Deflection 5 8 1 – Structural Beam Analysis

SLOPE DEFLECTION METHOD
A Term Paper Presented to: Department of Civil Engineering College of Engineering Mindanao State University – Iligan Institute of Technology
in Partial Fulfillment of the Requirements for CE 152 – Structural Theory II
Presented By:
Raizel Kae Reyes Queenie Ann Torralba Evon Diane Llanos Angela Liao Kea Nathasha Macalam Nikki Actub
May 2016
TABLE OF CONTENTS Page CHAPTER 1: INTRODUCTION
1
1.1
Introduction…………………………………………….
1
1.2
Objectives……………………………………………….
2
1.3
Scope and Limitations………………………………….
3
CHAPTER 2: REVIEW OF RELATED LITERATURE 2.1
4
Concept of Slope Deflection Method……………………
4
2.1.1 Kinematically Indeterminate………………………
5
2.1.2
Degree of Freedom……………………………….
6
2.2
Assumptions…………………………………………….
6
2.3
Advantages and Disadvantages………………………….
7
2.4
Sign Conventions……………………………………….
7
2.5
Applications and Derivations of Slope Deflection………
8
Method…………………………………………………. CHAPTER 3: METHODOLOGY
8 16
3.1
Sample Problems……………………………………….
17
3.2
Fixed-end Moments…………………………………….
18
3.3
Slope-deflection Equations………………………………
18
3.4
Procedure in Slope-deflection Method………………….
19
3.5
Supplementary Problems……………………………….
20
3.5.1
20
Continuous Beam……………………………….
3.5.2 Frames without Sidesway…………………….
22
3.5.3
25
Frames with Sidesway………………………….
CHAPTER 4: PRESENTATION AND ANALYSIS
29
CHAPTER 5: SUMMARY AND CONCLUSION
43
5.1
Summary……………………………………………….
43
TABLE OF CONTENTS
5.2
Conclusion………………………………………………
43
Appendix A
Fixed-end Moments…………………………………….
44
Appendix B
Calculations…………………………………………….
45
Appendix C
GRASP Checking……………………………………….
72
REFERENCES………………………………………………………….
114
CHAPTER 1 INTRODUCTION 1.1
Background about Slope Deflection Method Structural analysis is a broad subject given attention by researchers in order to
develop theories and provide methods that are most appropriate in solving determinate and indeterminate structures that could prove useful for practical calculations. Indeterminate structures are those which cannot be analyzed by the equations of static equilibriums alone. There are two distinct methods of analysis for statically indeterminate structures depending on how equations of equilibrium, load displacement and compatibility conditions are satisfied – force method of analysis and displacement method of analysis such as the slope-deflection method. The slope-deflection method is a structural system analysis method developed by Heinrich Manderla and Otto Mohr which was later presented by Professor G. A. Maney in 1915 as a method of analysis for rigid-jointed beam and frame structure. It is an equilibrium based method that primarily involves equations of equilibrium for the entire system as the name suggests; one equation is written for each kinematic degree of freedom while maintaining conditions of compatibility and the boundary conditions. By forming slope deflection equations and applying joint and shear equilibrium conditions, the rotation angles are calculated. Substituting them back into the slope deflection equations, member end moments are readily determined. As introduced earlier, the slope-deflection method can be used to analyze statically determinate and indeterminate beams and frames. In this method it is assumed that all deformations are due to bending only. In other words deformations due to axial forces are neglected. As discussed beforehand in the force method of analysis
2
compatibility equations are written in terms of unknown reactions. It must be noted that all the unknown reactions appear in each of the compatibility equations making it difficult to solve resulting equations. The slope-deflection equations are not that lengthy in comparison. 1.2
Objectives Slope-deflection method consists of series of simultaneous equations acting at the
end of each member considering it to be fixed-end and taking each moment in terms of slope and deflection. The solution of slope-deflection equations along with equilibrium equations gives the values of unknown rotations of the joints. In relation, this paper aims to comprehend the topic through the theory and derivation of slope-deflection hereby providing additional illustration and activity for the earnest learners that wish to grasp the whole of the lesson. More specifically, this paper will: 1.
Present the theories to strengthen our foundation of slope-deflection method and grasp the primary principles of the topic including the applications, sign conventions and assumptions related to the analysis.
2.
Derive the general formula of slope-deflection method for which can be used in structural analysis of the stiff connections between members of frames and beams in design.
3.
Provide examples to further understand the theory and additional problems to exercise one's understanding towards the topic.
4.
Compare results with Graphical Rapid Analysis of Structures Program (GRASP) for the provided problems.
3
1.3
Scope and Limitations This paper covers the discussion of frames made up of rectangular elements which
are kinematically indeterminate, the derivation of fundamental equations, the determination of moments, stresses, and deflections for a variety of typical structures, the analysis of slope-deflection method using different example problems, and the application of GRASP in validating the values obtained from the calculations. This method has the following advantage and disadvantage which limits its applicability to certain conditions. The analysis of slope-deflection is considered to be in elastic state, assuming it to have the same physical properties in all direction and stresses that can withstand are directly proportional to modulus of elasticity. Once approach to the problem begins, the unknowns, primarily, the displacements are automatically chosen and solved through equilibrium equations. For ease of calculations, it is bethought to be more suitable in computer implementations. However, it is advantageous only for the structures with small kinematic indeterminacy. Slope-deflection is also particularly tedious, time consuming and unwieldy for highly indeterminate structures especially when doing hand calculations. Hence this paper limits the application of slope-deflection method to beams and frames with four members.
