1.571 Structural Analysis and Control
Prof. Connor
Section 1: Straight Members with Planar Loading
Governing Equations for Linear Behavior
1.1 Notation
Y, v
V
a
M
F
X, u
Internal Forces
a
1.1.2 Deformation ­ Displacement Relations
a’
Y
a
β
θ
v
a’
b
a
Assume β is small
Displacements (u, v, β)
Longitudinal strain at location y :
For small β
Then
ε ( y ) = ∂ u( y )
∂x
u ( y ) ≈ u ( 0 ) – yβ
v(y ) ≈ v( 0)
ε ( y ) = u,x – yβ,x = ε a + ε b
ε a = u,x = stretching strain
ε b = –yβ, x = bending strain
1.571 Structural Analysis and Control
Prof Connor
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Shear Strain
a’
a
β
b’
θ
b
γ = decrease in angle between lines a and b
γ = θ–β
dv
= v, x
dx
γ = v,x – β
θ≈
1.3 Force ­ Deformation Relations
σ = Eε ⎫
⎬ stress strain relations for linear elastic material
τ = Gγ ⎭
τ
σ
V
M
F
X
F = ∫ σ dA
M = ∫ -yσ dA
V = ∫ τ dA
Consider initial strain for longitudinal actions
where
εσ + εo = εa + εb = εT
ε σ = strain due to stress
ε o = initial strain
Then
ε T = total strain = ε a + ε b
1
ε σ = ε T – ε o = --- σ
E
1.571 Structural Analysis and Control
Prof Connor
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σ = E(ε T + ε o ) = E(εa + ε b – ε o )
= E ( u,x – yβ, x – ε o )
∴
F = ∫ σdA
F = ∫ E ( u,x – yβ, x – ε o )dA
F = u,x ∫ E dA + β, x ∫ -yE dA + ∫ -ε o E dA
Also
M = ∫ -yσdA
M = ∫ -yE ( u, x – yβ, x – ε o )dA
2
M = u,x ∫ -yE dA + β, x ∫ y E dA + ∫ yε o E dA
If one locates the X ­axis such that
∫ yE dA
= 0
the equations uncouple to give:
F = u,x ∫ E dA + ∫ -εo E dA
2
Define
M = β, x ∫ y E dA + ∫ yε o E dA
D S = ∫ E dA = stretching rigidity
DB =
∫y
2
E dA = bending rigidity
F o = - ∫ εo E dA
Mo =
Then
∫ yε o E dA
F = D S u, x + Fo
M = D B β, x + M o
Consider no inital shear strain
τ = Gγ = G ( v, x – β)
V = ∫ GγdA = ∫ G(v, x – β)dA
V = (v, x – β) ∫ GdA
Define
DT =
∫ G dA
1.571 Structural Analysis and Control
Prof Connor
= transverse shear rigidity
Section 1
Page 3 of 17
then
V = D T (v, x – β)
1.4 Force Equilibrium Equations
V + ∆V
M + ∆M
M
C
F + ∆F
F
b x ∆x
m∆x
V
b y ∆x
Consider the rate of change of the internal force quantities over an interval ∆x
F
∑ x
= -F + F + ∆F + b x ∆x = 0
∆F + b x ∆x = 0
∆F
------- + b x = 0
∆x
∑ Fy
= -V + V + ∆V + b y ∆x = 0
∆V + b y ∆x = 0
∆V
------- + b y = 0
∆x
2
∆x
--------∑ M c = -M + M + ∆M + m∆x – b y 2 + V∆x = 0
2
∆x
∆M + m∆x + V∆x – b y --------- = 0
2
∆M
∆x
--------- + m + V–b y ------ = 0
∆x
2
Let ∆x → 0 (i.e. ∆x → dx )
∂F
+ bx = 0
∂x
∂V
+ by = 0
∂x
∂M
+V+m = 0
∂x
1.571 Structural Analysis and Control
Prof Connor
Section 1
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1.5 Summary of Formulation
Equations “uncouple” into 2 sets of equations; one set for “axial” loading and the
other set for “transverse” loading.
