Chapter 11 Displacement ethod of Analysis Slope Deflection ethod
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
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
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
Displacement ethod of Analysis
Slope-Deflection Equations Derivation
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
Slope-Deflection Equations Angular Displacement at B, B AB 4EI B L 2EI B L
Slope-Deflection Equations Relative Linear Displacement B ' 0. 1 2 1 L L L L 2 EI 3 2 EI 3 0. 6EI AB 2 L
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
Slope-Deflection Equations AB ( FE ) ( FE ) AB
Slope-Deflection Equations Fixed-End oment
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
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 2 3 0 F N ultiply the first equation by 2 and subtracting the second equation from it 3 EK ( FE )' N N N
Fixed End oment Table
Fixed End oment Table
Fixed End oment Table- Continue
Summary 2EK 2 3 ( FE ) N N F N 2EK 2 3 ( FE ) F F N F 3 EK ( FE )' N N N
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
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
Example 1 Draw the bending moment & shear force diagram. Fixed End oment 4 4 FE AB 2 t. m FE 2 t. m 8 2 2 6 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 6 6 3 2I 0 2 C CB 2E B 2 C 0 6 CB EIB 6 6 3
Joint Equilibrium Equations Joint B 0 4 1.7 EIB 2 EIB 6 0. B 3 EI Substituting in slope deflection equations 1.7 AB 0.5EI 2 1.15 t. m EI 1.7 EI 2 3.7 t. m EI 3.7 t. m 2 1.7 CB EI 6 7.1 t. m 3 EI 1.15 Computing The Reactions 4t 3.7 R R A B 1 1.15 4 2 3.7 1.36 t 4 4 1.36 2.64t R A R B1
R R B 2 C 263 3.7 7.1 5.43 t 6 2 6 5.43 6.57t 3.7 2t/m 7.1 5.43 R B2 R C 1.36 + - + 2.64-6.57 3.7 7.1 1.15 - + 1.57 - + 3.79 -
Determine the internal moments in the beam at the supports. Fixed End oment FE FE FE AB 2 6042 26.67 kn. m 2 6 2 6024 53.33 kn. m 2 6 2 30 6 135 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 0 135 EIB 135 6 2 A 0 AB 2 2 A B 0 26.67 AB B 26.67 A 0 2 A 2 B 0 53.33 B 53.33 4m 6m 30kN/m A B C I I 6m
Joint Equilibrium Equations Joint B 0 2 1 7 EIB 53.33 EIB 135 0 EIB 81.67 3 2 6 EI 70 B Substituting in slope deflection equations AB 1 (70) 26.67 3.33 kn. m 3 2 (70) 53.33 100 kn. m 3 1 (70) 135 100 kn. m 2
Example 3 Example 2: Determine the internal moments in the beam at the supports Support A; downward movement of 0.3cm & clockwise rotation of 0.001 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: 0.001 A AB 1.2 0.3 0.0015 600 0.6 1.2 0.001 600 Slope Deflection Equations AB 25000 2 0.001 B 3 0.0015 6 25000 2 B 0.001 3 0.0015 6 0. A B C I 6m I 6m
35000 B 6 Joint Equilibrium Equations Joint B 0 7 0.004 0 B B 0.00057rad 0.001
Substituting in slope deflection equations AB 25000 2 0.001 0.00057 3 0.0015 3.22 t. m 6 25000 2 0.00057 0.001 3 0.0015 3.93 t. m 6 35000 0.00057 0.001 3.93 t. m 6
Example 1b FE AB FE 0.0 A C 0.0
Slope Deflection Equations
Equilibrium Equations
Example 2b
Example 3b
Example 4
Analysis of Frames Without Sway The side movement of the end of a column in a frame is called SWAY.
