Chapter (3) SOPE DEFECTION ETHOD 3.1 Introduction:- The methods of three moment equation, and consistent deformation method are represent the FORCE ETHOD of structural analysis, The slope deflection method use displacements as unknowns, hence this method is the displacement method. In this method, if the slopes at the ends and the relative displacement of the ends are known, the end moment can be found in terms of slopes, deflection, stiffness and length of the members. In- the slope-deflection method the rotations of the joints are treated as unknowns. For any one member bounded by two joints the end moments can be expressed in terms of rotations. In this method all joints are considered rigid; i.e the angle between members at the joints are considered not-to change in value as loads are applied, as shown in fig 1. joint conditions:- to get θ B & θ C BA + BC + BD = 0. ( 1 ) CB + CE = 0. ( )
Figure (1) 3. ASSUPTIONS IN THE SOPE DEFECTION ETHOD This method is based on the following simplified assumptions. 1- All the joints of the frame are rigid, i.e, the angle between the members at the joints do not change, when the members of frame are loaded. - Distortion, due to axial and shear stresses, being very small, are neglected. 3..1 Degree of freedom:- The number of joints rotation and independent joint translation in a structure is called the degrees of freedom. Two types for degrees of freedom. In rotation:- For beam or frame is equal to D r. D r = j-f
Where: In translation:- D r j F = degree of freedom. = no. of joints including supports. = no. of fixed support. For frame is equal to the number of independent joint translation which can be give in a frame. Each joint has two joint translation, the total number or possible joint translation = j. Since on other hand each fixed or hinged support prevents two of these translations, and each roller or connecting member prevent one these translations, the total number of the available translational restraints is; f + h + r + m where f = no. of fixed supports. h = no. of hinged supports. r = no. of roller supports. m = no. of supports. The degree of freedom in translation, D t, is given by:- D t = j-(f+h+r+m) The combined degree of freedom for frame is:- D = D r + D t = j-f + j (f + h + r + m)
D = 3j 3j h r - m The slop 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. For example the frame shown in fig..a p (b) The frame (a) is nine times statically indeterminate. On other hand only tow unknown rotations, θ b and θ c i.e Kinematically
indeterminate to second degree- if the slope deflection is used. The frame (b) is once indeterminate. 3.3 Sign Conventions:- Joint rotation & Fixed and moments are considered positive when occurring in a clockwise direction. A. θ Al 3 EI A. 3 EI θ BI 1 3 A. EI A. 6 EI hence θb1 1 θa1 θa 1 3 B. EI B. 6 EI θb 3 B. EI B. 3 EI θ B1 + θ B = 0
Hence: A = B and θ A = θ A1 -θ A A. 3 E A. 1 E 3A. A 1EI A B 4EI.θA EI.θA Relation between Δ & R by moment area method or by conjugate beammethod. at B. ( ) 4EI 3. 6EI 6EI 6EI.R R (+ ve) when the rotation of member AB with clockwise. 3.4Fixed and moments: As given in the chapter of oment distribution method.
3.5 Derivation of slope deflection equation:- a1 4EI θ A b1 EI θ A a EI θ B b 4EI θ B Required ab & ba in term of (1) θ A, θ B at joint () rotation of member (R) (3) loads acting on member First assume:- Get f ab & f ba due to acting loads. These fixed and moment must be corrected to allow for the end rotations θ A,θ B and the member rotation R. The effect of these rotations will be found separately.