Axial (Stretching)
F,x + b x = 0
F = Fo + D S u, x
Boundary Condition
F or u prescribed at each end
Transverse (Bending)
V, x + b y = 0
M, x + V + m = 0
M = D B β, x + M o
V = D T (v, x – β)
Boundary Conditions
M or β prescribed at each end
and
V or v prescribed at each end
Note: These equations uncouple for two reasons
1. The location of the X ­axis was selected to eliminate the coupling term
∫ yEdA
2. The longitudinal axis is straight and the rotation of the cross­sections is
considered to be small. This simplification does not apply when:
i ­ the X ­axis is curved (see Section 2)
ii ­ the rotation, β, can not be considered small, creating
geometric non­linearity (see Section 4)
1.571 Structural Analysis and Control
Prof Connor
Section 1
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1.6 Fundamental Solution ­ Stretching Problem
bx
FB
B
A
L
x
Governing Equations:
∂F
+ bx = 0
∂x
F = Fo + D S
Boundary Conditions
F
B
= FB
u
A
= uA
From (i)
(i)
∂u
∂x
(ii)
F ( x ) = - ∫ b x dx + C 1
F(x)
∴
Then
L
= -( ∫ b x dx) + C 1 = F B
L
C 1 = F B + ( ∫ b x dx)
L
F ( x ) = - ∫ b x dx + ( ∫ b x dx) + FB
L
which can be written as
F(x) =
L
∫
x
b x dx + FB
Note: you could also obtain this result by inspection:
bx
F( x )
FB
B
x
F(x) =
L
L
∫
x
b x dx + FB
1.571 Structural Analysis and Control
Prof Connor
Section 1
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From (ii)
F – Fo
--------------- = u,x
DS
u(x) =
∫
F – Fo
--------------- dx + C 2
DS
F – Fo
u A = ⎛⎝ ∫ --------------- dx⎞⎠ + C 2
DS
0
F – Fo
C 2 = u A – ⎛ ∫ --------------- dx⎞
⎝ DS
⎠0
u(x) =
F – Fo
--------------- dx + u A
0 DS
x
∫
xF
u ( x ) = u A + ∫ -----B- dx + up ( x )
0 DS
FB x
u ( x ) = u A + --------- + u p ( x )
DS
where up ( x ) = particular solution due to b x and F o .
where
FB L
u B = u A + ---------- + u B, o
DS
u B, o = u p ( L )
1.571 Structural Analysis and Control
Prof Connor
Section 1
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1.7 Fundamental Solution: Bending Problem
V B, v B
VA
M B, β B
MA
B
A
L
x
V( x)
V B, v B
M B, β B
M( x)
B
L–x
Internal Forces
V ( x ) = VB
M ( x ) = M B + VB (L – x)
Governing Equations for Displacement
M – Mo
M = D B β, x + Mo → β, x = ----------------­ DB
V
V = D T (v, x – β) → v, x = β + ------­
DT
Integration leads to:
2
M B x VB
x
β ( x ) = β A + ----------- + ------- ⎛ Lx – -----⎞ + β o ( x )
DB DB ⎝
2⎠
2
MB L VB L
β ( L ) = βB = β A + ----------- + ------- ----- + β B, o
DB DB 2
VB
M B x 2 V B x2 x 3
v ( x ) = v A + β A L + ------------ + ------- ⎛⎝ L ----- – -----⎠⎞ + ------- x + v o ( x )
DB 2 DB 2 6
DT
M B L 2 VB L3 V B
v ( L ) = v B = v A + β A L + ------------ + ------- ----- + ------- L + v B, o
DB 2 DB 3 DT
1.571 Structural Analysis and Control
Prof Connor
Section 1
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1.8 Particular Solutions
Set
β i, o = end rotation at i due to span load
v i, o = end displacement at i due to span load
