a. Draw free body diagram b. Determine stress in the cable c. Determine support reactions A Wo = 10 Kn/M M 10 M

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Q: ction force at support I in kN (round to

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Q: The following beam is supported by three cables AD, BE, and CF. Cables AD and CF are made of steel,…

A: Same values is came once refer explanationExplanation:Step 1: Step 2: Step 3: Step 4:

Q: 2) The structure as shown supports the horizontal force F = 30 kN. Determine the safe diameter of…

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Q: The shear capacity v if a steel wide flange section, 600 mm deep whose web is 9 mm thick, is closest…

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Q: 2) The structure as shown supports the horizontal force F = 30 kN. Determine the safe diameter of…

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Q: 1.5 kN/m |B 2 m 3 m

A: Given:- w=1.5kN/m To find:- Maximum moment

Q: 2. The simple beam is subjected to loadings: w= 100 kN/m, F= 500 kN, M= 5000 kN-m. The lengths are:…

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Q: A) draw the free body diagram to prove its statically determinate and find the reaction forces at…

A: Find: A) draw the free body diagram to prove its statically determinate and find the reaction forces…

Q: For the beam below, (a) draw the shear and moment diagrams and (b) determine the maximum normal…

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Q: What is discharged condensate in condensing systems?

A: Discharged condensate in the condensing system:

Q: For the beam below, (a) draw the shear and moment diagrams and (b) determine the maximum normal…

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Q: The structure as shown supports the horizontal force F = 30 kN. Determine the safe diameter of pin…

A: GIVEN: The structure as shown in the attached figure, supports the horizontal force F = 30 kN. The…

Q: LA L2 a= 0.6 m ............... -2.0 m b= 1.0 m. Center of gravity of plate Steel plate (2.5 × 1.5 ×…

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Q: d carries a load P= 2,700 lb at joint B. The strut, which consists of two bars d to the beam by a…

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Q: Find the collapse load for the truss shown in the figure. All three bars have the same plastic axial…

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Q: Determine the maximum and minimum shearing stress in the web of the wide flange section in if V =…

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Q: 2) The structure as shown supports the horizontal force F = 30 kN. Determine the safe diameter of…

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Q: Determine sum of vertical support reaction at A and B

A: Given data : The value of point load P1=5 KNThe value of point load P2=4 KNTo find :Sum of vertical…

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Q: The rectangular beam ABC, 175 mm wide by 411 mm deep, is supported by a pin at A and the cable CD.…

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Q: 15 KN 4 m 5 kN/m 2 E1 B 3EI 5 m 5m 12 KN 3 EI 5m Question A For structure below, complete the…

A: Given An indeterminate frame Required  To find support reactions using force method.

Q: A IS 04 KN 15.VW KN B 6-55KN 6.XY KN 1.5 m 1.5 m 1.5 m 2.WX m 2.45m

A: ###Calculate reactions ofpoint A, B , C using equilibrium equations

Q: A steel cable of 2 cm diameter is used to lift a board 500 kg. Given that the length of the cable is…

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Q: The Channel section is used as a purlin placed on top of trusses with spacing of 4.9 m and slope 1V…

A: w = 24KN/ml = 4m

Q: The Channel section is used as a purlin placed on top of trusses with spacing of 4.9 m and slope 1V…

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Q: В C 4 m 3 m 6 m 3 m D E 4 kN 6 kN

A: To find the force in member BC -

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A: Given Fig : TO Find : a) Determine the shear and bending diagram  b) Determine the max. normal…

Q: Ceiling Pr. 4-47: A suspended walkway is modeled as a rigid bar and supported by steel rods (E…

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A: Draw the figure.

Q: H.W. Determine the horizontal and vertical components of force at pins A and D. D Ans: A = 12.0 kN =…

A: It is required to find horizontal and vertical components of pin at A & B.

Q: For the beam below, (a) draw the shear and moment diagrams and (b) determine the maximum normal…

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Q: answer number 3

A: To find maximum shear at mid span

Q: 4. Calculate the support reactions and plot the internal force diagrams for the following multi-span…

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Q: 2) The structure as shown supports the horizontal force F = 30 kN. Determine the safe diameter of…

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Q: For the beam below, (a) draw the shear and moment diagrams and (b) determine the maximum normal…

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Q: Where is neutral axis of the gutter section with reference from the bottom of the channel? What is…

A: area of C6X13 = 3.83 in2  Iy = 1.05 in4 but in the girder section it becomes Ix  x¯ = 0.514 in but…

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.