a1 b1 b a 4EI EI 4EI EI.θ A.θ A.θ B.θ B b3 6EI a 3. 6 EI.R by Superposition; f ab = f ab + a1 + a + a3 ab ab 4EI.θ f ab A EI θ EI B (θ 6EI.R A θ B 3R) In case of relative displacement between the ends of members, equal to zero (R = 0) The term ( (K) hence: ab ba f f ab ba EI EI (θ (θ a b θb) θ EI ) represents the relative stiffness of member say ab f ab K ab (θ A θ A b ) )
ba f ba K ba (θ B θ ) a Note: = R is (+ ve) If the rotation of member with clockwise. And ( ve) If anti clockwise. 6 EI =. (with ve R) 6 EI =. (with ve R) 3-5-1 Example 1 Draw B..D. S.F.D Solution:-.66 8 1- Relative stiffness:- K AB : K BC = : 1: - Fixed and oment:- 1 6 F BA 3 6 1 9 t. m. F BA 3 6 1 9, F BC 3 8 1 18 F CB 3 8 1 18 3- Two unknown θ B + θ C then two static equations are required. 1) B = 0 ) C = 0 Hence: BA + BC = 0 (1)
BC = 0. () But: AB = - 9 + θ B BA = 9 + 1 (θ B) BC = -16 + (θ B + θ C ) CB = +16 + (θ C + θ B ) From eqns. (1&) 9 + θ B + (- 16 + (θ B + θ C ) = 0 6θ B + θ C = 7.(3) and 4θ C + θ B = - 16 from 3 & 4 θ C + θ B = - 8.. (4) 5 θ B = 15 15 θ B = = 3.0 5 1.e. θ C = - 5.5 AB = - 9 + 3.4 = 5.6 t.m BA = 9 + 3.4 = 15.8 t.m BC = - 18 + ( 3.4) + (- 5.5) = - 15.0 t.m CB = 16 + (.3 5.7 + 3.4) = 0.0 (0.k)
1- Unknowns θ, A θ, B & θ C - Fixed end oment F AB = F BC = F CD = 3- Condition eqns. 6 1 = - 6 t.m etc AB = - 4 t.m, BA + BC = 0, & CB + CD = 0 4- Slope deflection equations From eqn.3 θ C = - AB = - 6 + (θ A + θ B ) = - 4 θ A + θ B =.(1) BA + BC = 0 + 6 + (θ B + θ A ) 6 + (θ B + θ C ) = 0 4θ B + θ A + θ C = 0 () CB + CD = 0 = 6 + θ C + θ B 6 + θ C = 0 4 θ C + θ B = 0.(3) θb 4 Substitute in eqn. () Hence: 3.75 θ B + θ A = 0...() 0.5θ B + θ A = 1 () 3.5θ B = - 1 θ B = - 1 θ A = 1.15 θ C = 0.077
Hence: AB = - 6 + xl.15 + (-.307) = - 4 t.m 0.K BA = 6 + x (-.307) + 1.15 = 6.536 t.m CB = 6 + x.77 + (-.307) = 5.85 t.m DC = 6 +.077 = 6.077 t.m Solution:- 1- Unknown displacements are θ B & θ D - Equations of equilibrium are:- DB = 0.(1) BA + BD + BC = 0 () 3- Relative Stiffness:- K AB : K BC : K BD = 35:31. 5: ; 51. 56:1. 4:1.0.
4- Fixed and oments: 9633 F AB 6 t. m 99 9636 F BA 1 99 t. m 3 7 F BD 1 1.5 t. m 3 7 F DB 1 1.5 t. m From the equations 1 & hence; DB = F DB + (θ D + θ B ) = 1.5 + 1 (θ D + θ B ) = 0 θ D + θ B + 1.5 = 0 ------- (3) and BA = 1 + 1.56 (θ B ) BD = 1.5 + 1.0 (θ B + θ D ) BC = 0 + 1.4 (θ B + 0) i.e. 1+1.56 (θ B ) 1.5 + θ B + θ D + 1.4 (θ B ) = o 7.9θ B + θ D.5 = 0 ------------- (4) 0.5θ B + θ D + 6.15 = 0 ------------- (3) i.e 7.4 θ B - 6.375 = o θ B = 0.86 θ D = - 6.55 Hence: BA = 1 + 1.56 (.86) = 14.68 t.m
