R-M Q1: Compute:- a) vertical distance AB. b) ABC area. M=R-cosR staff in A staff in B 6ố số AB Sinbo56 Sin591515 19.03 E59,515-

Q: Question 240K-ft where E = 1800 kst H=13,432 int 24 Ft Cal culate andl draw the de formed Shafe.…

A: Given  E = 1800 ksi  I= 18432 in 4

Q: 777 TTI

A: static indeterminacy : When the static equilibrium equations - force and moment equilibrium…

Q: Determine the reactions of each structure

A: Given:- A structure is given with two-point loads  The magnitude of the point load at F = 8 kips The…

Q: Determine the horizontal force Q applied to block B that would cause the system to impend its motion…

A: given the angles are 600, 150

Q: п sin?ydx + cos²xdy = 0; y () 4

A:

Q: 1. F1 = 324.56N @32.34 deg S of W F2 = 152.58N @18.33 deg N of W F3 = 500lbf @47.85deg. S of E

A: *Multiple questions*   As per our guidelines we are supposed to answer?️ only one question. Kindly…

Q: 100 1b/ft Fixed 50 1b.ft B 一 aft. 8ft -

A:

Q: N87°30'E 589.06 m (3 sOnknown. Bearin 686.66 m Unknown Bearing ---- -- 654.06 m S 42°45'E 656.98 m.…

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: Use the virtual work method to determine the horizontal deflection at joint H of the truss shown in…

A: Solution Draw the diagram with reaction

Q: The polar moment of inertia for the area is JC = 642 (106) mm4, about the z' axis passing through…

A: Given: The polar moment of inertia is 642 x 106 mm4. The moment of inertia about the y' axis is 264…

Q: * ски/т П вт B 4.5m 8KN m

A:

Q: Find the centroid along x- axis of the shaded area. x - axis = _mm. y ゥー 335 | 600 mm 345 243 O 249…

A: The centroid of give section is calculated as below

Q: LINE FORWARD BEARING BACKWARD BEARING N 15°00' W N 63°00' E S 14°00' W S 77°30' W S 13°30' E S…

A:

Q: Wo> pa WIOX 30 18' Find: Centerline spacing = a if p: .6 V : 18 Ksi

A:

Q: 6000√3³ (√3y-y²) dy

A: Answer We are given Density of water = 9000 N/m3  side of triangle = 2m We know that the Force due…

Q: E = 29,000 ksi I = 4,000 in.+ Do FIG. P6.16, P6.42 64 k B -9 ft- -18 ft- -9 ft El = constant E =…

A: VA+VD=64VD36=649VD=16VA=4843227=x18x=288 draw the M/EI diagram  to find slope at C then find area…

Q: Find the position of NA if 250*350 mm Ast is 1200 mm sq FE250. M25

A: This problem is related to RCC , Design concrete structure. In this Question given about Width and…

Q: Q-2-: In the following table: a. Check all calculations. b. What order of levelling does this…

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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: 0.0025

A: Area of the channel :A=120.25×2×0.25×tan60=0.108253m2Perimeter :P=0.25cos60=0.5m

Q: N87°30'E 589.06 m B 3 Unknown Bearing 654.06 m Unknown. Bearing 686.66 m .---- -- S 42°45'E 656.98…

A:

Q: Q3/ Resolve the force 130N into two non-rectangular components. One having a line of action along AB…

A: Intensity of force is given and it has been asked to resolve forces in AB and CD direction as given.…

Q: 2-7 Resolve the position vector A of the car (measured from fixed point O) into components parallel…

A: Introduction- A vector is a quantity that has both a direction and a magnitude, whereas a scalar…

Q: From the following set of lines whose magnetic bearings are given: OA, N 39° 25' E and OB, N 75° 50'…

A:

Q: Consider the vector V with V = 1 kN. Resolve V into components Vp along the p-axis and Vq along the…

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Q: 785 mm -19.7 mm 33.5 mm- 384 mm- Built-up Section Typical Section Ix = 4200 x 10° mm ly = 315 x 10°…

A:

Q: Q3) From the figure below Compute :- 1)Elevation of point B 2)Elevation of point C 3)Height of the…

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Q: 2P A SI 4I 2.5m 25m 1.25m 2.5m 1-25m

A:

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: -0.61 m B 356 N 0.61 m H 356 N 0.61 m G F 0.46 m 267 N 178 N 0.61 m

A:

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: dy + 2.xy = 0 dx 6.23.

A: As per our guidelines we only solve the first part of the question kindly resubmit the second…

Q: P.I.(Station 2 + 990) Elevation = 52.30m Elevation 52.30m H ga = -3.4% Elevation of HP- gi = +3%…

A: Elevation of PI = 52.30m g1 = +3% g2 = -3.4% L2 = 282.65m L = 563.30m

Q: 520 N/m 230 N/m Chinged 15m 6m BO 13m N

A:

Q: GIVEN: G.=-3.5% ELEV V = 71.50 L= 800 m G. = + 4.5 % STA V= 10 + 00

A: PVCSTATION=PVISTA-L2L=1000-L2=6+00PVCELV=PVIELV-g1L2PVCelv=71.5+0.035×8002PVCelv=85.5mEquation of…

Q: 450 mm 360 nim B 320 mm 500 mm 600 mm A 1165 N

A: Given:-

Q: %Y. 36 ull 0 1,0K/s Asiacell 3 Reciprocal ranging is employed when

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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: D. SM 3M 4M beam A I beam <105 200x MM A all = lo0 Mu? Eall= 200 GPm %3D → Find suppont ronctons ķ…

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Q: LINE FORWARD BEARING BACK BEARING AB S 37° 30' E N 37° 30' W N 44° 15' E S 43° 15' W N 73° 00' W BC…

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Q: Calcule la ubicación vertical del centroide para esta forma. y = 16 21 (3 48 24 1) 2 69 71 ref

A: The centroid is the object's center point. The centroid of a triangle is the point at where the…

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?…

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Q: TT 3.2 in. |B D 5.8 in.

A: Given:- P=7500lb Q=3000lb a=15ft b=8f c=10ft Iz=110in4 To find:- Magnitude of maximum tensile…

Q: dy * for y = x? - 8 Find dx x-1

A: answer = 2x + 8/(x-1)2

Q: Given: FB = (400î – 400j- 200k) Find: (a) Fc = ? (b) FR = FB + F and a = ? (c) Angle be tween Fg and…

A:

Q: Question #02 culate the Moment of Inertial (X-X & Y-Y) V X16 Steel plate en the centroid is (-0.159,…

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Q: Determine the distance between parallel planes 2x-4y+6z+20+0 and 2x-4y+6x-40=0

A:

Q: Given 2 planes x+4y-z+3=0 and x-12y+2z-7=0 . Det the angle value between 2 points

A: Solution: Given: Two planes: x+4y-z+3=0 x-12y+2z-7=0

Q: section 1, Q1= 0.11 m^3/s and p1= 138 kPa. d2=100mm 309 (2) water 2, d,=150mm 13 50 Od:375mm

A: Q1=0.11 cumec P1=138kPa

Asphalt Binders and Mixtures

Asphalt binders are subsidiary products procured during the refining process of petroleum while manufacturing gasoline, diesel, and similar petroleum products. Asphalt binders are obtained from the residue that remains after the distillation of petroleum, this residue is further subjected to steam distillation and oxidation to remove the traces of gas oil present in the residue. Certain additives or modifiers are also added to asphalt binders to alter them to suit different applications.