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 2 5248 80 kn. m 96 2 5248 80 kn. m 96 FE 0. DC Slope Deflection Equations I AB 2E 0. B 0. 0 12 AB I 2E 2 B 0 0. 0 12 A = 0. 0.1667EI A = 0. 0.333EI B B
I 2E 2B C 080 8 I CB 2E 2C B 0 80 8 I CD 2E 2 C 0 0. 0 12 I DC 2E 0 C 0. 0 12 Joint Equilibrium Equations Joint B 0 CB CD 0 0.333EI 0.5EI 0.25EI 80 0. C C B D = 0. D = 0. 0.5EI 0.25EI 80 B C 0.5EI 0.25EI 80 CB C B CD DC 0.333EI C 0.1667EI 0.333EI B 0.5EI B 0.25EI C 80 0. 0.833EI B 0.25EI C 80 Joint C C 0.833EI 0.25EI 80 C B
Two equation & two unknown B Substituting in slope deflection equations AB CB CD DC C 137.1 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 45.7 45.7 82.3 22.9 22.9
Example 6
Example 6b Draw the bending moment diagram Fixed End oment FE FE FE AB FE CB 2 2021 4.44 kn. m 2 3 2 2012 8.89 kn. m 2 3 2 48 4 64 kn. m 12 64 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 30 2 60 kn. m Slope Deflection Equations I AB 2E 2A B 0 4.44 3 A = 0. AB 2 EIB 3 4.44
I 2E A 2B 0 8.89 3 A = 0. 4 EIB 3 8.89 2I 2E 2B C 0 64 4 2I CB 2E B 2C 0 64 4 2EI EI 64 B C EI 2EI 64 CB B C I BE 2E 2B E 0 0 3 I EB 2E 2E B 0 0 3 E = 0. E = 0. BE EB 4 EIB 3 2 EIB 3 2I CF 2E 2C F 0 0 4 2I FC 2E 2F C 0 0 4 F = 0. F = 0. CF FC 2EI EI C C
Equilibrium Equations Joint B BE 0 4 4 EIB 8.89 2EIB EIC 64 EIB 0 3 3 Joint C CB CF CD EI 2EI 64 2EI 60 0 0 B C C EI B 4.67EI EI 55.11 4EI 4 C B C Two equation & two unknown EI B 12.70 EI C 4.18 Substituting in slope deflection equations AB 2 12.7 4.44 4.03 kn. m 3 212.7 4.18 64 42.77 kn. m 4 12.70 8.89 25.83 knm 3 CB 12.7 2 ( 4.18) 64 68.34 kn. m
BE EB 4 12.70 16.94 kn. m 3 2 12.70 8.47 kn. m 3 42.77 25.83 2 EI ( 4.18) 8.35 kn. m CF FC 4.18 kn. m 68.35 60 4.03 16.94 8.35 8.47 4.18
Slope Deflection (Frame with Sway) Analysis of Frames with Sway
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 3 0 12 12 1 1 EIB EI 6 24 I 2E 2 B 0. 3 0 12 12 1 1 EIB EI 3 24
0. 4 2 2 I E 2 B C 3 0 EIB EIC 15 15 15 15 2 4 2 I CB E 2 C B 0 0 EIB EIC 15 15 15 2 1 2 I CD E 2C 0 3 0. EIC EI 18 18 9 54 1 1 2 I DC E 0 C 3 0 EIC EI 18 18 9 54 Equilibrium Equations Joint B B = 0 0 1 1 4 2 EIB EI EIB EIC 0. 3 24 15 15 EI 14.4EI 3.2EI 0 1 Joint C C = 0 CB CD 0 B C
2 4 2 1 EIB EIC EIC EI 0. 15 15 9 54 EI 7.2EI B 26.4EI C 0 2 Three unknown & just two equations so we need another equilibrium equation. Let take F x = 0. 40 H H 0. H H A D A D CD 12 18 AB DC 4018 1.5 0. AB CD DC 1 1 1 1 720 1.5 EIB EI EI B EI 3 24 6 24 2 1 1 1 EIC EI EIC EI 0. 9 54 9 54 0.162EI 0.75EI 0.333EI 720 3 B C
