Structural AnalysisSTRUCTURAL– II ANALYSIS – II 10CV53 Subject Code : 10CV53 IA Marks : 25 No. of Lecture Hours/Week : 04 Exam Hours : 03 Total No. of Lecture Hours : 52 Exam Marks : 100
PART - A
UNIT - 1
ROLLING LOAD AND INFLUENCE LINES: Rolling load analysis for simply supported beams for several point loads and UDL. Influence line diagram for reaction, SF and BM at a given section for the cases mentioned in above uinit 1 6 Hours
UNIT - 2
SLOPE DEFLECTION METHOD: Introduction, Sign convention, Development of slope-deflection equations and Analysis of Beams and Orthogonal Rigid jointed plane frames (non-sway) with kinematic redundancy less than/equal to three. (Members to be axially rigid) 8 Hours
UNIT - 3 MOMENT DISTRIBUTION METHOD: Introduction, Definition of terms- Distribution factor, Carry over factor, Development of method and Analysis of beams and orthogonal rigid jointed plane frames (nonsway) with kinematic redundancy less than/equal to three. (Members to be axially rigid) 8 Hours
UNIT - 4 SWAY ANALYSIS: Analysis of rigid jointed plane frames (sway, members assumed to be axially rigid and kinematic redundancy £ 3) by slope deflection and moment distribution methods. 4 Hours
PART - B
UNIT – 5
KANIS METHODS: Introduction, Basic Concept, Analysis of Continuous beams and Analysis of rigid jointed non-sway plane frames. 6 Hours
UNIT - 6 FLEXIBILITY MATRIX METHOD OF ANALYSIS: Introduction, Development of flexibility matrix for plane truss element and axially rigid plane framed structural elements and Analysis of plane truss and axially rigid plane frames by flexibility method with static indeterminacy ≤ 3. 7 Hours
Dept. of Civil Engg., SJBIT Page 1 Structural Analysis – II 10CV53 UNIT - 7 STIFFNESS MATRIX METHOD OF ANALYSIS: Introduction, Development of stiffness matrix for plane truss element and axially rigid plane framed structural elements. And Analysis of plane truss and axially rigid plane frames by stiffness method with kinematic indeterminacy £3. 7 Hours
UNIT - 8 BASIC PRINCIPLES OF DYNAMICS: Basic principles of Vibrations and causes, periodic and aperiodic motion, harmonic and non- harmonic motion. Period and frequency, Forced and Free Vibration, Damping and Equations of Single Degree of Freedom System with and without damping 6 Hours
REFERENCE BOOKS: 1. Basic Structural Analysis- Reddy C.S. - Second Edition, Tata McGraw Hill Publication Company Ltd. 2. Theory of Structures Vol. 2 - S.P. Gupta, G.S. Pandit and R. Gupta, Tata McGraw Hill Publication Company Ltd. 3. Structural Dynamics-by M.Mukhopadhyay, 4. Structural Analysis-II -S. S. Bhavikatti – Vikas Publishers, New Delhi. 5. Basics of Structural Dynamics and Aseismic Design By Damodhar Swamy and Kavita PHI Learning Private Limited 6. Structural Analysis- D.S. Prakash Rao,, A Unified Approach, University Press 7. Structural Analysis, 4th SI Edition by Amit Prasanth & Aslam Kassimali, Thomson Learning.
Dept. of Civil Engg., SJBIT Page 2 Structural Analysis – II 10CV53
TABLE OF CONTENT
PAGE UNIT TOPIC NO Unit – 1 Rolling load and influence lines 4 Unit – 2 Slope deflection method 14 Unit – 3 Moment distribution method 31 Unit – 4 Sway analysis 43 Unit – 5 Kanis methods 57 Unit – 6 Flexibility matrix method of analysis 67 Unit – 7 Stiffness matrix method of analysis 77 Unit – 8 Basic principles of dynamics 86
Dept. of Civil Engg., SJBIT Page 3 Structural Analysis – II UNIT - 1 10CV53 ROLLING LOAD AND INFLUENCE LINES
1 Introduction: Variable Loadings So far in this course we have been dealing with structural systems subjected to a specific set of loads. However, it is not necessary that a structure is subjected to a single set of loads all of the time. For example, the single-lane bridge deck in Figure1 may be subjected to one set of a loading at one point of time (Figure1a) and the same structure may be subjected to another set of loading at a different point of time. It depends on the number of vehicles, position of vehicles and weight of vehicles. The variation of load in a structure results in variation in the response of the structure. For example, the internal forces change causing a variation in stresses that are generated in the structure. This becomes a critical consideration from design perspective, because a structure is designed primarily on the basis of the intensity and location of maximum stresses in the structure. Similarly, the location and magnitude of maximum deflection (which are also critical parameters for design) also become variables in case of variable loading. Thus, multiple sets of loading require multiple sets of analysis in order to obtain the critical response parameters.
Figure 1 Loading condition on a bridge deck at different points of time
Influence lines offer a quick and easy way of performing multiple analyses for a single structure. Response parameters such as shear force or bending moment at a point or reaction at a support for several load sets can be easily computed using influence lines.
Dept. of Civil Engg., SJBIT Page 4 StructuralFor example, Analysis we can– II construct influence lines for (shear force at B ) or10CV53(bending moment at) or (vertical reaction at support D ) and each one will help us calculate the corresponding response parameter for different sets of loading on the beam AD (Figure 2).
Figure 2 Different response parameters for beam AD
An influence line is a diagram which presents the variation of a certain response parameter due to the variation of the position of a unit concentrated load along the length of the structural member. Let us consider that a unit downward concentrated force is moving from point A to point B of the beam shown in Figure 3a. We can assume it to be a wheel of unit weight moving along the length of the beam. The magnitude of the vertical support reaction at A will change depending on the location of this unit downward force. The influence line for (Figure3b) gives us the value of for different locations of the moving unit load. From the ordinate of the influence line at C, we can say that when the unit load is at point C .
Figure 3b Influence line of for beam AB
Dept. of Civil Engg., SJBIT Page 5 StructuralThus, an Analysis influence – lineII can be defined as a curve, the ordinate to which10CV53 at any abscissa gives the value of a particular response function due to a unit downward load acting at the point in the structure corresponding to the abscissa. The next section discusses how to construct influence lines using methods of equilibrium.
2 Construction of Influence Lines using Equilibrium Methods The most basic method of obtaining influence line for a specific response parameter is to solve the static equilibrium equations for various locations of the unit load. The general procedure for constructing an influence line is described below. 1. Define the positive direction of the response parameter under consideration through a free body diagram of the whole system. 2..For a particular location of the unit load, solve for the equilibrium of the whole system and if required, as in the case of an internal force, also for a part of the member to obtain the response parameter for that location of the unit load.This gives the ordinate of the influence line at that particular location of the load. 3. Repeat this process for as many locations of the unit load as required to determine the shape of the influence line for the whole length of the member. It is often helpful if we can consider a generic location (or several locations) x of the unit load. 4. Joining ordinates for different locations of the unit load throughout the length of the member, we get the influence line for that particular response parameter. The following three examples show how to construct influence lines for a support reaction, a shear force and a bending moment for the simply supported beam AB .
Example 1 Draw the influence line for (vertical reaction at A ) of beam AB in Fig.1
Solution: Free body diagram of AB :
Dept. of Civil Engg., SJBIT Page 6 Structural Analysis – II 10CV53
So the influence line of :
Example 2 Draw the influence line for (shear force at mid point) of beam AB in Fig.2.
Solution:
Dept. of Civil Engg., SJBIT Page 7 Structural Analysis – II 10CV53
Example 3 Draw the influence line for (bending moment at ) for beam AB in Fig.3.
Solution:
Dept. of Civil Engg., SJBIT Page 8 Structural Analysis – II 10CV53
Dept. of Civil Engg., SJBIT Page 9 Structural Analysis – II 10CV53
Similarly, influence lines can be constructed for any other support reaction or internal force in the beam. However, one should note that equilibrium equations will not be sufficient to obtain influence lines in indeterminate structures, because we cannot solve for the internal forces/support reactions using only equilibrium conditions for such structures.
3 Use of Influence Lines In this section, we will illustrate the use of influence lines through the influence lines that we have obtained in Section 2. Let us consider a general case of loading on the simply supported beam (Figure 4a) and use the influence lines to find out the response parameters for their loading. We can consider this loading as the sum of three different loading conditions, (A), (B) and (C) (Figure 4b), each containing only one externally applied force.
Figure4: Application of influence lines for a general loading: (a) all the loads, and (b) the general loading is divided into single force systems
Dept. of Civil Engg., SJBIT Page 10 StructuralFor loading Analysis case (A),– II we can find out the response parameters using10CV53 the three influence lines. Ordinate of an influence line gives the response for a unit load acting at a certain point. Therefore, we can multiply this ordinate by the magnitude of the force to get the response due to the real force at that point. Thus
Similarly, for loading case (B):
And for case (C),
By the theory of superposition, we can add forces for each individual case to find the response parameters for the original loading case (Figure4a). Thus, the response parameters in the beam AB are:
One should remember that the method of superposition is valid only for linear elastic cases with small displacements only. So, prior to using influence lines in this way it is necessary to check that these conditions are satisfied. It may seem that we can solve for these forces under the specified load case using equilibrium equations directly, and influence lines are not necessary. However, there may be requirement for obtaining these responses for multiple and more complex loading cases. For example, if we need to analyse for ten loading cases, it will be quicker to find only three influence lines and not solve for ten equilibrium cases. The most important use of influence line is finding out the location of a load for which certain response will have a maximum value. For example, we may need to find the location of a moving load (say a gantry) on a beam (say a gantry girder) for which we get the maximum bending moment at a certain point. We can consider bending moment at point D of
Dept. of Civil Engg., SJBIT Page 11 Example3,Structuralwhere Analysis the beam–ABIIbecomes our gantry girder. Looking at the influence10CV53 line of one can say that will reach its maximum value when the load is at point D . Influence lines can be used not only for concentrated forces, but for distributed forces as well, which is discussed in the next section.
