200 mm B 300 mm 400 mm 100 D G 600 mm

Q: 200 mm 150 mm 100 mm 200 mm

A:

Q: 3 mm 600 N 6 mm 100 N/mm 3 mm Cable |1 mm 4 mm 4 mm

A: Drawing the FBD of the frame as shown below, Where, FT is the triangular load, FR is the…

Q: 3 m 2 m B 4 m F A 30 kN 18 kN

A:

Q: 80 mm 50 mm 2 1 7 15 mm 15 mm 3.

A: The Center of mass is a point where the sum of the moment of the weight of each particle is zero. It…

Q: X 300 mm D 800 mm B

A:

Q: E 600 mm B Rigid P. 80 mm 160 mm 80 mm

A:

Q: y 200 mm 200 mm + 150 mm- 125 mm 60 mm 125 mm 1.

A:

Q: 120 mm 600 mm 10-N 4N Fig. 5 600 mm 4N

A:

Q: 20 kN 50 kN 50 kN/m 25 kN/m B. 2 m 3 m

A: Consider the free body diagram. The cantiliver support will be replace by two forces FH and FV and a…

Q: 330 mm 220 100 mm mm 800 N 75 mm 200 N 10 1200 N

A:

Q: 125 mm 175 mm D -225 mm- 75 mm 150 N Fig.3

A: Answer: (a) The tension in cable ABD is 195 N. (b) The reaction at support C is 216.33 N at an angle…

Q: 75 mm D. 1.5 m E F B 2 m 150 mm 5 deg 2 deg 4 deg 1.5 deg 1 deg 3.5 deg

A: Write the twist angle of point A respect to point B.   φA/B=φA-φBφA-φB=3°…

Q: 1. 100 mm 225 mm 75 mm 125 mm 50 mm

A: According to the details provided in the question, we need to calculate centroid.

Q: 120 mm m x' 20 m

A: Data given-

Q: 120 mm 60 mm 90 mn D 80 mm C 45 mim 480 N

A:

Q: 5 m A 30 kN B 5 m 5 m 5 m 5 m 20 KN 5 m 5 m E

A:

Q: 3 cm4 cm cm 3 cm 3 cm 80 mm 2m 80 mm 60 mm

A:

Q: H G 60 E D 7 kN 7 kN 9 kN 6m 6m 6m 6m

A:

Q: A 150 mm B 240 mm 80 mm

A:

Q: Vertically 300lb/pulg2 and horizontally 150 lb/pulg2 I have only this data please answer.

A: Giver:- My=300 lb/pulg2 Mx=150 lb/pulg2   To find:- Resultant

Q: 200 N 250 N 630 mm 630 mm 630 mm

A: Total Distance = 630 +630+630 =1890 mmF1 = 200 NF2 = 250 NF3 = 355 N To-Determine:- Distance,d

Q: 100 kN 50 mm 50 mm 75 mm 75 mm 40 kN 30° 400 mm

A:

Q: 47 mm 47 mm Semiellipse. |26 mm 70 inm 2.

A: Given Data: The base of the triangle is b = 94 mm The height of the triangle is h = 70 mm The…

Q: 15 kN 25 kN 20 kN C В D 1 m 0.8 m 0.3 m 30 kN °0.4 m

A: Q7.35 Given: Load at A (P2)=15kNLoad at C (P1)=25kNLoad at D P3=30kNLoad at E (P4)=20kN

Q: 6 m 6 cm 12 om 6 cm 18 cm 12 cm 6 cm

A: "Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: 100 mm 100 mm 10 mm 10 mm. 200 mm 100 mm f 10 mm I II

A: For the selection of sections, we have to calculate the maximum bending stress in the beam. From the…

Q: F 1.0 1.0 cm cm 0.2 cm

A: The plate in the fluid medium subjected to not only  wall shear force but buoyant force also.  For a…

Q: 10 mm 20 mm 50 mm

A:

