Slope deflection method explained

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".

Introduction

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.

Slope deflection equations

The slope deflection equations can also be written using the stiffness factor

K=Iab
Lab
and the chord rotation

\psi=

\Delta
Lab
:

Derivation of slope deflection equations

When a simple beam of length

Lab

and flexural rigidity

EabIab

is loaded at each end with clockwise moments

Mab

and

Mba

, member end rotations occur in the same direction. These rotation angles can be calculated using the unit force method or Darcy's Law.

\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.

Equilibrium conditions

Joint equilibrium

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

Here,

Mmember

are the member end moments,

Mf

are the fixed end moments, and

Mjoint

are the external moments directly applied at the joint.

Shear equilibrium

When there are chord rotations in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account.

Example

The statically indeterminate beam shown in the figure is to be analysed.

L=10m

.

P=10kN

acts at a distance

a=3m

from the support A.

q=1kN/m

acts on BC.

P=10kN

.In the following calculations, clockwise moments and rotations are positive.

Degrees of freedom

Rotation angles

\thetaA

,

\thetaB

,

\thetaC

, of joints A, B, C, respectively are taken as the unknowns. There are no chord rotations due to other causes including support settlement.

Fixed end moments

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

Slope deflection equations

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

Joint equilibrium equations

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

Rotation angles

The rotation angles are calculated from simultaneous equations above.

\thetaA=

40.219
EI

\thetaB=

-6.937
EI

\thetaC=

5.785
EI

Member end moments

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

See also

References

Notes and References

  1. George A.. Maney. 1915. Studies in Engineering. University of Minnesota. Minneapolis.