Slope Deflection Equation - Displacement Method, Strength of Materials | Strength of Material Notes - Agricultural Engg - Agricultural Engineering PDF Download

Slope-Deflection Method

2.1 Sign Convention

For the development and application of Slope-Deflection Method we will use a new sign convention which different from the sign convention discussed in lesson 2.

Moment: At the end of a member clockwise moment is positive.

Transverse displacement: Transverse displacement in upward direction is positive.

Rotation: Rotation in anti-clockwise direction is positive.

The above sign convention is depicted in Figure 11.2.

Fig. 11.2.

2.2 Basic Concept

Slope-deflection equations are obtained by expressing moment at the end of a member as the superposition of end moment due to external loads on the member assuming ends are restrained and end moments caused by the actual end displacements and rotations. If a structure is composed of several members, which is the case with continuous beam, the above concept is applied to each member seperately. For instance, consider member BC in an arbitrarily loaded continuous beam as shown in Figure 11.3.

Fig. 11.3.

Let M_FBC and M_FCB are the end moments due to external loads assuming joint B and C are fixed as shown in Figure 11.4a. δ_B, θ_B and δ_C, _θC are the displacement, rotation at joint B and C respectively as shown in Fgiure 11.4b.

Fig. 11.4.

 The end moments at B and C are expressed as:

M_BC = M_FBC + Moment caused by δ_B, θ_B, δ_C, θ_C

M_CB = M_FCB + Moment caused by δ_B, θ_B, δ_C, θ_C

M_FBC/ M_FCB are often called as fixed end moment. An illustration of the above concept is discussed in the next sub-section.

2.3 Derivation of Slope-Deflection Equations

As mentioned above, in slope-deflection equations, moment at any end is expressed as the sum of fixed end moment and moments due to deflection/rotation.

Fixed end moment

Fixed end moment may be determined by any standard method for solving fixed beam subjected to transverse loading. For ready reference, the fixed end moment for a number of common loading cases are given bellow.

Moment due to rotation at B (= q_B)

Fig. 11.5.

 Let \[M_{BC}^{{\theta _B}}\] and \[M_{CC}^{{\theta _B}}\]  are the moment respectively at B and C due to rotation θ_B at B. Applying Moment-Area method as discussed in Lesson 5, we have,

B. Applying Moment-Area method as discussed in Lesson 5, we have,

\[{\theta _B} = {\rm{Area of M/EI diagram}}\]

\[{\Delta _B} = {\rm{Moment of M/EI diagram about C}}\]

Therefore,

        \[{\theta _B} =-{1 \over {2EI}}\left( {M_{BC}^{{\theta _B}} + M_{CB}^{{\theta _B}}} \right) \times {L_{BC}}\]                                  (1)

\[{\Delta _B}={\theta _B} \times {L_{BC}}=-{1 \over {2EI}}\left( {M_{BC}^{{\theta _B}} + M_{CB}^{{\theta _B}}} \right) \times {L_{BC}} \times \left( {{{{L_{BC}}} \over 2} + {{{L_{BC}}} \over 6}{{M_{BC}^{{\theta _B}} - M_{CB}^{{\theta _B}}} \over {M_{BC}^{{\theta _B}} + M_{CB}^{{\theta _B}}}}} \right)\]                           (2)

\[\Rightarrow {\theta _B} \times {L_{BC}} = {\theta _B} \times {L_{BC}} \times \left( {{1 \over 2} + {1 \over 6}{{M_{BC}^{{\theta _B}} - M_{CB}^{{\theta _B}}} \over {M_{BC}^{{\theta _B}} + M_{CB}^{{\theta _B}}}}} \right)\]

\[\Rightarrow M_{BC}^{{\theta _B}}=-2M_{CB}^{{\theta _B}}\]                                           (2)

From Equation (2) and Equation (1), we have,

\[M_{CB}^{{\theta _B}}={{2EI{\theta _B}} \over {{L_{BC}}}}\]  ; and  \[M_{BC}^{{\theta _B}}=-{{4EI{\theta _B}} \over {{L_{BC}}}}\]

Moment due to rotation at C (= θ_C)

Fig. 11.6.

 Following the similar procedure as in the previous case, \[M_{BC}^{{\theta _C}}\]  and \[M_{CB}^{{\theta _C}}\]  may be obtained as,

\[M_{BC}^{{\theta _B}}=-{{2EI{\theta _B}} \over {{L_{BC}}}}\]  and   \[M_{CB}^{{\theta _B}}={{4EI{\theta _B}} \over {{L_{BC}}}}\] .

Moment due to δ_B, δ_C

Fig. 11.7.

From the above figure we can write,

Change of slope between tangent at B and C = 0.

=>Area of M/EI diagram = 0

\[\Rightarrow M_{BC}^\delta=M_{CB}^\delta \]

Deviation of point C relative to the tangent drawn to B = Deviation of point Brelative to the tangent drawn to C = _δB - δ_C .

\[\Rightarrow-{1 \over 2}M_{BC}^\delta {{{L_{BC}}} \over 2}{{{L_{BC}}} \over 6} + {1 \over 2}M_{CB}^\delta {{{L_{BC}}} \over 2}\left( {{{{L_{BC}}} \over 2} + {{{L_{BC}}} \over 3}} \right) = \delta \]

\[\Rightarrow-M_{BC}^\delta {{L_{BC}^2} \over {24}} + M_{BC}^\delta {{5L_{BC}^2} \over {24}} = \delta \]

\[\Rightarrow M_{BC}^\delta=-{{6EI\delta } \over {L_{BC}^2}}\]

\[\Rightarrow M_{CB}^\delta=M_{BC}^\delta=-{{6EI\delta } \over {L_{BC}^2}}\]

Finally, the end moments at B and C are obtained by combining the contribution from fixed end moment and joint displacement/rotation as,

\[{M_{BC}}={M_{FBC}} + \left( { - M_{BC}^{{\theta _B}}} \right) + \left( { - M_{BC}^{{\theta _C}}} \right) + M_{BC}^\delta \]

\[{M_{CB}}={M_{FCB}} + M_{CB}^{{\theta _B}} + M_{CB}^{{\theta _C}} + M_{CB}^\delta\]

\[\left[ {\matrix{Moments are added according to the{\rm{}}}\cr{sign convention mentioned in 11.2.1}\cr}\right]\]

Substituting all the relevant terms, we have,

\[{M_{BC}}={M_{FBC}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right)\]

\[{M_{CB}}={M_{FCB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _C} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right)\]

The abover two equations are the Slope-Deflection equation.