Now solve the three equation 1 14.4 3.2 EI 0 1 7.2 26.4 EIB 0 0.162 0.75 0.333 EI C 720 EIC EIB 136.2 438.9 EI 6756.6 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
135 + 135 + - 95-95 208-110 +
Example 7 Draw the bending moment diagram. EI constant Fixed End oment FE FE FE FE AB CB CD FE 0. 2 2015 375 kn. m 12 2 2015 375 kn. m 12 FE 0. DC Slope Deflection Equations I AB 2E 0. B 3 0 12 12 1 1 EIB EI 6 24
I 2E 2 B 0. 3 0 12 12 1 1 EIB EI 3 24 I 0. 4 2 2E 2B C 3 375 15 15 EIB EIC 375 15 15 I 2 4 CB 2E 2C B 0 375 EIB EIC 375 15 15 15 I 2 1 CD 2E 2 C 0 3 0. 18 18 EIC EI 9 54 I 1 1 DC 2E 0 C 3 0 18 18 EIC EI 9 54 Equilibrium Equations Joint B B = 0 0 1 1 4 2 EIB EI EIB EIC 375 0. 3 24 15 15 EI 14.4EI 3.2EI 9000 1 B C
Joint C C = 0 CB CD 0 2 4 2 1 EIB EIC 375 EIC EI 0. 15 15 9 54 EI 7.2EI 26.4EI 20250 2 Three unknown & just two equations so we need another equilibrium equation. Let take F x = 0. 40 H H 0. A D AB H A 12 CD DC H D 18 4018 1.5 0. AB CD DC B 1 1 1 1 720 1.5 EIB EI EI B EI 3 24 6 24 2 1 1 1 EIC EI EIC EI 0. 9 54 9 54 0.162EI 0.75EI 0.333EI 720 3 B C C
Now solve the three equation 1 14.4 3.2 EI 9000 1 7.2 26.4 R1-R2 EIB 20250 0.162 0.75 0.333 EI 720 0.162R1-R3 C 1 14.4 3.2 EI 9000 0 7.2 23.2 0.22R2-R3 EIB 29250 0 1.58 0.185 EI C 2178 1 14.4 3.2 EI 9000 0. 7.2 23.2 EIB 29250 0. 0. 5.29 EI C 4257 Substituting in slope deflection equations AB 1 1 1467.2 9587.2 155 kn. m 6 24 1 1 1469.6 9587.2 90 kn. m 3 24 4257 EIC 804.7 5.29 EIB 1469.6 EI 9587.2
- CB 4 2 1469.6 ( 804.7) 375 90 kn. m 15 15 2 4 1469.6 ( 804.7) 375 356 kn. m 15 15 CD DC 2 1 ( 804.7) 9587.2 356 kn. m 9 54 1 1 ( 804.7) 9587.2 267 kn. m 9 54 90 90 - - 356 356 155 267
Example 8
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 5 5 2 6 EIB EI 1 5 25 1 2 0. B 3 0 I E 5 5 4 6 EIB EI 1 5 25 1 2 2 B 0 3 0
I 0 BE 2E 2B E 3 0 7 7 I 0 EB 2E 2E B 3 0 7 7 I 1 FE 2E 0. E 3 0 5 5 I 1 EF 2E 2 E 0 3 0 5 5 I E 5 5 I 2 CB 2E 2C B 3 0 5 5 2 2 2 B C 3 0 4 2 EIB EIE 7 7 2 4 EIB EIE 7 7 2 6 EIE EI 5 25 4 6 EIE EI 5 25 4 2 6 EIB EIC EI 5 5 25 2 4 6 EIB EIC EI 5 5 25 1 1 2 2 I 0 CD 2E 2C D 3 0 7 7 I 0 DC 2E 2D C 3 0 7 7 4 2 EIC EID 7 7 2 4 EIC EID 7 7
I E 5 5 I 2 ED 2E 2E D 3 0 5 5 2 DE 2 2 D E 3 0 4 2 6 EID EIE EI 5 5 25 2 4 6 EID EIE EI 5 5 25 2 2 Equilibrium Equations Joint B B = 0 BE 0 4 6 4 2 6 4 2 EIB EI 1 EIB EIC EI 2 EIB EIE 5 25 5 5 25 7 7 380EI 70EI 50EI 42EI 42EI 0 1 Joint E E = 0 B C E 1 2 EB ED EF 0 2 4 2 4 6 4 6 EIB EIE EID EIE EI 2 EIE EI 1 7 7 5 5 25 5 25 50EI 70EI 380EI 42EI 42EI 0 2 B D E 1 2 0 0
