Macaulay's method

Macaulay’s method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams. Use of Macaulay’s technique is very convenient for cases of discontinuous and/or discrete loading. Typically 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.

The first English language description of the method was by Macaulay. The actual approach appears to have been developed by Clebsh in 1862. Macaulay's method has been generalized for Euler-Bernoulli beams with axial compression, to Timoshenko beams, to elastic foundations, and to problems in which the bending and shear stiffness changes discontinuously in a beam

Method

Simply supported beam with a single eccentric concentrated load.

Consider a simply supported beam with a single eccentric concentrated load as shown in the adjacent figure. The reactions at the supports A and C are determined from the balance of forces and moments as

$R_{A}+R_{C}=P,~~LR_{C}=Pa$

Therefore $R_{A}=Pb/L$ and the bending moment at a point D between A and B ($0<x<a$) is given by

$M=R_{A}x=Pbx/L$

From the Euler-Bernoulli beam equation and the above expression for the bending moment, we have

$EI{\dfrac {d^{2}w}{dx^{2}}}={\dfrac {Pbx}{L}}$

Integrating the above equation we get, for $0<x<a$,

${\begin{aligned}EI{\dfrac {dw}{dx}}&={\dfrac {Pbx^{2}}{2L}}+C_{1}&&\quad \mathrm {(i)} ~\mathrm {(equation~for~slope)} \\EIw&={\dfrac {Pbx^{3}}{6L}}+C_{1}x+C_{2}&&\quad \mathrm {(ii)} ~\mathrm {(equation~for~deflection)} \end{aligned}}$

At $x=a_{-}$

${\begin{aligned}EI{\dfrac {dw}{dx}}(a_{-})&={\dfrac {Pba^{2}}{2L}}+C_{1}&&\quad \mathrm {(iii)} \\EIw(a_{-})&={\dfrac {Pba^{3}}{6L}}+C_{1}a+C_{2}&&\quad \mathrm {(iv)} \end{aligned}}$

For a point D in the region BC ($a<x<L$), the bending moment is

$M=R_{A}x-P(x-a)=Pbx/L-P(x-a)$

In Macaulay's approach we use the Macaulay bracket form of the above expression to represent the fact that a point load has been applied at location B, i.e.,

$M={\frac {Pbx}{L}}-P\langle x-a\rangle$

Ordinarily, when integrating $P(x-a)$ we get

$\int P(x-a)~dx=P\left[{\cfrac {x^{2}}{2}}-ax\right]+C$

However, when integrating expressions containing Macaulay brackets, we get the form

$\int P\langle x-a\rangle ~dx=P{\cfrac {\langle x-a\rangle ^{2}}{2}}+C_{m}$

with the difference between the two expressions being contained in the constant $C_{m}$.

Therefore the Euler-Bernoulli beam equation for this region has the form

$EI{\dfrac {d^{2}w}{dx^{2}}}={\dfrac {Pbx}{L}}-P\langle x-a\rangle$

Integrating the above equation, we get for $a<x<L$

${\begin{aligned}EI{\dfrac {dw}{dx}}&={\dfrac {Pbx^{2}}{2L}}-P{\cfrac {\langle x-a\rangle ^{2}}{2}}+D_{1}&&\quad \mathrm {(v)} \\EIw&={\dfrac {Pbx^{3}}{6L}}-P{\cfrac {\langle x-a\rangle ^{3}}{6}}+D_{1}x+D_{2}&&\quad \mathrm {(vi)} \end{aligned}}$

At $x=a_{+}$

${\begin{aligned}EI{\dfrac {dw}{dx}}(a_{+})&={\dfrac {Pba^{2}}{2L}}+D_{1}&&\quad \mathrm {(vii)} \\EIw(a_{+})&={\dfrac {Pba^{3}}{6L}}+D_{1}a+D_{2}&&\quad \mathrm {(viii)} \end{aligned}}$

Comparing equations (iii) & (vii) and (iv) & (viii) we notice that due to continuity at point B, $C_{1}=D_{1}$ and $C_{2}=D_{2}$. The above observation implies that for the two regions considered, though the equation for bending moment and hence for the curvature are different, the constants of integration got during successive integration of the equation for curvature for the two regions are the same.

The 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 $\langle x-a\rangle ^{n},\langle x-b\rangle ^{n},\langle x-c\rangle ^{n}$ 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.

Reverting back to the problem, we have
EI d2y/dx2 = W (b/L) x | – W(x-a)

It is obvious that the first term only is to be considered for x < a and both the terms for x > a.

EI dy/dx = W (b/L) (x2/2) +C1 | – W ((x-a) 2)/2 …………………………… (i)

EI y = W (b/L) (x3/6) +C1 x +C2 | – W ((x-a) 3)/6 ………………………… (ii)

Note: 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, a vertical line is placed after the discontinuity to remind the student that he should stop at that line when considering points with x < a.

