1.571 Structural Analysis and Control
Prof. Connor
Section 2: Linear Formulation for a General Planar Member
2.1 Geometric Relations: Plane Curve
t2
Y
t1
y
r
j
x
X
i
x = x(θ)
y = y(θ)
θ = Parameter to define the curve
r = position v ector for a point on the curve
r = xi + yj
2.1.1 Arc length
2
2 1⁄2
Differential arc length = (dx + dy )
ds =
dx⎞
⎛⎛ ----⎝ ⎝ dθ⎠
2
= ds
dy 2 1 ⁄ 2
+ ⎛⎝ ------⎞⎠ ⎞⎠ dθ
dθ
Define
α =
Then
dx-⎞
⎛ ⎛ ----⎝ ⎝ dθ⎠
2
dy 2 1 ⁄ 2
+ ⎛⎝ ------⎞⎠ ⎞⎠
dθ
ds = αdθ
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 1 of 24
2.1.2 Tangent and Normal Vectors
( x, y )
r1
∆r
( x + ∆x, y + ∆y )
r2
∆r = r 2 – r 1 = r(θ + ∆θ) – r ( θ )
∆r = ((x + ∆x)i + (y + ∆y)j) – (xi + yj)
∆r = (∆xi + ∆yj)
2
2 1⁄2
∆r = (∆x + ∆y )
= ∆s
Then for continuous and well­behaved functions
dr = (dxi + dyj)
drdx dy
dr
----- = ------= t1 = ⎛ ------ i + ------ j⎞
⎝ ds
ds
dr
ds ⎠
t 1 = un it tangent vector
1 dy
1 dx
t 1 = ⎛ ---------⎞ i + ⎛ ---------⎞ j
⎝ αdθ⎠
⎝ αdθ⎠
Define t 2 su ch that
t 1 × t 2 = k (u nit vector in z ­direction)
and enforce
t 1 • t2 = 0 ( perpendicular)
gives
1 dy
1 dx
t 2 = ⎛ – --- ------⎞ i + ⎛ --- ------⎞ j
⎝ α dθ⎠
⎝ α dθ⎠
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 2 of 24
2.1.3 Differentiation formulae for the tangent and normal vectors t 1 and t 2
t 1 • t1 = t1
2
t 2 • t2 = t2
2
= 1
= 1
dt
dt
dt
d
----- ( t1 • t 1 ) = t 1 • 1 + t 1 • 1 = 2 ⎛ t 1 • 1⎞ = 0
⎝
ds
ds
ds ⎠
ds
Therefore, dt 1
must be ⊥ to t 1 , and thus proportional to t 2
ds
The result is written as
dt 1
1
= --- t 2
ds
R
dt 2
1
= – --- t 1
ds
R
where
2
2
1- ⎛ d x dy d y dx ⎞
--1- = ----+ 2 ⎟
⎜–
3
2
R
α ⎝ dθ dθ dθ dθ⎠
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 3 of 24
2.1.4 Example: Curvilinear Formulation
Take the angle θ as the parameter defining the curve
Y
θ
R
t2
t1
θ
y
X
x
x = R sin θ
y = R (1 – cos θ)
α =
2
⎧ d
R ⎨ ⎛ sin θ⎞
⎝
⎠
⎩ dθ
+
⎛ d (1
⎝d θ
2
2
1⁄2
–
1⁄2
2⎫
⎞
cos θ) ⎬
⎠
⎭
α = R {cos θ + sin θ}
= R
1
t 1 = --- {R cos θi + R sin θj } = cos θi + sin θj
R
t 2 = – sin θi + cos θj
dt 1
1 dt
1 dt
1
1
= --- 1 = --- 1 = --- (– sin θi + cos θj ) = --- t 2
ds
αds
Rds
R
R
dt 2
1 dt
1 dt
1
1
= --- 2 = --- 2 = --- (– cos θi – sin θj ) = – --- t 1
αds
Rds
R
R
ds
This illustrates the equations on the previous page.
