The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam. M = N•m 4 15 mm ✓ 56 mm 28 mm 30 mm M

Q: C 105 22 B 118 37 Sta A C . D E 108°28′ E B 120 28 N30°00'W MERIDIAN 87°05' DEFLECTION ANGLE A ANGLE…

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Q: Та Tb 15° 30° Td Tc 60° 75° W

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Q: Complete the table by referring to the data below: - compute for the lat and dep - compute LEC and…

A: Solution: Solving by Transit rule: Corrected Lattitude: Clat=-Lat. ∑Lr∑L Corrected Departure:…

Q: 4 ) Calculate Lattitude and departure for length 77 m and Azimuth 23 Degree

A: given data  length of line = 77 meter azimuth of line = 23 degree we have to find  departure = ?…

Q: Data of a closed traverse are as follows: Compute the distance of line (5-1) a. 206.92 m b. 374.97 m…

A: Let distance 3-4 be x and distance 5-1 be y

Q: Two angles AOB and BOC and a single angle AOC were measured at the same point O. Adju angle AOC.…

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Q: 6-12. What is the angular misclosure of the closed traverse below? Ans.: A (a.) (b.) (C.) (d.) B85°…

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Q: Compute the DPD of each lines (AB, BC, CD, DE, EF, FA)

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Q: Given PI at 8+ 272.311, A = 24°24'20", and R = 500 m, compute E (external), M (midordinate), and the…

A: External Distance= E= RCos∆2-R= 500cos 12°12'10''-500= 11.558 m Midordinate = R1-cos∆2= 5001-cos…

Q: Course Adj. Lat. Adj. Dep. AB 490.71 m. 47.27 BC 587.12 608.89 CD -327.41 786.78 DE -1002.76 218.32…

A: Course  Adj lat Adj Dep DMD 2A AB 490.71 47.27 47.27 23195.86 BC 587.12 608.89 703.43…

Q: 6) B 11.7 с 7 Ө A

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Q: Given: Instrument @ "B", BS "C", FS "A" Northing B = 5000.000, Easting B = 10000.00, Elevation B =…

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Q: 10 - convert angle 1.497 rad to degrees a) O 850: 55:48" b) 269o: 48':55* C) 850: 48':S5“ d) 269o:…

A: The conversion of radian to the degree is done by multiplying the factor 180π,

Q: 5) Convert the following angles to the sexagesimal system: 23 grad 75.245 grad 151.83 grad

A: sexagesimal system 100 grades (or. 100g) = 90 degrees (or 90°)

Q: Given the following data, convert heading to azimuth or vice versa, leave procedure. 1) If the…

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Q: Given: Point Angle Bearing AB = N76-32-44E %3D 75°01'24" 41°19'20" 251°04'40" 54°06'24" 118°27'52"…

A: Given,  This is related to surveying.  We need to determine azimuths of each Line.

Q: find the elevation of reduced level for all point For sight Reduced level Back sight point Inter…

A: Given that:

Q: LINE DISTANCE BEARING CD 20 m S 850 20' E СА ? N 280 15' W СВ N 650 30' E DA ? N 700 W DB ? N 80°…

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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: STA BS HI FS IFS ELEV BMa 0.95 125.50m 0+ 00 3.0 + 10 2.3 TP1 3.13 0.64 0 + 22 + 30.5 + 34. 2 + 39.3…

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Q: STA BS HI FS IFS ELEV BMa 0.95 125.50m 0 + 00 + 10 3.0 2.3 TP1 3.13 0.64 0 + 22 2.7

A: The elevation is generally the vertical distance from a datum that is the sea level, to a point on…

Q: Corners Adj. Lat Adj. Dep АВ 80.20 -40.10 ВС -40.13 -36.82 CD 70.18 50.40 DE -30.14 30.36 EF 60.16…

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Q: Convert Azimuth to Bearing: 258° 34' 22" 128° 04' 44" 60° 21' 26" 309° 05' 38"

A: Quadrat -1.   N-E Quadrat -2   S-E Quadrat -3.   S-W Quadrat -4.   N-W

Q: COMPUTED CORRECTION BALANCED COURSE DISTANCE BEARING LAT DEP LAT DEP LAT DEP AB 495.85 N 05°30' E BC…

A: Latitude is Lcosθ Departure is Lsinθ    Latitude Correction=-LengthPerimeter × Latitude…

Q: Problem 3: Tunnel Surveying Objective: To prepare survey computations from preliminary survey of a…

A: Solution Data given is from the preliminary survey of the route of the traverse for the tunnel. When…

Q: Lines Latitude Departure AB -36.13 -25.77 ВС 74.56 -115.93 CD 12.82 0.39 DE 19.91 61.75 EA -68.4…

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Q: Given PLat 8+ 272.311, A = 24°24 20", and R = (midordinate), and the stations of the BC and EC. 500…

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Q: - A 500 mm x 500 mm square tied column shown is reinforced with 10- 26 mm diameter bars with a steel…

A: Given:- The dimension of the column = 500 mm × 500 mm  The steel reinforcement provided = 10-26 mm…

Q: 7777711

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Q: ree of static indeterminac

A: Given:- Beam To find:- degree of static indeterminacy

Q: Measured Interior Angle (2) N(Y) Point 610.24 (1) 100 45'37" 5000.00 N (Y) A B. 231 23'43" 10,000.00…

A: Azimuth is the angle between two points in a sphere. The angle between a projected vector and a…

Q: Convert the following to azimuth:  N64°21’18’’W S45°26’42’’E S79°42’16’’W

A: Convert the following to azimuth:  Part (a). N64°21’18’’W This is in quadrant 4. Azimuth  =   360°…

Q: COURSE LATITUDE DEPARTURE АВ 30 -70 BC 70 -30 CD Unknown Unknown DA -80 40 a) What is length of…

A: A closed traverse latitude and departure have been given in the tabulated form and it has been asked…

Q: 7. Determine whether the planes are parallel, orthogonal, or neither. 4x - 20y – 4z = 8 3x - 15y –…

A: If theta = 00 or 180⁰ : planes are parallel     theta = 90⁰          : planes are perpendicular

Q: Determine the adj. length of each path based on the observed data of a closed traverse.

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Q: A closed traverse has the following data: Course Bearing Distance 1-2 N9.27deg.E 58.7m. 2-3 S 88.43…

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Q: The length of a traverse line CD is 316.48 m and the magnetic azimuth from south of the line is 153…

A: Departure is given byD=length×sinbearingbearing=azimuth form northbearing=180°+153°54' = 333°54'

Q: The coordinates of control points A (1395.454, 1078.806), and L(1268.855, 1028.419). Calculate the…

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Q: SITUATION. Given the foliowing corrected data of a close traverse: Course Distance Latitude…

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Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.