4
CHAPTER 2 REVIEW OF RELATED LITERATURE 2.1
Concept of Slope Deflection Method The slope deflection method is a structural method for beams and frames. It was
first used for calculating secondary stresses in trusses but later it was applied to analyzing stresses in frames due to loads and variations of temperature. The first appearance of the method in the American press was in the article by C. Zunz in the Engineering Keys, Oct. 5, 1911, entitled 'Secondary Stresses' where he refers to Otto Mohr as the inventor of the method. In March 1915, it appeared in a bulletin by G. A. Maney, entitled 'Secondary Stress and Other Problems in Rigid Frames: a New Method of Solution' in University of Minnesota, Studies in Engineering where he gave a complete derivation of the fundamental deductions, including the case of any system of leads on the member. He used area-moments to find deflections and slopes. The equations derived in this bulletin are not in the form in which they are now used. However, they can be brought to the standard form by a simple transformation. In June 1915, it appeared in Bulletin Ko. 80, University of Illinois, Engineering Experiment Station, 'Wind Stresses in the Steel frames of Office Buildings' by W. Wilson and G. A. Llaney the theoretical work of Maney where the fundamental equations are derived in the form in which they are now used, but the case of loading on the member is not word out. The basic idea of the slope deflection method is to write the equilibrium equations for each node in terms of the deflections and rotations. Solve for the generalized displacements. Using moment-displacement relations, moments are then known. The structure is thus reduced to a determinate structure.
5
2.1.1. Kinematically Indeterminate Statical indeterminacy is difference of the unknown forces and the equations of equilibrium while kinematic indeterminacy is the number of possible relative displacements of the nodes in the directions of stress resultants. Redundants: 2 (By, Cy) Kinematics: 2 (θB, θC)
Redundants: 1 (Cy) Kinematics: 3 (θB, θC,
)
Redundants: 6 Kinematics: 1(θB)
Figure 2.1. Kinematically indeterminate structures vs. statically indeterminate structures
6
2.1.2. Degree of Freedom The concept degree of freedom refers to the ability of a joint to rotate or displace. Degree of freedom is the number of joints rotation and independent joint translation in a structure. It is found at the supports, joints and free ends of a member. For frames, it plays an important role in determining whether the frame has the ability to sway or not. A sidesway may occur if the degree of freedom of translation is greater than zero. 2.2.
Assumptions The slope defection method is applicable for beams and frames. It is useful for the
analysis of highly statically indeterminate structures which have a low degree of kinematical indeterminacy. Therefore, the analysis of this method is based on the following simplified assumptions. 1.
All the joints of the frame are rigid such that the angles between members at that joint are considered not to change in value when loads are applied. When beams are deformed, the rigid joints are considered to rotate only as a whole; in other terms, the angles between the tangents to the various branches of the elastic curve remain the same as in the original undeformed structure.
2.
The material of the structure is considered linearly elastic which shall be loaded within elastic limit. The material is of the same nature under having identical physical properties in all direction and stresses that can withstand are directly proportional to modulus of elasticity.
3.
Distortion, due to axial and shear stresses, being very small, are neglected and only flexural deformations are considered. In slope-deflection method, the rotations and displacements of joints are treated as unknowns that are expressed as
7
end moments. However, to satisfy equilibrium condition, the sum of end moments which any joints exerted on the ends of the union of the members must be zero, because the rigid joints in question are subject to the sum of these moments at the end; and so shall be the equation of sum in shear forces be zero.
2.3.
Advantages and Disadvantages In the force method of analysis, primary unknowns are forces and compatibility of
displacements is written in terms of pre-selected redundant reactions and flexibility coefficients using force displacement relations. Solving these equations, the unknown redundant reactions are evaluated. The remaining reactions are obtained from equations of equilibrium. As the name itself suggests, in the displacement method of analysis, slope-deflection method, the primary unknowns are displacements. Once the structural model is defined for the problem, the unknowns are automatically chosen unlike the force method. It is early on, when hand calculations were used that the force method were favored because it resulted in much smaller set of simultaneous equations, however, with the appearance of computers, slope-deflection method favored the ease of automation and low computational effort. In this method of analysis, first equilibrium equations are satisfied. The equilibrium of forces is written by expressing the unknown joint displacements in terms of load by using load displacement relations. These equilibrium equations are solved for unknown joint displacements. In the next step, the unknown reactions are computed from compatibility equations using force displacement relations.
2.4.
Sign Conventions Direction is critical when it comes to solving frames using slope-deflection
method. A simple error in one section of the method will relay to other parts of the
8
solution and the moment equilibrium at joints will not work out. Joint rotation is considered positive when occurring in counter-clockwise direction whereas fixed-end moments are positive when occurring in a clockwise direction.
Figure 2.2. Sign Conventions for Slope-Deflection Method
2.5.