Boundary Conditions:

As y = 0 at x = 0 C2 in the equation C2 = 0
And also, as y = 0 at x = L -->
W (b/L) (L3/6) +C1 L – W ((L-a) 3)/6 = 0
WbL2/6 +C1 L – W ((L-a) 3)/6 = 0
C1 L = -WbL2/6 + W ((L-a) 3)/6

C1   = -WbL/6 + W ((L-a) 3)/6L

      = -WbL/6 + W b 3/6L

      = - (Wb/6L)(L2 – b2)

Hence,
EI dy/dx = W (b/L) (x2/2) - (Wb/6L) (L2 – b2) | – W ((x-a) 2)/2 …………(iii)

EI y = W (b/L) (x3/6) - (Wb/6L) (L2 – b2) x | – W ((x-a) 3)/6 …………… (iv)

For y to be maximum,

dy/dx = 0

Assuming that this happens for x < a ………… (for a > b) Equation (iii) becomes

EI dy/dx = W (b/L) (x2/2) - (Wb/6L) (L2 – b2) (considering the equation till line)

W (b/L) (x2/2) - (Wb/6L) (L2 – b2) = 0 W (b/L) (x2/2) = (Wb/6L) (L2 – b2) x2/2 = (L2 – b2)/6

x = √((L2 – b2)/3) …………………………………………(v)

(v) in (iv) we get EI ymax = (Wb/6L) (((L2 – b2)/3)3/2) - (Wb/6L) (L2 – b2) ((L2 – b2)/3)1/2

            = (Wb/6L) (((L2 – b2)/3)3/2) - (Wb/6L) ((L2 – b2)3/2)/√3
            = (Wb/6L) (L2 – b2)3/2 [1/3√3 - 1/√3]
            = (1/√3) (Wb/6L) (L2 – b2)3/2 (1/3 – 1)
            = - (1/√3) (Wb/6L) (L2 – b2)3/2 (-2/3)
            = - (1/9√3) (Wb/L) (L2 – b2)3/2 

ymax = - (1/9√3) (Wb/LEI) (L2 – b2)3/2 ………………… (vi)

at x = a ie at C, from equation (iv)

EI yc = W (b/L) (a3/6) - (Wb/6L) (L2 – b2) a

        = (Wab/6L) [a2 - (L2 – b2)] 
        = (Wab/6L) (a2 - L2 + b2) 
        = (Wab/6L) (a2 + b2 – (a+b)2 ) ………………………(L= a+b)
        = (Wab/6L) (a2 + b2 – a2 - b2 -2ab) 
        = (Wab/6L) (-2ab) 
        = -W a2 b2 /3L …………………………………… (vii)

When a = b = L/2, for y to be maximum Equation (v) --> x = √((L2 – (L/2)2)/3)

x = √((L2 – L2/4)/3)
x = √(((3/4)L2 )/3)
x = L/2
(vi) --> ymax = - (1/9√3) (Wb/LEI) (L2 – (L/2)2)3/2
ymax = - (1/9√3) (W(L/2)/LEI) (L2 – L2/4)3/2
ymax = - (1/18√3) (W/EI) ((√3) 2L2 /22)3/2
ymax = - (1/18√3) (W/EI) ((√3) L /2)3
ymax = - (1/18√3) (W/EI) (3√3/8) L3
ymax = - W L3/48EI ………………………………… (viii)

also for a = b = L/2 equation (vii) -->

yc = -W (L/2)2 (L/2)2 /3LEI

   = - W L3/48EI  …………………………………… (ix)

Which is ymax as obtained from equation (viii). This is obvious because of symmetry in this case.

Note : It is instructive to work out the ratio of ymax / yx=L/2

at x = L/2 , from equation (iv)

EI yx=L/2 = (Wb/6L) (L/2)3 - (Wb/6L) (L2 – b2) (L/2)
EI yx=L/2 = (Wb/6L) (L3/8- L3/2 + b2L/2)
yx=L/2 = - (Wb/6L EI) (3L3/8- b2L/2)

ymax / yx=L/2 = [- (1/9√3) (Wb/L EI) (L2 – b2)3/2] / [- (Wb/6L EI) (3L3/8- b2L/2)]
ymax / yx=L/2 = (2/3√3) (1 – (b/L)2)3/2 / (3/8- b2L/2L3)
ymax / yx=L/2 = (2/3√3) (1 – (k)2)3/2] / (3/8- 0.5k2)

where k = b/L and for a < b; 0 < k < 0.5

Even when the load is as near as 0.05L from the support, the error in estimating the deflection is only 2.6%. Hence 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.

See also

• Bending
• Beam theory
• Timoshenko beam theory