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 4 of 24
2.2 Force Equilibrium Formulation: Curvilinear Formulation
t2 V
Y
t1
F
t3 = t1 × t2
j
X
i
F = Ft 1 + Vt 2
M =
Mt 3
Loading per unit arc length
b = b 1 t 1 + b2 t 2
m = mt 3
2.2.1 Vector Equations
F + ∆F
M + ∆M
∆s
M
F
F + ∆F – F + b∆s = 0
M + ∆M – M + m∆s – ∆st 1 × (– F) = 0
Then
dF
------- + b = 0
ds
dM
-------- + m + t 1 × F = 0
ds
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 5 of 24
2.2.2 Scalar Equations
F = Ft 1 + Vt 2
dt
dt
dF
dF
dV
------- = d ( Ft 1 + Vt 2 ) = F ------1- + t1 ------- + V ------2- + t 2 ------ds
ds
ds
ds
ds
ds
dF
dV
------- = --F- t 2 + t 1 dF
------- – V
--- t 1 + t 2 ------ds
R
ds R
ds
dF
--- t + t dF
------- + b = F
------- – --V- t + t dV
------- + b 1 t 1 + b2 t 2
ds
R 2 1 ds R 1 2 ds
Separating into components of t 1 a nd t 2 leads to
dF- V
-----– --- + b 1 = 0
ds R
dV F
------- + --- + b 2 = 0
ds R
For the moment equation
dM
-------- + m + t 1 × (Ft 1 + Vt 2 ) = 0
ds
Expanding and writing in scalar form
dM
-------- + m + V = 0
ds
Summary
dF
------- – V
--- + b 1 = 0
ds R
dV
------- + F
--- + b 2 = 0
ds R
dM
-------- + m + V = 0
ds
Note: F and V are coupled
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 6 of 24
2.3 Force Displacement Relations : Curvilinear Coordinates
( u – βξ )t 1
vt 2
ut 1
ξ
β
2.3.1 Strains : Displacement components defined wrt the local frame (t 1, t 2 )
∆l ( ξ ) = (R – ξ)∆θ
∆s
∆s = R∆θ → ∆θ = ----­ R
∆s
∆l ( ξ ) = (R – ξ) -----R
∆θ
R
∆l ( ξ )
ξ
ξ
∆l ( ξ ) = ⎛ 1 – ---⎞ ∆s
⎝
R⎠
∆s
u = ut 1 + vt 2 – ξβt1
ε = ε
γ
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 7 of 24
du
ε = -----dl
ξ
dl =
⎛⎝ 1 – ---⎞⎠ ds
R
1
du
ε = ------------------ -----­
ξ-⎞ ds
⎛ 1 – -⎝
R⎠
dv
β
du
v
------ = --u- t 2 + du
------ t 1 – --- t 1 + ------ t 2 – ξ --- t2 – ξ dβ
------ t 1
R
ds
ds
R
ds
R
ds
Separating t1 a nd t 2 terms
1 ⎧ du v
dβ ⎫
ε
= ------------------ ⎨ ------ – --­ – ξ
⎬
ds ⎭
ξ-⎞ ⎩ ds R
⎛ 1 – -⎝
R⎠
Define
du v
ε a = ------ – --­
ds R
dβ ⎫
1 ⎧
ε
= ------------------ ⎨ ε a – ξ
⎬
ds ⎭
ξ-⎞ ⎩
⎛ 1 –
-⎝
R⎠
⎫
1 ⎧ dv ξ
u
γ
= ------------------ ⎨ ------ – --- β + --- – β
⎬
R
ξ ⎞ ⎩ ds R
⎭
⎛ 1 –
-⎝
⎠
R
1
γ
= ------------------ γo
ξ⎞
⎛ 1 –
-⎝
R⎠
For most members ξ
« R then
ξ
--- ∼ 0
R
so
dv
u
γ
o ≅
------ –
β
+ --ds
R
Also if the member is thin
ξ
1 –
--- ∼ 1
R
Then, the strain distribution reduces to a linear distribution over the cross­section.