Applications and Derivations of Slope-Deflection Method
1.
Continuous beams with and without settlement Continuous beams are those that rest over three or more supports, thereby having
one or more redundant support reactions. Slope-deflection equations are derived considering the rotation of beam axis. In statically indeterminate beams, the rotation is due to support yielding and in turn induces reactions and stresses in the structure. Hence, in this case the beam end moments are related to rotations, applied loads and beam axes rotation.
a) Original beam
9
b) moment diagram
c) fixed-end support with displacement Figure 2.3. Derivation of Slope-Deflection Equation for Continuous Beam Using moment-area method, Theorem II:
( )
(
)
Theorem I:
[
[
]
(
)
]
10
Equilibrium equation,
(
)
For fixed-end support with linear displacement Δ,
Separate the beam with respect to the moments being applied such that FEMAB is the fixed-end moment of the beam as calculated in force method and is kinematically restrained structure. Superpose the fixed-end moments due to external load and end moments due to displacements, the end moments in the actual structure is obtained .Thus,
(
)
(
)
11
In such a case that the beam is without settlement, Δ=0, which gives
2.
Frames without sidesway A sidesway will not occur if the frame geometry and loading are symmetric, and
sidesway is prevented due to supports. For the frames without sidesway, the analysis is basically the same as that of continuous beams without settlement; however, beams could only have two members meeting at a joint whereas in the case of rigid frames two or more than two members meet at a joint.
Figure 2.4. Frame without Sidesway
12
Figure 2.5. Fixed-end Moments of the Frame
One would write as many equilibrium equations as the number of unknowns, and solving these equations joint rotations are evaluated. Substituting joint rotations in the slope–deflection equations member end moments are calculated. 3.
Frames with sidesway In general loading will never be symmetrical. Hence one could not avoid sidesway
in frames. A sidesway will occur if the frame geometry and loading are unsymmetrical, and sidesway is not prevented due to supports.
13
Figure 2.5. Frame undergoing Sidesway Assuming that rotation and displacement shown in Figure 2.5. are so small that
In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments MBC and MCB are not equal. If b is greater than a, then joint B and C are displaced toward the right by an amount Δ. The unknown joint rotations θB and θC are related to joint moments by the moment equilibrium equations. Similarly, when unknown linear displacement occurs, one needs to consider force-equilibrium equations. While applying slope-deflection equation to columns in the above frame, one must consider the column rotation φ as unknown. It is observed that in the column AB, the end B undergoes a linear displacement Δ with respect to end A. Hence the slopedeflection equation for column AB is similar to the one for beam undergoing support settlement.
14
Figure 2.6. Free body Diagram of Columns and Beams The horizontal shear force acting at A and B of the column AB is given by
Now, the required third equation is obtained by considering the equilibrium of member BC,
15
Substituting the values of beam end moments from the equations, we get three simultaneous equations in three unknowns, θB, θC, and Δ, solving which joint rotations and translations are evaluated.
16
CHAPTER 3 METHODOLOGY This chapter focuses on the application of slope-deflection method in analyzing the indeterminate beams and frames. Slope-deflection method requires less work both to write the equations and solve them. This method mainly aims to represent the end moments of the structure with respect to deflections (displacement or rotation). 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. Present the Sample Problem of Indeterminate Structure
Calculate Fixed end Moments
Apply Slope Deflection Equation
Compute the Support Reactions
Compare Results with GRASP
Figure 3.1. General Methodology
17
3.1.
Sample Problems
The sample problems for the indeterminate structures are provided below. 1. Determine the support moments for the continuous beam. Assume E = 200 GPa and I = 6 x 106 mm4. Label supports as A, B and C from left to right. Assume I for AB and 3I for BC.
Figure 3.1. Example 1
2. The indeterminate beam is loaded as shown as is fixed at A and D. The flexural rigidity EI for members AB, and BC are 2EI, 4EI, respectively. Determine the following: a. Internal moments at supports B and C b. Support reactions
Figure 3.2. Example 2
18
3. Determine only the moment at each joint of the gable frame. The roof load is transmitted to each of the purlins over simply supported sections of the roof decking. The center column, connected to the ridge point C, is incompressible keeping it from displacing vertically. EI is constant.
Figure 3.3. Example 3
3.2.
Fixed-end Moment The fixed-end moments for beams with specific type of loading is
provided in the Appendix A. 3.3.
Slope-deflection Equations The general equation for slope-deflection method is:
(
)
(
)
19
where:
3.4.
FEM
=
Fixed end moment due to the given loading
θ
=
Slope
=
Sinking of a given support with respect to a reference support
L
=
Span of the structure
E
=
Modulus of elasticity of the material
I
=
Moment of inertia of the structure
Procedure in Slope-deflection Method
In the slope-deflection method the individual equations are relatively easy to construct regardless of the number of unknowns. Hence, the slope deflection method is an easy process following the procedures which is the same whether it is applied to beams or frames and may be summarized as follows: 1.
Identify all kinematic degrees of freedom -
This can be done by drawing the deflection shape of the structure. All degrees of freedom are treated as unknowns in slope-deflection method.
2.
