1.571 Structural Analysis and Control
Prof. Connor
Section 5: Non-linear Analysis of Members
5.1 Deformation Analysis
Y
( 1 + ε y )dy
π
--- – γxy
2
dy
( 1 + ε x )dx
dx
X
5.1.1 Deformation of a fiber dx initially aligned with x .
u1
2
1
v1
v2
u 2 – u 1 = ∆u
dx + u 2 – u 1 = dx + d u
du
dx  1 + ------ = dx ( 1 + u, x )
dx
u2
dx + u 2 – u 1
v 2 – v 1 = ∆v
( 1 + εx )dx
v,x dx
dv
dv = ------ dx = v, x dx
dx
( 1 + u, x )dx
( 1 + ε x ) 2 = ( 1 + u, x ) 2 + v, x2
1 + 2ε x + ε x2 = 1 + 2u, x + u, x2 + v, x2
ε x ( 2 + ε x ) = u, x ( 2 + u,x ) + v, x2
εx
u,x
v,x2
ε x  1 + ---- = u,x  1 + ------- + ------2
2
2
For small strain ( ε x « 1 )
u, x
v,x2
ε x = u, x  1 + ------- + ------
2
2
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 1 of 17
If the member does not experience large relative rotations, then the non-linear u,x
term can be ignored.
Then for small relative rotations
1
ε x = u, x + --- v,x2
2
5.1.2 Deformation of a fiber dy initially aligned with y .
Proceed as done for dx to obtain:
For small strain
v,y
1
ε y = v,y  1 + ------ + --- u,y2
2
2
For small relative rotations
1
ε y = v,y + --- u, y2
2
5.1.3 Shear deformation
~ u, y
~ v,x
For small rotations
γ ≅ u,y + v, x
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 2 of 17
5.2 Straight Beams - Non-Linear Analysis for Plane Members
Y
B
y
X
A
dx
x
5.2.1 Axial Strain
For small rotations
u B = u A – y sin β
v B = v A – y ( cos β – 1 ) ≅ v A
β2 « 1
sin β ≅ β
cos β ≅ 1
εx
εx
εx
εx
5.2.2 Shear Strain
B
B
B
B
∂( uB ) 1 ∂ ( vB ) 2
= -------------- + ---  -------------
∂x
2 ∂x
1 ∂v A 2
∂
= ----- ( u A – y sin β ) + ---  --------
2 ∂x
∂x
1
= u A ,x – yβ, x + --- ( v A , x ) 2
2
= ε x – yβ, x = ε xo – yχ
A
∂u B ∂v B
∂
∂
γ B = --------- + -------- = ----- ( u A – y sin β ) + ----- ( v A )
∂x
∂y
∂x
∂y
γ B = – β + vA ,x
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 3 of 17
5.2.3 Summarizing
Define
ε x ( y ) = ε x'
γ = γ'
ε x' = ε x – yβ, x
1
ε x = u,x + --- v,x2
2
γ' = v, x – β
5.2.4 Apply the principle of virtual displacements
- Take a deformable body resisting some external loads
- Produce a perturbation (a virtual displacement)
- If body is in state of equilibrium, the first order work done by the stresses is equal to
the first order work done by the externally applied forces
i.e.
∫ (σ
⋅ ∂ε ) dVol =
∑ Forces
⋅ Virtual Displacements
For a beam: Internal virtual work is caused by stresses σ x, τ xy
∫ ∫ { ( σ x'δε x'
xA
+ τ xy'δγxy ) dA } dx =
∑ Forces
⋅ Virtual Displacements
ε x' = ε x – yβ, x = ε x – yχ
∂u 1 ∂v 2
1
ε x = u, x + --- v,x2 = ------ + ---  -----
∂x 2 ∂x
2
∂ ( u + δu ) 1 ∂ ( v + δv ) 2
( ε x + δε x ) = ----------------------- + ---  -----------------------
∂x
2
∂x
∂u ∂δu 1 ∂v ∂δv 2
( ε x + δε x ) = ------ + --------- + ---  ----- + ---------
∂x ∂x 2  ∂x ∂x 
∂u ∂δu 1 ∂v 2 ∂v ∂δv 1 ∂δv 2
( ε x + δε x ) = ------ + --------- + ---  ----- + ----- --------- + ---  ---------
∂x ∂x 2  ∂x
∂x ∂x 2  ∂x 
Then
∂δu ∂v ∂δv 1 ∂δv 2
δε x = --------- + ----- --------- + ---  --------- = δu, x + v, x δv, x
∂x ∂x ∂x 2  ∂x 
∂ ( β + δβ )
∂β
∂δβ
y ( χ + δχ ) = y ------------------------ = y ------ + y --------∂x
∂x
∂x
∂δβ
yδχ = y --------- = y∂β, x
∂x
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 4 of 17
Therefore
δε x' = δε x – yδβ,x
∂v
γ' = v,x – β = ----- – β = γ
∂x
∂ ( v + δv )
γ' + δγ' = ----------------------- – β – δβ = γ + δγ
∂x
∂δv
δγ' = --------- – δβ = δv, x – δβ = δγ
∂x
Right hand side of the virtual work equation
∫ ∫ { ( σ x'δε x'
xA
Giving
+ τ xy'δγxy ) dA } dx
∫ ∫ σ x'δε x' dA dx
xA
=
∫ ∫ ( σ x'δε x
xA
∫ ∫ σ x'δε x' dA dx
xA
=
∫ ( Fδε x
x
∫ ∫ τxy'δγ' xy dA dx
xA
=
– σ x yδβ,x ) dA dx
+ Mδβ,x ) dx
∫ Vδγ dx
x
RHS = ∫ ( Fδε x + Mδβ,x + Vδγ ) dx
x
Integrating by parts ∫ uv' = uv – ∫ u'v
RHS = ( Fδu + Fv,x δv + Mδβ + Vδv )
L
0
∂ ( Fv,x )
– ∫  δuF, x + δv  ------------------ + V, x + δβ ( M, x + V ) dx




