Q. 5le 3 30 KN 120

Q: q = 150 kN/m 92 55° 1.5m 3m 2.7m

A: To find out the magnitude of q2.

Q: SOVE FOR EQUILIBRIUM: Problem #3 lo00 300 # 3' 10' 4' Rz

A: This question is related to the Strength of Material or Mechanics. In this question we calculate…

Q: 文 1 m 7-190KN/m? Es KN/m2

A: Given, b = 1 m  L = 2 m  z = 1 m  μ = 0.3

Q: Reduce the set of profile notes in Table 2-7, and perform the arithmetic check.

A:

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…

A:

Q: 60 30 10 kN 60 10 kN 120 -Im 2m

A:

Q: Cable W=8ON smooth R 140

A:

Q: * ски/т П вт B 4.5m 8KN m

A:

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: Closed Traverse 65 13 95 19' B D 216 19 67 06' E 96 03 A Determine the following: 1. back azimuth of…

A:

Q: surfaces of CO 100 KN P F 30°. 30° 15 R

A:

Q: Cathedral Hill is on a line between Constable Hill and Aurora Hill. Cathedral Hill is 6.235 km from…

A: The elevation is generally the height above the ground or position of height. It is determined by…

Q: (2) jlubäi Change points in levelling are The instrument stations that are changed from one position…

A: Change point is also called as Turning point.

Q: YD D 7 m В 2 m 2 kN

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: Cable W=8ON smooth 3. 140

A:

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…

A:

Q: Determine the elevation of station BM1 using c = 0.65 and k = 100. Backsight Foresight DE ELEV…

A: When the height of a point and also its vertical distance is generally has above or below the…

Q: Given: PVI station=(8+70) and elevation=D140m g1=+3% , g2=-2% high point Station%=(8+92) and…

A:

Q: 120 KN 90 KN 4.00 m 4.00 m 4.00 m B 60 KN 6.00 m

A:

Q: 12KN/m of horizontal length d 5 4 9. a.

A: CALCULATE The Diagram for shear force , bending moment,  axial force ###

Q: Determine the T3, If y=18m. * 10 m 60 KN 40 KN 80 KN 50m 30 m 50m -30m O 158.678 KN O 179.013 KN O…

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: Upper reading = 1.64 m Middle reading = 1.57 m Lower reading = 1.51 m Vertical Middle Angle =…

A: From the given data :   Stadia interval factor K = 100  Stadia constant C = 0 Staff intercept (S) :…

Q: Akip/ft P=8 kip B 4' 60 61 ,2' 31

A: (2) The equivalent concentrated force of the UVL is calculated as, Equivalent concentrated…

Q: B) RA =, KN, Rg = KN/m %3D 45 KN/m 250 KN A 6.00 m Rg 1.50 m 4.00 m

A:

Q: Course Bearing Distance AB S 18°5' W 126 M BC S 08°51' E 90.2 M CD N 23° W 108 M DA

A:

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

A:

Q: 21(KN) 14 7(KN/m) 35ckN.m) 2, 4 2

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

Q: From the given traverse result: Lines Bearing Distance…

A: Closed traverse ABCDEA is given where bearing of line DE and EA is missing. It has been asking to…

Q: Least count of leveling staff is: O a. 0.5 b. 0.05 Ос. 0.0005 O d. 0.005

A:

Q: Hill B (elev. = 2,338.253 m) is in between hill A (elev. = 2,336.264 m) and hill C (elev. =…

A: Solution: Given Data: Elevation of hill A=2336.264 mElevation of hill B=2338.253 mElevation of hill…

Q: d=0:02 m 7m Lelom Uzo-2% find 7

A:

Q: 6 KNIM 45 KN/m コ行个1个才 3m 3.6m

A:

Q: direction of the AB, CD and EF. 65 14 12 m 20 m B. 25 105 23 A

A: an open traverse ABCDEF is given as shown in figure above, azimuth of DE is given and it has been…

Q: from point A in bet. B and C, the angles of elevation of B and C are 18deg30' and 8deg15'…

A:

Q: BC and BD.( 100 KN 150 KN 4.00 meters 5.00 m 4.00 meters 2.00 meters 2.00 meters

A: To find support reaction and force in member BC,BD

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: Hill B (elev. = 2,338.253 m) is in between hill A (elev. = 2,336.264 m) and hill C (elev. =…

A:

Q: Evaluate lim (3 +7¹) H18

A: Given  Limx-> ♾️ (3x+7x)1/x

Q: 12 kN/m d= 86 mm %D A B, 3 m 1.5 m –

A:

Q: tangent of each other at in tersect deflection angle of 60° The chainage The chainage of point Two…

A: Given that, Radius of curve R = 200m ∆ = 60°

Q: -kN, b- 39.2 kN, c = 2.7 m a" kN D H "c" m b" kN C K "c" m b" kN B J "c" m 4 m 5 m A = 40 x 10º mm?…

A:

Q: zimuth of 193°03'5 183.262 m with an

A:

Q: %3D v. 95 001 X :: 001X above thu liguid surface 11) Vs Volume above liquid= V- SL.0 V = 6.045 W=…

A:

Q: Q1 B) Find Q in 1/s and -V½= 2.5 m/s -D;= 300 mm find V3 in ft/min Q=? V, = 3 m/s D1 = 500 mm Vz = ?…

A: Discharge is the main quantity in any field used to transfer fluids from one place to another. It…

Q: An engineer's level with stadia constant of 0.30-m was set up on the line between A and B, the…

A: Given: The stadia constant is 0.30 m. The stadia interval factor is 100. The crosshair readings…

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…

Q: Determine the vesultant focce (R) and indieste its location from point (B)

A: Consider the free body diagram of the given loaded member 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.