Calculate Fixed-end moments -
Determine the fixed end moments at the end of each span due to applied loads acting on span by considering each span as fixed ended. Assign ± Signs w.r.t. above sign convention.
3.
Use slope deflection equations -
Express all end moments in terms of fixed end moments and the joint rotations by using slope – deflection equations.
20
4.
Summation of end moments of two members acting on a joint is equal to zero -
Establish simultaneous equations with the joint rotations as the unknowns by applying the condition that sum of the end moments acting on the ends of the two members meeting at a joint should be equal to zero.
5.
Rotations -
6.
Solve unknown joint rotations
Equate -
Substitute back the end rotations in slope – deflection equations and compute the end moments.
7.
Find the reactions -
3.5.
Determine all reactions and verify the answer using GRASP.
Supplementary Problems The supplementary problems below will serve as practice and recap for the
presented topic and its application to determine the support reactions of a beam and frame. 3.5.1. Continuous Beams 1.
A beam with rollers as supports at A and B and fixed support at C, it carries a a nodal load of 10 kN at the midspan of AB and a uniform load of 49.6 kN/m along span BC. EI is constant all throughout the beam. Assume E = 200 GPa and I = 6 x 106 mm4. a. Compute the slope at B. b. Find the MAB and MBC.
21
Figure 3.4. Problem 1
2.
Calculate the end moments and all support reactions using slope deflection method. Support answers using bending moment diagram and elastic curve of the beam. Assume E = 20 500 N/mm2 and I = 6.75 x 108 mm4.
Figure 3.5. Problem 2
3.
The indeterminate beam is loaded as shown. Assume constant flexural rigidity EI all throughout the beam span. Using slope-deflection method, determine the internal moments at point B and C as well as all the support reactions.
Figure 3.6. Problem 3
22
4.
Determine the member end moments and reactions for the beam using slope deflection method. Label supports as A, B and C from left to right. ΔB = 5 mm. Assume E = 20 500 N/mm2 and I = 6.75 x 108 mm4.
Figure 3.7. Problem 4
5.
For the beam shown below with E = 200 000 MPa and I = 1 x 106 mm4, determine the slope at joints B, C, and D if ΔC = 5 mm and ΔD = 10 mm.
Figure 3.8. Problem 5
3.5.2. Frames without Sidesway 6.
Calculate the end moments and all support reactions using slope deflection method. Support answers using bending moment diagram and elastic curve of the beam. Assume E = 20 500 N/mm2 and I = 6.75 x 108 mm4.
23
Figure 3.9. Problem 6
7.
Compute the reactions and the end moments for the rigid frame shown below. Draw bending moment diagram and sketch the elastic curve for the frame.
Figure 3.10. Problem 7
8.
Determine the member end moments and reactions for the frame shown by the slope‐deflection deflection method.
24
Figure 3.11. Problem 8
9.
Determine the moments at each joint and support of the battered-column frame. The joints are rigid. The supports are fixed connected. EI is constant.
Figure 3.12. Problem 9
25
10.
Find end member moments and draw the diagrams of the frame. Assume E = 200 GPa and I = 6 x 106 mm4. Label members as A, B and C from left to right.
Figure 3.13. Problem 10
3.5.3. Frames with Sidesway 11.
Determine the member end moments and reactions for the frame shown.
Figure 3.14. Problem 11
26
12.
For the frame shown, use the slope-deflection method to a. Determine the end moments of each member and reactions at supports. Assume 3EI for AC and 2EI for CD. b. Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.
Figure 3.15. Problem 12
13.
A frame is loaded as shown in the Figure below. Assuming the flexural rigidity EI of the frame is constant; compute the reactions at the joints using slope-deflection method.
27
Figure 3.16. Problem 13
14.
Determine the end member moments of the frame below.
Figure 3.17. Problem 14
28
15.
A frame is loaded as shown in the Figure Celow. Assuming the flexural rigidity EI of the frame is constant; compute the reactions at the joints using slope-deflection method. E = 200 000 MPa, I = 1 x 106 mm4
Figure 3.18. Problem 15
29
CHAPTER 4 PRESENTATION AND ANALYSIS In the previous chapter, a number of exercises where presented in order to understand the concept of slope-deflection method. Knowing joint rotations and translations, beam end moments are calculated from slope-deflection equations. The complete procedure is explained for the numerical examples in chapter 4. 1. Solution: In the given problem, joints B and C rotate. Hence, in this problem we have two rotational unknown displacements to be evaluated. Considering the kinematically determinate structure, fixed end moments are evaluated. Thus,
The end A is fixed and there is no translation in any joint of the given beam, hence, θA = Δ = 0. Now, writing the slope-deflection equations for the four beam end moments,
30
Now consider the equilibrium at B and C. Since the beam is continuous, the sum of the internal moments at a joint is equal to zero.
(
)
Equate (Eq. 4.11) and (Eq. 4.13),
Substituting the values of θB and θC in the slope-deflection equations, one could calculate beam end moments. Thus,
31
Joint
Segment
FEM (kN-m)
A
AB
-62.5
BA
62.5
Slope deflection equations
Slope
End member moment (kN-m)
Support reactions (kN)
-
-46.9
40.63
94.8
B
C
146.88 BC
-94.75
-94.8
CB
94.75
0
62.5
2. Solution: It is observed that the continuous beam is kinematically indeterminate to first degree as two joint rotations, θB and θC are unknown. By fixing the support or restraining the support at B and C against rotation, the fixed-end moments are obtained.