Joint C C = 0 CB CD 0 2 4 6 4 2 EIB EIC EI 2 EIC EID 0 5 5 25 7 7 70EI 240EI 50EI 42EI 0 3 B C D 2 Joint D D = 0 DC DE 0 2 4 4 2 6 EIC EID EID EIE EI 2 0 7 7 5 5 25 50EI 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
2 4 6 4 2 6 200 EIB EIC EI 2 EIB EIC EI 2 5 5 25 5 5 25 4 2 6 2 4 6 EID EIE EI 2 EID EIE EI 2 0 5 5 25 5 5 25 6EI 6EI 6EI 6EI 4.8EI 1000 B C D E 2 5 Bottom story F X = 0 40 80 H H 0 A F AB H A 5 EF FE H F 5 4 6 2 6 600 EIB EI 1 EIB EI 1 5 25 5 25 4 6 2 6 EIE EI 1 EIE EI 1 0 5 25 5 25 30EI 30EI 24EI 15000 6 B E 1 H A H F
6 unknown and 6 equation 380 70 0 50 42 42 EIB 0 50 0 70 380 42 42 EI 0 C 70 240 50 0 0 42 EID 0 0 50 240 70 0 42 EIE 0 6 6 6 6 0 4.8 EI 1 1000 30 0 0 30 24 0 EI 2 15000 Substituting in slope deflection equations AB 184.4 kn. m 115.6 kn. m BE EB 147.2 kn. m 147.2 kn. m FE EF EIB EIC EI D EIE 171.79 79.80 79.80 171.79 EI 1 1054.46 EI 2 837.29 184.4 kn. m 115.6 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
68.4 31.6-147.2 68.4 + + + + 115.6 - - - - + 31.6 68.4 68.4 147.2 115.6 184.4 184.4 +-
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
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 = 0.866 So in the slope deflection equations we will use; as the relative displacement of span AB. 0.5 as the relative displacement of span. 0.866 as the relative displacement of span CD. B 60 o 60 o CD AB =30 o D
Fixed End oment FE FE AB FE CB FE CD FE 2 4 12 2 2.667. m FE DC 0. 2.667 t. m 0. Slope Deflection Equations I AB 2E 0. B 3 0 3 3 I 2E 2 B 0 3 0 3 3 I 0.5 2E 2B C 3 2.667 4 4 I 0.5 CB 2E B 2C 3 2.667 4 4 2 2 EIB EI 3 3 4 2 EIB EI 3 3 1 3 EIB EIC EI 2.667 2 16 1 3 EIB EIC EI 2.667 2 16
I 0.866 CD 2E 2 C 0 3 0 6 6 I 0.866 DC 2E 0 C 3 0 6 6 2 0.866 EIC EI 3 6 1 0.866 EIC EI 3 6 Equilibrium Equations Joint B B = 0 0 4 2 1 3 EIB EI EIB EIC EI 2.667 0 3 3 2 16 112EI B 24EI C 23EI 128 1 Joint C C = 0 CB CD 0 1 3 2 0.866 EIB EIC EI 2.667 EIC EI 0 2 16 3 6 24EI B 80EI C 2.072EI 128 2
Third Equilibrium Equations (ethod 1) AB AB DC 3 6 8 2 0.0 DC CD 11 12.92 8.0 8 t 6.92m 21 22 12.92 11.92 96 0 AB CD DC 3.0 2.0 2.0 34.5EI 3.1EI 38.375EI 144 3 B C 3.0 6.0m 6.0
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 1.5 0 A AB A H A A 1.5m AB Free body diagram for Beam V 4 2 4 2 0 V B CB A V A CB V 4 B 4 1.5 AB CB H A 4 2.6 2.6 4 1.5 AB CB CD DC 4 0. 2.6 2.6 4 6 V B B 2t/m DC D C H D CB
24 9 10.4 144 AB CB CD DC 4 2 2 2 1 24 EIB EI EIB EI 9 EIB EIC 3 3 3 3 2 3 1 3 EI 2.667 EIB EIC EI 2.667 16 2 16 2 0.866 1 0.866 10.4 EIC EI EIC EI 144 3 6 3 6 34.5EI 3.1EI 38.375EI 144 3 B C Solving the three equation 112 24 23 EIB 128 24 80 2.072 EIC 128 34.5 3.1 38.375 EI 144 EIB EIC 3.11 2.71 EI 6.77
