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Chapter (3)
SLOPE DEFLECTION METHOD
3. 1 Introduction:-
The methods of three moment equation, and consistent deformation method are represent the FORCE METHOD of structural analysis, The slope deflection method use displace- ments 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 θB & θC MBA+MBC+MBD = 0 …………. ( 1 ) MCB+MCE = 0 …………. ( 2 )
Figure (1)
3 .2 ASSUMPTIONS IN THE SLOPE DEFLECTION METHOD 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. 2 - Distortion, due to axial and shear stresses, being very small, are neglected. 3 .2.1 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 Dr.
Dr = j-f
^
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. 2.a
The frame (a) is nine times statically indeterminate. On other hand only tow unknown rotations, θb and θc i.e Kinematically
D = 3j – 3j – 2h – r - m
^ ^
p
(b)
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.
3 EI
M (^) A.L 32 EI
θ^2 MA.L Al
6 EI
MA.L 2 EI
MA.L 3 θBI ^1 ^
hence (^) θ (^) B 1 ^12 θ (^) A 1
6 EI
MB.L 2 EI
MB.L 3 θA2 ^1
3 EI
MB.L 2 EI
MB.L 3 θB2 ^2
θB1 + θB2 = 0
^ ^ (^)
^
(^)
3.5 Derivation of slope deflection equation:-
Ma1 ^4 LEIθA
Mb1 ^2 LEIθA
Ma2 ^2 LEIθB
Mb2 ^4 LEIθB
Required Mab & Mba in term of (1) θA, θB at joint (2) rotation of member (R) (3) loads acting on member First assume:- Get Mfab & Mfba due to acting loads. These fixed and moment must be corrected to allow for the end rotations θA,θB and the member rotation R. The effect of these rotations will be found separately.
(^)
Ma1 ^4 LEI .θA
Mb1 ^2 LEI .θA
Mb2 ^4 LEI .θB
Ma2 ^2 LEI .θB
Mb3 M (^) a 3 L^62 EI .
^6 LEI.R
by Superposition; Mab = Mfab + Ma1 + Ma2 + Ma Mf (^) ab^4 LEI. θA^2 LEI θB^ ^6 LEI.R
M (^) ab Mfab^2 LEI( 2 θAθB 3 R)
In case of relative displacement between the ends of members, equal to zero (R = 0)
M (^) ab Mfab^2 LEI( 2 θaθb)
M (^) ba Mfba^2 LEI( 2 θbθA)
The term ( L
2 EI
) represents the relative stiffness of member say
(K) hence: M (^) ab MfabKab( 2 θAθb)
MBC = 0………………. (2)
But: MAB = - 9 + θB MBA = 9 + 1 (2θB) MBC = -16 + 2 (2θB + θC) MCB = +16 + 2 (2θC + θB) From eqns. (1&2) 9 + 2θB + (- 16 + 2 (2θB + θC) = 0 6 θB + 2θC = 7 …….(3) and 4 θC + 2θB = - 16 2 θC + θB = - 8 …….. (4) from 3 & 4 5 θB = 15 θB = 155 = 3. θC = - 5. 1.e. MAB = - 9 + 3.4 = 5.6 t.m MBA = 9 + 2 3.4 = 15.8 t.m MBC = - 18 + 2 (2 3.4) + (- 5.5) = - 15.0 t.m MCB = 16 + 2 (2.3 – 5.7 + 3.4) = 0.0 (0.k)
^ ^
(^) ^
^
^
^ ^
^
(^) ^
Hence: MAB = - 6 + 2xl.15 + (- .307) = - 4 t.m 0.K MBA = 6 + 2x (- .307) + 1.15 = 6.536 t.m MCB = 6 + 2x .77 + (- .307) = 5.85 t.m MDC = 6 + .077 = 6.077 t.m
Solution:- 1- Unknown displacements are θB & θD 2- Equations of equilibrium are:- MDB = 0 ……….(1) MBA + MBD + MBC = 0 ………(2) 3- Relative Stiffness:- KAB: KBC: KBD = 35:31. 5:22 ; 51. 56:1. 4:1.0.
^
^
(^) ^
^
(^)
4- Fixed and Moments: MFAB ^9 96 93 ^3 6 t. m
MFBA ^9 96 93 ^6 12 t. m
MFBD^312 712. 25 t. m
2
MFDB^312 712. 25 t. m
2
From the equations 1 & 2 hence; MDB = MFDB + (2θD + θB) = 12.25 + 1 (2θD + θB) = 0 2 θD + θB + 12.25 = 0 ------- (3) and MBA = 12 + 1.56 (2θB) MBD = 12.25 + 1.0 (2θB + θD) MBC = 0 + 1.4 (2θB + 0) i.e. 12+1.56 (2θB) – 12.25 + 2θB + θD + 1.4 (2θB) = o 7.92θB + θD – .25 = 0 ------------- (4) 0.5θB + θD + 6.125 = 0 ------------- (3) i.e 7.42 θB - 6.375 = o θB = 0. θD = - 6. Hence: MBA = 12 + 1.56 (2× .86) = 14.68 t.m
Two equilibrium eqns. MAB + MAA = 0 ………………… (1) MBB + MBA + 4= 0 ………………….(2)
Slope deflection eqns. MAB = o + 1.6 (2θA + θB) MAA = ^108 ^16 ( 2 θ A θA) MAA = - 20 + θA MBA = o + 1.6 (2θB + θA) MBB = - 42.67 + (2θB + θB) = - 42.67 + θB Hence: 3.2θA + 1.6θB + θA – 20 = o 4.2θA + 1.6θB = 20 ……………… (1)
- 42.67 + 4.2 θB + 1.6θA + 4 = 0 1.6θA + 4.2θB = 38.67 ……………….(2) 1.6θA + 0.61θB = 7.62 ……………….(1)
3.59θB = 31. θB = 8. θA = 1. MAB = -18.
MBA = 30
MBB = - 34
Example 5 Draw B.M.D for the shown frame Solution:-
- Two condition equations. MAA + MAB = 0 …….. …………...(1) MBA + MBB + 8 = 0 …………………..(2)
- Relative stiffness (^) 161 : (^) 101 = 1:1.
- Slope deflection equations: MAA = (2θA – θA) = θA MAB = (2θA – θB) × 1. MBA = (2θB – θA) × θA MBB = 42.67 + (2θB - θB)
Hence: θA + 3.2θA + 1.6VB = 0 4.2θA + 1.6θB = 0 .… (1) 3.2θB + 1.6θA + θB – 42.67 + 8 = 0 4.2θB + 1.6θA = 34.67… (2)
^
(^)
MBA + MBC + MBD = 0
MBA = MFBA + (2θB) MBA = 20 + 2θB ……………….. (1) MBC = - 5 + 2θB ……………….. (2) MBD = - 10 + 2θB ………………. (3)
Hence: 5 + 6θB = o θB = - o. Hence : MBA = 18.34 t.m , MBC = -6.67, MBD = -11.67 t.m MAB = - 20 = - 20.833 t.m MCB = 5 + θB = - 4.167 t.m MDB = 10 + θB = 9.167 t.m
Example 7:
Draw B.M.D for the shown frame Solution: “ 3 time statically ind.” θA , θB , & θC 1- Fixed end moments: MFAB = - 10 MFBA = + 10 MFBC = - 25
MFCD = MFDC = zero
2- Relative Stiffness 1:1: MAB = 0 …………………….(1) MBA + MBC = 0 …………………….(2) MCB + MCD = 0 …………………….(3)