4 Using Influence Lines for Uniformly Distributed Load Consider the simply-supported beam AB in Figure 6.5, of which the portion CD is acted upon by a uniformly distributed load of intensity w/unit length . We want to find the value of a certain response function R under this loading and let us assume that we have already constructed the influence line of this response function. Let the ordinate of the influence line at a distance x from support A be . If we consider an elemental length dx of the beam at a distance x from A , the total force acting on this elemental length is wdx . Since dx is infinitesimal, we can consider this force to be a concentrated force acting at a distance x .The contribution of this concentrated force wdx to R is:
Therefore, the total effect of the distributed force from point C to D is:
Figure 5 Using influence line for a uniformly distributed loading
Dept. of Civil Engg., SJBIT Page 12 StructuralThus, weAnalysis can obtain– II the response parameter by multiplying the intensity10CV53 of the uniformly distributed load with the area under the influence line for the distance for which the load is acting. To illustrate, let us consider the uniformly distributed load on a simply supported beam (Figure 6). To find the vertical reaction at the left support, we can use the influence line for that we have obtained in Example 1. So we can calculate the reaction as:
Figure 6.6 Uniformly distributed load acting on a beam
Similarly, we can find any other response function for a uniformly distributed loading using their influence lines as well. For non-uniformly distributed loading, the intensity w is not constant through the length of the distributed load. We can still use the integration formulation:
However, we cannot take the intensity w outside the integral, as it is a function of x .
Dept. of Civil Engg., SJBIT Page 13 Structural Analysis – II UNIT – 2 10CV53
SLOPE DEFLECTION METHOD
In this method the end moments or support moments expressed in terms of slopes, deflections, stiffness and length of the members. The unknown slope values (slopes) are determined from the condition of equilibrium of joints for moments that is
MBA + MBC = 0
MBA MBC A B C
Assumption
1. All the joints of the frame are rigid that is angle between the members do not change at a joint even after deformation. 2. The joints are assumed to rotate as a whole 3. Directions due to axial and shear stress are neglected because they are negligible or small
Sign Conventions
1. Moments:- Clockwise moment +ve Anticlockwise moment –ve
2. Rotation:- Clockwise Rotation +ve Anticlockwise Rotation –ve
3. Sinking of support If right support sinks down Δ is +ve If left support sinks down Δ is –ve
4. Bending Moments Sagging BM is +ve and Hogging BM is –ve
5. Shear Force Left side upward the SF is +ve Left side downward the SF is -ve Right side upward the SF is -ve Right side downwardward the SF is +ve
Dept. of Civil Engg., SJBIT Page 14 Structural1. Draw BMD, Analysis Elastic –curveII and SFD by slope deflection Method 10CV53
40kN 10kN/m
A B C 3m 2m 6m 1.5I 2I
Fixed End Moment:
MFAB = = = -19.2 kN-m
MFBA = = = 28.8 kN-m
MFBC = = = -30 kN-m
MFCB = = = 30 kN-m
Slope Deflection Equation
MAB = MFAB + (2θA + θB) 0
= -19.2 + (2θA + θB) (θA = 0 due to fixity at support A) .
MAB = -19.2+0.6EI θB
MBA = MFBA + (2θA + θB) 0
= 28.8 + (2θB + θA) .
MBA = 28.8 + 1.2EI θB
MBC = MFBC + (2θB + θC)
= -30 + (2θB + θC)
MBC = -30 +1.33EI θB + 0.667EI θC
MCB = MFCB + (2θC + θB)
= 30 + (2θC + θB)
MCB = 30 +1.33EI θC + 0.667EI θB
Dept. of Civil Engg., SJBIT Page 15 Structural Analysis – II 10CV53 Apply the condition of Equilibrium
MBA + MBC = 0 and MCB = 0
At ‘B’
MBA + MBC = 0
28.8 + 1.2EI θB -30 +1.33EI θB + 0.667EI θC = 0
2.533EI θB + 0.667EI θC = 1.2 1
At ‘c’
MCB = 0
30 +0.667EI θC + 0.667EI θB = 0
0.667EI θB+ 1.33EI θC = -30 2
Solving Eq 1 and 2
θB = θC = . . Substitute the above values in the S-D Equation
MAB = -19.2+0.6EI θB MAB = -14.76 kN-m
MBA = 28.8 + 1.2EI θB MBA = 37.68 kN-m
MBC = -30 +1.33EI θB + 0.667EI θC MBC = -37.68 kN-m
MCB = 30 +1.33EI θC + 0.667EI θB MCB = 0
Support Reaction 10kN/m 14.8 40kN 37.68 37.68 0
VA VB1 VB2 VC
Dept. of Civil Engg., SJBIT Page 16 StructuralVA = 11.424kN, Analysis VB1 =– 28.576kN,II VB2 = 36.28kN and VC = 23.72kN 10CV53 VB = VB1+ VB2 = 58.734kN 36.28
11.424 11.424
+ + A B C _
+ 23.72
28.576 28.576 SFD
48kN-m 40kN-m
37.68kN-m + +
_ 14.8kN-m _
BMD
Elastic Curve
2. Draw BMD, Elastic curve and SFD by slope deflection Method
40kN 20kN/m 10kN
A B C 1m 2m 4m 2m 2I 1.5I I
Fixed End Moment:
MFAB = = = -17.78 kN-m
MFBA = = = 8.89 kN-m
Dept. of Civil Engg., SJBIT Page 17 Structural Analysis – II 10CV53
MFBC = = = -26.67 kN-m
MFCB = = = 26.67 kN-m
MFCD = MCD = -10 X 2 = -20kN-m
Slope Deflection Equation
MAB = MFAB + (2θA + θB) 0
= -17.78 + (2θA + θB) (θA = 0 due to fixity at support A)
MAB = -17.78+1.33EI θB
MBA = MFBA + (2θA + θB) 0
= 8.89 + (2θB + θA)
MBA = 8.89 + 2.67EI θB
MBC = MFBC + (2θB + θC)
= -26.67 + (2θB + θC) .
MBC = -26.67 +1.5EI θB + 0.75EI θC
MCB = MFCB + (2θC + θB)
= 26.67 + (2θC + θB) .
MCB = 26.67 +1.5EI θC + 0.75EI θB
Apply the condition of Equilibrium
MBA + MBC = 0 and MCB+MCD = 0
At ‘B’
MBA + MBC = 0
8.89 + 2.67EI θB-26.67 +1.5EI θB + 0.75EI θC = 0
4.17EI θB + 0.75EI θC = 17.78 1
Dept. of Civil Engg., SJBIT Page 18 Structural Analysis – II 10CV53
At ‘c’
MCB+MCD = 0
26.67 +1.5EI θC + 0.75EI θB -20= 0
0.75EI θB+ 1.5EI θC = -6.67 2
Solving Eq 1 and 2
θB = θC = . . Substitute the above values in the S-D Equation
MAB = -17.78+1.33EI θB MAB = -10.3426 kN-m
MBA = 8.89 + 2.67EI θB MBA = 23.703 kN-m
MBC = -26.67 +1.5EI θB + 0.75EI θC MBC = -23.703 kN-m
MCB = 26.67 +1.5EI θC + 0.75EI θB MCB = 20kN-m
MCD = -20kN-m
Support Reaction 20kN/m 10.34 40kN 23.74 23.74 20 20
VA VB1 VB2 VC
VA = 22.21kN, VB1 = 17.79kN, VB2 = 40.935kN and VC = 49.065kN
VB = VB1+ VB2 = 58.734kN
Dept. of Civil Engg., SJBIT Page 19 Structural Analysis – II 10CV53 40.94
22.21 22.21 10 10 + + A B C + D
+ _
17.79 17.79
SFD 39.056
26.67kN-m 40kN-m
23.7kN-m + +
_ 10.38kN-m 20kN-m _ _
BMD
Elastic Curve
Dept. of Civil Engg., SJBIT Page 20 SINKINGStructural OF AnalysisSUPPORT – II 10CV53 A
δ
MAB = B
MBA =
B
δ MBA =
A
MAB =
Slope Deflection equations
MAB = MFAB + (2θA + θB - )
δ is +ve when right side support sinks δ is -ve when left side support sinks
1. Analyse the continuous beam by slope deflection method the support Bsinks by 5mm Draw BMD, EC and SFD. Take EI = 2 X 104 kN-mm2
60kN 80kN 20kN/m
A B C D 3m 2m 2m 2m 2m
Fixed End Moment:
MFAB = = = -28.8 kN-m
MFBA = = = 43.2 kN-m
Dept. of Civil Engg., SJBIT Page 21 StructuralMFBC = Analysis= – II= -40 kN-m 10CV53
MFCB = = = 40 kN-m
MFCD = MCD = -20 X 2 X 1 = -40kN-m
Slope Deflection Equation
MAB = MFAB + (2θA + θB – ) 0 ( )
= -28.8 + (2θA + θB - ) ) (θA = 0 due to fixity at support A) ( . ) 4 MAB = -52.8 + 0.8 x 10 θB
MBA = MFBA + (2θA + θB - ) 0
= 43.2 + (2θB + θA - ) ( . ) 4 MBA = 19.2 + 1.6 x 10 θB
MBC = MFBC + (2θB + θC - )
= -40 + (2θB + θC - ) ( . ) 4 4 MBC = -2.5 + 2 x 10 θB + 10 θC
MCB = MFCB + (2θC + θB - )
= 40 + (2θC + θB - ) . ( . ) 4 4 MCB = 77.5 + 2 x 10 θC + 10 θB
Apply the condition of Equilibrium
MBA + MBC = 0 and MCB+MCD = 0
At ‘B’
MBA + MBC = 0
4 4 4 19.2 + 1.6 x 10 θB-2.5 + 2 x 10 θB + 10 θC = 0
4 4 3.6 x 10 θB + 10 θC = -16.7 1
Dept. of Civil Engg., SJBIT Page 22 Structural Analysis – II 10CV53
At ‘c’
MCB+MCD = 0
4 4 77.5 + 2 x 10 θC + 10 θB -40= 0
4 4 10 θB+ 2 x 10 θC = -37.5 2
Solving Eq 1 and 2
-4 -4 θB = 0.66 x 10 θC = -19.1 x 10
Substitute the above values in the S-D Equation
4 MAB = -52.8 + 0.8 x 10 θB MAB = -52.275 kN-m
4 MBA = 19.2 + 1.6 x 10 θB MBA = 20.256 kN-m
4 4 MBC = -2.5 + 2 x 10 θB + 10 θC MBC = -20.256 kN-m 4 4 MCB = 77.5 + 2 x 10 θC + 10 θB MCB = 40kN-m
MCD = -40kN-m
Support Reaction 80kN 20kN/m 52.275 60kN 20.256 20.256 40 40
VA VB1 VB2 VC
VA = 30.4kN, VB1 = 29.6kN, VB2 = 35.064kN and VC = 84.94kN
VB = VB1+ VB2 = 64.664kN
Dept. of Civil Engg., SJBIT Page 23 40 Structural Analysis – II 35.064 35.064 10CV53 30.4 30.4 + + A B C + D
+ _
29.6 29.6
SFD 44.9 44.9
72kN-m 80kN-m
52.275kN-m + + 40kN-m 20.256kN-m
- _ _
BMD
Elastic Curve
Dept. of Civil Engg., SJBIT Page 24 AnalysisStructural of NON Analysis-SWAY Portal– II Frames 10CV53 A frame is a structure having both horizontal and vertical members, such as beams and columns. The joint between any two members is assumed to rotate has a whole when loads are applied (rigid) hence they are called rigid jointed frames.