For the fixed support A, the rotation is zero. Only two non-zero rotation is to be evaluated for this problem. Now, write slope-deflection equations for span AB and BC.
32
In the above four slope-deflection equations, the member end moments are expressed in terms of θB and θC unknown rotations. Now, the required equation to solve for the rotation θB and θC is the moment equilibrium equation at support B and C. For, moment equilibrium at support B and C, one must have,
Equate (Eq. 4.28) and (Eq. 4.30),
After evaluating θB and θC, substitute it in equations (Eq. 4.20) to (Eq. 4.23), to evaluate beam end moments. Thus,
33
Slope
End member moment (kN-m)
Support reactions (kN)
-
-2.3
4.14
Joint
Segment
FEM (kN-m)
A
AB
-4.6
BA
5.4
7.9
BC
-0.76
-7.9
CB
2.78
-15
Slope deflection equations
B
C
21.7
-1.83
3. Solution: In this problem, there are three unknowns - two rotations need to be determined i. e. θB and θD and one deformation, Δ. Thus the required equations to evaluate θB, Δ, and θD are obtained by considering the equilibrium of joint B and joint D. Now, calculate the fixed-end moments of the frame by considering the individual members as a beam and by fixing the supports.
For writing slope–deflection equations two spans must be considered, BC and CD. Since support A and E are fixed θA = θE = 0 and θC = 0.
34
(
)
(
(
)
(
)
(
)
(
)
(
)
(
Now consider the joint equilibrium of support B and C,
)
)
35
Equate (Eq. 4.44), (Eq. 4.46) and (Eq. 4.48),
Joint Segments A
FEM (kNm)
Slope deflection equations
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
37
22
94
-51
-
-
AB
-0
(
)
BA
0
(
)
BC
-48.75
51
-
-
CB
48.75
35
-
-
CD
-48.75
-35
-
-
DC
48.75
-51
-
-
DE
0
(
)
51
-
-
ED
0
(
)
-37
-22
94
B
C
D
E
-
Deflection
The supplementary problems serve as a practice to recap to the topic and do the same process in solving structures using slope-deflection method provided that one already understand the examples given above.
36
Problem 1 End member moments (kNm)
Support reactions (kN)
-12.5
0
-94.95
+12.5
999.48
Joint
Segments
FEM (kN-m)
A
AB BA
Slope deflection equations
Slope
B
C
551.91 BC
-1653.3
CB
+1653.3
-999.48 -
1980.21
545.04
Slope
End member moments (kNm)
Support reactions (kN)
-
-31.42
7.44
Problem 2 Joint
Segments
FEM (kN-m)
A
AB
-61.2
BA
+40.8
100.36
BC
-150
-100.36
CB
+150
Slope deflection equations
B
C
37.08
-
174.82
32.48
Problem 3 Joint
Segments
FEM (kN-m)
Slope deflection equations
Slope
End member moments (kNm)
Support reactions (kN)
A
AB
-
-
-
-
-
BA
-
-
20
B
0.95 BC
0
-20
CB
0
36.18
CD
-48.75
-36.18
DC
48.75
0
C
D
69.78
51.26
37
Problem 4 Joint
Segment s
FEM (kN-m)
A
AB
-72.29
BA
+72.29
Slope deflection equations
Slope
End member moments (kNm)
Support reactions (kN)
-
-73.76
90.95
52.75
B
C
100.03 BC
0
CB
0
-52.75 -
-34.68
-17.49
Problem 5 Joint
Segments
FEM (kN-m)
Slope deflection equations
Slope
End member moments (kNm)
Support reactions (kN)
A
AB
-
-
-
-
-
BA
-
-
BC
-15
CB
15
90
B
88.4 (
-90
)
(
-25
)
C
-45.4 CD
0
(
)
25
DC
0
(
)
56
D
E
83 DE
-84
(
ED
126
(
-56
) )
-
140
154
38
Problem 6 Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
19.43
14.53
17
0
38.86
-
-
BC
-45.33
-38.86
-
-
CB
45.33
-38.86
-
-
CD
0
38.86
-
-
DC
0
-
19.43
-14.53
17
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
3.4
0.45
6.51
Joint
Segments
FEM (kNm)
A
AB
0
BA
Slope deflection equations
B
C
D
Problem 7 Joint
Segment s
FEM (kNm)
A
AB
-3.92
BA
2.61
3.6
-
-
BC
-4.5
-4.6
-
-
BD
0
1
-
-
CB
4.5
2.2
-
-
CE
0
-2.2
-
-
D
DB
0
-
0.5
0.39
6.68
E
EC
0
-
-1.1
-0.83
2.61
B
Slope deflection equations
C
39
