Chapter 11. Displacement Method of Analysis Slope Deflection Method
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1 Chapter 11 Displacement ethod of Analysis Slope Deflection ethod
2 Displacement ethod of Analysis Two main methods of analyzing indeterminate structure Force method The method of consistent deformations & the equation of three moment The primary unknowns are forces or moments Displacement method The slope-deflection method & the moment distribution method The primary unknown is displacements (rotation & deflection) It is particularly useful for the analysis of highly statically indeterminate structures Easily programmed on a computer & used to analyze a wide range of indeterminate structures
3 Displacement ethod of Analysis Degree of Freedom The number of possible joint rotations & independent joint translations in a structure is called the degree of freedom of the structure In three dimensions each node on a frame can have at most three linear displacements & three rotational displacements. In two dimension each node can have at most two linear displacements & one rotational displacement. DOF = n J R - n number of possible joint s movements - J number of joints - R number of restrained movements
4 Displacement ethod of Analysis Degree of Freedom DOF = n J R - n number of possible joint s movements - n = 2 for two dimensional truss structures - n = 3 for three dimensional truss structures - n= 3 for two dimensional frame structures - n= 6 for three dimensional frame structures - J number of joints - R number of restrained movements Neglecting Axial deformation DOF = n J R m - m number of members
5 Displacement ethod of Analysis
6 Slope-Deflection Equations Derivation
7 Slope-Deflection Equations Angular Displacement at A, A 1 AB L 1 2L A' 0 L L 0 2 EI 3 2 EI 3 1 L 1 AB 2L B ' 0 L L AL 0 2 EI 3 2 EI 3 4EI 2EI A & B A A L L AB
8 Slope-Deflection Equations Angular Displacement at B, B AB 4EI B L 2EI B L
9 Slope-Deflection Equations Relative Linear Displacement B ' L L L L 2 EI 3 2 EI EI AB 2 L
10 Slope-Deflection Equations Fixed-End oment Fy 0. 1 PL 1 L 2 L 0. 2 EI 2 EI Fixed-End moment FE PL 8
11 Slope-Deflection Equations AB ( FE ) ( FE ) AB
12 Slope-Deflection Equations Fixed-End oment
13 Slope-Deflection Equations Slope-deflection equations The resultant moment (adding all equations together) 2 I E 2 3 ( FE ) L L AB A B AB 2 I E 2 3 ( FE ) L L B A Lets represent the member stiffness as k = I/L & The span ration due to displacement as = /L Referring to one end of the span as near end (N) & the other end as the far end (F). Rewrite the equations 2EK 2 3 ( FE ) N N F N 2EK 2 3 ( FE ) F F N F
14 Slope-Deflection Equations Slope-deflection equations for pin supported End Span If the far end is a pin or a roller support 2EK 2 3 ( FE )' N N F N 0 2EK F N ultiply the first equation by 2 and subtracting the second equation from it 3 EK ( FE )' N N N
15 Fixed End oment Table
16 Fixed End oment Table
17 Fixed End oment Table- Continue
18 Summary 2EK 2 3 ( FE ) N N F N 2EK 2 3 ( FE ) F F N F 3 EK ( FE )' N N N
19 Slope-Deflection Equations N, F = the internal moment in the near & far end of the span. Considered positive when acting in a clockwise direction E, k = modulus of elasticity of material & span stiffness k = I/L N, F = near & far end slope of the span at the supports in radians. Considered positive when acting in a clockwise direction = span ration due to a linear displacement / L. If the right end of a member sinks with respect the the left end the sign is positive
