Slope and Deflection Method ,The Moment Distribution Method ,Strain Energy Method

An introduction to Different method is given here. The strain energy method actually solves these equations by the method of successive approximations.

Problems are solved to illustrate the Strain energy method as applied element.

1. Noida Institute of Engineering and Technology, Greater Noida Slope and Deflection Method The Moment Distribution Method Strain Energy Method Aayushi Assistant Professor Civil Engg . Department 5/17/2020 1 Unit: 1 Aayushi RCE-502, DOS 1 Unit 1 Subject Name : Design of Structure I Course Details : B Tech 5 th Sem

2. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 2 Content Course Objective Course O utcome CO-PO & PSO Mapping Prerequisite & Recap Slope and Deflection Method Procedure for finding Slope and Deflection Numerical on Slope and Deflection The Moment Distribution Method Analyse structure by MDM Moment Distribution Method for Beam Numerical on MDM Syllabus of unit 1 Topic outcome and mapping with PO

3. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 3 Content You tube Video Links Daily Quiz Weekly Assignment MCQs Old Question Papers Expected Questions in University Examination Summary References Numerical on Strain Energy Strain Energy Method

4.   Objective 1 To impart the principles of elastic structural analysis and behavior of indeterminate structures. 2 To impart knowledge about various methods involved in the analysis of indeterminate structures. .. 3 To apply these methods for analyzing the indeterminate structures to evaluate the response of structures . 4 To enable the student get a feeling of how real-life structures behave. 5 To make the student familiar with latest computational techniques and software used for structural analysis. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 4 Course Objective

5.   Once the student has successfully completed this course, Students will be able CO1 To Identify and analyze the moment distribution in beams and frames by Slope Deflection Method, Moment Distribution Method and Strain Energy Method. CO2 To provide adequate learning of indeterminate structures with Muller’s Principle; Apply & Analyze the concept of influence lines for deciding the critical forces and sections while designing.. CO3 To learn about suspension bridge, two and three hinged stiffening girders and their influence line diagram external loading and analyze the same. CO4 To Identify and analyze forces and displacement matrix for various structural. CO5 To understand the collapse load in the building and plastic moment formed. CO6 Apply the concepts of forces and displacements to solve indeterminate structure. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 5 Course Outcome

6. CO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 RCE502.1 3 3 1 2 1 - - - 2 - - 3 RCE502.2 3 2 2 2 3 - - 1 2 - 2 2 RCE502.3 3 - 2 2 1 - - - 2 - 2 2 RCE502.4 3 1 2 2 1 - - - 2 - 2 2 RCE502.5 3 2 2 2 1 - - - 2 - 2 2 RCE502.6 3 2 3 2 1 - - - 2 - 2 3 AVG. 3 2 2 2 1 - - - 2 - 2 2 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 6 CO-PO and PSO Mapping COs PSO1 PSO2 PSO3 PSO4 RCE502.1 2 1 2 3 RCE502.2 2 1 2 2 RCE502.3 2 2 2 2 RCE502.4 2 1 2 3 RCE502.5 2 1 2 3 RCE502.6 2 1 2 3 Avg. 2 1 2 3

7. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 7 Prerequisite and Recap Basics of strength of material Basics of engineering mechanics

8. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 8 Syllabus of Unit 1 Analysis of fixed beams, Continuous beams and simple frames with and without translation of joint by 1 Slope-Deflection method, 2Moment Distribution method 3Strain Energy method.

9. 5/17/2020 9 Objective of Unit 1 Analysis of fixed beams, Continuous beams and simple frames with and without translation of joint by Different Method. Aayushi RCE-502, DOS 1 Unit 1

10. Derive slope-deflection equations for the case beam with yielding supports. Estimate the reactions induced in the beam due to support settlements. Analyse the beam undergoing support settlements and subjected to external loads. Write joint equilibrium equations in terms of moments. Relate moments to joint rotations and support settlements. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 10 Topic Objective

11. To Identify the moment distribution in beams and frames by Slope Deflection Method 5/17/2020 11 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different beams and frames. Apply the concepts of forces and displacements to solve indeterminate structure . Analyze moments to joint rotations and support settlements. Aayushi RCE-502, DOS 1 Unit 1

12. 5/17/2020 12 Objective of Topic Topic-1 Name The Slope Deflection Method Objective of Topic-1: To analysis of Different beam and frames by slope and deflection method Aayushi RCE-502, DOS 1 Unit 1