The frames is which the beams and columns are perpendicular to each other are called orthogonal frames. The moment of the joints in frames in the lateral direction is called Lateral Sway or Sway.
Δ Δ
The frames which do not sway in lateral direction are called non sway portal frames.
Examples:
Dept. of Civil Engg., SJBIT Page 25 Structural1. Analyse Analysis the frame– shownII in the figure and draw BMD 10CV53 40kN/m
B 6m, 2I C
4m I I
A D
Fixed End Moment:
MFAB = 0
MFBA = 0
MFBC = = = -120 kN-m
MFCB = = = 120 kN-m
MFCD = MFDC = 0
Slope Deflection Equation
MAB = MFAB + (2θA + θB) 0
= 0 + (2θA + θB) (θA = 0 due to fixity at support A)
MAB = 0.5EI θB
MBA = MFBA + (2θA + θB) 0
= 0 + (2θB + θA)
MBA = EI θB
MBC = MFBC + (2θB + θC)
= -120 + (2θB + θC)
MBC = -120+1.33EI θB + 0.667EI θC
Dept. of Civil Engg., SJBIT Page 26 StructuralMCB = MFCB Analysis+ (2θC + –θB)II 10CV53
= 120+ (2θC + θB)
MCB = 120+1.33EI θC + 0.667EI θB
MCD = MFCD + (2θC + θD) 0
= 0+ (2θC + θD)
MCD = EI θC
MDC = MFDC + (2θD + θC) 0
= 0+ (2θD + θC)
MDC = 0.5EI θC
Apply the condition of Equilibrium
MBA + MBC = 0 and MCB+MCD = 0
At ‘B’
MBA + MBC = 0
EI θB -120+1.33EI θB + 0.667EI θC = 0
2.33EI θB + 0.667EI θC = 120 1
At ‘c’
MCB+MCD = 0
120+1.33EI θC + 0.667EI θB+EI θC = 0
0.667EI θB+ 2.33EI θC = -120 2
Solving Eq 1 and 2
θB = θC = . .
Dept. of Civil Engg., SJBIT Page 27 SubstituteStructural the above Analysis values in– IIthe S-D Equation 10CV53
MAB = 0.5EI θB MAB = 36.14 kN-m
MBA = EI θB MBA = 72.28kN-m
MBC = -120+1.33EI θB + 0.667EI θC MBC = -71.6 kN-m
MCB = 120+1.33EI θC + 0.667EI θB MCB = 72.3kN-m
MCD = EI θC MCD = -72.26kN-m
MDC = 0.5EI θC MDC = -36.14kN-m 180
+
72 72
- -
72 72
- -
+ + 36.14 36.14 BMD
Dept. of Civil Engg., SJBIT Page 28 Structural2. Draw Analysis BMD and EC– II by slope deflection for the frame shown in figure. 10CV53
10kN/m 30kN 30kN A B C
3m 1m 2m 1m
3m
20kN 2m
D
Fixed End Moment:
MFAB = 0
MFBA = 10 x 3 x 1.5 = 45kN-m
MFBC = = = -22.5 kN-m
− − MFCB = = = 22.5 kN-m
+ + MFBD = = = 9.6kN-m
MFDB = = = -14.4kN-m
Slope Deflection Equation
MBC = MFBC + (2θB + θC) 0
= -22.5 + (2θB + θC)
MBC = -22.5+EI θB
MCB = MFCB + (2θC + θB) 0
= 22.5+ (2θC + θB)
MCB = 22.5+0.5EI θB
Dept. of Civil Engg., SJBIT Page 29 StructuralMBD = MFCD Analysis+ (2θB + –θDII) 10CV53 0
= 9.6+ (2θB + θD)
MBD = 9.6+0.8EI θB
MDB = MFDC + (2θD + θB) 0
= 0+ (2θD + θB)
MDB = -14.4+0.4EI θB
Apply the condition of Equilibrium
At ‘B’
MBA + MBC+ MBD = 0
45-22.5+EI θB+9.6+0.8EI θB = 0
1.8EI θB = -32.1 1
θB = . Substitute the above value in the S-D Equation
MBA = 45kN-m
MBC = -22.5+EI θB MBC = -40.33 kN-m
MCB = 22.5+0.5EI θB MCB = 13.585kN-m
MBD = 9.6+0.8EI θB MBD = -4.67kN-m
MDB = -14.4+0.4EI θB MDB = -21.52kN-m
Dept. of Civil Engg., SJBIT Page 30 Structural Analysis – II UNIT – 3 10CV53 MOMENT DISTRIBUTION METHOD
A MBA B MBC C D
Consider a continuous beam ABC as shown in the figure. The final moment developed at the intermediate support B should follow the condition MBA+ MBC = 0 but the fixed end moments (FEM) at that support will be MFBA+ MFBC = 0. The algebraic sum of the fixed end moments (FEM) thus obtained is the unbalanced moment at B. This unbalanced moment is balanced and redistributed to both MFBA and MFBC depending upon the stiffness of the member. Half of the distributed moment will be carried over to the far end. Hence we will again have unbalanced moments, so that we have to balance and distribute.
This iteration process will continue till we get balanced moments that is MBA+ MBC = 0, similarly at the last simple support MCB = 0. This iterative method of balancing and redistributing of unbalanced moments to obtain the final balanced moments is called moment distribution method. In order to distribute the moments we should calculate stiffness factors and distribution factors. Stiffness Factor (K) :- Stiffness factor depends upon the support condition at the far end.
A B C
If the far end is continuous or fixed K = If the far end is simply support or discontinuous K =
Ex: KAB = , KBA = , KBC = , KCB =
Distribution Factor(Δ):- The unbalanced moments are distributed only at the intermediate supports, hence distribution factors are calculated only at the intermediate supports.
ΔBA = or
ΔBC = or
Dept. of Civil Engg., SJBIT Page 31 Structural Analysis – II 10CV53 1. Analyse continuous beam shown in the figure by moment distribution method. Draw BMD, EC and SFD. 50kN 15kN/m 80kN A B C D 2m 2m 5m 2m 3m
Fixed End Moments:
MFAB = = = -25 kN-m
MFBA = = = 25 kN-m
MFBC = = = -31.25 kN-m
MFCB = = = 31.25 kN-m
MFCD = = = -57.6 kN-m
MFDC = = = 38.4kN-m
Stiffness Factor(K)
KBA = = = EI
KBC = = = 0.8EI
KCB = = = 0.8EI
KCD = = = 0.8EI
Distribution Factor(Δ)
ΔBA = = = 0.56 ΔCB = = = 0.5 . . . .
ΔBC = = = 0.44 ΔCD = = = 0.5 . . . . .
Dept. of Civil Engg., SJBIT Page 32 MomentStructural Distribution Analysis Table – II 10CV53 A B C D Δ 0.56 0.44 0.5 0.5 FEM -25 25 -31.25 31.25 -57.6 38.4 Balance +3.5 +2.75 +13.175 +13.175
Carry over 1.75 6.587 1.375 6.587 Balance -3.688 -2.89 -0.687 -0.687
Carry over -1.84 -0.343 -1.445 -0.343 Balance +0.192 +0.150 +0.722 +0.722
Carry over 0.096 0.361 0.075 0.361 Balance -0.202 -0.158 -0.0375 -0.0375
Carry over -0.10 -0.018 -0.079 -0.018 Balance +0.01 +0.008 +0.04 +0.04
Carry over 0.005 0.02 0.004 0.02 Balance -0.01 -0.008 -0.002 -0.002 Final -25 24.8 -24.8 44.39 -44.38 45 Moment
Support Reaction 15kN/m 25 50kN 24.8 24.8 44.39
VA VB1 VB2 VC1
44.39 80kN 45
VA VB1
VA = 25.05kN, VB1 = 24.95kN, VB2 = 33.6kN, VC1 = 41.42kN, VC2 = 47.878kN VB = VB1+ VB2 = 58.55kN VC = VC1+ VC2 = 89.20kN
Dept. of Civil Engg., SJBIT Page 33 Structural Analysis – 33.6II 47.87 47.87 10CV53
25.05 25.05
+ + +
A B C D - - -
24.95 24.95 32.12 32.12
41.42
SFD 96 50 46.9
45 + 44.39 + + 25 24.8
- _ - -
BMD
2. Analyse continuous beam shown in the figure by moment distribution method. Draw BMD, EC and SFD. 40kN 10kN/m A B C D 3m 2m 4m 5m
Fixed End Moments:
MFAB = = = -19.2 kN-m
MFBA = = = 28.8 kN-m
MFBC = = = -13.33 kN-m
MFCB = = = 13.33 kN-m
MFCD = = = -20.83 kN-m
Dept. of Civil Engg., SJBIT Page 34 StructuralMFDC = =Analysis= 20.83– IIkN-m 10CV53
Stiffness Factor(K)
KBA = = = 0.8EI
KBC = = = EI
KCB = = = EI
KCD = = = 0.6EI Distribution Factor(Δ)
ΔBA = = = 0.44 ΔCB = = = 0.625 . . .
