Unit III – Deflection in Beams 2. Macaulay’s method:
This is a convenient convenient method for
determining the deflections of a beam subjected to point loads
A
C
D
or in general discontinuous discontinuous loads. loads.
B
Example: In general at any section the B.M. is given by,
= The manner in which the above expression expression is written should be noted. Dr.P.V .Venkateswara enkateswara Rao, Associate Professor,
1
Unit III – Deflection in Beams 2. Macaulay’s method:
=
A
As the magnitude of goes on Increasing so that the law of Loading changes, additional expression appear.
C
D
B
For values of between = 0 and = , only the first term of the above expression should be considered. For values of between = and = , only the first two terms of the above expression should be considered. For values of between = and = , all the terms of the above expression should be considered. Dr.P.V .Venkateswara enkateswara Rao, Associate Professor,
2
Unit III – Deflection in Beams 2. Macaulay’s method: = ()
A
C
D
B
Integrating Equation (i) we get, the general expression for slope
=
+
−
−
()
It is very important to note the following two points (a) The constant of integration should written after the first term of the above expression. (b) The quantity ( ) should be integrated as
−
and not as
. Dr.P.Venkateswara Rao, Associate Professor,
3
Unit III – Deflection in Beams 2. Macaulay’s method:
=
−
+
2 ()
A
2
C
D
B
(c) Similarly the quantity ( ) should be integrated as −
and not as
.
Integrating Equation (ii), we get the deflection equation =
−
6
+ +
6
() Dr.P.Venkateswara Rao, Associate Professor,
4
Unit III – Deflection in Beams 2. Macaulay’s method: =
6
−
+ +
−
A
()
C
D
B
The constant and can be evaluated if end conditions are known.
Dr.P.Venkateswara Rao, Associate Professor,
5
Unit III – Deflection in Beams Problem 1:
A simply supported beam AB of span 10 m is subjected to an u.d.l. of 20 kN/m for a length of 6 m from the left support A and a concentrated load of 40 kN at 2 m from the right support B. Determine the slope at the support A and the deflection under the concentrated load interms of the flexural rigidity EI.
Dr.P.Venkateswara Rao, Associate Professor,
6
Unit III – Deflection in Beams Solution:
20 kN/m
Taking moments about A, ( × 10) = 40 × 8 +
A
6
=92
( 2 0 × 6 ×3) = 68 = 20 × 6 + 40 68 = 92 .
10
40 B C 2m 2m
=68
By using Macaulay’s method
The B.M. at any section distant from the end B is given by,
= 68 40 x 2 20
4 2
Dr.P.Venkateswara Rao, Associate Professor,
(i) 7
Unit III – Deflection in Beams Solution:
20
40
20 kN/m
= 68 40 x 2
4 2
A
6
=92
(i)
10
B C 2m 2m
=68
Integrating the above equation, we get slope equation,
= 68
2
+ 40
2
2
10
4
ii
3
Integrating the above equation, we get deflection equation, = 68
6
+ + 20
2 3
10
Dr.P.Venkateswara Rao, Associate Professor,
4 12
(iii)
Unit III – Deflection in Beams Solution:
= 68 20
20 kN/m
A
+ +
6 2
10
4
=92
(iii)
6
10
40 B C 2m 2m
=68
3 12 At the point B, i.e at = 0, = 0 From equation (iii), 0 = 0 + 0 + ∴ = 0 At the point A, i.e.,at = 10 , = 0 From equation (iii), 0 = (68 ×
20
102
3 ∴ = 684.
10
104
) + ( × 1 0 ) + 0
12 Dr.P.Venkateswara Rao, Associate Professor,
9
Unit III – Deflection in Beams Solution:
∴ = 684, = 0 substitute in equations (ii) and (iii).
= 34 684 20 2
A
6
=92
From equation (ii),
40
20 kN/m
10
At A, = 10 , = 34 × 10 684 20 10 2 716 =
−
10
10
B C 2m 2m
=68
ii slope eqn.
104
3
Dr.P.Venkateswara Rao, Associate Professor,
10
Unit III – Deflection in Beams Solution:
From equation (iii),
= 34 20
A
684
3 2 3
40
20 kN/m
10
=92
4 12
6
10
B C 2m 2m
=68
(iiia)
At point C, i.e., at at = 2 , 2 22 = 34 × 684 × 2 (20 × 3 3 1277.33 ∴ = Dr.P.Venkateswara Rao, Associate Professor,
11
Unit III – Deflection in Beams Problem 2:
A horizontal beam 4 m long is supported at the left end ‘A’ and at ‘B’ 1m from the right end C. It carries an uniformly distributed load of 10 kN/m over the span AB and A point load of 5 kN at the free end C. Determine the slope at the support A and the deflection at 1 m from left support A in terms of its flexural rigidity.
Dr.P.Venkateswara Rao, Associate Professor,
12
Unit III – Deflection in Beams Solution:
5
10 kN/m
Taking moments about A, ( × 3) = 5 × 4 +
C
B
A
3
=13.33
4 ( 1 0 × 3 ×1.5) = 21.67 = 10 × 3 + 5 21.67 = 13.33 .