BD = - 1.5 + (.86 6.55 1) = - 17.08 BC = 1.4 (.86) =.41 DB = 1.5 + ( -6.55) = zero 1 CB = BC = 1.05 AB = -6 + 1.56 (.86) = - 4.66
Two equilibrium eqns. AB + AA = 0 (1) BB + BA + 4= 0.() Slope deflection eqns. AB = o + 1.6 (θ A + θ B ) 10 16 8 AA = (θ θ ) A AA = - 0 + θ A BA = o + 1.6 (θ B + θ A ) BB = - 4.67 + (θ B + θ B ) = - 4.67 + θ B Hence: 3.θ A + 1.6θ B + θ A 0 = o 4.θ A + 1.6θ B = 0 (1) - 4.67 + 4. θ B + 1.6θ A + 4 = 0 1.6θ A + 4.θ B = 38.67.() 1.6θ A + 0.61θ B = 7.6.(1) A 3.59θ B = 31.05 θ B = 8.65 θ A = 1.46 AB = -18.5
BA = 30 BB = - 34 Example 5 Draw B..D for the shown frame Solution:- - Two condition equations. AA + AB = 0.....(1) BA + BB + 8 = 0..() - Relative stiffness 16 1 : 10 1 = 1:1.6 - Slope deflection equations: AA = (θ A θ A ) = θ A AB = (θ A θ B ) 1.6 BA = (θ B θ A ) θ A BB = 4.67 + (θ B - θ B ) Hence: θ A + 3.θ A + 1.6V B = 0 4.θ A + 1.6θ B = 0. (1) 3.θ B + 1.6θ A + θ B 4.67 + 8 = 0 4.θ B + 1.6θ A = 34.67 ()
By Solving 1 & θ A = - 3.68, θ B = 9.66 Hence AA = - 3.68, AB = 3.68 t.m BA = 5 BB = 33 Example 6: - Draw B..D for the given structure. Solution:- once statically indeterminate. 1- Fixed end moments F AB = - F BA = - F BC = - F CB = - F DB 8 0 8 8 0 8 4 10 8 10 8 8 10 0 t. m 0 t. m 5 t. m t. m = 10 t.m - From Static:- B = o
BA + BC + BD = 0 BA = F BA + (θ B ) BA = 0 + θ B.. (1) BC = - 5 + θ B.. () BD = - 10 + θ B. (3) Hence: 5 + 6θ B = o θ B = - o.833 Hence: BA = 18.34 t.m, BC = -6.67, BD = -11.67 t.m AB = - 0 = - 0.833 t.m CB = 5 + θb = - 4.167 t.m DB = 10 + θb = 9.167 t.m Example 7:
Draw B..D for the shown frame Solution: 3 time statically ind. θ A, θ B, & θ C 1- Fixed end moments: F AB = - 10 F BA = + 10 F BC = - 5 F CD = F DC = zero - Relative Stiffness 1:1:1 AB = 0.(1) BA + BC = 0.() CB + CD = 0.(3)
Equs. AB = - 10 + (θ A + θ B ) BA = 10 + (θ B + θ A ) BC CB CD = - 5 + θ B + θ C = 5 + θ C + θ B = θ C DC = θ C From 1, & 3 θ A + θ B = 10.(1) 4θ B + θ A + θ C = 15.() 4θ C + θ B = - 5 (3) By solving the three eqns. hence; θ A =.5 θ B = 5 θ C = - 7.5 Substitute in eqns of moments hence; AB = - 10 + 5 = zero (o.k) BA = 10 + 10 +.5 =.5 t.m BC = - 5 +10 7.5 = -.5 t.m CB = 5 15 + 5 = 15 t.m
CD = - 15 t.m DC = - 7.5 t.m 3-6 Frames with Translation Examples to frames with a single degree of freedom in translation.