13. 5/17/2020 13 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 1 Outcome of Topic-1: After the successfully competition of this topic students will be able to analysis of Different beam and Frames by slope and deflection method Aayushi RCE-502, DOS 1 Unit 1

14. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 14 Prerequisite and Recap Briefing about Fixed end moments. Basics of strength of material Basics of engineering mechanics Briefing about Indeterminate structure Diagram of bending moment and shear stress

15. The  slope-deflection method  is an alternate way to analyses indeterminate structures.  In the slope-deflection method, we will first identify unrestrained degrees-of-freedom (DOFs, which are rotations or deflections) and solve for them . In the slope-deflection method, we will find the deflections/rotations of the unrestrained DOFs in terms of the forces and moments at the DOF locations . For the slope-deflection method we will need to solve a system of equations with as many equations/unknowns as there are degrees-of-freedom (DOFs).  T he slope-deflection method will be good for analyzing highly-constrained structures, with a lot of degrees of indeterminacy, and not much freedom of movement (not many DOFs ). 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 15 Topic : The Slope-Deflection Method

16. A  degree-of-freedom  (or  DOF ) is a single way that a node is permitted to move or rotate.  In a 2D system, each node has three possible degrees-of-freedom: translation (movement) in one direction, translation in another direction perpendicular to the first one, and rotation. Usually, we consider the horizontal and vertical axes as the two perpendicular translational degrees-of-freedom. Although three DOFs are possible for each node, individual directions may be considered to be restrained, either by a support reaction or by one of the connected members. 5/17/2020 Aayushi RCE502, DOS 1 Unit 1 16 Topic: Degree of Freedom

17. The slope-deflection equations give us the moment at either end of each element within a structure as a function of both end rotations, the chord rotation, and the fixed end moments caused by the external loads between the nodes The slope-deflection method relies on the use of the  slope-deflection equation , which relate the rotation of an element (both rotation at the ends and rigid body rotation) to the total moments at either end. The ultimate goal is to find the end moments for each member in the structure as a function of all of the DOFs associated with both ends of the member. Apply equilibrium conditions at all of the joints to solve for the unknown rotations. This is the system of equations that we will have to solve, where the equations are the equilibrium equations for each node and the unknowns are the translations and rotations of the nodes. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 17 Topic : Slope and deflection of the beam

18. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 18 Continuous…

19. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 19 Continuous…. Equations of slope and deflection :-

20. 1. Find all of the unrestrained DOFs in the beam structure. 2. Define an equilibrium condition for each DOF (for rotations, the sum of all moments at each rotating node must equal zero). 3. Find the moments to put into the equilibrium conditions using the slope deflection equation. 4. For each element that has an external force between the nodes, find the fixed end moments at either end . If there are any support settlements or imposed displacements at node locations, find the chord rotations caused by that settlement/displacement. Construct each slope deflection equation. 5. Use the resulting equilibrium equations to solve for the values or the unknown DOF rotations (solving a system of equations). 6. Use the now-known DOF rotations to find the real end moments for each element of the beam. 7. Use the end moments and external loadings to find the shears and reactions. 8. Draw the resultant shear and bending moment diagrams . 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 20 Topic : Procedure for finding slope & deflection

21. 1.Find the bending moment for 4 degree indeterminate beam 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 21 Topic:- Numerical

22. Find the fixed end moment FEM AB =+81kNm (positive because it is counter-clockwise), and on the right side of the member FEM BA =−125kNm (negative because it is clockwise). Likewise , for member BC, which has a single point load, the fixed end moments FEM BC =+75kNm on the left (CCW ) FEM CB =−75kNm on the right (CW). 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 22 Numerical Continuous…

23. the moment at the end of member AB at point A, A is the near side and B is the far side, so: M AB =2EI/L(2θ A +θ B −3ψ AB )+ FEM M AB =2EI/9(2θ A +θ B −3(−0.00333))+ 81 For the moment at the end of member AB at point B, B is the near side and A is the far side, so: M BA =2EI/L(2 θ B + θ A −3 ψ AB )+FEM M BA =2EI/9(2 θ B + θ A −3(−0.00333))−121.5 Again , node A is fixed, so  θ A =0 M BA =2EI/9(2 θ B +0.0100)− 121.5 Moving on to member BC, for the moment at the end of member BC at point B, B is the near side and C is the far side, so: M BC =2EI/L(2 θ B + θ C −3 ψ BC )+FEM M BC =2EI/6(2 θ B + θ C −3(0.00500))+ 75 Since node C is also fixed, then we know that  θ C =0, so M BC =2EI/6(2 θ B −0.01500)+75 The last slope deflection equation is for the moment at the end of member BC at point C, where C is the near side and B is the far side, so: 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 23 Numerical Continuous…

24. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 24 Numerical Continuous… M CB =2EI/L(2 θ C + θ B −3 ψ BC )+FEM M CB =2EI/6(2 θ C + θ B −3(0.00500))−75 Again, node C is fixed, so  θ C =0: M CB =2EI/6( θ B −0.01500)−75 Now that we have all of the slope-deflection equations, we can apply our equilibrium condition at node B to solve for our only unknown rotation  θ B: M BA +M BC =0 2EI/9(2 θ B +0.0100)− 121.5+2EI/6(2 θ B−0.01500)+ 75=0 Since there is only one DOF in this problem, there is only one equilibrium equation and one unknown rotation, so we can solve directly for  θ B : 1.111θ B (EI)=0.00278(EI)+ 46.5 So far in the equations above, all of the units have been in 0kN and 0m, so we need to find an expression for EI in terms of 0kN and 0m: EI =(200000MPa)(127×106mm 4 ) EI =25.4×1012N/mm 2 =25400kNm 2

25. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 25 Numerical Continuous… We can now rearrange the previous equilibrium equation and solve for  θ B : 1.111θ B (EI )=0.00278(EI)+46.5 θ B =0.00278/1.111+46.5/1.111EI θ B =0.00278/1.111+46.5/1.111(25400) θ B =+ 0.00415rad Knowing the rotation at point B ( θ B ), we can sub that value back into the slope-deflection equations to get the actual end moments at either end of each member: M AB =2EI/9(0.00415+0.0100)+81=+ 160.9kNm likewise for the others: M BA =−18.2kNm M BC =+18.3kNm M CB =−166.9kNm At this point we should double check our calculations and see that  M BA +M BC =0  (which they do). The 0.1kN difference here is due to round off error. That could have been eliminated if we kept more significant figures along the way and then only rounded the number off at the end .

26. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 26 Numerical Continuous…

27. The slope-deflection equations are derived for the case of beam with yielding supports. Moments developed at the ends are related to rotations and support settlements . The equilibrium equations are written at each support. The continuous beam is solved using slope-deflection equations. The deflected shape of the beam is sketched. The bending moment and shear force diagrams are drawn for the examples solved in this lesson. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 27 Summary of Slope & Deflection

28. Calculate stiffness factors and distribution factors for various members in a continuous beam. Define unbalanced moment at a rigid joint. Compute distribution moment and carry-over moment. Derive expressions for distribution moment, carry-over moments Analyse continuous beam by the moment-distribution method. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 28 Topic Objective

29. To Identify the moment distribution in beams and frames by Moment Distribution Method 5/17/2020 29 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different beams and frames. Apply the concepts of forces and displacements to solve indeterminate structure . Analyze moments to joint rotations and support settlements. Aayushi RCE-502, DOS 1 Unit 1

30. 5/17/2020 30 Objective of Topic Topic-2 Name The Moment Distribution Method Objective of Topic-2: To analysis of Different beam and frames by Moment Distribution method Aayushi RCE-502, DOS 1 Unit 1

31. 5/17/2020 31 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 1 1 Outcome of Topic-2: After the successfully competition of this topic students will be able to analysis of Different beam and Frames by moment distribution method Aayushi RCE-502, DOS 1 Unit 1

32. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 32 Prerequisite and Recap Briefing about Fixed end moments. Basics of strength of material Basics of engineering mechanics Briefing about Indeterminate structure Diagram of bending moment and shear stress

33. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 33 Topic: The Moment Distribution Method The moment distribution method is also called the  Hardy Cross method  after its inventor, an engineering professor named Hardy Cross. The   moment distribution method   is the third method that we will learn for the analysis of indeterminate structures. Recall that the force method was good for structure with a small number of degrees of indeterminacy, regardless of how many degrees-of-freedom the system has. In contrast to that, the slope-deflection method was good for structures with a small number of degrees-of-freedom, regardless of how many degrees of indeterminacy the system has. The moment distribution method is an iterative method that gets around the problem of too many degrees of indeterminacy or too many degrees of freedom. Using moment distribution, we can analyses highly indeterminate structures with many degrees of freedom by hand . The moment distribution method is an exact method that relies on a set of successive better approximations to determine the bending moments for beams and frames