ΔBC = = = 0.56 ΔCD = = = 0.375 . . . Moment Distribution Table A B C D Δ 0.44 0.56 0.625 0.375 FEM -19.2 28.8 -13.33 13.33 -20.83 20.83 Release D -20.83
Carry over -10.41 Initial -19.2 28.8 -13.33 13.33 -31.24 0 Moment
Balance -6.8 -8.66 11.19 6.72
Carry over -3.4 -5.59 -4.33 0 Balance -2.56 -3.14 2.78 1.62
Carry over -1.28 1.39 -1.57 0 Balance -0.616 -0.765 0.98 0.59
Carry over -0.31 0.49 -0.383 0 Balance -0.23 -0.27 +0.24 +0.144
Carry over -0.005 0.02 -0.004 0
Dept. of Civil Engg., SJBIT Page 35 StructuralBalance Analysis – II-0.053 -0.067 +0.084 +0.051 10CV53
Carry over -0.027 0.042 -0.034 0 Balance -0.018 -0.023 -0.002 -0.013 Final -24.34 18.54 -18.61 22.15 -22.15 0 Moment
Support Reaction 15kN/m 24.34 50kN 18.54 18.61 22.15
VA VB1 VB2 VC1 10kN/m
22.15 0
VA VB1
VA = 17.15kN, VB1 = 22.85kN, VB2 = 19.1kN, VC1 = 20.9kN, VC2 = 29.5kN VB = VB1+ VB2 = 41.95kN VC = VC1+ VC2 = 50.40kN 19.1 29.5
17.15 17.15
+ + +
A B C D - - -
22.85 22.85 20.9 20.5 SFD 31.25 48 20
+ 22.15 + + 24.34 18.54
- _ -
BMD
Dept. of Civil Engg., SJBIT Page 36 Structural3. Analyse Analysis continuous –beamII shown in the figure by moment distribution method.10CV53 Draw BMD, EC and SFD. 20kN 60kN/m 20kN A B C D 8m, 2I 2m 2m,I 2m,I
Fixed End Moments:
MFAB = = = -106.67 kN-m
MFBA = = = 106.67 kN-m
MFBC = = = -30 kN-m
MFCB = = = 30 kN-m
MFCD = MCD = -20 x 2 = -40 kN-m
Stiffness Factor(K)
KBA = = = EI
KBC = = = 0.75EI
KCB = = = EI
KCD = 0
Distribution Factor(Δ)
ΔBA = = = 0.57 ΔCB = = = 1
.
ΔBC = = = 0.43 ΔCD = = 0 . .
Dept. of Civil Engg., SJBIT Page 37 MomentStructural Distribution Analysis Table – II 10CV53 A B C D Δ 0.57 0.43 1 0 FEM -106.67 106.67 -30 30 -40 BalanceOH +10
Carry over +5 Initial -106.67 106.67 -25 40 -40 Moment
Balance -46.55 -35.12
Carry over -23.28 0 Final -129.95 60.12 -60.12 40 -40 Moment
Support Reaction 60kN 20kN 129.95 20kN/m 60.12 60.12 40
VA VB1 VB2 VC1 VA = 88.73kN, VB1 = 71.27kN, VB2 = 35.03kN, VC = 106.3kN VB = VB1+ VB2 = 41.95kN 88.73
35.03 35.03 20 20 + + +
A B C D - - 24.94 24.94 71.27 160 SFD 129.95
+
-
60 60.12 + 40 - -
BMD
Dept. of Civil Engg., SJBIT Page 38 SINKINGStructural OF AnalysisSUPPORT – II 10CV53
1. Moments due to sinking
M = If right support sinks down δ is +ve If left support sinks down δ is –ve 2. Moments due to Rotation
Near end moment = For far end moment = Where θ is rotation, clockwise +ve and anticlockwise –ve The above should be added to the FEM
1. Analyse the continuous beam by moment distribution method. Support B yields by 9mm. Take EI = 1 x 1012 N-mm2, draw BMD and EC
60kN 20kN/m 20kN A B C D 2m 1m 6m 1m
Fixed End Moments:
MFAB = - = - = -19.33 kN-m .
MFBA = - = - = 20.67 kN-m .
MFBC = - = - = -58.5 kN-m ( ) ( . )
MFCB = - = - = 61.5 kN-m ( ) ( . )
MFCD = MCD = -20 x 1 = -20kN-m
Stiffness Factor(K)
KBA = = = 1.33EI
KBC = = = 0.5EI
KCB = = = 0.667EI
Dept. of Civil Engg., SJBIT Page 39 StructuralKCD = 0 Analysis – II 10CV53
Distribution Factor(Δ)
ΔBA = = = 0.725 ΔCB = = = 1 . . . . .
ΔBC = = = 0.275 ΔCD = 0 . . . Moment Distribution Table A B C D Δ 0.725 0.275 1 0 FEM -19.33 20.67 -58.5 61.5 -20 BalanceOH -41.5
Carry over -20.75 Initial -19.33 20.67 -79.25 20 -20 Moment
Balance 42.47 16.11
Carry over 21.235 0 Final 1.905 63.14 -63.14 20 -20 Moment
Support Reaction 20kN 20kN 1.905 60kN 63.14 63.14 20
VA VB1 VB2 VC1 VA = -1.8kN, VB1 = 61.8kN, VB2 = 67.19kN, VC = 72.81kN VB = VB1+ VB2 = 128.99kN 67.19
20 20 + + A B C D 1.8 1.8 - - SFD 52.81
61.8 61.8
Dept. of Civil Engg., SJBIT Page 40 Structural Analysis – II 10CV53
90 40 60.12 + 20 - + -
1.905 BMD
Analysis of Non – Sway frames
1. Draw BMD, EC and SFD for frame shown in the figure.
20kN/m 40kN
A B C 4m 2m 3m
2m
2m D
Fixed End Moments:
MFAB = = = -26.67 kN-m
MFBA = = = 26.67 kN-m
MFBC = = = -28.8 kN-m
MFCB = = = 19.2 kN-m
MFBD = = = 10 kN-m
MFDB = = = -10 kN-m
Stiffness Factor(K)
KBA = = = EI
KBD = = = EI
Dept. of Civil Engg., SJBIT Page 41 Structural Analysis – II 10CV53
KBC = = = 0.6EI
Distribution Factor(Δ)
ΔBA = = = 0.385 ΔCB = = = 0.385
. .