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 8 of 24
2.3.2 Stress Strain Relations
1
ε = --- σ + εo → σ = E (ε – ε o )
E
1
γ = ---- τ → τ = G γ
G
2.3.3 Force­Deformation Relations
F = ∫ σ dA
M = ∫ – ξσ dA
V = ∫ τ dA
Substituting for the strains leads to
F = ∫ E (ε – ε o ) dA =
F =
∫
⎧
⎪ E ⎛
------------ ε a
∫⎨
ξ⎝
⎪ 1 – -­
⎩
R
⎫
⎪
dβ
– ξ ⎠⎞ – E ε o ⎬dA
ds
⎪
⎭
E ­ du
E v
E
dβ
----------------dA – ∫ ------------------ --- dA – ∫ ------------------ ξ dA – ∫ E ε o dA
ξ⎞ R
ξ⎞ ds
ξ-⎞ d s
⎛ 1 – -⎛ 1 – -⎛ 1 – -⎝
⎝
⎝
R⎠
R⎠
R⎠
E
Eξ
F = ε a ∫ ------------------ dA – β, s ∫ ------------------ dA – ∫ E ε o dA
ξ⎞
ξ⎞
⎛ 1 – -⎛ 1 – -⎝
⎠
⎝
R
R⎠
Similarly
2
Eξ
Eξ
M = – ε a ∫ ------------------ dA + β,s ∫ ------------------ dA + ∫ ξE εo dA
ξ⎞
ξ-⎞
⎛ 1 – -⎛ 1 – -⎝
⎠
⎝
R⎠
R
G
V = γo ∫ ------------------ dA
ξ-⎞
⎛ 1 – -⎝
R⎠
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 9 of 24
Geometric Relations
1
R
ξ
1 ξ
------------ = ------------ = 1 + ----------- = 1 + --- -----------ξ
R
–
ξ
R
–
ξ
R
ξ
1 – --1 – --R
R
1
1
ξ
-----------dA = ∫ dA + --- ∫ ------------ dA
∫
ξ
ξ
R
A 1 – --A
A 1 – --­
R
R
Now
2
ξ
1
1 ξ
------------ = ξ ------------ = ξ + --------------ξ
ξ
R
ξ
1 – --1 – --1 – --R
R
R
ξ
--1- ∫ -----------dA =
R
ξ
A 1 – --R
2
1 -----------ξ
--ξ----dA
∫ R dA + 2 ∫
ξ
R
A
A 1 – --­
R
Making
1
------------ dA =
∫
ξ
A 1 – --R
2
ξ
1
ξ
-------- ------------ dA
∫ dA + ∫ R dA + 2 ∫
R A 1 – --ξ­
A
A
R
But
∫ ξ dA
A
= 0 i f ξ i s measured from the centroid of the section
Finally, one can write
1
------------ dA =
∫
ξ
A 1 – --R
2
1
ξ
I'------ ------------ dA = A + ----∫ dA + 2 ∫
2
R A 1 – --ξR
A
R
where
2
ξ
I' = ∫ ------------ d
A
ξ
A 1 – --­
R
We can use this expression to expand the force­deformation relations.
