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Structural Analysis (II) Chapter 5: Deflection
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Method (2) Conjugate beam Method ....بالكمرات. نقطه أي عند Slope(θ) Deflection (y)حساب في المستخدمه الطرق وإحدي
الكمرة. طول علي Slope(θ)Deflection (y) تحقق بحيث ( المعطي ) األصلية الكمرة من مستنتجه تخيليه كمرة هي
و
Conjugate beam:
قيم
(Conjugate beam)ب محمله ل( Elastic load)تكون ) مساوي -MEI )حيث ( M )شكل هو
: يكون الحاله هذه في و األصلية الكمرة علي العزم
Rotation or Slope(θ)or (y')= Shear force of elastic load = QelasticDeflection (δ)or (y)= Moment of elastic load =Melastic
Real(Original) beam Conjugate beam:
Load = d²M(x)dx² = w
Shear = dM(x)dx =Q
Moment = M(x)
Elastic load = d²(y)dx² =- M
EI
Rotation = d(y)dx =Qelastic
Deflection = y = Melastic
wt/m'
ملكات
ضلتفا
ملكات
ضلتفا
- M/EI
ملكات
ضلتفا
ملكات
ضلتفا
Rotation)
Deflection)
Structural Analysis (II) Chapter 5: Deflection
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θδ = 0 , θ ≠ 0 M = 0 , Q ≠ 0
Q
Free end Fixed supportδ ≠ 0 , θ ≠ 0 M ≠ 0 , Q ≠ 0
Fixed support Free end
δ = 0 , θ = 0
QM
δ = 0 , θ ≠ 0 M = 0 , Q ≠ 0
Interior support (Roller or Hinge) Internal hinge
θL
θR
(QL=QR)
Internal hinge
δ ≠ 0 , θL ≠ θR ≠ 0
Interior support (Roller or Hinge)θL
θR
M ≠ 0 , QL ≠ QR
من (Conjugate beam)(Original beam)التحويل إلي
M = 0 , Q = 0
End support (Roller or Hinge) End support (Roller or Hinge)
Conjugate SupportReal Support
Examples
Real Beam Conjugate Beam
Indetreminate Beam
(θR = θL)
(???)
Structural Analysis (II) Chapter 5: Deflection
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Steps of Solution :
بإستخدام نقطه أي عند (θ) (y)حساب عند التاليه الخطوات Conjugate beam method.أوتتبع
Using the Conjugate beam method, determine the rotation at points (a,b,c and d )and deflection at points(c and d ).
Ex(2)
4t 4t/m'
I I 2Ib
ca
d
Solution
bcad
4*6²8 =18.0 t.m
12t.m
bc
a
d
12
6t.m
18t.m
9t.m
63
91818
4.5
18
36
Elastic loads
adb
9 1818 4.5
1836
Elastic Reactionsa
d
b
9 18
18
4.5
1836
a
10.12
7.88 14.6212.36
Required rotation and deflection
Point (a)
θa = 1EI[+7.88] = + 7.88
EI (Clockwise)
Point (b)
θb =1EI[-14.62] = - 14.62
EI (Anti-Clockwise)
Point (c)
yc = 1EI[14.62(3) + 4.5(1.0) -18(1.125)]
= + 28.11EI (Downward)
θc = 1EI[18 - 14.62 - 4.5]
= - 1.12EI (Anti-Clockwise)
c
b
4.5
18
14.62
c
Point (d)
yd =1EI[+ 12.36] = + 12.36
EI (Downward)
θd =1EI[-10.12] = - 10.12
EI (Anti-Clockwise)
B.M.D
Conj.beam
I I 2I
I I I
Structural Analysis (II) Chapter 5: Deflection
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Ex(3) 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' 6ta
d bf
ec
Solution EI = Constant
2t/m'
6t
ad b
fe
c
3t4t
a d b f e c
9t.m
10t.m
16t.m1t.m
5t.m
40
85.33
1.3327
10
ad
b fe
c
j
6.75t
16t
16.25t
j
ad
b
fe
c
85.33
1.33 27
10
40 1629.33
24.13 10.2
16
271.3385.33
1040
Structural Analysis (II) Chapter 5: Deflection
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Required rotations and deflectionsPoint (a)
θa = 1EI[+29.33]
= + 29.33EI (Clockwise)
Point (b)
θb =1EI[-16] =-16
EI(Anti-Clockwise)
a
29.33
b
16
θf/L =1EI[-16+10-1.33] = - 7.33
EI (Anti-Clockwise)
yf =1EI[-16(2)+10(1.33)-1.33(1)] = - 20
EI (Upward)
Point (f)(Internal Hinge )
θf/R = 1EI[-16+10-1.33+24.13 ] = + 16.8
EI (Clockwise)
θf/rel = [θf/R - θf/L ]= + 16.8EI - -7.33
EI = + 24.13EI = Rf
EI
f
1.33
10
16
24.13
Point (d)
yd=1EI[+29.33(4)+10(1.33)-42.7(1.5)]
=+66.6EI (Downward)
a d
16t.m
5t.m
42.7
10
29.33
Point (e)
ye=1EI[+10.2(3)-13.5(1)] = + 17.1
EI (Downward)
e c
9t.m
13.5
10.2
Point (j)
M j= 6.75(3) - 6(1.5) = 11.25t.m
2t/m'
a
6.75t
j11.25t.m
29.33
Real beam Conj. beam
2.25 t.m
16.88
4.5
y j =1EI[+29.33(3)-16.88(1) -4.5(1.5)] = + 64.36
EI (Downward)
θd =1EI[29.33+10-42.7] = - 3.37
EI (Anti-Clockwise)
θe = 1EI[13.5-10.2 ] = + 3.3
EI (Clockwise)
6t
Structural Analysis (II) Chapter 5: Deflection
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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]