ΔBC = = = 0.23 . . Moment Distribution Table
AB BA BC CB BD DB Δ 0.385 0.23 0.385 FEM -26.67 26.67 -28.8 19.2 10 -10 BalanceOH -19.2
Carry over -9.6 Initial -26.67 26.67 -38.4 0 10 -10 Moment
Balance 0.67 0.4 0.67
Carry over 0.335 0 0.335 Final -26.34 27.33 -38 0 10.67 -9.67 Moment
48 48
38 + + 27.33
26.34 _ -
10.67 -
20 +
- 9.67
Dept. of Civil Engg., SJBIT Page 42 Structural Analysis – II UNIT – 4 10CV53 SWAY ANALYSIS
Sway analysis by slope deflection method
1. Analyse the frame shown in the figure by slope deflection method Draw BMD and EC 40kN/m
B 6m, 2I C
4m I 2I 6m
A
D
Fixed End Moment:
MFAB = 0
MFBA = 0
MFBC = = = -120 kN-m
MFCB = = = 120 kN-m
MFCD = MFDC = 0
Slope Deflection Equation
MAB = MFAB + (2θA + θB - ) 0
= 0 + (2θA + θB - ) (θA = 0 due to fixity at support A)
MAB = 0.5EI θB - 0.375EIδ
MBA = MFBA + (2θA + θB - ) 0
= 0 + (2θB + θA - )
MBA = EI θB – 0.375EIδ
Dept. of Civil Engg., SJBIT Page 43 StructuralMBC = MFBC Analysis+ (2θB + –θC-II ) 10CV53 0
= -120 + (2θB + θC- )
MBC = -120+1.33EI θB + 0.667EI θC
MCB = MFCB + (2θC + θB- ) 0
= 120+ (2θC + θB- )
MCB = 120+1.33EI θC + 0.667EI θB
MCD = MFCD + (2θC + θD- ) 0
= 0+ (2θC + θD - )
MCD = 1.33EI θC - 0.33EIδ
MDC = MFDC + (2θD + θC - ) 0
= 0+ (2θD + θC - )
MDC = 0.67EI θC - 0.33EIδ
Apply the condition of Equilibrium
MBA + MBC = 0 and MCB+MCD = 0
At ‘B’
MBA + MBC = 0
EI θB – 0.375EIδ-120+1.33EI θB + 0.667EI θC = 0
2.33EI θB + 0.667EI θC – 0.375EIδ = 120 1
At ‘c’
MCB+MCD = 0
120+1.33EI θC + 0.667EI θB+0.67EI θC - 0.33EIδ = 0
0.667EI θB+ 2.33EI θC - 0.33EIδ= -120 2
Dept. of Civil Engg., SJBIT Page 44 ShearStructural condition Analysis – II 10CV53 B C MBA MCD
4m 6m
HA A HD D MAB MDC
HA = HD =
HA = 0.375EIθB – 0.1875EIδ HD = 0.33EIθC - 0.11EIδ
ΣH = 0
HA + HD = 0
0.375EIθB – 0.1875EIδ + 0.33EIθC - 0.11EIδ = 0
0.375EI θB+ 0.33EI θC - 0.296EIδ= 0 3
Solving Eq 1 , 2 and 3
θB = θC = δ = . . . Substitute the above values in the S-D Equation
MAB = .5EI θB - 0.375EIδ MAB = 27.31 kN-m
MBA = EI θB – 0.375EIδ MBA = 63.67kN-m
MBC = -120+1.33EI θB + 0.667EI θC MBC = -63.65 kN-m
MCB = 120+1.33EI θC + 0.667EI θB MCB = 88.64kN-m
MCD = 1.33EI θC - 0.33EIδ MCD = -88.1kN-m
MDC = 0.67EI θC - 0.33EIδ MDC = -48.29kN-m
Dept. of Civil Engg., SJBIT Page 45 180 Structural Analysis – II 88.64 10CV53 + - 63.65 -
63.67 88.1
+ -
_
27.3
+
48.29
2. Analyse the frame shown in the figure by slope deflection method Draw BMD and EC 80kN
1m 3m, 2I 3m 40kN
3m I I 4m
A
D
Fixed End Moment:
MFAB = = = -7.5 kN-m
MFBA = = = 22.5 kN-m
MFBC = = = -60 kN-m
MFCB = = = 60 kN-m
Dept. of Civil Engg., SJBIT Page 46 Structural Analysis – II 10CV53 MFCD = MFDC = 0
Slope Deflection Equation
MAB = MFAB + (2θA + θB - ) 0
= -7.5+ (2θA + θB - ) (θA = 0 due to fixity at support A)
MAB = -7.5 + 0.5EI θB - 0.375EIδ
MBA = MFBA + (2θA + θB - ) 0
= 22.5 + (2θB + θA - )
MBA = 22.5 + EI θB – 0.375EIδ
MBC = MFBC + (2θB + θC- ) 0
= -60 + (2θB + θC- )
MBC = -60+1.33EI θB + 0.667EI θC
MCB = MFCB + (2θC + θB- ) 0
= 60+ (2θC + θB- )
MCB = 60+1.33EI θC + 0.667EI θB
MCD = MFCD + (2θC + θD- ) 0
= 0+ (2θC + θD - )
MCD = EI θC - 0.375EIδ
MDC = MFDC + (2θD + θC - ) 0
= 0+ (2θD + θC - )
MDC = 0.5EI θC - 0.375EIδ
Dept. of Civil Engg., SJBIT Page 47 Structural Analysis – II 10CV53 Apply the condition of Equilibrium
MBA + MBC = 0 and MCB+MCD = 0
At ‘B’
MBA + MBC = 0
22.5 + EI θB – 0.375EIδ-60+1.33EI θB + 0.667EI θC = 0
2.33EI θB + 0.667EI θC – 0.375EIδ = 37.5 1
At ‘c’
MCB+MCD = 0
60+1.33EI θC + 0.667EI θB+EI θC - 0.375EIδ = 0
0.667EI θB+ 2.33EI θC - 0.375EIδ= -60 2
Shear condition B C MBA MCD
1m
4m 3m
HA A HD D MAB MDC
HA = HD = –( )
HA = 0.375EIθB – 0.1875EIδ-6.25 HD = 0.33EIθC - 0.185EIδ
ΣH = 0
HA + HD+40 = 0
0.375EIθB – 0.1875EIδ-6.25 + 0.33EIθC - 0.185EIδ+40 = 0
0.375EI θB+ 0.375EI θC - 0.375EIδ= -33.75 3
Dept. of Civil Engg., SJBIT Page 48 SolvingStructural Eq 1 , 2Analysisand 3 – II 10CV53
θB = θC = δ = . . Substitute the above values in the S-D Equation
MAB = -7.5+0.5EI θB - 0.375EIδ MAB = -29.07 kN-m
MBA =22.5+ EI θB – 0.375EIδ MBA = 20.61kN-m
MBC = -60+1.33EI θB + 0.667EI θC MBC = -20.62 kN-m
MCB = 60+1.33EI θC + 0.667EI θB MCB = 60.62kN-m
MCD = EI θC - 0.33EIδ MCD = -60.61kN-m
MDC = 0.5EI θC - 0.33EIδ MDC = -50.93kN-m 180 60.62 + - 20.5 -
20.5 60.61
+ -
29.07
+
50.93
Dept. of Civil Engg., SJBIT Page 49 SwayStructural analysis Analysisby Moment –DistributionII method 10CV53
In moment distribution method of sway frames, we have to carry out two analysis to get final moments. They are
1. Non sway analysis 2. Sway analysis
Final Moment = Non sway moment + (C x Sway moment)
1. Analyse the frame by moment distribution method.
80kN B 4m C I δ
6m 2I 1.5I 4m
D
A
Non Sway Analysis
The sway in the frame is prevented by the force δ. As there are no loads on the members, there are no FEM’s developed. Hence the final moments for non sway analysis will be absent. Apply the shear conditions to the vertical members
ΣH = 0,
HA + HD+80-δ= 0, δ = 80kN [HA = HD = 0]
Sway Analysis
Fixed End Moments:
MFAB = MFBA = =
MFCD = MCD = = .
Dept. of Civil Engg., SJBIT Page 50 Structural Analysis – II 10CV53 = =
. Stiffness Factor(K)
KBA = = = 1.33EI
KBC = = = EI
KCB = = = EI
KCD = = = 1.5EI .
Distribution Factor(Δ)
ΔBA = = = 0.57 ΔCB = = = 0.4 . . .
ΔBC = = = 0.43 ΔCD = = = 0.6 . . . Moment Distribution Table A B C D Δ 0.57 0.43 0.4 0.6 FEM -16 -16 0 0 -27 -27 Balance +9.12 +6.88 +10.8 +16.2
Carry over 4.56 5.4 3.44 8.1 Balance -3.08 -2.32 -1.38 -2.06
Carry over -1.54 -0.68 -1.16 -1.03 Balance +0.39 +0.29 0.464 +0.696
Carry over 0.194 0.232 0.146 0.348 Balance -0.13 -0.099 -0.0584 -0.088
Carry over -0.066 -0.029 -0.05 -0.044 Balance -0.0165 -0.0125 -0.02 -0.03 Final -12.77 -9.68 -9.68 12.25 -12.25 -19.58 Moment
Dept. of Civil Engg., SJBIT Page 51 StructuralShear Conditions Analysis – II 10CV53
ΣH = 0, HA + HD + δ’= 0
HA = HD =
δ’ = 11.7kN
Sway correction factor, C = = = 6.84
. Final Moment = Non sway moment + (C x Sway moment)
MAB = 0 + (6.84 X -12.77) = -87.35kN-m
MBA = 0 + (6.84 X -9.68) = -66.21kN-m
MBC = 0 + (6.84 X 9.68) = 66.21kN-m
MCB = 0 + (6.84 X 12.25) = -83.79kN-m
MCD = 0 + (6.84 X -12.25) = -83.79kN-m
MDC = 0 + (6.84 X -19.58) = -133.93kN-m
83.79
+ 66.2 83.77
+ - 66.2
+
- 139.9
87.3
Dept. of Civil Engg., SJBIT Page 52 Structural2. Analyse Analysis the frame by– momentII distribution method. 10CV53
20kN/m B C 8m, 2I δ
4m I I 4m
A D
Non Sway Analysis
Fixed End Moments:
MFAB = MFBA = 0
MFCD = MCD = 0
MFBC = = -106.67kN-m
MFCB = = 106.67kN-m
Stiffness Factor(K)
KBA = = = EI
KBC = = = EI
KCB = = = EI
KCD = = = 0.75EI
Distribution Factor(Δ)
ΔBA = = 0.5 ΔCB = = 0.57
ΔBC = = 0.5 ΔCD = = 0.43
Dept. of Civil Engg., SJBIT Page 53 Structural Analysis – II 10CV53 Moment Distribution Table A B C D Δ 0.5 0.5 0.57 0.43 FEM 0 0 -106.67 106.67 0 0 Balance 53.33 53.33 -60.8 -45.87
Carry over 26.67 -30.4 26.69 0 Balance 15.2 15.2 -15.2 -11.47
Carry over 7.6 -7.6 7.6 0 Balance 3.8 3.8 -4.33 -3.27
Carry over 1.9 -2.17 1.9 0 Balance 1.09 1.09 -1.083 -0.82
Carry over -0.54 -0.54 0.54 0 Balance 0.27 0.27 -0.31 -0.133
Carry over -0.135 -0.15 0.135 0 Balance 0.077 0.077 -0.077 -0.058
Carry over -0.039 -0.03 0.039 0 Balance 0.019 0.019 -0.02 -0.0167 Final 36.85 73.76 -73.76 61.72 -61.72 0 Moment
Apply the shear conditions to the vertical members
ΣH = 0,
HA + HD - δ= 0,
HA = HD =
δ = 12.22kN
Dept. of Civil Engg., SJBIT Page 54 Structural Analysis – II 10CV53 Sway Analysis
Fixed End Moments:
MFAB = MFBA =
MFCD = MFCD =
= = = Moment Distribution Table A B C D Δ 0.5 0.5 0.57 0.43 FEM -10 -10 0 0 -10 -10 Release D 10
Carry over 5 Initial -10 -10 0 0 -5 0 Moment Balance 5 5 2.85 2.15
Carry over 2.5 1.425 2.5 0 Balance -0.72 -0.72 -1.43 -1.1
Carry over -0.36 -0.713 -0.36 0 Balance 0.36 0.36 0.21 0.15
Carry over 0.18 0.11 0.18 0 Balance -0.055 -0.055 -0.103 -0.077
Carry over -0.028 -0.0513 -0.028 0 Balance 0.026 0.026 0.016 0.012 Final -7.71 -5.34 5.34 3.84 -3.84 0 Moment
Shear Conditions
ΣH = 0, HA + HD + δ’= 0
HA = HD =
Dept. of Civil Engg., SJBIT Page 55 Structural Analysis – II 10CV53 δ’ = 4.23kN
Sway correction factor, C = = = 2.89 . . Final Moment = Non sway moment + (C x Sway moment)
MAB = 36.85 + (2.89 X -7.71) = 14.66kN-m
MBA = 73.76 + (2.89 X -5.34) = 58.15kN-m
MBC = -73.76 + (2.89 X 5.34) = -58.14kN-m
MCB = 61.70 + (2.89 X 3.84) = 72.84kN-m
MCD = -61.72 + (2.89 X -3.84) = -72.82kN-m
MDC = 0 + (2.89 X 0) = 0 160
+ -
- 72.8 58.14
- -
_ 14.6
BMD
Dept. of Civil Engg., SJBIT Page 56 Structural Analysis – II UNIT – 5 10CV53 KANI’S METHOD
It is a method based on the rotation contributions at each joints. It is one of the oldest and most convenient methods for the analysis of multi storied frames. This method is also an iterative method like moment distribution method.
