Derivation of the Slope Deflection Equation PDF

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ivation of the Slope Deflection Equation...

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UNIVERSITY OF GUYANA FACULTY OF ENGINEERING & TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING CIV 3115 – STRUCTURAL ANALYSIS I DERIVATION OF THE SLOPE-DEFLECTION EQUATION In order to develop the general form of the slope-deflection equations, consider the typical span AB of a continuous beam as shown. The beam is subjected to an arbitrary loading and has a constant 𝐸𝐼. The objective is to relate the beam’s internal end moments 𝑀𝐴𝐵 and 𝑀𝐵𝐴 in terms of its degree of freedom, 𝜃𝐴 and 𝜃𝐵 , and the linear displacement ∆ which is caused by a relative settlement between the supports and the given loadings. Moments and angular displacements are considered positive when acting clockwise.

Figure 1

Consider node A of the member shown. There is a rotation at A, 𝜃𝐴 while its far-end B is held

fixed. The method of the conjugate beam will be utilized to determine the moment 𝑀𝐴𝐵. The end shear at 𝐴’ acts downward on the beam because 𝜃𝐴 is clockwise. The deflection on the real beam at A and B are both zero. Therefore, the sum of moments at each end 𝐴’ and 𝐵’ of the beam

must also be zero.

Figure 2

𝐿 1 𝑀 2𝐿 1 𝑀 ↺ +, ∑ 𝑀𝐴′ = 0, [ ( 𝐴𝐵 ) 𝐿] − [ ( 𝐵𝐴 ) 𝐿] =0 3 2 𝐸𝐼 3 2 𝐸𝐼 1 𝑀𝐴𝐵 𝐿 2𝐿 1 𝑀𝐵𝐴 ) 𝐿] − [ ( ) 𝐿] + 𝜃𝐴 𝐿 = 0 ↺ +, ∑ 𝑀𝐵′ = 0, [ ( 3 2 𝐸𝐼 2 𝐸𝐼 3 Simplifying both equations:

1

6𝐸𝐼 1 𝑀𝐵𝐴 2𝐿 𝐿 𝐿 ([ 𝑀𝐴𝐵 = 0) × ( ) 𝐿] ) 𝐿] − [ ( 3 3 2 𝐸𝐼 2 𝐸𝐼 1 𝐿 ([ (𝑀𝐵𝐴 ) 𝐿] 𝐿 − [1 (𝑀𝐴𝐵) 𝐿] 2𝐿 + 𝜃 𝐿 = 0) × 6𝐸𝐼 𝐴 2 𝐸𝐼 3 2 𝐸𝐼 3 [1]

(𝑀𝐴𝐵 𝐿 − 2𝑀𝐵𝐴 𝐿 = 0) × 2 𝑀𝐵𝐴 𝐿 − 𝑀𝐴𝐵 2𝐿 + 6𝜃𝐴 𝐸𝐼 = 0 2𝑀𝐴𝐵 − 4𝑀𝐵𝐴 𝐿 = 0 𝑀𝐵𝐴 𝐿 − 𝑀𝐴𝐵 2𝐿 + 6𝜃𝐴 𝐸𝐼 = 0 Adding the two equations: −3𝑀𝐵𝐴 𝐿 + 6𝜃𝐴 𝐸𝐼 = 0 𝑴𝑩𝑨 =

𝟐𝑬𝑰 𝑳

𝜽𝑨 … 𝑬𝒒. 𝟏

𝑀𝐴𝐵 𝐿 − 2𝑀𝐵𝐴 𝐿 = 0 (𝑀𝐵𝐴 𝐿 − 𝑀𝐴𝐵 2𝐿 + 6𝜃𝐴 𝐸𝐼 = 0) × 2 𝑀𝐴𝐵 𝐿 − 2𝑀𝐵𝐴 𝐿 = 0 2𝑀𝐵𝐴 𝐿 − 𝑀𝐴𝐵 4𝐿 + 12𝜃𝐴 𝐸𝐼 = 0 Adding the two equations: −3𝑀𝐴𝐵 𝐿 + 12𝜃𝐴 𝐸𝐼 = 0 𝑴𝑨𝑩 =

𝟒𝑬𝑰 𝜽 … 𝑬𝒒. 𝟐 𝑳 𝑨

For the angular displacement at B, 𝜃𝐵 . In similar manner, if end B of the beam rotates to its final

position 𝜃𝐵 , while end A is held fixed, we can relate the applied moment 𝑀𝐵𝐴 to the angular

displacement 𝜃𝐵 and the reaction moment 𝑀𝐴𝐵 at the wall. The results are: 𝑴𝑩𝑨 = 𝑴𝑨𝑩 =

𝟒𝑬𝑰 𝑳

𝜽𝑩 … 𝑬𝒒. 𝟑

𝟐𝑬𝑰 𝜽 … 𝑬𝒒. 𝟒 𝑳 𝑩

Figure 3

The relative displacement, ∆, can be determined if the far node B of the member is displaced relative to A, so that the chord of the member rotates clockwise but the ends do not rotate, then equal but opposite moment and shear reactions are developed in the member. Utilizing the conjugate beam method to relate the moment to the linear displacement. The displacement of

the real beam at B, the moment at the end 𝐵’ of the conjugate beam must have a magnitude of ∆ as indicated. Summing moments about 𝐵’: [2]

Figure 4

1𝑀 1 1𝑀 2 𝐿 ( 𝐿)] − ∆= 0 𝐿 ( 𝐿)] − [ ↺ +, ∑ 𝑀𝐵′ = 0, [ 2 𝐸𝐼 3 2 𝐸𝐼 3 𝑀𝐿2 𝑀𝐿2 − − ∆= 0 3𝐸𝐼 6𝐸𝐼 −

𝑀𝐿2 =∆ 6𝐸𝐼

𝑴𝑨𝑩 = 𝑴𝑩𝑨 = 𝑴 = −

𝟔𝑬𝑰 𝑳𝟐

∆ … 𝑬𝒒. 𝟓

The final component is the fixed-end moments. In general, linear or angular displacements of the nodes are caused by loadings acting on the span of the member, not by members acting at its nodes. The span loadings must be transformed into equivalent moments acting at the nodes and then use the load-displacement relationships just derived. This can be done by finding the reaction moment that each load develops at the nodes. Consider the fixed-supported member subjected to a concentrated load P at its centre. The conjugate beam method will be utilized again. The slope at each end must be equal to zero:

Figure 5

1 𝑀 1 𝑃𝐿 ) 𝐿] − 2 [ ( ) 𝐿] = 0 ↑ +, ∑ 𝐹𝑦 = 0, [ ( 2 4 𝐸𝐼 2 𝐸𝐼 𝑃𝐿2 /8𝐸𝐼 𝑀= 𝐸𝐼 𝐿

𝑴=

𝑷𝑳 … 𝑬𝒒. 𝟔 𝟖

This moment is called a fixed-end moment (FEM). Based on the sign convention, at node A, the moment is negative while at B, it is positive. For the FEM for various loading have been developed in texts books. We have in general: 𝑀𝐴𝐵 = (𝐹𝐸𝑀)𝐴𝐵

𝑀𝐵𝐴 = (𝐹𝐸𝑀)𝐵𝐴...