34. For example, the structure shown is typical for a concrete building frame The structure has rigid connections between all of the members, and this results in 27 degrees of static indeterminacy and 15 degrees of freedom. This system would clearly be all but impossible to analyse using the force method or the slope-deflection method. Prior to the development of the moment distribution method in 1932, there was simply no practical way to analyse such a system. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 34 Continuous…

35. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 35 Topic: How to Analyse Structure

36. The Sign-Convention The sign-convention for the moment distribution method is the same as the sign convention for the slope-deflection method All counter-clockwise moments are considered positive All clockwise moments are considered negative. There is no consideration of the internal moment sign convention. All moments that are considered are point moments at member ends. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 36 Continuous…

37. 1. Determine the stiffness for each member. - For a member that is fixed at both ends ,. k AB =4EI/L - For a member that has a pin at one end, k AB =3EI/L 2. Determine the distribution factors for each member at each node based on relative stiffness of the members DF AB = k AB /∑ k i Use a distribution factor of zero for a fixed support and 1.0 for a pinned support with only one connected member. 3. Determine the fixed end moments for all members that have external loads applied between the end nodes. 4. For each node in turn: -Determine the unbalanced moment on the node. -Distribute the unbalanced moment to each member connected to the node in proportion to the distribution factors  in the reverse direction of the unbalanced moment . 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 37 Topic: Moment Distribution Method for Beam

38. - For each member that the moment has been distributed to, carry over some of the moment to the opposite end of the member CO=1/2 CO=0 For a member with a fixed end opposite (a regular locked node), carry over half of the moment that was applied by the distribution. For a member with a pinned end opposite (where there are no other members connected to that pin) do not carry over any moment. 5. Repeat the previous step for each node, multiple times as necessary until the carry over moments are a small fraction of the total moments at each member end. 6. Sum all of the moments in each member end from all previous steps (including the original fixed end moments). This sum gives the total moment at each member end in the real system. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 38 Continuous…

39. Find the bending moment diagram for the indeterminate beam. Take EI constant. Step 1   Find the stiffness of each member . k AB =4EI/L =4EI/5 =0.8EI Member BC has a pin at the right side node C with only one member (BC) connected to that node . k BC =3EI/L =3EI/4 =0.75EI 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 39 Topic: Numerical 50 KN 5 m 2 m 2 m

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41. Step 2 find all of the distribution factors for all the members at each joint DF AB = k AB / k AB +k wal l =0.8EI/0.8EI + ∞ =0 At node B there are two members that are connected, members AB and BC DF BA = k AB / k AB +k BC =0.8EI/0.8EI+0.75EI =0.516 For the pin end at node C, there is only one member attached, member BC DF BC =K BC / k AB +k BC =0.75EI/0.8EI+0.75EI=0.484 DF CB = k BC / k BC =1.0 Step 3 The next step is to determine the fixed end moments FEM BC =+ 25kNm, FEM CB =−25kNm 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 41 Continuous…

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44. An introduction to the moment-distribution method is given here. The moment distribution method actually solves these equations by the method of successive approximations. Various terms such as stiffness factor, distribution factor, unbalanced moment, distributing moment and carry-over-moment are defined in this lesson. Few problems are solved to illustrate the moment-distribution method as applied to continuous beams with unyielding supports. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 44 Summary of MDM

45. Deflection by strain energy method. Evaluation of strain energy in member under different loading. Application of strain energy method for different types of structure. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 45 Topic Objective

46. To Identify Evaluation of strain energy in member under different loading. 5/17/2020 46 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different beams and frames. Apply the concepts of to strain energy method solve indeterminate structure . Analyze moments to joint rotations and support settlements. Aayushi RCE-502, DOS 1 Unit 1

47. 5/17/2020 47 Objective of Topic Topic-3 Name The Strain Energy Method Objective of Topic-3: To Identify & analysis Evaluation of strain energy in member under different loading. Aayushi RCE-502, DOS 1 Unit 1

48. 5/17/2020 48 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 Outcome of Topic-3: After the successfully competition of this topic students will be able to Identify & analysis Evaluation of strain energy in member under different loading. Aayushi RCE-502, DOS 1 Unit 1

49. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 49 Prerequisite and Recap Deflection by strain energy method. Evaluation of strain energy in member under different loading. Application of strain energy method for different types of structure.