Procedure Calculation of FEM of each span Calculation of stiffness factors and rotations factors rotation factor U = -1/2( ) Computation of rotation contribution M’ = U [ΣFEM + Σ far end contribution] The rotation contributions are calculated at intermediate joints by kani’s table. The iteration process is carried out till we get the values almost same as previous cycle. Calculation of final support moments
MAB = MFAB + 2 M’AB + M’BA & MBA = MFBA + 2 M’BA + M’AB
1. Draw BMD and EC by kani’s method
30kN/m 180kN 20kN/m A B C D 8m, 1.5I 4m 2I 8m 6m, I
Fixed End Moments:
MFAB = = = -160 kN-m
MFBA = = = 160 kN-m
MFBC = = = -320 kN-m
MFCB = = = 160 kN-m
MFBD = = = -60 kN-m
MFDB = = = 60 kN-m
Dept. of Civil Engg., SJBIT Page 57 StructuralStiffness Factor(K) Analysis – II 10CV53
KBA = = = 0.75EI .
KBC = = = 0.67EI
KCB = = = 0.67EI
KCD = = = 0.67EI
Distribution Factor(Δ)
UBA = -1/2 = -1/2[ = -0.265 . [ ] . . ] UBC = -1/2 = -1/2 = -0.235 . [ ] [ . . ] UCB =-1/2 =-1/2 = -0.25 . [ ] [ . . ] UCD =-1/2 =-1/2 = -0.25 . [ ] [ . . ]
B C
A -0.265 -0.235 -0.25 -0.25 D -160 160 -160 -320 160 100 -60 60
M’BA M’BC M’CB M’CD 42.4 37.6 -34.4 -34.4 51.52 45.684 -36.42 -36.42 52.05 46.16 -36.54 -36.54 52.08 46.19 -36.55 -36.55 52.08 46.19 -36.55 -36.55
M’BA = 52.08 M’BC = 46.19 M’CB = -36.55 M’CD = -36.55
Final Moments
MAB = MFAB + 2 M’AB + M’BA
MAB = -160 + 2 X (0) + 52.08 = -107.92kN-m
MBA = 160 + 2 X (52.08) + 0 = 264.16kN-m
Dept. of Civil Engg., SJBIT Page 58 Structural Analysis – II 10CV53 MBC = -320 + 2 X (46.19) – 36.42 = -264.17kN-m
MCB = 160 + 2 X (-36.42) + 46.19 = 133.1kN-m
MCD = -60 + 2 X (-36.55) – 0 = -133.1kN-m
MDC = 60 + 2 X (0) – 36.55 = 23.45kN-m 480
264 240 + 133.1 107 - 90 + - - -
A B C D
2. Draw BMD and EC by kani’s method
30kN/m 180kN 20kN/m A B C D 8m, 1.5I 4m 2I 8m 6m, I
Fixed End Moments:
MFAB = = = -160 kN-m
MFBA = = = 160 kN-m
MFBC = = = -320 kN-m
MFCB = = = 160 kN-m
MFCD = = = -60 kN-m MFDC = = = 60 kN-m
= -30kN-m CO = -60kN-m
MFCD = -90kN-m MFDC = 0kN-m
Dept. of Civil Engg., SJBIT Page 59 StructuralStiffness Factor(K) Analysis – II 10CV53
KBA = = = 0.75EI .
KBC = = = 0.67EI
KCB = = = 0.67EI
KCD = = = 0.5EI
Distribution Factor(Δ)
UBA = -1/2 = -1/2[ = -0.265 . [ ] . . ] UBC = -1/2 = -1/2 = -0.235 . [ ] [ . . ] UCB =-1/2 =-1/2 = -0.286 . [ ] [ . . ] UCD =-1/2 =-1/2 = -0.214 . [ ] [ . . ]
B C
A -0.265 -0.235 -0.286 -0.214 D -160 160 -160 -320 160 70 -90 0
M’BA M’BC M’CB M’CD 42.4 37.6 -30.77 -23.026 50.55 44.83 -32.84 -24.66 51.1 45.32 -32.98 -24.68 51.1 45.35 -32.99 -24.68
M’BA = 51.1 M’BC = 45.35 M’CB = -32.99 M’CD = -24.68
Final Moments
MAB = MFAB + 2 M’AB + M’BA
MAB = -160 + 2 X (0) + 51.1 = -108.9kN-m
MBA = 160 + 2 X (51.1) + 0 = 262.28kN-m
MBC = -320 + 2 X (45.35) – 32.99 = -262.29kN-m
Dept. of Civil Engg., SJBIT Page 60 Structural Analysis – II 10CV53 MCB = 160 + 2 X (-32.99) + 45.35 = 139.36kN-m
MCD = -90 + 2 X (-24.68) – 0 = -139.36kN-m
MDC = 0 480
262.28 240 + 139.36 108.9 - 90 + - -
A B C D
3. Draw BMD and EC by kani’s method
30kN/m 180kN 20kN/m A B C D 8m, 1.5I 4m 2I 8m 6m, I
Fixed End Moments:
MFAB = = = -160 kN-m
MFBA = = = 160 kN-m
MFBC = = = -320 kN-m
MFCB = = = 160 kN-m
MFCD = MCD = -20 x 6 x 3 = -360 kN-m
MFBC = = = -320 kN-m MFCB = = = 160 kN-m
= 100kN-m CO = 200kN-m
MFCD = -220kN-m MFCB = 360kN-m
Dept. of Civil Engg., SJBIT Page 61 StructuralStiffness Factor(K) Analysis – II 10CV53
KBA = = = 0.75EI .
KBC = = = 0.5EI
KCB = = = 0.67EI
Distribution Factor(Δ)
UBA = -1/2 = -1/2[ = -0.3 . [ ] . . ] UBC = -1/2 = -1/2 = -0.2 . [ ] [ . . ] UCB =-1/2 =-1/2 = -0.5 . [ ] [ . ]
B C
A -0.3 -0.2 -0.5 0 D -160 160 -60 -220 360 0 -360 0
M’BA M’BC 18 12
M’BA = 18 M’BC = 12
Final Moments
MAB = MFAB + 2 M’AB + M’BA
MAB = -160 + 2 X (0) + 18 = -142kN-m
MBA = 160 + 2 X (18) + 0 = 196kN-m
MBC = -220 + 2 X (12) + 0 = -196kN-m
MCB = 360kN-m
MCD = -360kN-m
Dept. of Civil Engg., SJBIT Page 62 480 Structural Analysis – II 360 10CV53 + 196 240 + - 142 -
-
A B C D
Sinking of Support
1. Analyse the continuous beam by Kani’s method, support B sinks by 10mm. Support A rotates by 0.002rad. Draw the BMD and EC. Given E = 200 x 106 kPa and I = 100 x 106 mm4
40kN 10kN 20kN/m A B C D 2m 2m 6m 2m
Fixed End Moments:
MFAB = + - = + - = -55 kN-m . .
MFBA = + - = + - = -35 kN-m . .
MFBC = - = + = 3.33 kN-m ( ) .
MFCB = - = + = 63.33 kN-m ( ) .
