200 lb - 3 ft 3 ft - 50 Ib 30 -3 ft 100 lb

Q: 3 ft- Wo G 4 ft 60% OD MG 2 ft FG (NG B G E B.

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Q: (M=450 lb-ft (M)s = 300 lb-ft 3.5 ft 1.5 ft 2 ft 2 ft (Mh = 250 lb-ft

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Q: 150 lb/ft A В -6 ft -8 ft

A: Let us draw the reaction at the supports  Weight of the rectangular load = 150 lb/ft(8ft)=1200 lb…

Q: y 800 lb - 40° Calculater for Fr 35°

A:

Q: In figure 2 a) what is the total weight of all bodies? b) Which one of the following is the free…

A: Total weight will be the sum of weight of all bodies.

Q: 40 kN E 2 m B F 3 m 50

A: to find reaction component at B and C

Q: 600 lb-ft B 350 lb-ft 3ft 1 ft 500 lb-ft 2 ft

A: For solution refer below images.

Q: free-body diagrams. 6 ft 3 kip/ft B. 2 kip/ft -3 ft C

A: As per bartleby guidelines we are allowed to solve only 1st question if multiple questions are…

Q: 500 lb 600 lb · ft A B -5 ft - -5 ft -5 ft

A: Forces and moments generated in any member depend upon the type of support and type of loading…

Q: The reaction at C 120 lb 300 lb lb www 120 lb

A: Given

Q: Give an example of Lifting of a large weight by a small force by the application of Pascal’s law. A…

A: According to Pascal law pressure on the enclosed fluid is transmitted throughout the fluid with…

Q: 100 lb B 12а++ - 4 ft - 80 lb - 3 1 -

A: To find : The shear stress in the beam. Free body diagram : The free body diagram is shown below:…

Q: 8'-0" 900 lb 91 006 10'-0" B 400 lb 68'-0" 48'-0" 12'-0|

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Q: D 50 lb-ft B 4 in. 30 6 in. A.

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Q: 350 lb 30 250 lb

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Q: Neglecting the mass of the beam, compute the reactions at A and B

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Q: The diameter of the large piston of a hydraulic jack is found to be 75 mm. A force of 58 N is…

A: Given that, Diameter of large piston, D2=75 mm,Force applied of small piston, F1=58 N,Load lifted by…

Q: Fi = 549 lb F2= 839 lb F - 806 lb and has the following geometry: w- 20 ft F2 F1 F3 B

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Q: 300 lb/ft 1500 lb-ft B A 16'- 6' 1200 lb

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Q: 80 lb/ft В - 5 ft – 2-5 5 ft A 60 lb/ft 12 ft

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Q: =150 le 120 F = 80 lb.

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Q: 3 kN 12 kN.m 3 kN / m 5 kN / m 7 3 m 3 m 2 m

A: given:

Q: F = 250 lb - 30°, 45 F2 = 375 lb

A:

Q: 120° F= 400 lb F = 600 lb

A: Here for force F1, Angle with x-Axis (alpha) = 60o Angle with Y-Axis (beta) = ? Angle with Z-Axis…

Q: y 800 lb/ft A B –- 2 ft →- -4 ft - 2 ft→

A: Q. To determine the reaction forces for the beam as shown.

Q: In each case, draw the shear and moment diagrams for the beam.

A: The free body diagram of the first beam is as follows: Calculate the force in the vertical…

Q: In the Figure 3, the weight component parallel to the inclined surface can be calculated as: *…

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Q: - 2- 45 100lbs/ft 300lb 30 200lbs

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Q: 150 Ib/ft 100 lb

A: Consider the member AB and drawing FBD as shown below, Balancing vertical…

Q: F = 120 lb | 30° 12 kN 1.25"- 30 4 m 1.5" 5 m 11"

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Q: 100 KN 6m T

A: GIVEN DATA- A SIMPLY SUPPORTED BEAM GIVEN SUBJECTED TO POINT LOAD WE HAVE TO DRAW SFD AND BMD

Q: 17 4-m 180 lbs 3-m 2-m 5-m 160 lbs

A:

Q: 180 lb · ft В. C 9 ft 4.5 ft

A: Now draw the free body diagram for the above problem. Convert the triangular varying load into the…

Q: car is parked i. ii. The rear wheels of the car in Figure 2 is kept locked by the brakes whiles the…

A: Given data: Weight of the car, W = 3300 lb. Part (a) Draw the free-body diagram for the upslope…

Q: 300, lb 300 lb 800, lb 300, lb 800, lb 9 ft G B C D E F 700 Ib. 700 lb 6 panels e 5 rt/panet

A: GIVEN DATA - Truss Given   We have to find force in given member using method of joint

Q: RB 50KN bm 100 KN 4m B 4m

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Q: 6 kN 2.5m

A: Write the equilibrium condition for the above problem.

Q: 400 lb 108 in

A: Calculate the reaction of the support, RA+RB=400   .............. (1)Taking moment about…

Q: Determine the x and y components of each of the forces shown.

A: To find out the x component of the forces Resolve all the force in the x-axis…

Q: 40cm 4000 N Determine pressure? Diameter force —

A: To Find : The pressure applied on the piston. Given :  The Force applied to the piston is F=2000 N.…

Q: 10 KN/m 1a KN (m Ra 1.5 m 4m 3 m 1.5m

A: For solution refer below images.

Q: 30 2f- 6 ft-

A:

Q: Define the following terms: Weight Tension Friction Equilibrium

A: GIVEN DATA- SOME TERMS GIVEN WE HAVE TO DEFINE IT

Q: 400 lb F1 = 400 F2 = -600 k F3 = 500 sin 45 i-s00 cos 45k 7t Y600 lb 45 M- rxF,. M,- r2x , M,- rgxf,…

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

Q: 600 lb · ft -5 ft – -5 ft -5 ft

A: The beam can said to be in static equilibrium, when the vector sum of all the forces on the beam is…

Q: 100 kN I 120 kN/m 60 kN/m Ic D 315 kN-m 3 m 3 m 3 m 3 m

A: Draw the free-body diagram of the beam.

Q: 20 KN 40 kN/m 150 kN m 3m

A: According to the company's guidelines when multiple questions are asked in a single request,  only…

Q: 15 ft 150 Ib 35 5ft 2 ft 250 Ib 25

A:

Bending Moment

In general, we can define the bending moment as the reaction induced when a certain amount of load or force is applied to an element which makes it bend.