20 Slope-Deflection Equations Steps to analyzing beams using this method Find the fixed end moments of each span (both ends left & right) Apply the slope deflection equation on each span & identify the unknowns Write down the joint equilibrium equations Solve the equilibrium equations to get the unknown rotation & deflections Determine the end moments and then treat each span as simply supported beam subjected to given load & end moments so you can workout the reactions & draw the bending moment & shear force diagram
21 Example 1 Draw the bending moment & shear force diagram. Fixed End oment 4 4 FE AB 2 t. m FE 2 t. m FE 6 t. m FE CB 6 t. m 12 4t 4m 2t/m A I B 2I C 6m Slope Deflection Equations I A 0 AB 2E 2 A B 0 2 AB 0.5EIB 2 4 I A 0 2E A 2 B 0 2 EIB 2 4 2I 0 4 C 2E 2 B C 0 6 EIB I 0 2 C CB 2E B 2 C 0 6 CB EIB 6 6 3
22 Joint Equilibrium Equations Joint B EIB 2 EIB 6 0. B 3 EI Substituting in slope deflection equations 1.7 AB 0.5EI t. m EI 1.7 EI t. m EI 3.7 t. m CB EI t. m 3 EI 1.15 Computing The Reactions 4t 3.7 R R A B t t R A R B1
23 R R B 2 C t t 3.7 2t/m R B2 R C
24 Determine the internal moments in the beam at the supports. Fixed End oment FE FE FE AB kn. m kn. m kn. m 8 Slope Deflection Equations Example 2 60kN I 1 E EI 6 3 I 2 E EI 6 3 I 1 3E B EIB A 0 AB 2 2 A B AB B A 0 2 A 2 B B m 6m 30kN/m A B C I I 6m
25 Joint Equilibrium Equations Joint B EIB EIB EIB EI 70 B Substituting in slope deflection equations AB 1 (70) kn. m 3 2 (70) kn. m 3 1 (70) kn. m 2
26 Example 3 Example 2: Determine the internal moments in the beam at the supports Support A; downward movement of 0.3cm & clockwise rotation of rad. Support B; downward movement of 1.2cm. Support C downward movement of 0.6cm. EI = 5000 t.m 2 Fixed End oment FE AB FE FE FE CB Displacements: A AB Slope Deflection Equations AB B B A B C I 6m I 6m
27 35000 B 6 Joint Equilibrium Equations Joint B B B rad 0.001
28 Substituting in slope deflection equations AB t. m t. m t. m 6
29 Example 1b FE AB FE 0.0 A C 0.0
30 Slope Deflection Equations
31 Equilibrium Equations
32 Example 2b
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35 Example 3b
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37 Example 4
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40 Analysis of Frames Without Sway The side movement of the end of a column in a frame is called SWAY.
41 Example 5 Determine the moment at each joint of the frame. EI is constant Fixed End oment FE FE 0. AB FE FE FE CB CD kn. m kn. m 96 FE 0. DC Slope Deflection Equations I AB 2E 0. B AB I 2E 2 B A = EI A = EI B B
42 I 2E 2B C I CB 2E 2C B I CD 2E 2 C I DC 2E 0 C Joint Equilibrium Equations Joint B 0 CB CD EI 0.5EI 0.25EI C C B D = 0. D = EI 0.25EI 80 B C 0.5EI 0.25EI 80 CB C B CD DC 0.333EI C EI 0.333EI B 0.5EI B 0.25EI C EI B 0.25EI C 80 Joint C C 0.833EI 0.25EI 80 C B
43 Two equation & two unknown B Substituting in slope deflection equations AB CB CD DC C EI 22.9 kn. m 45.7 kn. m 45.7 kn. m 45.7 kn. m 45.7 kn. m 22.9 kn. m
44 Example 6
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48 Example 6b Draw the bending moment diagram Fixed End oment FE FE FE AB FE CB kn. m kn. m kn. m kn. m FE BE FE EB FE CF FE FC 0 A 1m 20kN I 3m B I E 48kN/m 2I 3m 4m C 2I F 4m 2m 30kN I D CD kn. m Slope Deflection Equations I AB 2E 2A B A = 0. AB 2 EIB
49 I 2E A 2B A = 0. 4 EIB I 2E 2B C I CB 2E B 2C EI EI 64 B C EI 2EI 64 CB B C I BE 2E 2B E I EB 2E 2E B E = 0. E = 0. BE EB 4 EIB 3 2 EIB 3 2I CF 2E 2C F I FC 2E 2F C F = 0. F = 0. CF FC 2EI EI C C