1m
=21.67
By using Macaulay’s method The B.M. at any section distant from the end C is given by,
= 5 + 21.67 x 1 10
1
Dr.P.Venkateswara Rao, Associate Professor,
2
(i) 13
Unit III – Deflection in Beams Solution:
10
= 5 + 21.67 x 1 1
5
10 kN/m
(i)
A
=13.33
C
B
3 4
1m
=21.67
2 Integrating the above equation, we get slope equation, 1 1 = 5 + + 21.67 10 ii 2 2 6 Integrating the above equation (ii), we get deflection equation, 1 1 = 5 + + + 21.67 10 6 6 24 (iii)
Dr.P.Venkateswara Rao, Associate Professor,
14
Unit III – Deflection in Beams Solution:
= 5 +21.67
5
10 kN/m
A
+ +
6 1
10
1
C
B
3
=13.33 (iii)
4
1m
=21.67
6 24 At the point B, i.e at = 1 , = 0
From equation (iii), 0 = 5( ) + (1) +
∴ + = 0.833 (A) At the point A, i.e.,at = 4 , = 0 From equation (iii), 0 = (5 ×
+21.67
41
10
41
) + ( × 4 ) +
6 24 ∴ 4 + = 10.435 () Dr.P.Venkateswara Rao, Associate Professor,
15
Unit III – Deflection in Beams Solution:
5
10 kN/m
4 + = 10.435 () + = 0.833 (A)
A
=13.33
C
B
3
4
1m
=21.67
3 = 11.268 B (A) = 3.756 = 0.833 3.756 = 4.589 ∴ = 4.589 Substitute and values in equation (ii) and (iii) 1 = 5 3.756 + 21.67 2 2
10
Dr.P.Venkateswara Rao, Associate Professor,
1 6
iia
16
Unit III – Deflection in Beams Solution: = 5 +21.67
5
10 kN/m
6 1 6
A
3.756 + 4.589
10
1 24
3
=13.33 ()
1 = 5 3.756 + 21.67 2 2
C
B
10
4
1 6
1m
=21.67
iia
At A, i.e. at x= 4m, from Equation (iia),
= 5(
4 2
) 3.756 + 21.67
=
41 2
10
41 6
iia
.
Dr.P.Venkateswara Rao, Associate Professor,
17
Unit III – Deflection in Beams Solution: = 5 +21.67
5
10 kN/m
D
A
3.756 + 4.589
6 1
10
6
1 24
=13.33 ()
C
B
3
4
1m
=21.67
At D, i.e. at x= 3m, from Equation (iiia),
= 5
4 2
3.756 3 + 4.589 + 21.67
=
31 2
10
31 6
.
Dr.P.Venkateswara Rao, Associate Professor,
18
Unit III – Deflection in Beams Problem 3:
A simply supported beam AB 9 m span is subjected to a vertical point load of 30 kN at 3 m from the left support A and a clockwise moment of 20 kNm at 3 m from the right support B. Using Macaulay’s method, determine the slope at A and the deflection under the point load. E= 2 × 10 MPa and = 12 × 10 .
Dr.P.Venkateswara Rao, Associate Professor,
19
Unit III – Deflection in Beams Solution:
30 kN
Taking moments about A,
× 9 = (30 × 3)+20 = 12.22 = 30 12.22
20 kN m D 3m 3m B
C A
3m
17.78
x
= 17.78 Using macaulay’s method, the B.M. at any section from A is given by,
= 17.78 30 3 + 20 6 Dr.P.Venkateswara Rao, Associate Professor,
20
Unit III – Deflection in Beams Solution:
30 kN
C
= 17.78 30 3
+20 6 By integrating ,
= 17.78
2
+ 30
A
20 kN m D 3m 3m B
3m x
3
2
+ 20 6
By integrating, = 17.78
6
+ + 30
3 6
+ 20
6
2
At x=0, y=0, from the above deflection equation, 0 = 0 + 0 + ∴ = 0 Dr.P.Venkateswara Rao, Associate Professor,
21
Unit III – Deflection in Beams Solution:
= 17.78 30
3 6
6
30 kN
+ +
+ 20
6
20 kN m D 3m 3m B
C A
3m
2
x
At x=9 m, y=0, from the above deflection equation, 0 = 17.78
9 6
+ 9 + 30
93 6
+ 20
96
2
∴ = 130.03 ∴
−
= 17.78 130.03 30
+ 20 6 --(i) slope equation
Dr.P.Venkateswara Rao, Associate Professor,
22
Unit III – Deflection in Beams Solution:
= 17.78 130.03 6 − − ---(ii) 30 + 20
30 kN C A
3m
x
deflection equation.
∴
20 kN m D 3m 3m B
−
= 17.78 130.03 30
+ 20 6 --(i) slope equation
Slope at A, i.e., i.e. at x=0,
=
130.03
=
130.03
− = 5.418 × 10 2.4 × 10
Dr.P.Venkateswara Rao, Associate Professor,
23
Unit III – Deflection in Beams Solution:
= 17.78 130.03 6 − − ---(ii) 30 + 20
30 kN
20 kN m D 3m 3m B
C A
3m
deflection equation.
x
Deflection at C, i.e., at x= 3m, from (ii), 3 33 = 17.78 130.03(3) 30 6 6 310.08 310.08 − = = = 1.292 × 10 2.4 × 10 ∴ = 1.292 () Dr.P.Venkateswara Rao, Associate Professor,
24
Unit III – Deflection in Beams Problem 4 (H.W.):
A beam ABC is loaded as shown in Figure. If E= 2 × 10 N/ and = 9 × 10 , determine (i) slope at the end ‘A’ (ii) deflection at the free end ‘C’ (iii) Maximum deflection. 15 kN
A
6m
4m
2 kN/m
B 2m
Dr.P.Venkateswara Rao, Associate Professor,
25
Unit III – Deflection in Beams Problem 5 (H.W.):
A beam of uniform section 10 m long is simply supported at its ends and is subjected to 100 kN and 60 kN at distance of 2 m and 5 m respectively, from the left end. Compute the deflection under each load using Macaulay’s method. Assume E=200 Gpa and I=0.0118 .
Dr.P.Venkateswara Rao, Associate Professor,
26
