Solve the following problem on drawing S.F.D. and B.M.D. by hand. -2 m- B 4 m- 1.5 kN/m

Q: 8a kN a kN/m A -4.00- -6.00–

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Q: I0 KN 15 KN 10 KN E. 0.9 m C. 0.9m G IF 1.2m 1.2m 1.2m 1.2m

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Q: What is Deflection (DF) for BC? a. 0.753 b. 0.553 c. 0.981 d. 0.417   What is DF for CB? a. 0.583…

A: Given: Required : Distribution factor for BC, CB, CD.

Q: = (5t3/2)m/s, where t is in seconds.

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Q: P=10 KN 3.75m 1.25 m 5m 71. Find the deflection under point of application of concentrated load P.…

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Q: 10 N 10 N 20 N C E 10 N B R 6.0m 4.0m A J 8 BAYS @ 3.0 m EACH 20 N The vertical reaction at S is…

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Q: 4K/ft loft a). horizontal displacement @ C b). slope and displacement using unit load method @ 3ft…

A: Solution Find the support reactions ΣFx = 0 HA = 4×10        40 k ΣMA= 0 VC ×8 = 4×10×5 VC  = 25…

Q: Solve the following problem completely using Area moment method and Conjugate beam method for each…

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Q: efficient of static friction at all contacting surfaces is 0.3, determine the inclination ∅ (theta)…

A: Given :- μ=0.3   To find:- θ

Q: A car is on a slope of 20 degrees. The handbrake can deliver 9 kN. What is the maximum mass that can…

A: Assume mass of the car = m Free body diagram of the given problem:

Q: The flat roof of the steel-frame building shown in the photo is intended to support a total load of…

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Q: 5 KNIM 5KN 5m A

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Q: Find roller support reaction

A: Given data : The data of point load are: Point load=P Point load=Q Point load=R   Required : Finding…

Q: LEASE SHOW COMPLETE FBD AND COMPLETE SOLUTIO

A: MAKE FBD OR EQULIBRIUM LETS MAKE NO ACCERLERATION

Q: I- Moment Distribution Method 1. Find the RA, R3 Rc and R, using the Moment Distribution Method.

A: Fixed End Moments: MFAB=-wl8=-10×88=-10 kN.mMFBA=wl8=10×88=10 kN.mMFBC=-wl212=-5×8212=-26.67…

Q: ASSUME w = 12 kN/m P = 46 kN El = constant w kN/m A 6m 3m

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Q: Determine the magnitude and directione of the minimum force P needed to pull the 50-kg roller over a…

A: Given: The mass of the roller is 50 kg. Consider the figure,

Q: Given the following beam loadings a) Derive the shear and moment equations for the beam shown b)…

A: Solution:-  Support reactions, Ra+Re=7 kN Taking moment about A; ∑MA=0Re×8-7×1-5-10=0Re=228…

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: Draw S.F and B.M diagrams for the beam loaded as shown below. Use graphical method. -1.5 kN/m 0.5…

A: Solution  Given diagram

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: 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: 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: kN 20 kN 18 kN 4 kN/m A B C. 2m 6m 2m 2m 4m 2m E one: -0.60 0.40 1.00 0.50

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Q: Problem 1: Draw the bending moment diagram and calculate 0 at B using slope- deflection equations. A…

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Q: Calculate shear force and draw SFD and BMD using Moment Distribution Method

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Q: Solve for the maximum shear and maximum moment of the simply supported beam. Show the shear and…

A: Using equilibrium conditions, find out reactions at the support. Shear force at a section =…

Q: 10 N 10 N 20 N E 10 N B 6.0m 4.0m D F H J L Q 8 BAYS @ 3.0 m EACH 20 N • The Vertical reaction at A…

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Q: The man picks up the 52-ft chain and holds it just high enough so it is completely off the ground.…

A: The horizontal tension in the cable is calculated as, The height if the chain than can be lifted by…

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: BLOCK A WITH WEIGHT EQUAL TO 100 N, RESTS ON BLOCK B OF NEGLIGIBLE WEIGHT. THE COEFFICIENT OF STATIC…

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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: Find support reaction

A: Given data in question  Span length  Simply supported beam  UDL UVL To find out  Reactions