∂x
x
Left hand side of the virtual work equation
LHS = ∫ b y δv dx + ∫ b x δu dx + ∫ mδβ dx + ∑ P ⋅ δp
x
x
x
Differentiating both RHS and LHS wrt x , recognizing that the first RHS term is a constant and that both the second RHS term and the LHS are integrated over the same
interval, we get
( F, x + b x )δu + ( V, x + ( Fv, x ), x + b y )δv + ( M, x + V + m )δβ = 0
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 5 of 17
Since δu , δv , and δβ are independent, each term becomes and independent expression
F, x + b x = 0
V, x + ( Fv, x ), x + b y = 0
M, x + V + m = 0
These are the governing equations of equilibrium for a non-linear member.
Note: in V, x + ( Fv, x ), x + b y = 0 , ( Fv,x ), x = source of P-δ effect.
5.3 Compatibility Equations
F =
∫ σ x' dA
A
F = u, x ∫ E dA
A
2
1

--∫ Eε x' dA = ∫ E  u, x + 2 v, x
A
A
1 2
+ --- v,x ∫ E dA – β, x ∫ yE dA
2
A
A
=
– yβ, x dA
If y is measured from the mechanical centroid ∫ yE dA = 0
A
Then
1 2
F = D S  u, x + --- v, x
2
where
DS =
∫ E dA
A
1 2
M = – ∫ yσ x' dA = –  u, x + --- v, x  ∫ Ey dA + β, x ∫ Ey 2 dA