Example 8: Draw B..D for the shown frame. 1- Unknowns: θb, θc, - Relative stiffness K AB : K BA : K CD 1 1.5 : : 4 8 6 1: 1 : 1 3- Fixed end moments F AB = o F BC = CB = zero F CD = - 6 t.m F DC = + 6 t.m BA = o
4- From Statics the equilibrium eqns BA + BC = 0 (1) CB + CD = 0.() 5- Shear equation (In direction of X, = o) 6 + X A + X D 8 = o BA AB CD DC 6 + 40 4 6 (3) hence X A BA AB 4 and xd 6- Slope deflection eqns: CD DC 6 4 BA 0 + 1 (θ B 3 4 ), AB = 0 + 1 (θ B 3. 4 )
BC = 0 + 1 (θ B + θ C ) Hence: 4θ B 0.75 + θ C = 0 (1) CB = 0 + 1 (θ C + θ B ) CD = - 6 + 1 (θ C 3 ), 6 Hence: DC = + 6 + 1 (θ C -3 4 ) 1 4 θ C + θ B - = 6 () ( B.75 ) (1 B.75 ) ( 6 C ) (61 C + =0 4 6 + 0.75 θ B -.375 + 1 θc 0.1667 = o θ B +.67 θ C - 07 = -.66 (3) Subtract (3) from () 3.33 θ 1 + 0.88 = 8.33 θ C 0.067 =.6 (4) Subtract (1) from () (4) 15 θ C 1.5 = 4 θ C 0.08 = 1.6 (5) From (4) & (5) 0.147 = 1 = 6.80 θ C =.149 θ B = 0.799
BA = - 3.5 t.m, AB = - 4.301 t,m, BC = 3.79 CB = 5.1 t.m, CD = - 5.1 t.m, DC = 4.744 Example 9:- Write the shear & condition eqns for the following frame. Solution:- Three unknowns: θb, θc, Condition equations: BA + BC = o (1) CB + CD = o () Shear eqn. X A +X B + P1 + P = o P1 AB BA ( ) + ( h1 Example 10: CD h Find the B..D for the shown structure. DC ) P1 + P) = o (3)
Solution:- θ D = θ E = o θ C = - θ C θ B = - θ B 1- Unknown displacements are: θb, θc, - Relative Stiffness: AB : BE : BC : CD : ED 1 : 5 3 1 1 1 : : : 5 3 3 3 : 10: 3 : 5: 5 3- Fixed end moment:-
F BE = - 436 1 F EB = + 1 t.m F CD = 1.536 1 F DC = + 4.5 1 t. m 4.5 t. m 4- Equilibrium equations:- 1- CD + CB = o - BC + BA + BE = o 3- Shear condition:(33 16.5)+ CD = - 4.5 + 5 (θ C + θ D 3R) CB = 0 3 (θ C + θ B ) BC = 0 + 3 (θ B + θ C ) BA = 0 + 3 (θ B ) BE = - 1 + 10 (θ B 3R) CD 6 DC DE 6 ED Hence - 4.5 + 10 θ C 15 R + 6 θ C + 3 θ B = 0 16θ C + 3θ B 15R 4.5 = 0 (1) And 16θ B + 3θ C + 6θ B 1 + θ B - 3θ R = 0 3θ C + 3θ B 30R 1 = 0 () and 15θC 30R 30 θb 60R 16.5( ) 0 6 6
.5 θ C + 5θ C + 17R + 16.5 = 0 (3) by solving equation 1, & 3 get AB = + 6.66 t.m BA = + 13.3 t.m BC = + 19.0 t.m CB = + 18 t.m BE = - 3.3 t.m EB = - 30.53 t.m CD = - 18 t.m DC = - 18.43 t.m 3-7 Frame with multiple degree of freedom in translation. Example 11: Write the shown equations and condition eqns for the given frame. Solution Unknowns: θb, θc, θd, θe, 1, 1
Condition eqns BE + BA + BC = 0 (1) CB + CD = 0 () DC + DE = 0 (3) EB + EF + ED = 0 (4) Shear eqns : Equilibrium of the two stories. At sec (1) (1) :- (evel CD) P + X c + X E = 0 P + CB BC DE h h At sec. () ():- (evel BE) or x = 0 P 1 + P + x A + x F = 0 P 1 + P + BA ED = 0 AB EF h1 h1 FE = 0
Example 1:- Draw B..D for the given structure. S olution:- 1- Relative Stiffness:-
- Equilibrium equations:- AB + AC = 0 (1) BA + BD = 0 () CA + CD + CE = 0 (3) DB + DF + DC = 0 (4) Σx = 0 at evel A-B + (6-3) + AC 6 CA BD 6 DB = 0 (5) Σx = 0 at evel CD 11 + CE 6 EC DF 6 FD = 0 (6)