Problem 8 Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
-185.8
-37.13
57.4
185
183.3
-
-
BC
-200
-183.3
-
-
CB
200
235.8
-
-
CD
0
36.7
-
-
CE
-200
-272.5
-
-
D
DC
0
-18.3
2.74
136.2
E
EC
200
0
-39.61
46.4
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
244.15
381.4
447.6
Joint
Segment s
FEM (kNm)
A
AB
-185
BA
Slope deflection equations
B
C
-
Problem 9 Joint
Segments
FEM (kNm)
A
AB
0
BA
0
488.3
-
-
BC
-895.2
-488.3
-
-
CB
895.2
488.3
-
-
CD
0
-488.3
-
-
DC
0
-244.15
-381.4
447.6
Slope deflection equations
B
C
D
-
40
Problem 10 Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
14.82
-4.72
41.76
10
67.64
-
-
BC
-74.5
-67.64
-
-
CB
31.5
0
-26.29
8.24
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
4.86
1.94
7.87
8.71
-
-
Joint
Segments
FEM (kNm)
A
AB
-15
BA
Slope deflection equations
B
C
Problem 11 FEM Joint Segments (kNm) A
Slope deflection equations
AB
0
(
)
BA
0
(
)
BC
-13.12
-8.71
-
-
CB
9.84
7.14
-
-
CD
0
(
)
-7.14
-
-
DC
0
(
)
-2.56
-1.94
5.52
B
C
D
Deflection
-
41
Problem 12 Joint Segments A
FEM (kNm)
Slope deflection equations
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
131.6
-40
83.79
11.6
-
-
AC
-30
CA
30
CD
-70.2
-21.6
-
-
DC
70.2
0
0
66.61
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
-64.1
-36
5.41
31.9
-
-
( (
) )
C
D
Deflection
Problem 13 Joint Segments A
FEM (kNm)
Slope deflection equations
AB
-53.33
BA
53.33
BC
0
-31.9
-
-
CB
0
0
-44
7.98
( (
) )
B
C
Deflection
42
Problem 14 FEM Joint Segments (kNm) A
Slope deflection equations (
)
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
-
3.33
1.29
0.21
-3.33
-
-
AB
0
BA
0
BC
-9.25
-3.33
-
-
CB
9.25
-5.33
-
-
CD
0
5.33
-
-
DC
0
-
-4.3
3.67
8.79
Slope deflection equations
Slope
End member moments (kNm)
Rx (kN)
Ry (kN)
(
-
1.68
14.53
74
-15
-
-
√ √
√
(
)
√
B
C
D
√ √
(
)
√
(
)
√
Deflection
Problem 15 FEM Joint Segments (kNm) A
AB
-37
BA
37
BC
0
15
-
-
CB
0
22
-
-
CD
0
(
)
-49
-
-
DC
0
(
)
0
-
16.3
)
(
)
B
C
D
Deflection
-
43
CHAPTER 5 SUMMARY AND CONCLUSION 5.1
Summary In this paper the slope-deflection equations are derived for beams and frames with
unyielding supports. The kinematically indeterminate structures are analyzed by slopedeflection equations. After the equations have been written for each member, by equating the sum of the moments at each joint to zero and employing one equation of statics, a number of equations are obtained which contain values of θ and Δ as the only unknowns. From these equations can be found values of θ and Δ, which, when substituted in the original slope-deflection equations, give values of the various moments. The advantages of displacement method of analysis over force method of analysis are also brought out here. A couple of examples are solved to illustrate the slope-deflection equations. 5.2
Conclusion The general form of the fundamental equation of slope-deflection method is easily
memorized, and the equations may be written for all members of a structure with little effort. It is frequently possible to simplify the equations through noting where values of θ and Δ must be equal to zero from the conditions of the problem. No integrations need be performed and there is little danger of the omission of the effect of a single indeterminate quantity. The method has been explained in sufficient detail to enable the designing engineer to use it in the solution of his particular problems. It is believed that the fundamental principles can be quickly coordinated with the ordinary principles of mechanics so that the more complex problems and even the simpler ones may be studied from a new viewpoint.