MFCD = MCD = -20 x 2 x 1 = -40 kN-m
MFBC = = = 3.33 kN-m MFCB = = = 63.33 kN-m
= -6.665kN-m CO = -13.33kN-m
MFCD = -3.335kN-m MFCB = 40kN-m
Stiffness Factor(K)
KBA = = = EI
KBC = = = 0.5EI
Dept. of Civil Engg., SJBIT Page 63 StructuralKCB = = Analysis= 0.67EI– II 10CV53
Distribution Factor(Δ)
UBA = -1/2 = -1/2[ = -0.33
[ ] . ] UBC = -1/2 = -1/2 = -0.17 . [ ] [ . ] UCB =-1/2 =-1/2 = -0.5 . [ ] [ . ] B C
A -0.33 -0.17 -0.5 0 D -55 -35 -38.35 -3.35 40 0 -40 0
M’BA M’BC 12.66 6.4
M’BA = 12.66 M’BC = 6.4
Final Moments
MAB = MFAB + 2 M’AB + M’BA
MAB = -55 + 2 X (0) + 12.66 = -42.34kN-m
MBA = -35 + 2 X (12.66) + 0 = -9.68kN-m
MBC = -3.33 + 2 X (6.4) + 0 = 9.47kN-m
MCB = 40kN-m
MCD = -40kN-m 45 40 +
40 + - 42.34
-
A B C D 9.68
Dept. of Civil Engg., SJBIT Page 64 AnalysisStructural of Non Analysis-Sway Frames– II 10CV53
1. Draw the BMD and EC by Kani’s method
30kN 30kN 10kN/m
A B C 1m 2m 1m 4m
3m
20kN 2m D
Fixed End Moments:
MFAB = - = - = -22.5 kN-m
MFBA = - = + = 22.5 kN-m
MFBD = = = 9.6 kN-m
MFDB = = = -14.4 kN-m
MFBC = MBC = -10 x 4 x 2 = -80 kN-m
Stiffness Factor(K)
KBA = = = 0.75EI .
KBC =0
KBD = = = 0.8EI
Distribution Factor(Δ)
UBA = -1/2 = -1/2[ = -0.278 UBC = 0
[ ] . ] UCB =-1/2 =-1/2 = -0.22 . [ ] [ . ]
Dept. of Civil Engg., SJBIT Page 65 Structural Analysis – II 10CV53
B
A -0.278 0 C -22.5 22.5 -47.5 -80
M’BA -0.22 M’BC 13.32 9.6 M’BD 10.54
14.4 D
M’BA = 13.32 M’BC = 10.54
Final Moments
MAB = MFAB + 2 M’AB + M’BA
MAB = -22.5 + 2 X (0) + 13.32 = -9.18kN-m
MBA = 22.5 + 2 X (13.32) + 0 = 49.14kN-m
MBC = -80
MBD = 9.6 + 2 X (10.54) + 0 = 30.68kN-m
MDB = -14.4 + 2 X (0) + 10.54 = -3.86kN-m
Dept. of Civil Engg., SJBIT Page 66 Structural Analysis – II UNIT-6 10CV53 FLEXIBILITY MATRIX METHOD
The systematic development of consistent deformation method in the matrix form has lead to flexibility matrix method. The method is also called force method. Since the basic unknowns are the redundant forces in the structure. This method is exactly opposite to stiffness matrix method. The flexibility matrix equation is given by
[P] [F] = {[Δ] – [ΔL]}
-1 [P] = [F] {[Δ] – [ΔL]}
Where, [P] = Redundant in matrix form [F] = Flexibility matrix [Δ] = Displacement at supports [ΔL]= Displacement due to load
1. Analyse the continuous beam shown in the figure by flexibility matrix method, draw BMD
60kN/m 100kN A B C 4m 1.5m 1.5m
Static Indeterminacy SI = 2 (MA and MB) MA and MB are the redundant Let us remove the redundant to get primary determinate structure
60kN/m 100kN
A B B C 4m 1.5m 1.5m
120/EI 75/EI
A’ B’ B’ C’
Dept. of Civil Engg., SJBIT Page 67 Structural Analysis – II 10CV53 Δ1L [ΔL] = Δ2L
Δ1L = Rotation at A = SF at A’
Δ1L = ½ [2/3 X 4 X ]
Δ1L =
Δ2L = Rotation at A = SF at B’
= VB1’ + VB2’
Δ2L = ½ [2/3 X 4 X ] + ½ [1/2 X 3 X ]
Δ2L = . 160
[ΔL] = 216.25
Note: The rotation due to sagging is taken as positive. The moments producing due to sagging are also taken as positive.
To get Flexibility Matrix Apply unit moment to joint A
1kN-m A B B C
δ11 δ12 [F] =
δ21 δ22
Dept. of Civil Engg., SJBIT Page 68 Structural Analysis – II 10CV53
δ11 = = = .
δ21 = = = . Apply unit moment to joint A
1kN-m 1kN-m A B B C
δ12 = = = .
δ22 = + = + = .
δ11 δ12 1.33 0.67 [F] = =
δ21 δ22 0.67 1.33
Apply the flexibility equation
-1 [P] = [F] {[Δ] – [ΔL]}
0 [Δ] = 0
-1 1.33 0.67 0 160 [P] = EI - 0.67 1.33 0 216.25
MAB -86.00 [P] = = kN-m
MBA -68.08
Dept. of Civil Engg., SJBIT Page 69 Structural Analysis120 – II 10CV53
86 + 75 68 - - +
BMD
2. Analyse the continuous beam shown in the figure by flexibility matrix method, draw BMD
40kN/m 120kN 20kN/m A B C D 12m 4m 8m 12m
Static Indeterminacy SI = 2 (MB and MC) MB and MC are the redundant Let us remove the redundant to get primary determinate structure
40kN/m 120kN
A B B C 12m 4m 8m
720/EI 320/EI
A’ B’ B’ C’
Dept. of Civil Engg., SJBIT Page 70 Structural Analysis – II 10CV53 20kN/m
C D 12m
360/EI
C’ D’
Δ1L [ΔL] = Δ2L
Δ1L = Rotation at B = SF at B’
= VB1’ + VB2’
Δ1L = .
Δ2L = Rotation at C = SF at C’
= VC1’ + VC2’
Δ2L = . 3946.67
[ΔL] = 2293.33
Dept. of Civil Engg., SJBIT Page 71 Structural Analysis – II 10CV53 To get Flexibility Matrix Apply unit moment to joint A
1kN-m 1kN-m A B B C
δ11 δ12 [F] =
δ21 δ22
δ11 = + = + =
δ21 = = =
Apply unit moment to joint A
1kN-m 1kN-m B C C D
δ12 = = =
δ22 = + = + =
δ11 δ12 8 2 [F] = =
δ21 δ22 2 8
Apply the flexibility equation
-1 [P] = [F] {[Δ] – [ΔL]}
0 [Δ] = 0
Dept. of Civil Engg., SJBIT Page 72 Structural Analysis – II 10CV53 -1 8 2 0 3946 [P] = EI - 2 8 0 2293
MAB -449.97 [P] = = kN-m
MBA -174.22
120
86 + 75 68 - - +
BMD
SINKING OF SUPPORT
1. Analyse the continuous beam by flexibility method, support B sinks by 5mm. Sketch the BMD and EC given EI = 15 X 103 kN-m2
30kN/m 120kN A B C 6m, 2I 2m I 2m
NOTE: In this case of example with sinking of supports, the redundant should be selected as the vertical reaction.
Static indeterminacy is equal to 2. Let VB and VC be the redundant, remove the redundant to get the primary structure.
A’ B’ C’
120/EI
240/EI 480/EI
Dept. of Civil Engg., SJBIT Page 73 StructuralA’ Analysis – II B’ C’ 10CV53
270/EI
Δ1L [ΔL] = Δ2L
Δ1L = Displacement at B in primary determinate structure = BM at B’ in conjugate beam
Δ1L = [ X 6 X X (2/3 X 6) ] + (6 X X 6/2) + [ X 6 X X (3/4 X 6) ]
Δ1L =
Δ2L = Displacement at C in primary determinate structure = BM at C’ in conjugate beam
Δ2L = [ X 6 X X (2/3 X 6 + 4) ] + (6 X X 6/2 + 4) + [ X 6 X X (3/4 X 6 +4) ]
Δ2L =
8910
[ΔL] = 19070
To get Flexibility Matrix Apply unit Load at B
A B C 6m 4m 1kN 3/EI
A’ B’ C’
Dept. of Civil Engg., SJBIT Page 74 Structural Analysis – II 10CV53 δ11 δ12 [F] =
δ21 δ22
δ11 = - X 6 X X (2/3 X 6) =
δ21 = - X 6 X X (2/3 X 6 + 4) =
Apply unit load at C
A B C 6m 4m 1kN 5/EI 4/EI
2/EI
A’ B’ C’
δ12 = - X 6 X X (2/3 X 6) – [6 X X (6/2)] =
δ22 = - X 6 X X (2/3 X 6 + 4) – [6 X X (6/2 + 4)] - X 4 X X (2/3 X 4) = .
δ11 δ12 -36 -72 [F] = =
δ21 δ22 -72 -177.33
Apply the flexibility equation
-1 [P] = [F] {[Δ] – [ΔL]}
0.005 [Δ] = 0
Dept. of Civil Engg., SJBIT Page 75 Structural Analysis – II 10CV53 -1 -36 -72 0.005 8910 [P] = EI - -72 -177.33 0 19070
VB 161.43 [P] = = kN-m
VC 41.98
Support Reaction 120kN
MA 30kN/m MB MB 0
VA VB1 VB2 VC
VA = 96.64kN, VB1 = 83.36kN, VB2 = 78.07kN, VC = 41.98kN
VB = VB1+ VB2 = 161.43kN
MA 112.48 = kN-m
MB 72.28
135
112.48 + 120 72.28 - - +
BMD
Dept. of Civil Engg., SJBIT Page 76 Structural Analysis – II UNIT-7 10CV53 STIFFNESS MATRIX METHOD
The systematic development of slope deflection method in the matrix form has lead to Stiffness matrix method. The method is also called Displacement method. Since the basic unknowns are the displacement at the joint.
The stiffness matrix equation is given by
[Δ] [K] = {[P] – [PL]}
-1 [Δ] = [K] {[P] – [PL]}
Where, [P] = Redundant in matrix form [F] = Stiffness matrix [P] = Final force at the joints in matrix form
[PL]= force at the joints due to applied load in matrix form
1. Analyse the continuous beam by Stiffness method Sketch the BMD
20kN/m 120kN A B C 6m 3m 3m
Kinematic Indeterminacy KI = 1 (θB)
20kN/m 120kN A B C 6m 3m 3m
[PL] = MFBA + MFBC
= + ( ) = -
− [PL] = -30kN-m Apply unit displacement at joint B.