Then
β i = β i, e + β i, o
v i = v i, e + v i, o
where
β i, e = end rotation at i due to end actions
v i, e = end displacement at i due to end actions
Concentrated Moment
b
a
B
A
M*
L
M* a
β B, o = ---------­DB
2
Concentrated Force
M* a
M* a
v B, o = -------------- + ----------- ( L – a)
2 DB 2 DB
P*
b
a
B
A
L
2
P* a
β B, o = -----------­
2 DB
3
v B, o
2
P* a P* a
P* a
= ---------- + ------------ + ------------ ( L – a)
3 DB 2 DB
DT
1.571 Structural Analysis and Control
Prof Connor
Section 1
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Distributed Loading
bdx
B
A
dx
L
Replace P* with bdx and integrate from x = 0 to x = L
β B, o
2
L
x
= ∫ ---------- b dx
0 2 DB
3
L
v B, o =
2
L
L
x
x
x
----------------------d
d
b
x
+
b
x
+
∫ D
∫ 3D
∫ 2 D b dx( L – x)
0 T
0
0
B
B
for b constant (ie uniformly distributed loading)
3
bL
β B, o = ---------­
6 DB
2
4
bL
bL
v B, o = ---------- + ---------­
2 DT 8 DB
1.571 Structural Analysis and Control
Prof Connor
Section 1
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1.9 Summary
FB L
u B = u B, o + ---------- + u A
DS
MB L 2 VB L3 VB
v B = v B, o + ------- ----- + ------- ----- + ------- L + v A + β A L
DB 2 DB 3 DT
2
MB L VB L
β B = β B, o + ----------- + ------- ----- + β A
DB D B 2
These equations can be written as
uB
u B, o
L
-----DS
0
0
0
L - -----L
--------+ 3DB DT
L--------­
2 DB
0
L
---------2 DB
3
v B = v B, o +
βB
β B, o
2
2
1
-----­ DB
FB
VB
MB
1 0 0 uA
+ 0 1 L vA
0 0 1 βA
Rigid body transformation from A to B
Also
F A = F A, o – F B
V A = V A, o – V B
M A = MA, o – MB – LV B
FA
F A, o
VA =
V A, o
MA
M A, o
1.571 Structural Analysis and Control
Prof Connor
1 0 0 FB
– 0 1 0 VB
0 L 1 M
B
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1.10 Matrix Formulation ­ Straight Members
Define:
uB
u B = v B Displacement Matrix
βB
FB
FB =
V B End Action Matrix
MB
General Force Displacemnt Relation
Express displacement at B as:
u B = u B, o + f B F B + TAB u A
u B, o : Due to applied loading ⎫
⎪
f B F B : Due to forces at B ⎬ Based on cantilever model
⎪
T AB u A : Effect of motion at A ⎭
Interpret
f B = Member flexibility matrix
T AB = Rigid body transformation from A to B
For the prismatic case
L
-----DS
0
0
0
L
L
---------- + ------3 DB DT
L
---------­
2 DB
0
L --------2 DB
3
fB =
2
1.571 Structural Analysis and Control
Prof Connor
2
1
-----­ DB
Section 1
Page 12 of 17
Force Displacement Relations
Define k B = f –1
= Member stiffness matrix
B
Start with
u B = u B, o + f B F B + TAB u A
Solve for F B
f B FB = u B – T AB uA – u B, o
Define
Then
F B = k B uB – k B T AB u A – k B u B, o
F B, i = – k B uB, o
F B = k B uB – k B T AB u A + FB, i
Next, determine F A
T
F A = FA, o – T AB FB
T
where
T
F A = ( – TAB k B )u B + ( TAB k B T AB )u A + FA, i
T
F A, i = FA, o – T AB F B, i
Note FA, i and F B, i are the initial end actions with no end displacements
Finally, rewrite as
FB = k BB u B + k BA u A + FB, i