50 Equilibrium Equations Joint B BE EIB EIB EIC 64 EIB Joint C CB CF CD EI 2EI 64 2EI B C C EI B 4.67EI EI EI 4 C B C Two equation & two unknown EI B EI C 4.18 Substituting in slope deflection equations AB kn. m kn. m knm 3 CB ( 4.18) kn. m
51 BE EB kn. m kn. m EI ( 4.18) 8.35 kn. m CF FC 4.18 kn. m
52 Slope Deflection (Frame with Sway) Analysis of Frames with Sway
53 Example 7 Draw the bending moment diagram. EI constant Fixed End oment As there is no span loading in any of the member FE for all the members is zero Slope Deflection Equations I AB 2E 0. B EIB EI 6 24 I 2E 2 B EIB EI 3 24
54 I E 2 B C 3 0 EIB EIC I CB E 2 C B 0 0 EIB EIC I CD E 2C EIC EI I DC E 0 C 3 0 EIC EI Equilibrium Equations Joint B B = EIB EI EIB EIC EI 14.4EI 3.2EI 0 1 Joint C C = 0 CB CD 0 B C
55 EIB EIC EIC EI EI 7.2EI B 26.4EI C 0 2 Three unknown & just two equations so we need another equilibrium equation. Let take F x = H H 0. H H A D A D CD AB DC AB CD DC EIB EI EI B EI EIC EI EIC EI EI 0.75EI 0.333EI B C
56 Now solve the three equation EI EIB EI C 720 EIC EIB EI Substituting in slope deflection equations AB CB CD DC 208 kn. m 135 kn. m 135 kn. m 95 kn. m 95 kn. m 110 kn. m
57
58 Example 7 Draw the bending moment diagram. EI constant Fixed End oment FE FE FE FE AB CB CD FE kn. m kn. m 12 FE 0. DC Slope Deflection Equations I AB 2E 0. B EIB EI 6 24
59 I 2E 2 B EIB EI 3 24 I E 2B C EIB EIC I 2 4 CB 2E 2C B EIB EIC I 2 1 CD 2E 2 C EIC EI 9 54 I 1 1 DC 2E 0 C EIC EI 9 54 Equilibrium Equations Joint B B = EIB EI EIB EIC EI 14.4EI 3.2EI B C
60 Joint C C = 0 CB CD EIB EIC 375 EIC EI EI 7.2EI 26.4EI Three unknown & just two equations so we need another equilibrium equation. Let take F x = H H 0. A D AB H A 12 CD DC H D AB CD DC B EIB EI EI B EI EIC EI EIC EI EI 0.75EI 0.333EI B C C
61 Now solve the three equation EI R1-R2 EIB EI R1-R3 C EI R2-R3 EIB EI C EI EIB EI C 4257 Substituting in slope deflection equations AB kn. m kn. m EIC EIB EI
62 - CB ( 804.7) kn. m ( 804.7) kn. m CD DC 2 1 ( 804.7) kn. m ( 804.7) kn. m
63 Example 8
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67 Draw the bending moment diagram. EI constant Fixed End oment Example 9 As there is no span loading in any of the member FE for all the members is zero Slope Deflection Equations I AB E EIB EI B 3 0 I E EIB EI B 0 3 0
68 I 0 BE 2E 2B E I 0 EB 2E 2E B I 1 FE 2E 0. E I 1 EF 2E 2 E I E 5 5 I 2 CB 2E 2C B B C EIB EIE EIB EIE EIE EI EIE EI EIB EIC EI EIB EIC EI I 0 CD 2E 2C D I 0 DC 2E 2D C EIC EID EIC EID 7 7
69 I E 5 5 I 2 ED 2E 2E D DE 2 2 D E EID EIE EI EID EIE EI Equilibrium Equations Joint B B = 0 BE EIB EI 1 EIB EIC EI 2 EIB EIE EI 70EI 50EI 42EI 42EI 0 1 Joint E E = 0 B C E 1 2 EB ED EF EIB EIE EID EIE EI 2 EIE EI EI 70EI 380EI 42EI 42EI 0 2 B D E
70 Joint C C = 0 CB CD EIB EIC EI 2 EIC EID EI 240EI 50EI 42EI 0 3 B C D 2 Joint D D = 0 DC DE EIC EID EID EIE EI EI 240EI 70EI 42EI 0 4 C D E Top story F X = 0 40 H H 0 H H B E B E CB DE 5 5 ED 2 H B H E
71 EIB EIC EI 2 EIB EIC EI EID EIE EI 2 EID EIE EI EI 6EI 6EI 6EI 4.8EI 1000 B C D E 2 5 Bottom story F X = H H 0 A F AB H A 5 EF FE H F EIB EI 1 EIB EI EIE EI 1 EIE EI EI 30EI 24EI B E 1 H A H F
72 6 unknown and 6 equation EIB EI 0 C EID EIE EI EI Substituting in slope deflection equations AB kn. m kn. m BE EB kn. m kn. m FE EF EIB EIC EI D EIE EI EI kn. m kn. m 31.6 kn. m CD 68.4 kn. m DE 68.4 kn. m CB 68.4 kn. m DC 68.4 kn. m ED 31.6 kn. m