50. Castigliano’s Theorem : The displacement corresponding to any force applied to an elastic stricture and collinear with that force is equal to the partial derivative of the total strain energy with respect to that force. δi = where δi is the displacement at the point of application of force Fi in the direction of Fi.   5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 50 Topic: Strain Energy Method

51. When external loads are applied on an elastic body they deform. The work done is transformed into elastic strain energy U that is stored in the body. We will develop expressions for the strain energy for different types of loads. Axial Force : Consider a member of length L and axial rigidity AE subjected to an axial force P applied gradually as shown in the Figure. The strain energy stored in the member will be equal to the external work done by the axial force i.e 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 51 Topic: Calculation of Strain Energy

52. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 52 Continuous… Bending Moment : Consider a beam of length L and flexural rigidity EI subjected to a general loading as shown in Figure. Consider a small differential element of length, dx . The energy stored in the small element is given by The total strain energy in the entire beam will be

53. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 53 Continuous… Shear Force: The strain energy stored in the member due to shearing force is expressed by where V is the shearing force; and GAs is the shearing rigidity of the member. Twisting Moment: The strain energy stored in the member due to twisting moment is expressed by

54. The truss members are steel rods with a 50 mm diam. The load F is 4 kN. Find the deflection at the point A Rod area = π x .052 /4 = 0.001963 m 2 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 54 Topic: Numerical Find the forces : FAB = 0.75 F = 3 kN (tension ) FAC = -1.25 F = 5 kN (compression )

55. Total energy U = =5.902 ×F Nm δ a = =1.18×10 -8 F Nm Giving δa =0.047 mm Deflection of curved members: Consider the curved frame shown in (a) below. We want to find the deflection of the frame due to force F, in the direction of F and at its point of application.   5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 55 Continuous…

56. Consider the strain energy in the element defined by the angle dθ . The force F is resolved into components F r and F θ . There are three parts of the strain energy : 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 56 Continuous…

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58. An introduction to the strain energy method is given here. The strain energy method actually solves these equations by the method of successive approximations. Various Forces such as Axial, Shear, moment and Twisting moment are defined in this lesson. Few problems are solved to illustrate the Strain energy method as applied element . 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 58 Summary of Strain Energy Method

59. Youtube /other Video Links 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 59 Youtube Video Links Topic Links Process Slope Deflection Method https://www.youtube.com/watch?v=76MtP7WAGr4&t=2033s Click on the link Moment Distribution Method https://www.youtube.com/watch?v=SGObzuMBCpc Click on the link Strain Energy Method https://www.youtube.com/watch?reload=9&v=WigWnQuj98k Click on the link

60. If in a pin-jointed plane frame (m + r) > 2j, then the frame is (Where ‘m’ is number of members, ‘r’ is reaction components and ‘j’ is number of joints ) (A) Stable and statically determinate (B) Stable and statically indeterminate (C) Unstable (D) None of the above   The Castigliano's second theorem can be used to compute deflections (A) In statically determinate structures only (B) For any type of structure (C) At the point under the load only (D) For beams and frames only 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 60 Daily Quiz

61. Discuss the slope deflection method to analyze an indeterminate structure with its necessary assumptions and sign convention. Write down the step involved in the analysis of indeterminate beams by slope deflection method . Using slope deflection equations, explain what are the effects of support settlements on the indeterminate structures ? Use neat diagrams to explain your point. Discuss the situations with the help of neat diagrams where in sway will occur in portal frames. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 61 Daily Quiz

62. A beam ABC, 10 m long, fixed at ends A and C is continuous over joint B and is loaded as shown in Fig Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. El is constant for both spans . 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 62 Daily Quiz

63. Analyze the beam given in figure by slope deflection method. Analyze the frame shown in fig, by moment distribution method. Take EI as constant. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 63 Daily Quiz

64. Write the assumption made in slope and deflection method? Write the assumption made in Moment distribution method? Define degree of freedom? Define Hardy Cross method ? Define the Castigliano's second theorem ? Find the horizontal deflection at joint C of the pin-jointed frame as shown. AE is constant for all members . 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 64 Weekly Assignment

65. Q Analyze the frame shown in fig, by moment distribution method. Take EI as constant. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 65 Weekly Assignment