T
F A = k BA u B + k AA uA + FA, i
Notice that there are two fundamental matrices: k B and TAB
1.571 Structural Analysis and Control
Prof Connor
Section 1
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Matrices for Prismatic Case
uB
vB
βB
uA
vA
βA
FB
DS
-----L
0
0
DS
– -----L
0
0
VB
0
MB
0
FA
D
– ------S
L
VA
0
MA
0
( –6 )D*
12 D*B
-------------------------------------B3
2
L
L
( –6 )D* (4 + a)D*
---------------------B- ---------------------------B2
L
L
0
0
(–12 )D*
6 D* B
-------------------------------------B3
2
L
L
( –6 )D* (2 + a)D*
---------------------B- ---------------------------B2
L
L
12 D B
a = -----------2
L DT
1.571 Structural Analysis and Control
Prof Connor
0
0
(–12 )D*
( –6 )D*
------------------------B- --------------------­B3
2
L
L
6 D* B
(2 + a)D* B
--------------------------­ -------------2
L
L
DS
-----L
0
0
0
0
12 D* B
----------------3
L
6 D* B
-------------2
L
6 D* B
------------­2
L
(4 + a)D* B
--------------------------­ L
DB
D* B = ---------------­ (1 + a)
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1.11 Transformation Relations
Rigid Body Displacement Transformation
uA
A
uB
B
r AB
ωB
ωA
ωB = ω A
u B = u A + ω A × r AB
uB = uA
⎫
⎪
v B = v A + ω A L ⎬ in two dimensions
⎪
ωB = ωA ⎭
uB
vB
ωB
1 0 0 uA
= 0 1 L vA
0 0 1 ωA
u B = T AB u A
Statically Equivalent Force Transformation
Translate force system acting at B to point A
PB
B
PA
r AB
A
mB
mA
PA = P B
m A = m B + r AB × PB
T
F A = FA, o – T AB FB
1.571 Structural Analysis and Control
Prof Connor
Section 1
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Coordinate Transformation
y’ y
x’
x
z,z’
ax
ax'
a = ay
a' = ay'
az
az'
a' = Ra
cos θ sin θ 0
R = – sin θ cos θ 0
0
0 1
The inverse is
Take
cos θ = cos (–θ) ⎫
T
–1
sin θ = –sin ( –θ ) ⎪
⎬ R = R
⎪
–1
R = R(–θ) ⎭
(x, y, z) = Global frame
(x', y', z') = Local frame
F
R
(l )
( gl )
= R
( gl ) ( g )
F
= R
Given k in local frame ( k ( l ) ), transform to global frame
F
F
(l )
(g )
( l) ( l )
= k u
= R
( l ) ( gl ) ( g )
= k R
( lg ) ( l )
F
= R
u
( lg ) ( l ) ( gl ) ( g )
k R
u
If
F
(g )
= k
(g) (g )
u
Then
k
(g )
= R
( lg ) ( l ) ( gl )
1.571 Structural Analysis and Control
Prof Connor
k R
= (R
( gl ) T ( l ) ( gl )
) k R
Section 1
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1.12 Structural Stiffness Matrix assembly
g
g
g
g
g
g
F B = k BB u B + k BA uA + FB, i
g
g
T g
g
g
g
F A = ( k BA ) u B + k AA u A + FA, i
g
T l
F i = R Fi
Use direct stiffness method to generate the system equations referred to the global
frame.
Take B as the positive end and A as the negative end.
B → n+
A → n-
for member n
Write system equation as
E
= PI + KU
Work with the partitioned form of system stiffness matrix K .
k BB in n+,n+
k AA in n­,n­
T
k BA in n+,n­
T
with
k BA in n­,n+
F B, i in n+ of PI
F A, i in n­ of PI
1.571 Structural Analysis and Control
Prof Connor
Section 1
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