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

Q: 900 lb 60 lb/ft 400 lb D A |B - 8 ft 6 ft→ 4 ft

A: following is the solution-

Q: 2 ft F D E -1.5 ft 600 lb 300 lb -3 ft -3 ft-

A:

Q: F2 = 81 lb F = 100 lb 4 ft 3 ft 7 ft 40° 4 ft

A:

Q: 2 kN 4m A 2m 2m MA=2 kN.m O MA=-8 kN.m MA=4 kN.m O None of them

A: Given data: Magnitude of the force = 3 kN Need to determine the moment about the point A.

Q: 25 lb 40 lb 24 in. 0.5 in. 4 in. a B Section a – a

A:

Q: 1 kN 2 kN 2 kN 2 kN 1 kN H 0.46 m F D 2.62 B I A C| E G| 2.4 m 2.4 m 2.4 m 2.4 m Figure 2

A:

Q: Draw isometrie

A: Isometric projection Isometric projection is a method for representing three-dimensional objects…

Q: Figure VC 0.3 m 0.2 m B 1 of 1

A: Given: VB = 3 m/s VC = -1.7 m/s

Q: -5 ft 3 ft- 2 ft 4 ft B В 5 3 12 13 4 5 500 lb 200 lb 260 lb

A:

Q: -C-((3/4)*D)))-A)/2 C-((3/1)*D)-A)/2

A: Given, A+2B = 5C-3/4D A= [20 22;4 1;6 1]C= [10 10;5 5 ;2 2]D = [1 1;1 1;1 1]B=…

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

A: For solution refer below images.

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: 100 lb 800 lb · ft A |C B 5 ft 5 ft

A:

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

A: Data given-

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

A:

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

A:

Q: 1000 Ib -12 ft 12 ft- B 2000 lb 8 ft D. ton -12 ft- 6 ft 6 ft

A:

Q: 2. 1000 lb M= 4500 lb.ft B A RA 3 ft 6 ft 3ft Ro

A:

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

A:

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

A:

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

A:

Q: 20 kN 12 kN A В M = 10 kN-m %3D 4m 4m 4m Figure Q1 (b)

A:

Q: Find the force of the rods AB, AC, BC and BD. 20 N 30 N В 4m 4m 4m 4m 4m 4m A 4m E

A:

Q: 300 lb 6. TE

A: Given that, Calculate the reaction at the support at C and D.

Q: 2 kN 50 mm 5 kN• m D 80 mm B C 1.0 m 1.0 m 1.0 m

A:

Q: 3000 lb 90 Ib /ft 9 ft 200 lb/ft

A: For solution refer below images.

Q: 13378.2898 lbm to lbf

A:

Q: Determine the maximum shear stress acting at section a–a in the beam.

A: Given beam with reactions at 'A' and 'D' are as shown below: Applying equilibrium conditions: (i)…

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: 15 kN 10 kN/m 20 kN · m A 2 m 1m 1m 2 m What is the reaction force at A?

A: Given: Point load=15kN Moment=20kNm Uniform load=10kN/m

Q: 18 kN 3 B 25 kN 8m DE 6m D 10 kN Figure 1

A:

Q: 100 lb 30° 200 lb 2 ft 300 ft-lb 2 ft 2 ft 2 ft-

A: First we resolved 100 lb force, and calculated horizontal force, vertical force and moment due to…

Q: |1500 lb 4500 lb • ft 2.5 in. A E NA B |D I = 96 in. 8 in. 6 ft 6 ft +3 ft

A: To find: The shear and moment diagram. Calculation: The force balance equation. R1+R2=1500 The…

Q: 4 m A) 12 kN.m B) 16 kN m C) 20 kN m D) 22 kN.m 4 KN co 2 m 10 kN 2 m 2 m 2 kN/m E

A:

Q: < 1 of 1 Figure E 6 m D 4 m 4 m 4 m Pi P2

A:

Q: Force in the cable CD in Newton

A: Draw the free body diagram of the joint C.

Q: 80 kN 40 kN 60 kN -10 kN/m A B 2 mim. 3 m 3 m 2m 1 m (b) 10 k I k/ft B 12 8' (b)

A:

Q: In the Figure 6, the absolute force value in the bar AB is (in kN): Figure 6: 5 m 5 m A 20 KN 5 m B

A:

Q: 22: Moment of Q? 500 lb 700 Kg*m/s² O 10 m 300 Kg

A:

Q: A 0.8 ft 4.8 ft 60 lb 3.5 ft E 48 lb 1.8 ft B |C D 3.6 ft

A:

Q: Define the following terms: Weight Tension Friction Equilibrium

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

Q: Q.7 Degree of freedom at triple point is A 1 B 2 D -1

A: This concept is related with the basic thermodynamics.

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: 2 kN/m 10 kN t. 5 m 3 m 2 m Figure 2

A:

Q: 1500 lb 300 lb/ft 300 lb/ft A B 3 ft- 3 ft- HINGE ROLLER (2)

A:

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: F3 = 52 lb 13 12 F1 = 80 lb 25°

A: Given that, The component of forces in the x-direction,

Types of loading

In mechanical terminology, the term load indicates that external force/weight, either constant or variable, acts on a mechanical member/structure. The weight of an object of a specific material is also an external load, and the external force that works on a structure comes under the categories of loading. The external load generates its effects as stress and strain in the structure.