Q: 50 mm 150 mm 150 mm 300 mm *

A: A1=300×50A1=15000 mm2y¯1=3002=150 mm   from bottom,A2=300×50A2=15000 mm2y¯2=300+502=325 mm,

Q: F B 4 m 4 m 4 m 4 m

A: A beam is defined as a structural member subjected to transverse shear load during its…

Q: y -9 cm- -3 cm 2 cm 19 cm 2 cm C 15 cm 9 cm- 3 cm→

A: Divide given figure in 3 section/shape,

Q: a) 10mm IG mm IG mm 1050mm IG mm 16mm 250 mm 300 mm

A:

Q: 9 cm cm S cm 7 cm

A: Answer- 5.9

Q: 300 mm 300 mm Figure Q2

A: Given data:- Area =1950 mm^2 Maximum deflection of point C = 0.6mm

Q: Two holes have been drilled through a long steel bar that is subjected to a centric axial load as…

A: Solution: Given Data: P=32 kNdA=20 mmdB=50 mmw=100 mmt=15 mm   To Find: (a) Maximum stress at A. (b)…

Q: 100 Ib 100 Ь 100 b

A: Once applied, force is described as an external cause that alters or tends to modify the…

Q: F=20 lb a Q6 25 mm 70 mm 100 mm- 70 25 mm mm

A:

Q: L = 3 m Lotfen birini seçin: a 98.48 mm D. 120.04 mm C 150.92 mm d. 175.14 mm . 142.02 mm

A:

Q: H, = 0.40 Hx = 0.30 D 150 mm 150 mm B E 300 mm 300 mm \F 150 mm 150 mm 250 mm

A:

Q: 20 cm 30 cm

A: This is the case of the forced vortex motion. here radius = 0.3 m distance between the two limb del…

Q: 1.0 m 2.0 m 05 m 05 m 20 kN 4 KN 10 kN 4.0 m

A: Given Data: Force (F) = 10 kN

Q: (1) (2) (3) 50 N 10 mm 10 mm 10 mm A:=50 mm? A2=20 mm? A:=10 mm? E=200 GPa

A:

Q: C 110 mm F В 150 mm E 300 mm A D 200 mm 150 mm

A: There is a circumferential contact force between gears B and EF=TABrB=TEFrETEF=TABrErBTAB=170…

Q: The volume of a 45’ container is 87.16 m3. The internal measures are: Length: 44’4” ft Height: 8’4”…

A: Volume of container, V=87.16m3 Length, L=44'4"=(44*12)+4=532" (inches) Height, H=8'4"=(8*12)+4=100"…

Q: З KN 8 kN A 25 kN m 2 m 2 m 2 m 5 kN

A: Note: As per bartleby guidelines for more than one question asked only one question to be solved…

Q: 40 mn 40 mm 40 mm 40 mm N.A N.A 80 mm 80 mm 40 mm 40 mm 100 mm 100 mm 100 mm 100 mm case 2 case 1

A: For case 1:let:b=40  d=160b1=100    d1=40y=40+20 =60mmy'=80…

Q: 24 mm 5 mm 30 mm 8 mm 5 mm 30 mm Section AA 12,000 N

A: The correction factor is rc=3015=2 The correction factor for 2 is k=1.41. The moment of inertia is…

Q: C 30 kN 50 kN 70 kN 50 kN d=20 mm d=40 mm 1m 2 m 1m

A: draw FBD of each section and find stresses.for solution refer below image.

Q: D C 2 mm 5 mm

A: GivenAD=15mmAB=15mm

Q: 50 kN 500 40 kN 30 kN 50 mm. 200 mm 200 mm 50 mm 250 mm 30 kN

A:

Center of Gravity and Centroid

The term center of gravity indicates the position or location of a point about which the whole weight of a body or system is supposed to be concentrated, which means the whole weight of the body or system works through this point. It is closely related to mass distribution. For all positions of the body, the center of gravity would be a single point.