66. A continuous beam ABCD is simply supported at A, B, C and is fixed at D. The span AB, BC and CD are 3 m, 4 m and 2 m long. The beam carries a point load of 12 kN on AB at 2 m from A, a point load of 20 kN at the middle of BC and a point load of 6 kN at middle of CD. If I(AB) : I(BC) : I(CD) =1:2:2. Find the supports moments using moment distribution method. Draw the bending moment diagram for the continuous beam shown in using moment distribution method. El is constant. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 66 Weekly Assignment

67. Which of the following methods of structural analysis is a force method? (A) Slope deflection method (B) Column analogy method (C) Moment distribution method (D) None of the above The deflection at any point of a perfect frame can be obtained by applying a unit load at the joint in (A) Vertical direction (B) Horizontal direction (C) Inclined direction (D) The direction in which the deflection is required 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 67 MCQ s

68. The maximum negative bending moment in fixed beam carrying udl occurs at ____ a) Mid span b) 1/3 of the span c) Supports d) Half of the span A fixed beam of the uniform section is carrying a point load at the center, if the moment of inertia of the middle half portion is reduced to half its previous value, then the fixed end moments will __ a) Increase b) Remains constant c) Decrease d) Change their direction 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 68 MCQ s

69. A propped cantilever beam carrying total load “W” distributed evenly over its entire length calculate the vertical force required in the prop. a) 3/4 W b) W c) 5/8 W d) 3/8 W ___ is a small opening made in the bottom or sides of a tank. a) Mouthpiece b) Orifice c) Sill d) Sluice 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 69 MCQ s

70. Calculate the actual velocity of jet if the coefficient of velocity is 0.97. The head of water on the orifice of diameter 2 cm is 6 m. a) 11 m/s b) 12 m/s c) 10.5 m/s d) 13 m/s The value of Cv varies ___ to ____ a) 0.95 – 0.99 b) 0.93 – 0.95 c) 0.97 – 1 d) 0.94 – 0.96 The relation between hydraulic coefficients is Cd = Cc × Cv. a) False b) True 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 70 MCQ s

71. The principle of virtual work can be applied to elastic system by considering the virtual work of (A) Internal forces only (B) External forces only (C) Internal as well as external forces (D) None of the above Which of the following methods of structural analysis is a displacement method ? (A) Moment distribution method (B) Column analogy method (C) Three moment equation (D) None of the above 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 71 MCQ s

72. In moment distribution method, the sum of distribution factors of all the members meeting at any joint is always (A) 1 (B)  less than 1 (C)  greater than 1 (D) None of the above 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 72 MCQ s

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82. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 82 Old Question Papers

83. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 83 Old Question Papers

84. For the beams shown below determine the reaction forces and draw the shear and moment diagrams using the slope-deflection method. All elements have the same  EI  unless otherwise indicated . For the frames shown below determine the reaction forces and draw the shear and moment diagrams using the slope-deflection method. All elements have the same EIEI unless otherwise indicated 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 84 Expected Questions for University Exam

85. A beam ABC of length 2L rests on the three supports equally spaced and is loaded with UDL 'w/unit length throughout the length of the beam as shown in Fig. Plot the BM and shear force diagrams . Draw the bending moment diagram for the continuous beam shown in using moment distribution method. El is constant. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 85 Expected Questions for University Exam

86. Using moment distribution method analyze the frame shown in Fig Draw the bonding moment diagram. The comparative moment of inertia is mentioned against each member of the frame. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 86 Expected Questions for University Exam

87. Draw the bending moment diagram for the continuous beam ABCD loaded as shown.The relative moment of inertia of each span of the beam is also shown in the figure. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 87 Expected Questions for University Exam

88. An introduction to Different method is given here. The strain energy method actually solves these equations by the method of successive approximations. P roblems are solved to illustrate the Strain energy method as applied element. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 88 Summary

89. Jain, A. K., “Advanced Structural Analysis “, Nem Chand & Bros., Roorkee . Hibbeler , R.C., “Structural Analysis”, Pearson Prentice Hall, Sector - 62, Noida-201309 C . S. Reddy “Structural Analysis”, Tata Mc Graw Hill Publishing Company Limited,New Delhi. Timoshenko , S. P. and D. Young, “ Theory of Structures” , Tata Mc- Graw Hill BookPublishing Company Ltd., New Delhi. Dayaratnam , P. “ Analysis of Statically Indeterminate Structures”, Affiliated East- WestPress . Wang, C. K. “ Intermediate Structural Analysis”, Mc Graw -Hill Book PublishingCompany Ltd. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 89 References

90. 5/17/2020 Aayushi RCE-502, DOS 1 Unit 1 90 References Thank You