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# The moment area theorem (10.01.03.131)

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### The moment area theorem (10.01.03.131)

1. 1.  Topic : Solving Statically Indeterminate Structure : Moment-Area Theorem . Ikramul Ahasan Bappy ID:10.01.03.131
2. 2.  Definition: A member of any type is classified statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations, e.g. a continuous beam having 4 supports
3. 3.  Assumptions: Contributions for the theorem .  Beam is initially straight,  Is elastically deformed by the loads, such that the slope and deflection of the elastic curve are very small, and  Deformations are caused by bending. Otto Mohr.(1868) Charles E. Greene.(1873)
4. 4.  Moment-curvature relationship:  Sign convention:
5. 5. B BA A M dx EI
6. 6. *12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD Theorem 2 The vertical deviation of the tangent at a pt (A) on the elastic curve w.r.t. the tangent extended from another pt (B) equals the moment of the area under the ME/I diagram between these two pts .  This moment is computed about pt (A) where the vertical deviation 6 (tA/B) is to be determined. 
7. 7. M/EI Diagram  Determine the support reactions and draw the beam’s M/EI diagram.  If the beam is loaded with concentrated forces, the M/EI diagram will consist of a series of straight line segments.
8. 8.  If the loading consists of a series of distributed loads, the M/EI diagram will consist of parabolic or perhaps higher-order curves.
9. 9.    An exaggerated view of the beam’s elastic curve. Pts of zero slope and zero displacement always occur at a fixed support, and zero displacement occurs at all pin and roller supports. When the beam is subjected to a +ve moment, the beam bends concave up, whereas -ve moment gives the reverse .
10. 10.    An inflection pt or change in curvature occurs when the moment if the beam (or M/EI) is zero. Since moment-area theorems apply only between two tangents, attention should be given as to which tangents should be constructed so that the angles or deviations between them will lead to the solution of the problem. The tangents at the supports should be considered, since the beam usually has zero displacement and/or zero slope at the supports.
11. 11. # Since axial load neglected, a there is a vertical force and moment at A and B. Since only two eqns of equilibrium are available, problem is indeterminate to the second degree. # Let By and MB are redundant, #By principle of superposition, beam is represented as a cantilever. # Loaded separately by distributed load and reactions By and MB, as shown.
12. 12. 0 B 'B ' 'B 1 0 B 'B ' 'B 2 wL3 48EI B 7 wL4 384EI B 42 EI 9 kN/m 4 m 48EI 3 7 9 kN/m 4 m 384EI 12 EI 4
13. 13. 2 3 'B 'B ' 'B PL 2 EI By 4 m By 4 m 8By 2 EI PL 3EI ML EI 2 ' 'B 2 ML 2 EI EI 3 21.33B y 3EI MB 4 m EI MB 4 m 2 EI EI 4M B EI 2 8M B EI
14. 14. Substituting these values into Eqns (1) and (2) and canceling out the common factor EI, we have 0 12 8B y 0 4M B 42 21.33B y 8M B Solving simultaneously, we get By 3.375 kN MB 3.75 kN m
15. 15. Thank you for your attention
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