The slope deflection method is a structural analysis method for beams and frames introduced in 1914 by George A. Maney.[1] The slope deflection method was widely used for more than a decade until the moment distribution method was developed. In the book, "The Theory and Practice of Modern Framed Structures", written by J.B Johnson, C.W. Bryan and F.E. Turneaure, it is stated that this method was first developed "by Professor Otto Mohr in Germany, and later developed independently by Professor G.A. Maney". According to this book, professor Otto Mohr introduced this method for the first time in his book, "Evaluation of Trusses with Rigid Node Connections" or "Die Berechnung der Fachwerke mit Starren Knotenverbindungen".
By forming slope deflection equations and applying joint and shear equilibrium conditions, the rotation angles (or the slope angles) are calculated. Substituting them back into the slope deflection equations, member end moments are readily determined. Deformation of member is due to the bending moment.
The slope deflection equations can also be written using the stiffness factor
K= | Iab |
Lab |
\psi=
\Delta | |
Lab |
When a simple beam of length
Lab
EabIab
Mab
Mba
\thetaa-
\Delta | |
Lab |
=
Lab | |
3EabIab |
Mab-
Lab | |
6EabIab |
Mba
\thetab-
\Delta | |
Lab |
=-
Lab | |
6EabIab |
Mab+
Lab | |
3EabIab |
Mba
Rearranging these equations, the slope deflection equations are derived.
Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. be in equilibrium. Therefore,
\Sigma\left(Mf+Mmember\right)=\SigmaMjoint
Mmember
Mf
Mjoint
When there are chord rotations in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account.
The statically indeterminate beam shown in the figure is to be analysed.
L=10 m
P=10 kN
a=3 m
q=1 kN/m
P=10 kN
Rotation angles
\thetaA
\thetaB
\thetaC
Fixed end moments are:
MABf=-
Pab2 | |
L2 |
=-
10 x 3 x 72 | |
102 |
=-14.7kNm
MBAf=
Pa2b | |
L2 |
=
10 x 32 x 7 | |
102 |
=6.3kNm
MBCf=-
qL2 | |
12 |
=-
1 x 102 | |
12 |
=-8.333kNm
MCBf=
qL2 | |
12 |
=
1 x 102 | |
12 |
=8.333kNm
MCDf=-
PL | |
8 |
=-
10 x 10 | |
8 |
=-12.5kNm
MDCf=
PL | |
8 |
=
10 x 10 | |
8 |
=12.5kNm
The slope deflection equations are constructed as follows:
MAB=
EI | |
L |
\left(4\thetaA+2\thetaB\right)=
4EI\thetaA+2EI\thetaB | |
L |
MBA=
EI | |
L |
\left(2\thetaA+4\thetaB\right)=
2EI\thetaA+4EI\thetaB | |
L |
MBC=
2EI | |
L |
\left(4\thetaB+2\thetaC\right)=
8EI\thetaB+4EI\thetaC | |
L |
MCB=
2EI | |
L |
\left(2\thetaB+4\thetaC\right)=
4EI\thetaB+8EI\thetaC | |
L |
MCD=
EI | |
L |
\left(4\thetaC\right)=
4EI\thetaC | |
L |
MDC=
EI | |
L |
\left(2\thetaC\right)=
2EI\thetaC | |
L |
Joints A, B, C should suffice the equilibrium condition. Therefore
\SigmaMA=MAB+
f | |
M | |
AB |
=0.4EI\thetaA+0.2EI\thetaB-14.7=0
\SigmaMB=MBA+
f | |
M | |
BA |
+MBC+
f | |
M | |
BC |
=0.2EI\thetaA+1.2EI\thetaB+0.4EI\thetaC-2.033=0
\SigmaMC=MCB+
f | |
M | |
CB |
+MCD+
f | |
M | |
CD |
=0.4EI\thetaB+1.2EI\thetaC-4.167=0
The rotation angles are calculated from simultaneous equations above.
\thetaA=
40.219 | |
EI |
\thetaB=
-6.937 | |
EI |
\thetaC=
5.785 | |
EI |
Substitution of these values back into the slope deflection equations yields the member end moments (in kNm):
MAB=0.4 x 40.219+0.2 x \left(-6.937\right)-14.7=0
MBA=0.2 x 40.219+0.4 x \left(-6.937\right)+6.3=11.57
MBC=0.8 x \left(-6.937\right)+0.4 x 5.785-8.333=-11.57
MCB=0.4 x \left(-6.937\right)+0.8 x 5.785+8.333=10.19
MCD=0.4 x -5.785-12.5=-10.19
MDC=0.2 x -5.785+12.5=13.66