The slope deflection method is a structural analysis method for beams and frames introduced in 1914 by George A. Maney. 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".
Now let us consider following example(take lengths of all beams=L)
As in the above case of statically indeterminate beams let us consider there is no sinking of support anywhere.Now we will construct the slope deflection equations(assuming that you know what is fixed end moment or refer to the image below).
(Example of fixed end moment calculation:- M(fbc)=PL/8)
For beam section AB
As end A is fixed there would be no immediate deflection there that is ɵ(A)=0
but however as end B is a roller support a ɵ(B) would be existing.
Now
Moment about A:(Mab)= M(fab) + 2EI/L(2ɵ(A)+ɵ(B))............................(1)(predefined equation)
In above cast M(fab)=0(because there is no loading in section AB)
∴ M(AB) = 2EI/L (ɵ(B)).....................................................(2)
similarly
M(BA) = 2EI/L (2ɵ(B))...................................................(3)
For beam section BA
In this case both ends are roller and also a load of P KN is applied at mid point of beam thus there would be fixed end moments at both ends which can be obtained from above image.
Now
M(BC)= -M(fbc) + 2EI/L(2ɵ(B)+ɵ(C))..........................(4)( minus sign is introduced because fixed end moment about b would be anti-clockwise)
Similarly
M(CB)= M(fbc) + 2EI/L(2ɵ(C)+ɵ(B))..............................(5)
For beam section CD
In this case both the ends are roller with no loading.
Now
M(CD)= -M(fcd) + 2EI/L(2ɵ(C)+ɵ(D))..............................(6)
Similarly
M(DC)= M(fdc) + 2EI/L(2ɵ(D)+ɵ(C)).................................(7)
Now let us consider following example(take lengths of all beams=L)
ANALYSIS
As in the above case of statically indeterminate beams let us consider there is no sinking of support anywhere.Now we will construct the slope deflection equations(assuming that you know what is fixed end moment or refer to the image below).
For beam section AB
As end A is fixed there would be no immediate deflection there that is ɵ(A)=0
but however as end B is a roller support a ɵ(B) would be existing.
Now
Moment about A:(Mab)= M(fab) + 2EI/L(2ɵ(A)+ɵ(B))............................(1)(predefined equation)
In above cast M(fab)=0(because there is no loading in section AB)
∴ M(AB) = 2EI/L (ɵ(B)).....................................................(2)
similarly
M(BA) = 2EI/L (2ɵ(B))...................................................(3)
For beam section BA
In this case both ends are roller and also a load of P KN is applied at mid point of beam thus there would be fixed end moments at both ends which can be obtained from above image.
Now
M(BC)= -M(fbc) + 2EI/L(2ɵ(B)+ɵ(C))..........................(4)( minus sign is introduced because fixed end moment about b would be anti-clockwise)
Similarly
M(CB)= M(fbc) + 2EI/L(2ɵ(C)+ɵ(B))..............................(5)
For beam section CD
In this case both the ends are roller with no loading.
Now
M(CD)= -M(fcd) + 2EI/L(2ɵ(C)+ɵ(D))..............................(6)
Similarly
M(DC)= M(fdc) + 2EI/L(2ɵ(D)+ɵ(C)).................................(7)
SOLVING THE EQUATIONS
It is a known fact that moment about a roller support is zero thus moment about B,C and D must be zero. Hence
M(BA)+M(BC)=0..............................................................(8)
And
M(CB)+M(CD)=0...............................................................(9)
Also
M(DC)=0............................................................................(10)
Now using equations 8,9 and 10 ɵ(A),ɵ(B),ɵ(C),ɵ(D) can be obtained. Thus after obtaining them we could use them to obtain all the moments about the end A,B,C and D and thus enabling us to draw BENDING MOMENT DIAGRAM
I HOPE THAT YOU HAVE UNDERSTOOD THE BASICS OF SLOPE DEFLECTION METHOD. IN NEXT BLOG I WILL BE COVERING MORE DIVERSITY IN BEAMS AND FRAMES.
FOR ANY QUERY OR SUGGESTION PLEASE COMMENT.
HAVE A NICE DAY.....................😇😇😇😇