A B C
[K] = + = + = 1.33EI (θ=1)
Dept. of Civil Engg., SJBIT Page 77 ByStructural condition of Analysisequilibrium –atII joint B 10CV53
[P] = 0
-1 [Δ] = [K] {[P] – [PL]}
= {[P] – [PL]}
θB = {[0] – [-30]} = . . Slope deflection equation
MAB = MFAB + (2θA + θB) 0
= -60 + (2θA + ) (θA = 0 due to fixity at support A) .
MAB = -52.5kN-m
MBA = MFBA + (2θB + θA) 0
= 60 + (2 X + θA) .
MBA = 75.04kN-m
MBC = MFBC + (2θB + θC)
= -90 + (2 X + 0) .
MBC = -75kN-m
MCB = MFCB + (2θC + θB)
= 90 + (0 + ) .
MCB = 97.52kN-m 120 90 + 97.5
+ 75 52.5 _ - -
BMD
Dept. of Civil Engg., SJBIT Page 78 Structural2. Analyse Analysis the continuous– II beam by Stiffness method Sketch the BMD 10CV53
60kN/m 100kN A B C 4m 1.5m 1.5m
Kinematic Indeterminacy KI = 2 (θB & θC)
60kN/m 100kN A B C 4m 1.5m 1.5m
[P1L] = MFBA + MFBC
= + ( ) = - = 42.5kN-m
− [P2L] = MFCB = = = 37.5kN-m
P1L 42.5 [PL] = = kN-m P2L 37.5
Apply unit displacement at joint B.
A B C
K11 = + = + = 2.33EI (θ=1)
K21 = = = 0.67EI
Apply unit displacement at joint B.
A B C
K12 = = = 0.67EI
K22 = = = 1.33EI
Dept. of Civil Engg., SJBIT Page 79 ByStructural condition of Analysis equilibrium –atII joint B 10CV53 [P] = 0 -1 [Δ] = [K] {[P] – [PL]} -1 2.33 0.67 0 42.5 [Δ] = - 0.67 1.33 0 37.5
θB -11.88 =
θC -22.19
Slope deflection equation
MAB = MFAB + (2θA + θB) 0
= -80 + (2θA - ) (θA = 0 due to fixity at support A) .
MAB = -85.94kN-m
MBA = MFBA + (2θB + θA) 0
= 80 + (2 X + θA) .
MBA = 68.12kN-m
MBC = MFBC + (2θB + θC)
= -37.5 + (2 X + ) . .
MBC = -68.6kN-m
MCB = 0
120 75 + 68.6 86 - _ +
BMD
Dept. of Civil Engg., SJBIT Page 80 SinkingStructural of support Analysis – II 10CV53
1. Analyse the continuous beam shown in figure by stiffness method. Support B sinks by 300/EI units and support C sinks by 200/EI units
100kN 60kN A B C 4m 4m 4m 4m
Kinematic Indeterminacy KI = 2 (θB & θC)
100kN 60kN A B C 8m 4m 4m
[P1L] = MFBA + MFBC - -
= - – + = 21.25kN-m
[P2L] = MFCB - = + = 69.38kN-m
P1L 21.25 [PL] = = kN-m P2L 69.38
Apply unit displacement at joint B.
A B C
K11 = + = + = EI (θ=1)
K21 = = = 0.25EI Apply unit displacement at joint B.
A B C
K12 = = = 0.25EI
Dept. of Civil Engg., SJBIT Page 81 Structural KAnalysis22 = =– II= 0.50EI 10CV53 By condition of equilibrium at joint B [P] = 0 -1 [Δ] = [K] {[P] – [PL]} -1 1 0.25 0 21.25 [Δ] = - 0.25 0.50 0 69.38
θB 15.36 =
θC -146.44
Slope deflection equation
MAB = MFAB + (2θA + θB – ) 0 ( )
= -100 + (2θA + - ) (θA = 0 due to fixity at support A) . ( / )
MAB = -124.29kN-m
MBA = MFBA + (2θA + θB - ) 0
= 100 + (θA + - ) . ( / ) 2 MBA = 79.55kN-m
MBC = MFBC + (2θB + θC - )
= -60 + ( + - ) . . ( / ) 2 MBC = -79.55kN-m
MCB = 0
200 120 + 79.5 124.3 - _ +
BMD
Dept. of Civil Engg., SJBIT Page 82 AnalysisStructural of frames Analysis – II 10CV53 1. Analyse the frame by stiffness method
100kN 30kN /m A B C 2m 2I 3m I, 3m
3m I
D
Kinematic Indeterminacy KI = 2 (θB & θC)
100kN 30kN /m A B C 2m 2I 3m I, 3m
3m I
[P1L] = MFBA - MFBC + MFCD
= - = 25.5kN-m
[P2L] = MFCB = = 22.5kN-m
P1L 25.5 [PL] = = kN-m P2L 22.5
Dept. of Civil Engg., SJBIT Page 83 Structural Analysis – II 10CV53 Apply unit displacement at joint B.
A B C
K11 = + + = + + = 4.267EI (θ=1)
K21 = = = 0.67EI Apply unit displacement at joint C.
A B C
K12 = = = 0.67EI
K22 = = = 1.33EI By condition of equilibrium at joint B [P] = 0
-1 [Δ] = [K] {[P] – [PL]}
Dept. of Civil Engg., SJBIT Page 84 Structural Analysis – II -1 10CV53 4.267 0.67 0 25.5 [Δ] = - 0.67 1.33 0 22.5
θB -3.604 =
θC -15.01
Slope deflection equation
MAB = MFAB + (2θA + θB) 0
= -72 + (2θA + ) (θA = 0 due to fixity at support A) .
MAB = -74.88kN-m
MBA = MFBA + (2θA + θB) 0
= 72 + (θA + ) . 2 MBA = 42.23kN-m
MBC = MFBC + (2θB + θC)
= -22.5 + ( + ) . . 2 MBC = -37.37kN-m
MBD = MFBD + (2θB + θD)
= 0 + ( + 0) . 2 MBD = -4.81kN-m
MDB = MFDB + (2θD + θB)
= 0 + ( + ) . 2 0 MDB = -2.402kN-m
MCB = 0
Dept. of Civil Engg., SJBIT Page 85 Structural Analysis – II UNIT – 8 10CV53 STRUCTURAL DYNAMICS Terminology:- 1. Periodic motion: - periodic motion is a motion that repeats itself at equal interval of time.
Vibration Parameter
Time
2. Time Periodic: - The time required for motion of one cycle is called time period.
Vibration Parameter
Time 3. Cycle : - The motion during one time period is called cycle.
Vibration Parameter
Time 4. Frequency: - Frequency is a no of cycle per unit time.
Vibration Parameter
Time 5. Amplitude (A):- peak displacement from the mean position is called amplitude.
A Vibration Parameter
Time
Dept. of Civil Engg., SJBIT Page 86 Structural6. Free Vibration Analysis: - It –is IIthe vibration under no external vibration but occurs 10CV53with the initial displacement.
7. Natural frequency (fn): - It is the frequency of a body under free vibration.
Circular Natural frequency (wn): - wn = 2π fn 8. Damping: - It is the interval resistance to the motion of a body 9. Degree of freedom: - It is the no of independent co-ordinate required to determine (define) the displacement configuration at any time is called the degree of freedom. x y
X (1DOF)
Y (1DOF)
Simple harmonic Motion: -
Displacement
Mean Position
Is a periodic reciprocating motion is accelerate always directed towards mean position and is proportion to the displacement from mean position.
Vector representation of SHM:- Y
P X X
Y
Dept. of Civil Engg., SJBIT Page 87 StructuralLet us consider Analysis a displacement– II OP as y = Asinwnt 10CV531 Where, y = displacement vector A = Amplitude
wn = natural circular frequency Differentiate 1 w.r.t ‘t’ Y1 = A wn sinwnt = A wn [ sin(wnt + π/2)] 2
Differentiate 2 w.r.t ‘t’ 2 Y2 = -A wn sinwnt 2 = A wn [ sin(wnt + π)] 3
Characteristic 1. Velocity vector leads the displacement vector by π/2 2. Displacement, velocity and acceleration vector will rotate in counter clockwise direction with same frequency
Vibration analysis 1. Mathematical modelling 2. Formulation of governing equation of motion 3. Solution of equations of motion 4. Analysis and interpretation of results
Model :- A Mathematical representation of a physical system/a physical process
Mass
Spring Dashpot
Physical system Model
Dept. of Civil Engg., SJBIT Page 88 DStructuralifferential equation Analysis of motion– II by Simple harmonic equation 10CV53
Approach:- consider a single degree of freedom spring mass system as shown in figure which oscillates about the mean position as shown in the figure.
K
Mass
Let the y(t) = Asinwnt 1
The velocity is
Y1(t) = Awncoswnt = Awnsin(wnt + π/2) 2 The velocity is
2 Y2(t) = -Awn sinwnt 2 = Awn sin(wnt + π) 3
2 Y2(t) = -wn y(t)
2 Hence Y2(t) = -wn y(t)
Acceleration is proportional to the displacement
2 Y2(t) + wn y(t) = 0 But 2 wn =
Y2(t) + y(t) = 0
Y2(t) m + k y(t) = 0
m y2(t) = inertia force k y(t) = Spring force
Dept. of Civil Engg., SJBIT Page 89 Structural1. Determine Analysis the natural– frequencyII and time period of the system as shown in10CV53 fig. Take E = 2.1 x 105 N/mm2 and I = 13 x 106 mm4
4m
K = 1200kN/mm
500kg
K = 120kN/mm
500kg
To find the natural frequency
f =
T =
= 2π
m = = =
= 0.05096.
T = 2π
. T = 0.004 sec
Dept. of Civil Engg., SJBIT Page 90