Macaulay's method
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Macaulay’s method is a technique used in structural analysis. 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.Though the method uses Double integrationbut still it is not called as " Double Integration method". The Double Integration Methodfor solving the deflection and slope involves some similar steps but is actually different. The three Methods for solving these problems are - Double Integration Method. Moment Area Method. Macaulay's Method.
Method
Consider a simply supported beam with a single eccentric concentrated load.
With A as origin, For the region AC M = W (b/L) x We know, from Euler's equation,
EI d2y/dx2 = W (b/L) x
Integrating the above equation, we get
EI dy/dx = W (b/L) (x2/2) +C1 ……………………………………… (i) (equation for slope)
EI y = W (b/L) (x3/6) +C1 x +C2 …………………………………… (ii) (equation for deflection)
at x=a :
EI (dy/dx) c = W (b/L) (a2/2) +C1 ………………………………… (iii)
EI yc = W (b/L) (a3/6) +C1 a +C2 ………………………………… (iv)
For the region CB M = W (b/L) x – W(x-a) As already we know,
d2y/dx2 = M/EI
EI d2y/dx2 = W (b/L) x – W(x-a)
Integrating the above equation, we get
EI dy/dx = W (b/L) (x2/2) – W ((x-a)2)/2 +C’1 …………………… (v)
EI y = W (b/L) (x3/6) – W ((x-a)3)/6 +C’1 x +C’2 ………………… (vi)
at x=a -->
EI (dy/dx) c = W (b/L) (a2/2) – W ((a-a)2)/2 +C’1
= W (b/L) (a2/2) +C’1 ……………………………………………… (vii)
EI yc = W (b/L) (a3/6) – W ((a-a)3)/6 +C’1 a +C’2
= W (b/L) (a3/6) +C’1 a +C’2 ……………………………………… (viii)
Comparing equations (iii) & (vii) and (iv) & (viii)
C’1 = C1 and C’2 = C2
The above observation implies that for the two regions considered, though the equation for B.M. 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. This is true only when in the equations for curvature and the integration there from, the term for the second region are kept in the form W(x-a), W(x-a)2/2 & W(x-a)3/6.
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 (x-a)n, (x-b)n, (x-c)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.