Structural Analysis (II)
Chapter Chapter 5: Deflectio Deflection n
Method (2) Conjugate beam Method ....
Slope( θ) Deflection ( y) Conjugate beam:
Deflection ( y) (M)
-M ) ( EI
(Elastic load )
Slope( θ) (Conjugate beam)
Rotation or Slope( θ)or (y')= Shear force of elastic load = Qelastic Deflection ( δ )or (y)= Moment of elastic load = Melastic δ )or Real(Original) beam Load = d²M(x) = w dx² Shear = dM(x) = Q dx
Moment = M(x) wt/m'
Conjugate beam: M Elastic load = d²(y) = EI dx²
Rotation = d(y) = Qelastic dx Deflection = y = Melastic
- M/EI
Rotation)
Deflection)
Structural Analysis (II)
Chapter 5: Deflection
(Conjugate beam) Real Support
(Original beam) Conjugate Support
Fixed support
Free end
M≠0,Q≠0
δ≠0,θ≠0
M
Q
Free end
Fixed support
M=0,Q=0
δ=0,θ=0 End support (Roller or Hinge) δ=0,θ≠0
End support (Roller or Hinge) Q M=0,Q≠0
θ Interior support (Roller or Hinge)
Internal hinge
θR δ=0,θ≠0
M=0,Q≠0 θL
(QL=QR )
(θR = θL )
Interior support (Roller or Hinge)
Internal hinge θL
δ ≠ 0 , θL ≠ θR ≠ 0
θR
M ≠ 0 , QL ≠ QR
Examples Real Beam
Conjugate Beam
(???) Indetreminate Beam
Structural Analysis (II)
Chapter 5: Deflection
Steps of Solution :
( y)
( θ)
.Conjugate beam method
FOR Real beam Reactions Bending Moment Diagram
( B.M.D)
Examples 6t
(1)
12t.m
3t/m`
12t.m
B.M.D
= WL² 8
WL² 8
= 24m.t
= 24m.t
(2) 4 t/m`
6t
2 t/m` C
15 t
14 t
21 t WL²/8=32m.t
8 m.t
WL²/8=9m.t
12 m.t
B.M.D
8 m.t
12 m.t
B.M.D WL²/8=32m.t
WL²/8=9m.t
Structural Analysis (II)
(
Chapter 5: Deflection
)
(
) 8t
8t 2 C
8t 12mt 6mt
B.M.D 12mt
12mt 6mt
3mt
12mt
6mt
Or
2
6mt 12mt
12mt
Modified B.M.D
24mt
FOR Conjugate beam (Conjugate Beam) .(Modified B.M.D)
(Given)
(Elastic loads) (Conjugate Beam)
.(Conjugate Beam)
(Elastic loads)
R .( B.M.D)
R
(C.g) .( B.M.D) ( Elastic loads) 1 ML 2
( B.M.D) ML
Datum
Datum 4
Structural Analysis (II)
Chapter 5: Deflection 1 M L 2 1
M1
Datum
1 M L 2 2
M2 Datum 2 3ML
M1
1 M L 2 1
M1
Datum
Datum
=
M2
M2
1 M L 2 2
Datum
M1 Datum
M1
1 M L 2 1
=
1 M L 2 2
M2
+
M2 Datum
2 ML 3 1 M L 2 1
M1 Datum
M1 Datum
=
+
1 M L 2 2
M2
M2
Datum
2 3ML
Structural Analysis (II)
Chapter 5: Deflection
(Rotation) (Deflection ) Qelastic
Rotation ( θ )=
Deflection (y) =
Melastic
Qelastic Melastic
Moment
Shear force Negtive )
Positive ) Q
M
Q
Q
M
- v e
Q
( θ )= + ve
Negtive )
Positive )
+v e
M
+ v e
( θ )= - ve
- v e
M
( y )= - ve
( y )= + ve -
θ
θ
+
Solved Examples Ex(1)
Using the Conjugate beam method, determine the rotation at points (a,b,c and d) and deflection at points(c,d and e).
10t
2t/m' a
c
10t
3t
b
d
e
Solution
EI = Constant
Reactions and Bending moment diagram (Original beam) 10t
2t/m'
10t
18.8t a
3t 16.2t 6.0t.m
c
d b
2*3²=2.25 t.m 8
47.4t.m
46.8t.m 2*3²=2.25 t.m 8
e
Structural Analysis (II)
Chapter 5: Deflection
Elastic loads( Area of bending moment)
12
6.0 6.0
a
c
d e
b
71.1
2.25 t.m
47.4
46.8 71.1 70.2
M) Modified B.M.D( EI 93.6
4.5
4.5
Conjugate beam b
a
e
b
a
e
Elastic loads on Conjugate beam a
6 . 3 9
2 1 . . 5 . 0 1 7 4 7
1 5 . . 1 7 4
b
c
e
d
0 . 6
2 1
Elastic Reactions a
6 . 3 9
2 1 . . 5 . 0 1 7 4 7
1 5 . . 1 4 7
M) ( EI
b
c d
164.3
2 1
138.7 b
269.4 e
0 . 6
132.7 7
Structural Analysis (II)
Chapter 5: Deflection
Required rotation and deflection Rotation ( θ )=
Qelastic
Deflection (y) = Left
Melastic
Right
+ve Sign summary
+ ve
Point (a)
a
1 [164.3] = + 164.3 θa = EI EI
164.3 Point (b) 1 [-138.7] = - 138.7 θb = EI EI
138.7 b 1 5 . . 1 4 7
Point (c) 1 [164.3-4.5-71.1] = + 88.7 θc = EI EI 1 [164.3(3)-4.5(1.5)-71.1(1)] = + 415.05 yc = EI EI
c
164.3 6 . 3 9
Point (d) 1 [ -138.7 -12 +93.6] = - 57.1 θd = EI EI
b d 2 1
Point (e) 1 [ -132.7] = - 132.7 θe = EI EI 1 [ -268.8] = - 269.4 ye = EI EI
269.4 e
132.7
7 . 8 3 1
Structural Analysis (II)
Ex(2)
Chapter 5: Deflection
Using the Conjugate beam method, determine the rotation at points (a,b,c and d ) and deflection at points(c and d ).
