03 Slopedeflection

DISPLACEMENT METHOD OF ANALYSIS:
SLOPE DEFLECTION EQUATIONS
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General Case
Stiffness Coefficients
Stiffness Coefficients Derivation
Fixed-End Moments
Pin-Supported End Span
Typical Problems
Analysis of Beams
Analysis of Frames: No Sidesway
Analysis of Frames: Sidesway

1

Slope – Deflection Equations
P

i

w

j

k

Cj

settlement = ∆j

Mij

P

i

w

j

Mji

θi ψ θj

2

Degrees of Freedom
M

θΑ

A

B

1 DOF: θΑ

C

2 DOF: θΑ , θΒ

L

θΑ

A

P
B
θΒ

3

Stiffness

kAA

kBA

1

B

A
L
k AA =

4 EI
L

k BA =

2 EI
L

4

kBB

kAB

A

B

1
L
k BB =

4 EI
L

k AB =

2 EI
L

5

Fixed-End Forces
Fixed-End Moments: Loads
P
L/2

PL
8

L/2

PL
8

L
P
2

P
2

w wL2 12 wL 2

L

wL2
12
wL
2

6

General Case

P

i

w

j

k

Cj

settlement = ∆j

Mij

P

i

w

j

Mji

θi ψ θj

7

Mij

P

i

j

w

Mji

θi
L

θj

ψ
4 EI
2 EI θi + θj = M ij L
L

Mji =

2 EI
4 EI θi + θj L
L

θj

θi
(MFij)∆

+
(MFji)∆
settlement = ∆j

+
P
(MFij)Load

M ij = (

settlement = ∆j

w
(MFji)Load

4 EI
2 EI
2 EI
4 EI
)θ i + (
)θ j + ( M F ij ) ∆ + ( M F ij ) Load , M ji = (
)θ i + (
)θ j + ( M F ji ) ∆ + ( M F ji ) Load 8
L
L
L
L

Equilibrium Equations i P

w

j

k

Cj

Cj M

Mji
Mji

jk

Mjk j + ΣM j = 0 : − M ji − M jk + C j = 0

9

Stiffness Coefficients
Mij

i

j

Mji

L

θj

θi

kii =

4 EI
L

k ji =

2 EI
L

×θ i

k jj =

4 EI
L

×θ j

1

kij =

2 EI
L

+
1

10

Matrix Formulation

M ij = (

4 EI
2 EI
)θ i + (
)θ j + ( M F ij )
L
L

M ji = (

2 EI
4 EI
)θ i + (
)θ j + ( M F ji )
L
L

 M ij  (4 EI / L) ( 2 EI / L) θ iI   M ij F 
M  = 
 θ  +  M F 
EI
L
EI
L
(
2
/
)
(
4
/
)
  j   ji 
 ji  

 kii
k ji

[k ] = 

kij  k jj 

Stiffness Matrix
11

P

i

Mij

w

j

Mji

θi

L

[ M ] = [ K ][θ ] + [ FEM ]

θj

ψ

∆j

([ M ] − [ FEM ]) = [ K ][θ ]

[θ ] = [ K ]−1[ M ] − [ FEM ]

Mij

Mji

θj

θi

Fixed-end moment
Stiffness matrix matrix

+
(MFij)∆

(MFji)∆

[D] = [K]-1([Q] - [FEM])
+
(MFij)Load

P

w

(MFji)Load

Displacement matrix Force matrix

12