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 10 of 24
Summarizing
F = D S ε a + D SB β, s + F o
M = D SB ε a + D B β,s + Mo
V = D T γo
where
DS =
∫
E ----------------dA
ξ⎞
⎛ 1 – -⎝
R⎠
ξE
D SB = – ∫ ------------------ dA
ξ-⎞
⎛ 1 – -⎝
R⎠
DT =
∫
G
------------------ dA
ξ⎞
⎛ 1 – -⎝
R⎠
2
DB =
ξ E
------------------ dA
∫
ξ-⎞
⎛ 1 – -⎝
R⎠
For the homogeneous case E and G ar e constant, and the above simplify to
I'
D S = EA + E -----2
R
I'
D SB = – E --­
R
D B = EI'
I'D T = GA + G-----2
R
Additionally, for a thin curved member, one can neglect ξ ⁄ R wrt 1, obtaining the simplified form
EI
D S = EA + -----2­
R
EI
D SB = – ----­R
D B = EI
GI
D T = GA + -----2
R
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 11 of 24
2.3.4 Example for a Rectangular Section
2
ξ
I' = ∫ ------------ dA
ξ
A 1 – --­
R
dA = bdξ
ξ
CL
d
d⁄2
2
bξ
-----------dξ
I' = ∫
ξ
–d ⁄ 2 1 – -R
b
⎧
⎫
3 d 4
3 d 2
I' = I ⎨ 1 + ------ ⎛ ---⎞ + --------- ⎛⎝ ---⎞⎠ + ... ⎬
112 R
20 ⎝ R⎠
⎩
⎭
3
bd
I = ------­12
d
d n
and for --- < 1 , ⎛ ---⎞ « 1 for n > 1
⎝ R⎠
R
Summary of Equations
du v
dβ
F = D S ⎛ ------ – ---⎞ + D SB ------ + Fo
⎝ ds R⎠
ds
du v
dβ
M = D SB ⎛⎝ ------ – ---⎠⎞ + D B ------ + M o
ds
ds R
dv
u
V = D T ⎛ ------ – β + ---⎞
⎝ ds
R⎠
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 12 of 24
2.3.5 Flexibility Matrix ­ Circular Member
A
X
Note: k into page Y
R
θB
B
V B, v B
}
F B, u B
M B, β B
End Actions and displacements
referred to the local frame at B
Assumptions
1. “Thin” linear elastic member
2. Neglect stretching and transvers shear deformations. This is valid for non­
sha llow cases only
uB
uB = vB
FB
FB =
βB
VB
MB
u B = f B FB
2 3
1
2
1 2
R ⎛⎝ --- – 2 sin θ B + --- sin θ B cos θ B⎞⎠ R ⎛⎝ 1 – cos θ B – --- sin θ B⎞⎠ R(θ B – sin θ B )
2
2
2
fB =
1 2
2
R ⎛ 1 – cos θ B – --- sin θ B⎞
⎝
⎠
2
R (θ B – sin θ B cos θ B )
R(1 – cos θ B )
R(θ B – sin θ B )
R(1 – cos θ B )
θB
1.571 Structural Analysis and Control
Prof Connor
2
Section 2
Page 13 of 24
2.4 Cartesian Formulation ­ Governing Equations
Consider the case where the centroidal axis is defined by y = y ( x ) . The i ndependent variable is x, the position coordinate on the x­axis. In the angular formulation, we worked with displacement components and loads re ferred to the local curvilinear
frame defined by the unit vectors t 1 and t 2 . In this section, we are going to work with quantities referred to the global cartesian reference frame. This approach was originally suggested by Marguerre.
Y, v g
t1
vg j
t2
ψ
x
ds
α = ------ → ds
dx
dy
sin ψ = ------ =
ds
dx
cos ψ = ------ =
ds
ug i
= αdx
dy­
--------αdx
--1­
α
y
s
X, u g
Ny
V
M
F
ψ
Nx
by dx
b x dx
m g dx
dx
The above figure defines the “cartesian” notation. The distributed loading ( b x, b y, m g )
is defined with respect the the x and y directions and the loading per unit projected length ( dx ). Similarly, the displacement components ( u g, v g ) are referred to the
cartesian global reference directions.
