC and DA-

Q: 10 KN 15 KN 10 KN E 0.9 m 0.9m A G B IF 1.2m 1.2m 1.2m C.

A:

Q: 160.00 kN 2 m 10.00 kN 10 -2.67 m @ 3

A: Find the internal member force of member 6,7 and 9 using appropriate method

Q: 25 kN A B 45° 5m 4m F E 30 kN 5m →-2.5m→– 2.5m

A:

Q: 5m 2E1 (2) 60 kN/m ΖΕΙ | 1m El -12 kN 1m -12 kN

A: Slope Deflection Method   Fixed end moments MfAB=-wL212=-60×5212=-125 kN.m MfBA=wL212=60×5212=125…

Q: B) RA = KN, Rg = KN/m 45 KN/m 200 50 KN A 9.00 m Rg 3.50 m 150 KN-m 7.00 m

A: Given simply supported beam with over hang at the right end .

Q: C 8 kN/m 4 3 А В 3 m- 2 m-

A: HA = -1.8 KNVA = 9.6 KNT = 3 KNexplanation is given in below step

Q: resultant couple moment. 10 kN 4 3. 4 m- 1 m B 1 m 3 45 10 kN

A: Couple moment is the Force multiplied by perpendicular distance between them.

Q: D. E. C B. -4 m 4 m 15 kN 20 kN 00

A:

Q: 30 kN 40 kN 28 kN 2m 20 kN/m 1m C 4m 1.5 m

A:

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

A:

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

A:

Q: Problem #2: 10 + A (kN) 3+B (kN/m) |A |B C |D -E - 3 m --- 4 m – 4 C (m) – 3 m -| 2 m

A: Given A=1 B=5 C=5 To find RA, HA, RC, RE

Q: -60 kN 3 m 50 kN 50 kN 50 kN 50 kN 4 at 4 m = 16 m

A: Using Method of Section , a section is cut passing through HI , CH , BC and equilibrium of RHS is…

Q: o (kN/m) 0.92 0.77 150 300 ©Cengage Leaning 2014

A: Given data Pre-consolidation Pressure=100kN/m2 Pressure=450kN/m2

Q: A 3.6 m 40 kN/m 5m 88 KN B 56 kN-m 1.5m ● 1.5m 91 KN 2.4m 60 KN 1.4m. 7.3 m 74 kN/m D

A: Given- a continuous beam with different loading system.

Q: 10 kN 50 kN - m C -B 2m- 3 m

A:

Q: Determine the reactions at the supports for the structures shown

A: Given data:

Q: 20 KN 2m Jom RI R2 y at point B

A: Calculate deflection at point B for given beam.

Q: 30 kN 60 kN 6 kN/m d. d /3 . R 36 kN R= 27.9 kN R 80.1 kN -3 m -3 m--5 m- -4 m-

A: To determine: Shear force diagram and Bending moment diagram of the given loading diagram.

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

A:

Q: 20 kN 20 kN 15 KN 15 kN B 13 5 12 26 kN -3 m--3m-3 m- -3m- -3 m-

A:

Q: 8 kN-m 2 kN B C トー2m→ ー 2m ー 2m

A:

Q: 20 kN 40 kN/m -8 m 150 kN-m Prob. 7–55

A: Note: As you have nothing mentioned in this problem so, we are solving it for reaction at support A…

Q: 25 kN 20 kN 30 kN B D А +1.5 m→- 3 m -3 m

A: Given:- A=600mm2 To find:- Maximum normal stress in bar

Q: 10 kN 3. A 4kN/m 1 m 1 m

A:

Q: Determine the resultant af the system and its position prom point A. 10 KN to ku /m 10 kN-m

A:

Q: ass of plane: ension, T: 22,000 kg 400 N

A:

Q: 30 kN/m -6.5 m- B + -10 m nea 45 kN/m

A:

Q: ITI 20 KN 50 KN 30 KN 40 KN 30 KN 2m 2.5m 4m 2m 3.5m

A: solution  first find the resultant force = sum of upward forces - sum of down ward forces  R = 50+30…

Q: 12P kN 8P kN 4 m D -3 m -3 m -3 m -3 m

A: Third highest digit of your student nnumber ; 3  (assume)

Q: 0 kN 10 kN/m A B -7 m -7 m

A:

Q: 20 kN 30 kN 16 k N/ m 10 k N/ m 40 kN-m A В |C D E F -3 m - 3 m 2 m 8 m 2 m

A: Consider the free body diagram of the beam as shown below.

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.