{"history":[],"article":{"id":"cb196b47-002c-49c1-b088-890971641548","title":"Macaulay's method","slug":"macaulay-s-method","content":":::summary\n:::\n\n# \n**Macaulay's method (the double integration method)** is a technique used in [structural analysis](https://bliptext.com/articles/structural-analysis) to determine the [deflection](https://bliptext.com/articles/deflection-engineering) of [Euler-Bernoulli beams](https://bliptext.com/articles/beam-theory).\nUse of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading.\nTypically partial uniformly distributed loads (u.d.l.) and uniformly varying loads (u.v.l.) over the span and a number of concentrated loads are conveniently handled using this technique.\nThe first English language description of the method was by [Macaulay](https://bliptext.com/articles/william-herrick-macaulay).\nThe actual approach appears to have been developed by [Clebsch](https://bliptext.com/articles/alfred-clebsch) in 1862.\nMacaulay's method has been generalized for Euler-Bernoulli beams with axial compression, to [Timoshenko beams](https://bliptext.com/articles/timoshenko-beam-theory), to [elastic foundations](https://bliptext.com/articles/elastic-foundation), and to problems in which the bending and shear stiffness changes discontinuously in a beam.\n\n# Method\nThe starting point is the relation from [Euler-Bernoulli beam theory](https://bliptext.com/articles/beam-theory)\nWhere w is the deflection and M is the bending moment.\nThis equation is simpler than the fourth-order beam equation and can be integrated twice to find w if the value of M as a function of x is known.\nFor general loadings, M can be expressed in the form\nwhere the quantities represent the bending moments due to point loads and the quantity is a [Macaulay bracket](https://bliptext.com/articles/macaulay-bracket) defined as Ordinarily, when integrating P(x-a) we get\nHowever, when integrating expressions containing Macaulay brackets, we have\nwith the difference between the two expressions being contained in the constant C_m.\nUsing these integration rules makes the calculation of the deflection of Euler-Bernoulli beams simple in situations where there are multiple point loads and point moments.\nThe Macaulay method predates more sophisticated concepts such as [Dirac delta functions](https://bliptext.com/articles/dirac-delta-function) and [step functions](https://bliptext.com/articles/step-function) but achieves the same outcomes for beam problems.\n\n# Example: Simply supported beam with point load\nAn illustration of the Macaulay method considers a simply supported beam with a single eccentric concentrated load as shown in the adjacent figure.\nThe first step is to find M. The reactions at the supports A and C are determined from the balance of forces and moments as Therefore, R_A = Pb/L and the bending moment at a point D between A and B (0 < x < a) is given by Using the moment-curvature relation and the Euler-Bernoulli expression for the bending moment, we have\nIntegrating the above equation we get, for 0 < x < a, At x=a_{-}\nFor a point D in the region BC (a < x < L), the bending moment is In Macaulay's approach we use the [Macaulay bracket](https://bliptext.com/articles/macaulay-bracket) form of the above expression to represent the fact that a point load has been applied at location B, i.e.,\nTherefore, the Euler-Bernoulli beam equation for this region has the form\nIntegrating the above equation, we get for a < x < L At x=a_{+}\nComparing equations (iii) & (vii) and (iv) & (viii) we notice that due to continuity at point B, C_1 = D_1 and C_2 = D_2.\nThe above observation implies that for the two regions considered, though the equation for [bending moment](https://bliptext.com/articles/bending-moment) and hence for the [curvature](https://bliptext.com/articles/curvature) are different, the constants of integration got during successive integration of the equation for [curvature](https://bliptext.com/articles/curvature) for the two regions are the same.\nThe above argument holds true for any number/type of discontinuities in the equations for curvature, provided that in each case the equation retains the term for the subsequent region in the form etc. It should be remembered that for any x, giving the quantities within the brackets, as in the above case, -ve should be neglected, and the calculations should be made considering only the quantities which give +ve sign for the terms within the brackets.\nReverting to the problem, we have\nIt is obvious that the first term only is to be considered for x < a and both the terms for x > a and the solution is Note that the constants are placed immediately after the first term to indicate that they go with the first term when x < a and with both the terms when x > a. The Macaulay brackets help as a reminder that the quantity on the right is zero when considering points with x < a.\n# [Simply supported beam with a single eccentric concentrated load. | upload.wikimedia.org/wikipedia/commons/2/2c/SimpSuppBeamPointLoadUnsymm.svg]\n\n## Boundary Conditions\nAs w = 0 at x = 0, C2 = 0.