44
Appendix A Fixed-End Moments
45
Appendix B Calculations Problem 1 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
46
Appendix B (cont'd)
(
)
47
Appendix B (cont'd) Problem 2 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
48
Appendix B (cont'd) Problem 3 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
49
Appendix B (cont'd) Problem 4 Fixed-end moments:
Chord equations:
Slope-deflection equations:
(
)
(
)
(
)
(
)
Equilibrium equations:
50
Appendix B (cont'd) Problem 5 Fixed-end moments:
Chord equations:
Slope-deflection equations:
(
)
(
)
(
)
(
)
(
)
(
)
51
Appendix B (cont'd)
(
)
(
)
(
)
(
)
(
(
Equilibrium equations:
(
)
(
)
)
)
52
Appendix B (cont'd) Problem 6 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
53
Appendix B (cont'd)
54
Appendix B (cont'd) Problem 7 Fixed-end moments:
Slope-deflection equations:
55
Appendix B (cont'd)
Equilibrium equations:
56
Appendix B (cont'd) Problem 8 Fixed-end moments:
Slope-deflection equations:
57
Appendix B (cont'd)
Equilibrium equations:
58
Appendix B (cont'd) Problem 9 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
59
Appendix B (cont'd)
60
Appendix B (cont'd) Problem 10 Fixed-end moments:
Slope-deflection equations:
Equilibrium equations:
61
Appendix B (cont'd)
Equate (Eq. A – 10.1) and (Eq. A – 10.2),
62
Appendix B (cont'd) Problem 11 Fixed-end moments:
Slope-deflection equations:
(
)
(
(
)
(
)
(
)
(
)
(
)
(
)
)
63
Appendix B (cont'd) Equilibrium equations:
64
Appendix B (cont'd) Problem 12 Fixed-end moments:
Slope-deflection equations:
(
)
(
(
)
(
Equilibrium equations:
)
)
65
Appendix B (cont'd)
66
Appendix B (cont'd) Problem 13 Fixed-end moments:
Slope-deflection equations:
(
)
(
(
)
(
⁄ (
)
⁄ (
Equilibrium equations:
)
)
)
67
Appendix B (cont'd)
68
Appendix B (cont'd) Problem 14 Fixed-end moments:
Slope-deflection equations:
(
)
(
)
(
)
(
)
√
√
(
(
√
√
)
)
√
√
√
√
(
√
(
(
(
)
)
√
)
√
√
)
69
Appendix B (cont'd) Equilibrium equations:
√
√
70
Appendix B (cont'd) Problem 15 Fixed-end moments:
Slope-deflection equations:
(
)
(
(
)
(
(
)
(
(
)
(
(
(
)
)
)
)
)
)
71
Appendix B (cont'd) Equilibrium equations:
72
Appendix C GRASP Checking Example 1
Figure C-1.1 Support Reactions of Particular Indeterminate Beam
Figure C-1.2 Shear Diagram of Particular Indeterminate Beam
Figure C-1.3 Moment Diagram of Particular Indeterminate Beam
73
Appendix C (cont'd)
Figure C-1.4 Elastic Curve of Particular Indeterminate Beam
74
Appendix C (cont'd) Example 2 (Assumption: E = 20 500 MPa, I = 1 x 106 mm4)
Figure C-2.1 Support Reactions of Particular Indeterminate Beam
Figure C-2.2 Shear Diagram of Particular Indeterminate Beam
Figure C-2.3 Moment Diagram of Particular Indeterminate Beam
75
Appendix C (cont'd)
Figure C-2.4 Elastic Curve of Particular Indeterminate Beam
76
Appendix C (cont'd) Example 3
Figure C-3.1 Support Reactions of Particular Indeterminate Frame
Figure C-3.2 Axial Diagram of Particular Indeterminate Frame
77
Appendix C (cont'd)
Figure C-3.3 Shear Diagram of Particular Indeterminate Frame
Figure C-3.4 Moment Diagram of Particular Indeterminate Frame
78
Appendix C (cont'd)
Figure C-3.4 Elastic Curve of Particular Indeterminate Frame
79
Appendix C (cont'd) Problem 1
Figure C-4.1 Support Reactions of Particular Indeterminate Beam
Figure C-4.2 Shear Diagram of Particular Indeterminate Beam
Figure C-4.3 Moment Diagram of Particular Indeterminate Beam
Figure C-4.4 Elastic Curve of Particular Indeterminate Beam
80
Appendix C (cont'd) Problem 2
Figure C-5.1 Support Reactions of Particular Indeterminate Beam
Figure C-5.2 Shear Diagram of Particular Indeterminate Beam
Figure C-5.3 Moment Diagram of Particular Indeterminate Beam
Figure C-5.4 Elastic Curve of Particular Indeterminate Beam
81
Appendix C (cont'd) Problem 3 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-6.1 Support Reactions of Particular Indeterminate Beam
Figure C-6.2 Shear Diagram of Particular Indeterminate Beam
82
Appendix C (cont'd)
Figure C-6.3 Moment Diagram of Particular Indeterminate Beam
Figure C-6.4 Elastic Curve of Particular Indeterminate Beam
83
Appendix C (cont'd) Problem 4
Figure C-7.1 Support Reactions of Particular Indeterminate Beam
Figure C-7.2 Shear Diagram of Particular Indeterminate Beam
Figure C-7.3 Moment Diagram of Particular Indeterminate Beam
Figure C-7.4 Elastic Curve of Particular Indeterminate Beam
84
Appendix C (cont'd) Problem 5
Figure C-8.1 Support Reactions of Particular Indeterminate Beam
Figure C-8.2 Shear Diagram of Particular Indeterminate Beam
Figure C-8.3 Moment Diagram of Particular Indeterminate Beam
Figure C-8.4 Elastic Curve of Particular Indeterminate Beam
85
Appendix C (cont'd) Problem 6
Figure C-9.1 Support Reactions of Particular Indeterminate Frame
Figure C-9.2 Axial Diagram of Particular Indeterminate Frame
86
Appendix C (cont'd)
Figure C-9.3 Shear Diagram of Particular Indeterminate Frame
