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

Q: Bar Member Magnitude , KN Type of Bar Force AB ВС BD DE DF 150 KN 3 4 D 200 KN 6.00 m 8.00 m A E F…

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Q: 120 KN 90 KN 4.00 m 4.00 m 4.00 m E 1.50 m F G 60 KN 2.00 m 6.00 m 1.00 m 5.00 m B.

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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: SITUATION 5. For Items 12-14. 12. Find the force acting on member FE. a. 1.803 kN "T" 1.Şm H b.…

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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: Solve FG only

A: Support reaction: Due to symmetry:AY=12×(2×(0.6+1.24+1.04+1)+1.2)=4.48 KN

Q: Determine the magnitude and type of bar forces for members BE, CF, DG, and FG.

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Q: 8 kN 48 kN/m В A 1 m – -1 m –

A: Calculate the support reactions at A and B:…

Q: a. 9.30 kN b. 18.61 KN c. 37.21 AN d. 74.42 kN Compute the value of eccentricity e. 0.026 m 0.062 m…

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Q: Find the reaction Ey?

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Q: 7. What is the support reaction force (kN) at B? 20 kN AA 3m 15m y a) 10 b) 5 C) d) 20 Bos birak

A: Free body diagram of given beam:-

Q: Untitled Section F1= 10 kN F2 = 26 kN 8- 45 45 F3 =8 KN B 4 m F4= 44 kN E La-45 FS = 8 kN 1] What is…

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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: 4m 6.5 kN/m - 4 m— Draw SFD and BMD

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

A: Finding out reaction at support C.

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: What is the internal bending moment at point B (including the sign based on the sign convention for…

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Q: Circle all zero force members.

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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: A force of 10 kN acts through point A(2, 4) and B(8, 6). Resolve this force into two components,…

A: Step 1:Lets find vectors along AC and along AB :    Vector AC =  AC=C−A=(5−2,−2−4)=(3,−6)Vector CB =…

Q: What is the normal force exerted by the roller? 25 KNim 150 KN 10 m Select one: a. 149.30 kN b.…

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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: 10, KN A 1m 10 kN/m 1m 5 kN/m 3m Find the resultant Find the location of the resultant from pt. A…

A: Objective: To determine resultant of the forces system, location of the resultant from points A and…

Q: Determine the magnitude of resultant force

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Q: P = 10 1.5P KN- P KN- F 1.5 m GH -3 m E 1.5 m, IC D 2m 6 m

A: To find force in member EF

Q: Calculate the magnitude of the tension T and the angle 0, for which the eye bolt will be under a…

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Q: Determine the internal force system acting on section 1 by analyzing the FBD o (a) segment AD and…

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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 in member CD. H 3 m C E -2 m 2 m 2 m- 2 m 2 m 2 m- 4 kN 5 kN 6 kN 8 kN O 12.0 kN…

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Q: B1

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Q: Q3 50 kN 5 m 5 m. A'nalyze the above symmetrical otructure under the given symmetrical looding by…

A: Solution Draw the diagram with reaction

Q: Find roller support reaction

A: Given:- A truss is loaded with various point loads The magnitude of the point load at points B, D,…

Q: 3. Draw AFD, SFD, BMD of the below structure (Figure:02). 50KN 3 m 80KN 3 m 5m 3 m Figure: 02

A: Assuming that the heap on a segment is applied through the focal point of gravity of its cross area,…

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: F1 = 3 kN F2 - 11 kN C. F3= 2 kN e-60 30 2m 7 m F4 = 20 kN D ElJa- 30 FS = 2 kN 1] What is the…

A: Given data  Length of the span Point loads Angles of these point loads.

Q: 30. Bending moment equation for segment AB. O a. 0.58x(KN)-0.16x3(KN/m2) O b.…

A: Draw the free body diagram. Equilibrium of horizontal forces, ∑Fx=0Ax=0

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: Problem 2 For the given frame shown, it is subjected to a concentrated load of 100 kN at the midspan…

A: Fixed end Moment (Mf) MfAB = MfBA = 0 MfBC = -(PL/8) = - (100×6/8) = - 75KNm MfCB = +(PL/8)=…

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.…

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.