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Slope Deflection Method

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slope deflection
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  Slope Deflection Method 1.   Relationship between end moments and joint rotations and displacements The moment diagram of a beam with distinct but, equal sign moments at its ends, and no loads along its span is depicted in figure 1 : Figure 1: Moment Diagram, of a beam subjected to Moments at its edge. According to first Moment -Area moment method, difference between slopes (and consequently, rotations) at the ends is given by the net area of the Moment diagram, divided by EI, which can  be calculated by superimposition of triangles dAc and ABc.    AB BA AB BAdAC  ABc  A B  M  M  EI  L L M  L M   EI  EI  A A  2221     (1) According to the second Moment-Area theorem, moment of area of whole M/EI diagram along  point B, is equal to the distance between a tangent at point A and point B. As the tangent at point A has a slope equal to  a, the following expression can be proposed:    AB BA AB BAdAC  ABc  A AB  M  M  EI  L L M  L L M  L  EI  EI  A L A L  L  26232231323  2      (2) If Equation 2 is algebraically manipulated, moment at end B can be expressed in terms of moment at end A:    BA AB A B  M  M  L EI      2  (3) After replacing this expression in Equation 2, the following is obtained      23226 22  AB A B AB AB A B A AB  M  L EI  EI  L M  M  L EI  EI  L L         (4) Equation 4 explicitly relates end moments, their rotations and their relative displacements. A B c d e   M AB M BA        23 2  AB A B A AB  M  L EI  L L EI          22333 2  AB B A AB A B A AB  M  L EI  L EI  L L EI  L EI  L EI  L EI  L EI            (5) Finally, arranging similar terms, Moment at end A is given by:    AB B A  M  R L EI    322     (6) Where R is called chord rotation, and is equal to the ratio of deflection and element length. After replacing Equation 6 into Equation 3, moment at end B can be expressed in terms of nodal deflections and slopes:        BA B A A B B A A B  M  R L EI  R L EI  L EI    3223222           (7) Finally, Moment at the other end becomes:    BA A B  M  R L EI    322     (8)   2.   Example, slope deflection method: Figure 2: Problem definition (on top left), Element orientation (Top, Right) and Ratio between relative flexural inertia and element length (Bottom Right). Taking into account the problem outlined, the following expressions for moment equilibrium at the joints can be proposed Joint C: 0   V  AB II  AB I  BA  M  M  M   (1) Joint D 0   IV  AB III  BAV  BA  M  M  M   (2) Joint A 0   VI  AB II  BA  M  M   (3) 16’   16’   8’   12’   A B   C   D   E   F   3I   3I   2I   2I   4I     2I   3klbf    6klbf    A   B   C   D   E   F   I   II   III   IV   V   VI   A   B   C   D   E   F   3I/12=I/4   I/8   2I/16=I/8   4I/16=I/4   2I/8=I/4   I/4  Problem statement Element Orientation EI/L  Joint B 0   VI  BA IV  BA  M  M   (4) Taking into account the orientations defined in Figure 1, (Arrows are at end B on each element) following expressions can be proposed for each joint equilibrium equation. In particular, for node A, this is observed:        028232823242 21    DC  AC C  E   EI  R EI  R EI          (5) It must be stressed that rotation and joint E is zero, because it is a support. Consequently Equation 4 can be written in the following manner:          022232322 4  21    DC  AC C   R R EI                 0243264 21    DC  AC C   R R         (6) Collecting similar terms, Equation 5 becomes 036210 21    R R  DC  A       (7) For joint D, the following is observed:      03282213242242 21    R EI  R EI  EI   B D D F  DC           (8) If the EI term is factored,      0322132222 4  22   R R EI   B D D F  DC           (9) Recalling that rotation at end F is zero, because it is a fixed support:        03262222 21    R R  B D D DC           03212442 21    R R  B D D DC          (10) Finally, after arranging and collecting similar terms, Equation 9 becomes: 0312102 21    R R  DC  B       (11) For joint A, the following moment-rotation-displacement relationships can be proposed:      02423282 2    B A AC   EI  R EI        (12) After collecting all flexural stiffness terms, Equation 11 can be reformulated into:

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