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 14 of 24
2.4.1 Force Equilibrium Equations
b y F + ∆F
o
M
bx
M + ∆M
dy
F
dx
∑F
= 0 = F + ∆F + ∆x ( b x i + by j) – F = 0
∆F-----+ bx i + byj = 0
∆x
then
dF
------- + bx i + b y j = 0
dx
Knowing
F = (F cos ψ – V sin ψ)i + ( Fsin ψ + V cos ψ)j
we get
d----(( F cos ψ – V sin ψ)i + (F sin ψ + V cos ψ)j) + b x i + by j = 0
dx
Separating
d----(F cos ψ – V sin ψ) + b x = 0
dx
d
----(F sin ψ + V cos ψ) + b y = 0
dx
Similarly
1
= 0 = –(∆xi + ∆yj) × (–F) + (M + ∆M – M + m g ∆x) – --- (∆xi + ∆yj) × (b x i + b y j)∆x
2
Second order term goes to 0 in the limit ∆x → 0
∑ Mo
(∆xi + ∆yj) × (( F cos ψ – V sin ψ)i + (F sin ψ + V cos ψ)j) + (∆M + m g ∆x)k = 0
(∆x ( F sinψ + V cos ψ)–∆y ( F cos ψ – V sin ψ) + ∆M + m g ∆x)k = 0
∆y
∆y
∆M
F sinψ + V cos ψ– ------ F cosψ – ------ V sinψ + --------- + m g = 0
∆x
∆x
∆x
dy
dy
dM
F sinψ + V cos ψ– ------ F cos ψ – ------ V sin ψ + -------- + m g = 0
dx
dx
dx
But
dy
------ = α sin ψ
dx
1.571 Structural Analysis and Control
Prof Connor
1
cos ψ = --­
α
Section 2
Page 15 of 24
Then
1
dM
F sinψ + V cos ψ–α sin ψF --- – α sin ψV sin ψ + -------- + m g = 0
α
dx
dM
2
V cosψ + αsin ψV + -------- + m g = 0
dx
1
Multiply both sides by --α
dM
--1- ⎛ V cos ψ + αsin 2 ψV + -------- + m g⎞ = 0
⎠
dx
α⎝
1 dM 1
2
2
Vcos ψ + Vsin ψ + --- -------- + --- m g = 0
α dx α
Therefore
1 dM
1
--- -------- + V + --- m g = 0
α dx
α
d
------ (F cos ψ – V sin ψ) + b x = 0
dx
d----(F sin ψ + V cos ψ) + b y = 0
dx
N x = F cosψ – V sin ψ
N y = F sin ψ + V cos ψ
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 16 of 24
4.2 Deformation ­ Displacement Relations
e2
s
β o
vg
e 1 = s tretching deformation
e1
ug
e 2 = tr ansverse shear deformation
dx
dx ------ = cos ψ → ds = -----------­
ds
cos ψ
u = (u g i + v g j)
du
du------ = ----cos ψ
ds
dx
du
dv
du
------ = ⎛ --------g i + -------g- j⎞
⎝ dx
dx
dx ⎠
dv
du
du
------ = cos ψ ⎛ --------g i + -------g- j⎞
⎝ dx
dx ⎠
ds
du g dv g
du
e 1 = ------ ⋅ t1 = cos ψ ⎛ -------- i + -------- j⎞ ⋅ ( cos ψi + sin ψj)
⎝
ds
dx
dx ⎠
dv g
2 dug
e 1 = cos ψ -------- + sin ψ cos ψ -------­
dx
dx
du g dv g
du
e 2 = ------ ⋅ t2 – β = cos ψ ⎛ -------- i + -------- j⎠⎞ ⋅ (– sin ψi + cos ψj) – β
⎝ dx
dx
ds
du g
2 dv g
e 2 = – sin ψ cos ψ -------- + cos ψ -------- – β
dx
dx
dβ
dβ
k = bending deformation = ------ = cos ψ ----­ds
dx
note: e 1 and e 2 a re referred to the centroidal axes (s)
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 17 of 24
2.4.3 Force Deformation Relations