4t
a
4t/m'
I
d
b
I
2I
c
Solution 12t.m 6t.m d
Required rotation and deflection
I
1 [+7.88] = + 7.88 θa = EI EI
2I
c
B.M.D
Point (a)
9
18
18
12
1 [-14.62] = - 14.62 θb = EI EI
a
8 1
c
d
5 . 4
1 [18 - 14.62 - 4.5] θc = EI = - 1.12 EI
6 3
c
I
I
9t.m
Conj.beam
b
14.62
6 3
d
8 1
1 [-10.12] = - 10.12 θd = EI EI 1 [+ 12.36] = + 12.36 yd = EI EI
10.12
d
I
8 1
b 5 . 4
9 8 1 6 3
a
12.36
b
a
Elastic loads
Point (d)
4.5
18 18t.m
36
1 [14.62(3) + 4.5(1.0) -18(1.125)] yc = EI = + 28.11 EI
b
4*6²=18.0 t.m 8
Point (b)
Point (c)
I
a
8 1
b
c
7.88
9 8 1
5 . 4
a 8 1
Elastic Reactions
14.6
Structural Analysis (II)
Ex(3)
Chapter 5: Deflection
Using the Conjugate beam method, determine : * the rotation at points (a ,b,d and e ), * the relative (change in) rotation at point( f ), * the deflection at points(d ,e ,f and j).
2t/m' a
6t
f j
e
b
d
c
Solution
6t
f
EI = Constant c
e
a
4t 3t
16t
2t/m' j
b
d
6.75t
16.25t 40
10 10t.m
5t.m b
d
a
f
1t.m
16t.m
e
c
9t.m
1.33 27 85.33 85.33
1.33
27
b
a
e
d
85.33 a
40
10 b
d
29.33
c
f
40
16
16
1.33
27
f
c
e
10
24.13
10.2
Structural Analysis (II)
Chapter 5: Deflection
Required rotations and deflections Point (a)
Point (b) a
1 [+29.33] θa = EI = + 29.33 EI
29.33 Point (f)(Internal Hinge )
16
1 [-16+10-1.33] = - 7.33 θf/L = EI EI 1 [-16+10-1.33+24.13 ] = + 16.8 θf/R = EI EI -7.33 24.13 Rf θf/rel = [θf/R - θf/L ]= + 16.8 EI - EI = + EI = EI 1 [-16(2)+10(1.33)-1.33(1)] = - 20 yf = EI EI
16
3 3 . 1
f
24.13
0 1
10
Point (d) 1 [29.33+10-42.7] = - 3.37 θd = EI EI
5t.m d
a
29.33
1 [+29.33(4)+10(1.33)-42.7(1.5)] yd=EI =+66.6 EI
16t.m
Point (e)
42.7
1 [13.5-10.2 ] = + 3.3 θe = EI EI 1 [+10.2(3)-13.5(1)] = + 17.1 ye=EI EI
e
c
9t.m
10.2 13.5
Point (j) Real beam 6t 2
Conj. beam 16.88
t/m'
a
b
1 [-16] =-16 θb =EI EI
j
6.75t
29.33
2.25 t.m
11.25t.m 4.5
M = 6.75(3) - 6(1.5) = 11.25t.m j
1 [+29.33(3)-16.88(1) -4.5(1.5)] = + 64.36 y = EI EI j
11
Structural Analysis (II)
Ex(4)
Chapter 5: Deflection
Using the Conjugate beam method, determine : * the rotation at points (c and e ), * the relative (change in) rotation at point( d ), * the deflection at points(c ,e and d ).
[take EI = 6000 t/cm²]
4t
2t/m' b
a
I
Solution
I
d
2
I
2I
2I
e
7t
c
13t
a
I
d
7t 7t.m
8t.m
4t.m e
a
b
d
B.M.D(Original beam)
4.0
8.0
7t.m a
8.0 4t.m
4t.m 2t.m
d
e
42.67
4.0 c
b
16t.m 21.33 21.33
42.67 a
c
16t.m
3.5
Elastic Reactions
4t
b
Reactions(Original beam)
Conjugate beam
c
16t
t/m'
7.0t.m
2I
2I
e
d e
3.5
24.2
8.0
4.0
8.0
16.29
27.25 c
4.0
12.29
Structural Analysis (II)
Chapter 5: Deflection
Required rotations and deflections Point (c)
c
27.25
12.29 1 [-12.29] = - 0.002rad θc = 6000 1 [-27.25] = - 0.454 cm yc = 6000
Point (d)(Internal Hinge )
d
a 5 . 3
24.2
1 [+ 3.5] = 5.833*10 -4 rad θd/L = 6000 1 [+ 3.5 + 24.2 ] = + 4.6167*10 -3 θd/R = 6000 24.2 = 4.033*10-3 rad θd/rel = │θd/R - θd/L│= Rd = EI 6000 1 [3.5(0.67)] = + 0.039 cm yd = 6000
21.33
Point (e) e
4.0
8.0
16.29
1 [-16.29-8-4+21.33] = - 0.00116 rad θe = 6000 1 [16.29*4 + 8*2.67+ 4*1.33-21.33*1.5] = + 0.997 cm ye = 6000