AB = - 8 + 1 (θ A + θ B ) AC = 3 + (θ A + θ C 3R 1 ) AC = - 3 + (θ C + θ A 3R 1 ) CA = 16 + (θ B + θ A ) BD = 0 + (θ B + θ D 3R 1 ) DB = 0 + (θ D + θ B - 3R 1 ) DF = 0 + (θ D + 0-3R ) FD = 0 + (θ D - 3R ) CD = - 48 + (θ C + θ C ) CD = + 48 + (θ D + θ C ) CE = - 8 + (θ C - 3R ) EC = + (θ C - 3R ) 3- Fixed end moment:- 9484 11 F AB = 8t.m 98 4 F BA = 16 t.m 1 16 1 F AC = 3t.m
16 1 F CA = 3t.m F CD = 48 41 1 t.m F DC = + 48 t.m 4- Unknown displacement: θa, θb, θc, θd, 1, by Solving the six equations one can get; AB = - 3.84 t.m BA = + 18.39 t.m AC = 3.84 t.m CA = + 7.9 t.m BD = - 18.39 t.m DB = -.97 t.m CD = - 11.15 t.m DC = - 53.44 t.m CE = 3.87 t.m EC = - 13.44 t.m DF = - 30.47 t.m FD = - 6.15 t.m
Example (13):- Write the shear equations & equilibrium equations for the shown frame. Solution: Shear eqns: X CE + X BA + P 1 EC h = 0...(1) CE AB BA 1 h1 h x D + x G + x E + P = 0 --- () + P 1 = o Or: DE h ED GF FG EC h h1 X A + X D + X G + P 1 + P = 0 h AB DE ED GF 1 h h h FG CE + P 1 + P = 0 + P = 0
Example 14:- a- Write the equations of equilibrium including the shear equations for the frame. b- Write the slope deflection equations in matrix for members CE & GH. c- By using the slope - deflection method; sketch elastic curve. d- Sketch your expected B..D Solution:-
(Unknowns = θ C, θ D, θ E, θ F, θ G, θ A +θ K, θ, 1,, 3, 4 Relative stiffness: 1 : 1 a- equilibrium equations K + KG = 0 (1) K + H = 0 () GK + GH + GE = 0 (3) HG + H + HF = 0 (4) EG + EC + FF = 0 (5) FE + FD + FH = 0 (6) CE + CD + CA = 0 (7) DC + DB + DF = 0 (8) Shear equations:- a- at evel GH 5 + 10 + (X G 5) + X H = 0 (9)
Where: X G = X H = GK KG 5 H H 5 b- at evel EF Where: 5 + 10 + 0 + (X E 5) + X F = 0 30 + X E + X F = 0 (10) EG X E = GE 5 X F = FH HF 5
c- at evel CD Where: 5 + 10 + 10 + 30 + (X C 5) + X D = 0 50 + X C + X D = 0 (11) X C = X D = CE EC 5 DF FD 5 d- at Sec AB:- 5 + 10 + 10 + 10 + 40 + (X A 5) + X B = 0
70 + X A + X B = 0..(1) X A = X B = AC CA 5 CE EC 5 3-8 Slope deflection eqns in matrix form: 1- ember CE Where: CD = F CE + EC = F EC + F CE = - 5 1 EI (θ C + θ E 3 5 EI (θ E + θ C 3 5 = - 4.16 t.m F EC = + 4.16 t.m In atrix form: CE = - 4.16 EI 5 1 5 1 5 ) ) 1 3 θ C EC 4.16 1 3 R θ E Where: R = 1 5
- member GH GH = - 16.67 6I 5 1 θ G HG + 16.67 1 θ H d- B..D Example 15:- By using slope deflection method; 1- Draw B..D for the shown frame. - Sketch elastic curve. Solution: 1- Relative stiffness 1: 1 - unknowns: θ B = - θ B (From symmetry) 3- Equilibrium eqns BA + BC + BD + BB = o (1)
4-Fixed end moments F AB = 46 1 = - 1 t.m F BA = = + 1 F BC = F CB = F BD = F DE = o F BB = - 1 1 81 8 = - 36 t.m 4- Slope deflection eqns AB = - 1 +(θb) BA = 1 + θb
BC = θb BD = θb BB = - 36 + θb CB = θb DB = θb From eqn (1) (1 + θ B ) + (θ B ) + (θ B ) + (- 36 + θ B ) = 0 7θ B - 4 = 0 θ B = 3.486 hence AB = - 8.57 t.m BA = 18.86 t.m BC = 6.86 t.m BB = - 3.58 t.m CB = 3.48 t.m
DB = 3.48 t.m BD = 6.86
The Free Body Diagram to find the S. F. & N. F. SHEET (3) 1) Draw S.F.D. and B..D. for the statically indeterminate beams shown in figs. From 1 to 10.
) Draw N.F.D., S.F.D. for the statically indeterminate frames shown in figs. 11 to 17. Using matrix approach 1.