STRUCTURAL ANALYSIS VBB 2053
LECTURE 2
SLOPE-DEFLECTION METHOD 2005 Pearson Education South Asia Pte Ltd
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Learning Objectives
Learn to analyse statically indeterminate structures using slope-deflection method. Knowing the slope-deflection equations. Discuss the application of slope-deflection method to beams and frames.
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Outline
Introduction to slope-deflection method. General procedure of slope-deflection method of analysis. Derivation of slope-deflection equations. Work examples on slope-deflection method of analysis: beams and frames.
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Statically Indeterminate Structures: Method of Analysis
Analyzing any indeterminate structure, it is necessary to satisfy:
Equilibrium – Equilibrium is satisfied when the reactive forces hold the structure at rest. Compatibility – Compatibility is satisfied when the various segments of the structure fit together without intentional breaks or overlaps. Force-displacement – This requirements depend upon the way the material responds – linear elastic response.
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Statically Indeterminate Structures: Method of Analysis
There are two ways to analyze a statically indeterminate structure
Force Method Displacement Method
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Force Method of Analysis
Identifying unknown redundant forces. Satisfying structure’s compatibility equations. This is done by expressing the displacements in terms of the loads by using the loaddisplacement relations. The solution of the resultant equations yields the redundant reactions, and then the equilibrium equations are used to determine the remaining reactions on the structure.
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Displacement Method of Analysis
Satisfying equilibrium equations for the structure. Unknown displacements are written in terms of the loads by using the load-displacement relations, then solved for the displacements. Once the displacement obtained, the unknown loads are determined from the compatibility equations using the loaddisplacement relations.
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Degree of Freedom
When a structure is loaded, the node will undergo unknown displacements. These displacements are referred to as the “degree of freedom”. Specify degree of freedom is a necessary 1st step when apply displacement method. In 3D, each node on a beam or frame can have at most 3 linear displacements & 3 rotational displacements. In 2D, each node can have at most 2 linear displacements & 1 rotational displacement.
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Degree of Freedom
The number of these unknowns are referred to as the degree in which the structure is kinematically indeterminate. Any load applied to the beam will cause node A to rotate. Node B is completely restricted from moving. Hence, the beam has only one unknown degree of freedom. Therefore kinematically indeterminate to 1st degree.
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Degree of Freedom The beam has nodes at A, B & C There are 4 degrees of freedom A, B, C and C. It is kinematically indeterminate to 4th degree.
The frame has 3 degrees of freedom, B, C and B. It is kinematically indeterminate to 3rd degree.
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Slope-Deflection Method
The slope-deflection method uses displacements as unknowns and is referred to as a displacement method. In the slope-deflection method, the moments at the ends of the members are expressed in terms of displacements and end rotations of these ends. An important characteristic of the slope-deflection method is that it does not become increasingly complicated to apply as the number of unknowns in the problem increases. In the slope-deflection method the individual equations are relatively easy to construct regardless of the number of unknowns.
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Derivation of Slope-Deflection Equations
To derive the general form of the slope-deflection equation, let us consider the typical span AB of the continuous beam shown below when subjected to arbitrary loading.
The slope-deflection equation can be obtained using the principle of superposition By considering separately the moments developed at each support due to θA, θB, Δ, P. Assume clockwise is positive.
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Derivation of Slope-Deflection Equations
Moment due to angular displacement at A, θA
To determine the moment MAB needed to cause this displacement, we will use the conjugate beam method. MB' 0
MA ' 0
1 MAB L 1 MBA 2 EI L 3 2 EI
2L 0 L 3
1 MBA 2 EI
L 1 MAB 2L AL 0 L L 3 2 EI 3
From which we obtain the following relationships.
MAB
4EI A L
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MBA
2EI A L 13
Derivation of Slope-Deflection Equations
Moment due to angular displacement at B, θB
The applied moment MBA to the angular displacement B & the reaction moment MAB at the wall can be related as:
MBA
4EI B L
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MAB
2EI B L
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Derivation of Slope-Deflection Equations
Moment due to relative linear displacement, Δ
If the far node B of the member is displaced relative to A, so that the cord of the member rotates clockwise & yet both ends do not rotate, then equal but opposite moment and shear reactions are developed in the member. Moment M can be related to the displacement using conjugate beam method.
MB' 0 2L 1 M L 1 M L 0 L 2 EI 3 2 EI 3 6EI MAB MBA M 2 L 2005 Pearson Education South Asia Pte Ltd
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Derivation of Slope-Deflection Equations
Moment due to loading (P or w)
These moments are called Fixed End Moment (FEM). In general, linear and angular displacement of the nodes are caused by loadings acting on the span of the member. To develop the slope-deflection equation, we must transform these span loadings into equivalent moment acting at the nodes and then use the load-displacement relationships just derived.
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Derivation of Slope-Deflection Equations
Slope-Deflection Equation
The end moments due to each displacement and loadings are added together, the resultant moments at the ends can be written as: MAB = MθA + MθB + MΔ + Mload
I Δ MAB 2E 2θA θB 3 FEMAB 0 L L I Δ MBA 2E 2θB θA 3 FEMBA 0 L L 2005 Pearson Education South Asia Pte Ltd
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Derivation of Slope-Deflection Equations
Slope-Deflection Equation
The slope-deflection equations can be generalized as follows:
MN 2Ek2θN θF 3ψ FEMN
MN = internal moment in the near end of the span, E = modulus of elasticity k = span stiffness, k = I/L N, F = near- and far-end slopes or angular displacements of the span at the supports = span rotation of its cord due to a linear displacement, = /L FEMN = fixed end moment at the near end support.