Figure C-9.4 Moment Diagram of Particular Indeterminate Frame
Figure C-9.4 Elastic Curve of Particular Indeterminate Frame
87
Appendix C (cont'd) Problem 7 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-10.1 Support Reactions of Particular Indeterminate Frame
Figure C-10.2 Axial Diagram of Particular Indeterminate Frame
88
Appendix C (cont'd)
Figure C-10.3 Shear Diagram of Particular Indeterminate Frame
Figure C-10.4 Moment Diagram of Particular Indeterminate Frame
89
Appendix C (cont'd)
Figure C-10.4 Elastic Curve of Particular Indeterminate Frame
90
Appendix C (cont'd) Problem 8 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-11.1 Support Reactions of Particular Indeterminate Frame
Figure C-11.2 Axial Diagram of Particular Indeterminate Frame
91
Appendix C (cont'd)
Figure C-11.3 Shear Diagram of Particular Indeterminate Frame
Figure C-11.4 Moment Diagram of Particular Indeterminate Frame
92
Appendix C (cont'd)
Figure C-11.4 Elastic Curve of Particular Indeterminate Frame
93
Appendix C (cont'd) Problem 9 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-12.1 Support Reactions of Particular Indeterminate Frame
Figure C-12.2 Axial Diagram of Particular Indeterminate Frame
94
Appendix C (cont'd)
Figure C-12.3 Shear Diagram of Particular Indeterminate Frame
Figure C-12.4 Moment Diagram of Particular Indeterminate Frame
95
Appendix C (cont'd)
Figure C-12.4 Elastic Curve of Particular Indeterminate Frame
96
Appendix C (cont'd) Problem 10 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-13.1 Support Reactions of Particular Indeterminate Frame
Figure C-13.2 Axial Diagram of Particular Indeterminate Frame
97
Appendix C (cont'd)
Figure C-13.3 Shear Diagram of Particular Indeterminate Frame
Figure C-13.4 Moment Diagram of Particular Indeterminate Frame
98
Appendix C (cont'd)
Figure C-13.4 Elastic Curve of Particular Indeterminate Frame
99
Appendix C (cont'd) Problem 11 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-14.1 Support Reactions of Particular Indeterminate Frame
Figure C-14.2 Axial Diagram of Particular Indeterminate Frame
100
Appendix C (cont'd)
Figure C-14.3 Shear Diagram of Particular Indeterminate Frame
Figure C-14.4 Moment Diagram of Particular Indeterminate Frame
101
Appendix C (cont'd)
Figure C-14.4 Elastic Curve of Particular Indeterminate Frame
102
Appendix C (cont'd) Problem 12 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-15.1 Support Reactions of Particular Indeterminate Frame
Figure C-15.2 Axial Diagram of Particular Indeterminate Frame
103
Appendix C (cont'd)
Figure C-15.3 Shear Diagram of Particular Indeterminate Frame
Figure C-15.4 Moment Diagram of Particular Indeterminate Frame
104
Appendix C (cont'd)
Figure C-15.4 Elastic Curve of Particular Indeterminate Frame
105
Appendix C (cont'd) Problem 13 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-16.1 Support Reactions of Particular Indeterminate Frame
Figure C-16.2 Axial Diagram of Particular Indeterminate Frame
106
Appendix C (cont'd)
Figure C-16.3 Shear Diagram of Particular Indeterminate Frame
Figure C-16.4 Moment Diagram of Particular Indeterminate Frame
107
Appendix C (cont'd)
Figure C-16.4 Elastic Curve of Particular Indeterminate Frame
108
Appendix C (cont'd) Problem 14 (Assumption: E = 20 500 MPa, I = 6.75 x 108 mm4)
Figure C-17.1 Support Reactions of Particular Indeterminate Frame
Figure C-17.2 Axial Diagram of Particular Indeterminate Frame
109
Appendix C (cont'd)
Figure C-17.3 Shear Diagram of Particular Indeterminate Frame
Figure C-17.4 Moment Diagram of Particular Indeterminate Frame
110
Appendix C (cont'd)
Figure C-17.4 Elastic Curve of Particular Indeterminate Frame
111
Appendix C (cont'd) Problem 15
Figure C-18.1 Support Reactions of Particular Indeterminate Frame
Figure C-18.2 Axial Diagram of Particular Indeterminate Frame
112
Appendix C (cont'd)
Figure C-18.3 Shear Diagram of Particular Indeterminate Frame
Figure C-18.4 Moment Diagram of Particular Indeterminate Frame
113
Appendix C (cont'd)
Figure C-18.4 Elastic Curve of Particular Indeterminate Frame
114
REFERENCES
Chapter (3) Slope Deflection Method. http://www1.zu.edu.eg/zusted/Advanced%20 Stractural%20analysis%20Part%201/chapter%203%20(slope%20Deflection).pdf. Chapter 5: Indeterminate Structures – Slope-Deflection Method. Structural Design II. http://user.engineering.uiowa.edu/~design1/StructuralDesignII/Chapter5-Slope-de fl_Method.pdf Structural Analysis. IIT, Kharagpur. 2008. https://archexamacademy.com/download/ Structural%20Systems/beam%20diagrams/Structural%20Analysis%20e-Book.pdf.
Wilson, W. M, Richart, F. E., and Weiss, C. (1918). Analysis Of Statically Indeterminate Structures By The Slope Deflection Method. University of Illinois Bulletin, vol. XVI, no. 10. https://engineering.purdue.edu/~ce474/Docs/engineeringexper v00000i00108_slopedeflection.pdf.