5kN AD 3m

Q: 4 kN Draw F.B.D then find reactions at supports A and B take beam AB wieght 20K g Ax B Rgsin30 30°…

A: Given that: FBD of a beam as shown:

Q: 15 kN/m 5 kN/m 9 m

A: Given data  Simply supported beam  UVL Span length  I will break the given loading into UDL and UVL

Q: support B? a. 68.145 kN-m b. -68.145 kN-m c. 28.43 kN-m d. -28.43 kN-m   What is the final moment at…

A: A indeterminate beam is given with loading as mentioned.   To determine The final moment at A The…

Q: rent Attempt ln Progress Determine the shear force Vin the beam at a point located 0.70 m to the…

A: Given data, By = 300 kN Cy = 125 kN

Q: 5 kN 5 kN 5 kN 1.5 kN/m 1.5 kN/m A B 1.5m RA Rc

A:

Q: Using MDM, what is the reaction at support D? 4 kN 20 kN 18 kN 4 kN/m A B C. D 2m 6m 2m 2m 4m Select…

A: Finding out reaction at support D.

Q: 13.66 kN

A:

Q: / Replace the loading b- e moment acting at point C | Replace the loading by ly the location of the…

A: Replace the loading by an equivalent force Draw the free body diagram

Q: What is the vertical reaction at support A?  A. 35.81 kN B. 46.31 kN C. 50.90 kN D. 51.33 kN   What…

A: Detailed solution is attached below as

Q: Find: a. b. c. Situation 4: Considering the beam provided below, fixed support at point A, an…

A: As per guidelines, first three sub parts have been solved.

Q: 1.50 m COMPUTE THE HORIZONTAL AND VERTICAL COMPONENTS OF THE h,-4.50 m RESULTANT FORCE ON THE CURVED…

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Q: -5.37 5 KN 6.62- -4.29- -5.02- 4 KN 3.73 3.24 E 1.04 6.2 KN B. 1.96 0.43 C 4 KN 2.4 KN -2.59 5.77-…

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Q: 600 KN 240 KN E 16m A B 12m 12m 12m 300 KN 12) Determine the reaction at B. * 715.00 KN 751.20 KN…

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Q: Determine the resultant moment about points A and B, if Fc = 17 kN, L=8 ft and M = 63 kN-ft A L Fe L…

A: Given data:

Q: SITUATION. For the given truss. 1 m. 2 m. B. C D 2 m. 2 m. 2 m. - 2 m. 10 kN 25 kN 15 kN Determine…

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Q: Derive the shear force equation,bending moment equation for the beam own. Neglect the weight of the…

A: Uniformly varying load with 2.1kN/m at 7 m span. To calculate shear force and bending moment…

Q: Refer to figure below of a combined footing P2 a 35. If Pi = 1200 kN, P2 = 800 kN, b = 6, and c = 3,…

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Q: For W1=8.9 KN/m, W2=5 KN/m, the resultant of applied loads is WI KN/m W2 KN/m Select one: OA 36.70…

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Q: Determine the force BD? 400 kN 3 m 500 kN 50° D. 4 m

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Q: F7-4. Determine the normal force, shear force, and moment at point C. |12 kN 9 kN/m |C -1.5 m-1.5 m…

A: F7-4. Draw the free-body diagram of the given beam. Write the equilibrium equation for the sum of…

Q: 2. The 6-ft-wide Tainter gate in the form of a quarter circular arc is used as a sluice gate.…

A: Radius of OB, OA, and ODAngle, Width, To find: Moment about point O.

Q: The t-component of the force F is known to be 75 N. Determine only the n-component. 10 A. 119.5 N B.…

A: Finding out n component force.

Q: PROBLEM 4 The beam shown is loaded with the triangular, rectangular and the 10kN point load. Find…

A: Give data  Find resultant of loading diagram.    Concept

Q: 7. Get the force in CD and CE. 30 kN 1.5 m A 20 kN B C D F E G -2 m2 m2 m 2m

A: Given: Required : Force in CD and CE

Q: a. Slope 8A b. Slope Oc c. Deflection Ac d. Deflection AE e. Maximum deflection between support 10…

A: Given:Required:Slope at ASlope at CDeflection at CLet us solve this by moment area method as not…

Q: 20 EN 20LN 15 KN 15KN SEN 10 EN SEN 301m - 3 m D - 60 3m = 18

A:

Q: Derive the shear force equation,bending moment equation for the beam own. Neglect the weight of the…

A: Determine the support reactions. The sum of moments of all the forces about Point A is zero.…

Q: 8 kN 1 m -2.5 m DRAW M N T DIAGRAMS 8 kN -2.5 m 16 kN/m -4 m 4 kN/m -2 m

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Q: 20 KN 20 KN ACTIVITY 3 20 KN 10 KN/m 10 KN/m A PROBLEM 1 20 KN/m 10 KN/m C Compute the 10 KN-m 10…

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Q: support

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Q: ALI TITT 2 kN/m 3 m B 2 m 2 m D

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Q: Problem Number 5 1 kN 1 kN 1 h=8m B A F E G 5 kN 5 kN 5 kN 1 kN D 1 kN K -6 panels @ 5 m = 30 m- 12.…

A: To Determine Force at member GH,     FGH

System of Forces

A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum changes. If the body is at rest, under the action of force, its state of rest changes to motion. In other words, the magnitude of the momentum of the body changes from zero to a measurable value. Similarly, if the body possesses some initial velocity, it has some associated momentum, the effect of force can cause the body to halt or come to rest, that is, the momentum reduces to zero.