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Slope-Deflection Equation
Pin supported end span
Sometimes an end span of a beam or frame is supported by a pin or roller at its far end.
The moment at the roller or pin is zero provided the angular displacement at this support does not have to be determined. Use the generalised slope-deflection equation, we have:
MN 2Ek2θN θF 3ψ FEMN 0 2Ek2θN θF 3ψ 0
Since far end is pinned, FEMF = 0 FEMN can be obtained
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Slope-Deflection Equation
Simplifying, we get:
MN 3EkθN ψ FEMN
Only applicable for end span with far end pinned or roller supported
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Work Example 1
Draw the shear & moment diagrams for the beam shown below.
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Work Example 2
Draw the shear & moment diagrams for the beam shown below. 40 kN/m A
60 kN 1m
1m
C
6m B
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Analysis of Frames: No Sidesway
Slope-deflection method can also be used to analyse frames. A frame will not sidesway to the left or right provided it is properly restrained.
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Analysis of Frames: No Sidesway
No sidesway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry.
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Work Example 6: Frame With No Sidesway
Determine the moments at each joint of the frame shown below. EI is constant.
3 spans must be considered in this case: AB, BC & CD
Note that θ A θD 0 and ψAB ψBC ψCD 0
Step 1: FEM
FEMBC FEMCB
5 wL2 80kN.m 96 5 wL2 80kN.m 96
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Work Example 6: Frame With No Sidesway
Step 2: Slope-Deflection Equations
I MN 2E 2θN θF 3ψ FEMN L MAB 0.1667EIθB
MBA 0.333EIθB MBC 0.5EIθB 0.25EIθC 80 MCB 0.5EIθ C 0.25EIθB 80 MCD 0.333EIθC MDC 0.1667EIθC 2005 Pearson Education South Asia Pte Ltd
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Work Example 6: Frame With No Sidesway
Step 3: Equilibrium Equations
Moment equilibrium at joints B & C, we have:
M M
B
0;
MBA MBC 0
C
0;
MCB MCD 0
Substitute MBA, MBC and MCB, MCD into above equations, we get:
0.833EIθB 0.25EIθC 80 0.833EIθC 0.25EIθB 80
137.1 Solving simultaneously yields θB θC EI
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Work Example 6: Frame With No Sidesway
Step 4: Moment Calculation MAB 22.9kNm MBA 45.7kNm
Using these results, the reactions at the ends of each member can be determined using equations of equilibrium. The moment diagram for the frame can be drawn.
MBC 45.7kNm MCB 45.7kNm MCD 45.7kNm MDC 22.9kNm 2005 Pearson Education South Asia Pte Ltd
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Work Example 7: Frame With No Sidesway
Determine the internal moments at each joint of the frame shown below. The moment of inertia for each member is given in the figure. Take E = 200 Gpa.
Take as self study at home….Refer to Hibbeler 7th edition, Example 11-6 50 kN/m
30 kN
160 (106) mm4
B
2.4 m
2.4 m C
E
320 (106) mm4
3.6 m 260 (106) mm4
4.5 m
A
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80 (106) mm4
D
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Analysis of Frames: Sidesway
A frame will sidesway when it or the loading acting on it is nonsymmetric. Consider the frame shown here. The loading P causes an unequal moments at joint B and C. MBC tends to displace joint B to the right. MCB tends to displace joint C to the left. Since MBC > MCB, the net result is a sidesway of both joint B & C to the right.
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Analysis of Frames: Sidesway
When applying the slope-deflection equation to each column, we must consider the column rotation ( = /L) as an unknown in the equation. As a result, an extra equilibrium equation must be included in the solution. The techniques for solving problems for frames with sidesway is best illustrated by work examples.
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Work Example 8: Frame With Sidesway
Determine the moments at each joint of the frames shown below.
Ends A & D are fixed. Sidesway occurs here. AB = /4 (+ ve) DC = /6 (+ ve) AB = (6/4)DC
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Work Example 8: Frame With Sidesway
Slope-Deflection Equations
MAB EI(0.5θB 2.25ψDC )
(1)
MBA EI(1.0θC 2.25ψDC )
(2)
MBC EI(0.8θB 0.4θC )
(3)
MCB EI(0.8θC 0.4θB )
(4)
MCD EI(0.667θC 1.0ψDC )
(5)
MDC EI(0.333θC 1.0ψDC )
(6)
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Work Example 8: Frame With Sidesway
Equilibrium Equation
9 unknowns and 6 equations. Thus, we need 3 equilibrium equations. 2 moment equilibrium equations for joints B & C.
M M
B
0;
MBA MBC 0
(7)
C
0;
MCB MCD 0
(8)
F
0;
Another equilibrium equation obtained from summation of forces (entire frame) in x-direction, since there is horizontal displacement . X
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200 VA VD 0
(9) 35
Work Example 8: Frame With Sidesway
Equilibrium Equation
Considering the free-body diagram of each column separately.
MAB MBA MB 0 VA 4 MDC MCD MC 0 VD 6 MAB MBA MDC MCD 200 0 4 6
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(9)
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Work Example 8: Frame With Sidesway
Equilibrium Equation
Substitute equation (2) and (3) into equation (7). Substitute equation (4) and (5) into equation (8). Equation (1), (2), (5) and (6) into equation (9).
Solve simultaneously, we have
EIθB 243.78 EIθC 75.66 EIψDC 208.48 2005 Pearson Education South Asia Pte Ltd
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Work Example 8: Frame With Sidesway
Moment calculation
Using these results & solving for equation (1) to (6)
MAB 347kNm MBA 225kNm MBC 225kNm MCB 158kNm MCD 158kNm MDC 183kNm 2005 Pearson Education South Asia Pte Ltd
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