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74 Example 10 Draw the bending moment diagram. EI constant Degree of freedom DOF = 3x4-6-3 = 3 That means we got three unknown & we need three equations Before we start let us discus the relative displacement () of each span
75 The relative displacement () for span AB is equal AB 0. = AB (clockwise) B CD C The relative displacement () for span is equal 0. = (counterclockwise) The relative displacement () for span CD is equal 0. ( CD )= CD (clockwise) A 60 o AB Let us build a relationship between AB, & CD take AB = = AB sin30 = 0.5 CD = AB cos30 = So in the slope deflection equations we will use; as the relative displacement of span AB. 0.5 as the relative displacement of span as the relative displacement of span CD. B 60 o 60 o CD AB =30 o D
76 Fixed End oment FE FE AB FE CB FE CD FE m FE DC t. m 0. Slope Deflection Equations I AB 2E 0. B I 2E 2 B I 0.5 2E 2B C I 0.5 CB 2E B 2C EIB EI EIB EI EIB EIC EI EIB EIC EI
77 I CD 2E 2 C I DC 2E 0 C EIC EI EIC EI 3 6 Equilibrium Equations Joint B B = EIB EI EIB EIC EI EI B 24EI C 23EI Joint C C = 0 CB CD EIB EIC EI EIC EI EI B 80EI C 2.072EI 128 2
78 Third Equilibrium Equations (ethod 1) AB AB DC DC CD t 6.92m AB CD DC EI 3.1EI EI B C m 6.0
79 2.6m Third Equilibrium Equations (ethod 2) Third equation F X = 0 H H 0 A D From the free body diagram for column CD H D CD 6 DC B C CD Free body diagram for column AB H 2.6 V A AB A H A A 1.5m AB Free body diagram for Beam V V B CB A V A CB V 4 B AB CB H A AB CB CD DC V B B 2t/m DC D C H D CB
80 AB CB CD DC EIB EI EIB EI 9 EIB EIC EI EIB EIC EI EIC EI EIC EI EI 3.1EI EI B C Solving the three equation EIB EIC EI 144 EIB EIC EI 6.77
81 Solving the three equation EIB EIC EI 144 EIB EIC EI 6.77
82 Substituting in slope deflection equations AB 2.44 t. m 0.36 t. m 2.78 CB 0.36 t. m 2.78 t. m B _ C _ 2.78 CD DC 2.78 t. m 1.88 t. m 2.44 A D
83 Example Draw the bending moment diagram. EI constant Before we start let us discus the relative displacement () of each span
84 The relative displacement () for span AB is equal AB 0. = AB (clockwise) The relative displacement () for span is equal ( 2 ) 1 = ( ) = (counterclockwise) The relative displacement () for span CD is equal 0. ( CD )= CD (clockwise) A = B C 2 1 AB CD D Let us build a relationship between AB, & CD take AB = = 2( AB cos) = 2 5/8.6 = CD = AB = So in the slope deflection equations we will use; as the relative displacement of span AB as the relative displacement of span. as the relative displacement of span CD. B AB CD = 2 & CD = AB because of the symmetry in the geometry
85 Fixed End oment FE FE AB FE CB FE CD FE kn. m FE DC kn. m 0. Slope Deflection Equations I AB 2E 0. B I 2E 2 B I E 2B C I CB 2E 2C B EIB EI EIB EI EIB EIC EI EIB EIC EI
86 I CD 2E 2 C I DC 2E 0 C EIC EI EIC EI Equilibrium Equations Joint B B = EIB EI EIB EIC EI EI 28.57EI 6.13EI CB B Joint C C = 0 CD 0 C EIB EIC EI EIC EI EI EI 6.13EI B C
87 7m 7m Third equation F X = 0 6 H H 0 A D Free body diagram for column AB H 7 V 5 0 A AB A Free body diagram for Beam V B CB H A AB CB V A V B 14 7 AB 5 CB H A From the free body diagram for column CD H 7 V 5 0 D CD DC D V V C CB D CB VC 14 7 A V A 5m V B B B 4kN/m C V C C CB CD DC D 5m H V D D
88 5 CD DC CB H D H H 0 A D AB CB CD DC CB AB CD DC CB EIB EI EIB EI 7 EIC EI EIC EI 10 EIB EIC EI EIB EIC EI EI 3.688EI 5.12EI Solving the three equation B EIB EIC EI 294 C EIB EIC EI
89 Substituting in slope deflection equations AB 0.6 kn. m 3.8 kn. m CB 3.8 kn. m kn. m B _ C CD kn. m 14.4 DC kn. m 0.6 A D
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