\nAlso, as w = 0 at x = L, or,\nHence,\n\n## Maximum deflection\nFor w to be maximum, dw/dx = 0.\nAssuming that this happens for x < a we have\nor\nClearly x < 0 cannot be a solution.\nTherefore, the maximum deflection is given by or,\n\n## Deflection at load application point\nAt x = a, i.e., at point B, the deflection is or\n\n## Deflection at midpoint\nIt is instructive to examine the ratio of.\nAt x = L/2\nTherefore,\nwhere k = B/L and for.\nEven when the load is as near as 0.05L from the support, the error in estimating the deflection is only 2.6%.\nHence in most of the cases the estimation of maximum deflection may be made fairly accurately with reasonable margin of error by working out deflection at the centre.\n\n## Special case of symmetrically applied load\nWhen a = b = L/2, for w to be maximum\nand the maximum deflection is","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-26T09:47:51.149Z","updated_at":"2025-01-26T09:47:51.149Z","search_vector":"'macaulay':1 'method':3","search_content":" \n \n\n# \n**Macaulay's method (the double integration method)** is a technique used in [structural analysis](https://bliptext.com/articles/structural-analysis) to determine the [deflection](https://bliptext.com/articles/deflection-engineering) of [Euler-Bernoulli beams](https://bliptext.com/articles/beam-theory).\nUse of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading.\nTypically partial uniformly distributed loads (u.d.l.) and uniformly varying loads (u.v.l.) over the span and a number of concentrated loads are conveniently handled using this technique.\nThe first English language description of the method was by [Macaulay](https://bliptext.com/articles/william-herrick-macaulay).\nThe actual approach appears to have been developed by [Clebsch](https://bliptext.com/articles/alfred-clebsch) in 1862.\nMacaulay's method has been generalized for Euler-Bernoulli beams with axial compression, to [Timoshenko beams](https://bliptext.com/articles/timoshenko-beam-theory), to [elastic foundations](https://bliptext.com/articles/elastic-foundation), and to problems in which the bending and shear stiffness changes discontinuously in a beam.\n\n# Method\nThe starting point is the relation from [Euler-Bernoulli beam theory](https://bliptext.com/articles/beam-theory)\nWhere w is the deflection and M is the bending moment.\nThis equation is simpler than the fourth-order beam equation and can be integrated twice to find w if the value of M as a function of x is known.\nFor general loadings, M can be expressed in the form\nwhere the quantities represent the bending moments due to point loads and the quantity is a [Macaulay bracket](https://bliptext.com/articles/macaulay-bracket) defined as Ordinarily, when integrating P(x-a) we get\nHowever, when integrating expressions containing Macaulay brackets, we have\nwith the difference between the two expressions being contained in the constant C_m.\nUsing these integration rules makes the calculation of the deflection of Euler-Bernoulli beams simple in situations where there are multiple point loads and point moments.\nThe Macaulay method predates more sophisticated concepts such as [Dirac delta functions](https://bliptext.com/articles/dirac-delta-function) and [step functions](https://bliptext.com/articles/step-function) but achieves the same outcomes for beam problems.\n\n# Example: Simply supported beam with point load\nAn illustration of the Macaulay method considers a simply supported beam with a single eccentric concentrated load as shown in the adjacent figure.\nThe first step is to find M. The reactions at the supports A and C are determined from the balance of forces and moments as Therefore, R_A = Pb/L and the bending moment at a point D between A and B (0 a and the solution is Note that the constants are placed immediately after the first term to indicate that they go with the first term when x a. The Macaulay brackets help as a reminder that the quantity on the right is zero when considering points with x < a.\n# [Simply supported beam with a single eccentric concentrated load. | upload.wikimedia.org/wikipedia/commons/2/2c/SimpSuppBeamPointLoadUnsymm.svg]\n\n## Boundary Conditions\nAs w = 0 at x = 0, C2 = 0.\nAlso, as w = 0 at x = L, or,\nHence,\n\n## Maximum deflection\nFor w to be maximum, dw/dx = 0.\nAssuming that this happens for x < a we have\nor\nClearly x < 0 cannot be a solution.\nTherefore, the maximum deflection is given by or,\n\n## Deflection at load application point\nAt x = a, i.e., at point B, the deflection is or\n\n## Deflection at midpoint\nIt is instructive to examine the ratio of.\nAt x = L/2\nTherefore,\nwhere k = B/L and for.\nEven when the load is as near as 0.05L from the support, the error in estimating the deflection is only 2.6%.\nHence in most of the cases the estimation of maximum deflection may be made fairly accurately with reasonable margin of error by working out deflection at the centre.\n\n## Special case of symmetrically applied load\nWhen a = b = L/2, for w to be maximum\nand the maximum deflection is","revisionCount":0},"hasMore":false,"nextCursor":null}

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