Solving the three equation 112 24 23 EIB 128 24 80 2.072 EIC 128 34.5 3.1 38.375 EI 144 EIB EIC 3.11 2.71 EI 6.77
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 0.36 0.36 + _ C _ 2.78 CD DC 2.78 t. m 1.88 t. m 2.44 A 2.8 1.88 + D
Example Draw the bending moment diagram. EI constant Before we start let us discus the relative displacement () of each span
The relative displacement () for span AB is equal AB 0. = AB (clockwise) The relative displacement () for span is equal ( 2 ) 1 = ( 1 + 2 ) = (counterclockwise) The relative displacement () for span CD is equal 0. ( CD )= CD (clockwise) A = 1 + 2 B C 2 1 AB CD D Let us build a relationship between AB, & CD take AB = = 2( AB cos) = 2 5/8.6 = 1.163 CD = AB = So in the slope deflection equations we will use; as the relative displacement of span AB. 1.163 as the relative displacement of span. as the relative displacement of span CD. B AB CD 2 1 1 = 2 & CD = AB because of the symmetry in the geometry
Fixed End oment FE FE AB FE CB FE CD FE 47 12 2 0. 16.33 kn. m FE DC 16.33 kn. m 0. Slope Deflection Equations I AB 2E 0. B 3 0 8.6 8.6 I 2E 2 B 0 3 0 8.6 8.6 I 1.163 2E 2B C 3 16.33 7 7 I 1.163 CB 2E 2C B 3 16.33 7 7 2 6 EIB EI 8.6 73.96 4 6 EIB EI 8.6 73.96 4 2 6.978 EIB EIC EI 16.33 7 7 49 2 4 6.978 EIB EIC EI 16.33 7 7 49
I CD 2E 2 C 0 3 0 8.6 8.6 I DC 2E 0 C 3 0 8.6 8.6 4 6 EIC EI 8.6 73.96 2 6 EIC EI 8.6 73.96 Equilibrium Equations Joint B B = 0 0 4 6 4 2 6.978 EIB EI EIB EIC EI 16.33 0 8.6 73.96 7 7 49 103.65EI 28.57EI 6.13EI 1633 1 CB B Joint C C = 0 CD 0 C 2 4 6.978 4 6 EIB EIC EI 16.33 EIC EI 0 7 7 49 8.6 73.96 28.57EI 103.65EI 6.13EI 1633 2 B C
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 7 473.5 0 B CB H A AB CB V A V B 14 7 AB 5 CB H A 14 7 7 7 From the free body diagram for column CD H 7 V 5 0 D CD DC D V 7 473.5 0 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
5 CD DC CB H D 14 7 7 7 6 H H 0 A D AB CB CD DC CB 649 7 5 7 5 0 7 7 10 294 AB CD DC CB 2 6 4 6 2 6 7 EIB EI EIB EI 7 EIC EI 8.6 73.96 8.6 73.96 8.6 73.96 4 6 4 2 6.978 EIC EI 10 EIB EIC EI 16.33 8.6 73.96 7 7 49 2 4 6.978 EIB EIC EI 16.33 294 7 7 49 3.688EI 3.688EI 5.12EI 294 3 Solving the three equation B 103.65 28.57 6.13 EIB 1633 28.57 103.65 6.13 EIC 1633 3.688 3.688 5.12 EI 294 C EIB EIC 18.897 24.603 EI 61.532
Substituting in slope deflection equations AB 0.6 kn. m 3.8 kn. m CB 3.8 kn. m 16.43 kn. m 3.8 3.8 B + 16.43 _ C 16.43 CD 16.43 kn. m 14.4 DC 10.71 kn. m 0.6 A 10.71 D