2.4.3.1 Thin slightly curved member: use relations developed for the prismatic case
Assume
ε = e 1
γ = e 2
F =
∫ E ( e1
A
DS =
V =
∫
– e 1, o ) dA = (e 1 – e 1, o )D S
E dA A
∫ Ge 2 dA
A
DT =
M =
∫
= e2 DT
G dA A
∫ E (e 1
A
– e 1, o )ξ dA =
M = (k – k 0 )D B
DB =
∫ E( k
A
2
2
– k o )ξ dA = ( k – ko ) ∫ Eξ dA
A
2
∫
Eξ d
A
A
Summarize as follows
F
e 1 = e 1, o + ----­DS
V
e 2 = ------­
DT
M
k = k o + -----­ DB
2.4.3.2 Thick Case ( ξ ≠ 0 wrt R) : change in l o as a function of ξ must be considered
Use equations developed in section 2.3.3 with
e a
= e 1
γ o = e 2
dβ
k = ----­ds
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 18 of 24
4.4 Approximate Formulation ­ Shallow Car tesian
2
­ Sha llow assumption ψ «
1
Then
cos ψ ≅ 1
dy
sin ψ ≅ tan ψ ≅ ------ = y, x
dx
α = 1
(ds ≅ dx)
­ M arguerre neglects V terms in the x equilibrium equation
Nx = F
N y = V + Fy, x
So, resulting equations (shallow and Marguerre)
Equilibrium
dF
------- + b x = 0
dx
dV- ----d dy
-----+ - ⎛ F ------⎞ + b y = 0
dx dx ⎝ dx⎠
dM------+V+m = 0
dx
Deformation ­ Displacement
du
dy dv
e 1 = --------g + ------ -------g- = ε a
dx dx dx
dv g
e 2 = -------- – β = γ o
dx
dβ
k = ----­dx
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 19 of 24
For the Force­Displacement Relations, thin and thick should be considered separately
­ Thi n, Shallow, Marquerre
du
dy dv
F = D S ⎛ --------g + ------ -------g-⎞ – D S e, o
⎝ dx dx dx ⎠
DS =
∫
DT =
∫
DB =
∫
E dA
A
dv
V = D T ⎛ -------g- – β⎞
⎝ dx
⎠
G dA
A
dβ
M = D B ------ – D B k o
dx
2
Eξ dA
A
­ Thi ck, Shallow, Marguerre
du
dy dv
dβ ⎫
F = D S ⎛ --------g + ------ -------g-⎞ + D SB ------ ⎪
⎝ dx
dx dx ⎠
ds ⎪
⎬
du g dy dv g
dβ ⎪
M = D SB⎛ -------- + ------ --------⎞ + D B ------ ⎪
⎝ dx
dx dx ⎠
ds ⎭
ds ≅ dx
dv g
V = D T ⎛ -------- – β⎞
⎝ dx
⎠
E
D S = ∫ ------------------------ dA
(1 – ξ ⁄ R)
ξE
D SB = – ∫ ------------------------ dA
(1 – ξ ⁄ R)
G
D T = ∫ ------------------------ dA
(1 – ξ ⁄ R)
2
ξ E
D B = ∫ ------------------------ dA
(1 – ξ ⁄ R)
Boundary Conditions
in general, the B.C.’s are
u g or N x ⎫
⎪
v g or N y ⎬ pre scribed at each end
⎪
β or M ⎭
for the Marguerre formulation
ug or F
v g or V +
β or M
⎫
⎪
y, x F ⎬
⎪
⎭
1.571 Structural Analysis and Control
Prof Connor
pres cribed at each end
Section 2
Page 20 of 24
Example
Y
h
X
L
1 2
y = --- ax
2
y, x = ax
y, xx
= a
2 H
a = -----­-2
L
x
H x
y, x = aL --- = ⎛ 2 ----⎞ --⎝ L⎠ L
L
x goes from 0 to 1; then 2 H ⁄ L provides a measure of y, x .
The maximum value of y, x is 2 H ⁄ L .
α =
⎧ ⎛ dx ⎞ 2
⎨ ⎝ ------⎠
⎩ dθ
+
1⁄2
dy-⎞ 2 ⎫
⎛ ----⎝ dθ⎠ ⎬
⎭
using θ = x
2 1 ⁄
2 α = {1 + (y, x ) }
α max =
⎧
⎨1
⎩
+
1⁄2
2 H⎞ 2 ⎫
⎛ -----⎝ L ⎠ ⎬
⎭
Suppose H = L ⁄ 10 then α max = 1.02 . Thus, it is reasonable to assume α is constant
2
and equal to 1. In gen eral, when (y, x ) is small wrt 1, one can simplify these equations. This is what we call the “shallow cartesian formulation.” The r esulting approximate equations are:
1
--- ≅ y, xx
α≅1
R
t 1 ≅ i + y, x j
t 2 ≅ – y, x i + j
cos ψ ≅ 1
sin ψ ≅ y, x
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 21 of 24
2.4.5 Shallow Cartesian Solution ­ Parabolic Geometry
b
M B, β B
V B, v B
Y
B F B, u B
1
y = --- ax 2
2
A
X
Use Marguerre formulation, assume thin and shallow
b x = 0 , b y = constant = b , m g = 0
y, x = ax
at x = 0
u g = v g = β = 0 (fi xed support)
at x = L
N x = N Bx
N y = N By
M = MB
Governing equations
Equilibrium
F, x = 0
V, x + (Fax),x + b = 0
M, x + V = 0
Force Deformation (set u g = u ; v g = v )
F
-----= u,x + axv, x
DS
V-----= v, x – β
DT
M-----= β,x
DB
Boundary Conditions
at x = L
F = N Bx
at x = 0
u = v = β = 0
V = N By – aLF
M = MB
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 22 of 24
Solution for the internal forces
F, x = 0
F = C 1 = N Bx
V, x + (N Bx ax), x + b = 0
V = – bx – ( ax
)N Bx + C 2
N By – aLF = – bx – ( ax
)N Bx + C 2
C 2 = N By + bL
V = – bx – ( ax
)N Bx + N By + bL
V = N By + b ( L – x) – ( ax
)N Bx
M = N By x +
b ⎛ Lx
⎝
2
2
MB
2
x
x
– -----⎠⎞ – ⎛⎝ a -----⎠⎞ N Bx + C 3
2
2
2
L
L
= N By L + b ----- – a ----- N Bx + C 3
2
2
2
2
L
L
C 3 = M B – N By L – b ----- + a ----- N Bx
2
2
b
2 aN Bx 2
2
M = --- (L –x) – ------------ (L – x ) + (L – x)N By + MB = D B β, x
2
2
Solution for displacements
3
2
b
x
3 aN Bx
2 x
D B β = – --- (L – x) + ------------ ⎛ – xL + -----⎞ + ⎛⎝ Lx – -----⎠⎞ N By + xM B + C 4
6
2 ⎝
3 ⎠
2
b 3
C 4 = --- L
6
V
v, x = ------- + β
D T
2
2
⎫
1 ⎧
x
x
v = ------- ⎨ xN By + b ⎛ Lx – -----⎞ – ⎛ a -----⎞ N Bx ⎬
⎝
D T ⎩
2⎠ ⎝ 2⎠
⎭
2
3
4
2 2
4
2
3
2
aN Bx x
⎫
x
1 ⎧
x x
x
x
x
2x
+ -------⎨ b ⎛⎝ L ----- – L ----- + ------⎠⎞ + ------------ ⎝⎛ – ----- L + ------⎠⎞ + ⎛⎝ L ----- – -----⎠⎞ N By + ----- MB ⎬ + C 5
6 24
12
2 6
2
DB ⎩
4
2
2
⎭
C 5 = 0
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 23 of 24
F
u, x = ------ – axv, x
D S
V
F
u,x = ------ – ( ax ) ⎛ ------- + β⎞
⎝D
⎠
DS
T
2
2 3
2
3
N Bx
aLx
ax ⎫
a x
1 ⎧ ------ax
- x + ------ ⎨ – - N By + ---------- N Bx + b ⎛ – ------------ + --------⎞ ⎬
u = -------⎝
DS
3
2
3 ⎠⎭
D T ⎩ 2
2
3
4
5
3
5
3
a N Bx x 3 2 x 5
⎫
2x
x
x
x
ax
1 ⎧
x
+ -------⎨– ab ⎛ L ----- – L ----- + ------⎞ + --------------- ⎛ ----- L – ------⎞ + aN By ⎛ – L ----- + -----⎞ + -------- M B ⎬ + C 6
⎝
⎝
⎝
⎠
⎠
⎠
6
2
3
8 30
3
15
3 8
D B ⎩
⎭
C6 = 0
Flexibility matrix ­ cartesian formulation
Set
u gB
N Bx
u B = v gB
F B = N By
βB
MB
u B = u B, o + T AB u A + f B FB
T AB
1 0 –h
= 01 L
00 1
N Ax = N Ax, o – N Bx
N Ay = N Ay, o – N By
M A = MA, o –M B – LN By + hN Bx
1.571 Structural Analysis and Control
Prof Connor
Section 2
Page 24 of 24