2 
A
A
A
M = D B β, x
where
DB =
M =
∫ Ey
A
2 dA
∫ τ xy dA
A
=
∫ Gγ' dA
A
=
∫ G ( v, x
A
– β ) dA
V = D T ( v,x – β )
DT =
∫ G dA
A
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 6 of 17
Summarizing
F, x + b x = 0
V, x + ( Fv, x ), x + b y = 0
M, x + V + m = 0
1 2
F = D S  u, x + --- v, x
2
M = D B β, x
Boundary conditions
V = D T ( v,x – β )
u or Fx
β or M
v or V + Fv,x = Py
Note Py = effective shear
5.3.1 Examples of boundary conditions
a
kr
P
ke
x
L
@x = 0
u = v = β = 0
@x = L
F = –P
M = – k r β ( L ) ; M = 0 without spring
V – v,x P = – k e v ( L )
b
P
@x = 0
u = v = β = 0
@x = L
F = –P
β = 0
V – v, x P = 0
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 7 of 17
Example
bx = m = 0
D B, D S, D T = constant
by
P
A
B
F,x + b x = 0 → F ( x ) = constant = – P
1 2
D S  u, x + --- v,x = – P
2
P 1 2
u,x = – ------ – --- v, x
DS 2
Px 1 x 2
u ( x ) – u ( A ) = – ------ – --- ∫ v, x dx
DS 2 0
M = D B β, x
M, x = D B β, xx
M,x + V + m = 0
D B β, xx + V = 0
V = –D B β, xx
D
V
V
------- = v, x – β → v, x = β + ------- = β – ------B- β, xx
DT
DT
DT
V,x + ( Fv, x ), x + b y = 0
– D B β,xxx – Pv, xx + b y = 0
DB
– D B β,xxx – P  β, x – ------- β, xxx + b y = 0
DT
P
D B β, xxx  ------- – 1 – Pβ, x + b y = 0
DT
by
P
β, xxx + ----------------------------- β, x = ---------------------------P-
P-


----------DB  1 – 
DB 1 – 
DT
DT
Define
P
µ 2 = ----------------------------P
D B  1 – -------
DT
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 8 of 17
Solving for β and then v , starting with
µ2
β,xxx + µ 2 β,x = ------ b
P y
P
β = C 2 + µ  1 – ------- ( – C 4 sin µx + C 5 cos µ x ) + β part

D T
v =

∫ β
D
– ------B- β, xx dx
DT
v = C 2 x + C 3 + C 4 cos µ x + C 5 sin µ x + v part
M = D B β, x = P ( – C 4 cos µ x – C 5 sin µ x ) + D B ( β part ), x
V = –D B β, xx
V = µP ( – C 4 sin µx + C 5 cos µ x ) – D B ( β part ),xx
For b y = 0
@x = 0
v = 0 → C3 + C4 = 0
M=0 →
C4 = 0



C4 = 0
C3 = 0
@x = L
v = 0 → C 2 L + C 5 sin µL = 0
M=0 →
C 5 sin µL = 0



C2 = 0
For a non-trivial solution sin µL = 0 (ie µL = nπ , n = 1, 2, … )
For n = 1
So
π
µL = π → µ = --L
π2
P
----- = ---------------------------2
P
L

D B  1 – -------
DT
P cr
π 2 DB
------------L2 = ---------------------π 2 DB
1 + -----------L2 DT
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 9 of 17
5.4 Member Relations Geometrically Non-Linear Case
Y, v
βB
uA
βA
vB – vA
ρ = ----------------L
ρ
vB
vA
X, u
A
B
L
uB
VB
MB
P
VA
FB
FA
vB – vA
L
PL 2
( µL ) 2 = --------- = λ 2
DB
D B = EI
D = 2 ( 1 – cos µL ) – µL sin µL
Dφ 1 = µL ( sin µL – µL cos µ L )
Dφ 2 = µL ( µL – sin µL )
φ3 = φ1 + φ2
Assumption: Neglect transverse shear deformation wrt bending deformation
uA
uB
uA = vA
uB = vB
βA
βB
FA
FB
FA =
VA
MA
1.571 Structural Analysis and Control
Prof Connor
FB =
VB
MB
Section 5
Page 10 of 17
5.4.1 Member Equations
PL
1
u B = u A – ------- – ∫ --- v,x2 dx
DS
2
1 1
e r = --- ∫ --- v, x2 dx
L 2
PL
u B = u A – ------- – e r L
DS
Set
DS
F B = ------ ( u B – u A + e r L ) = – P (member in compression)
L
D
M B = ------B- ( φ 1 β B + φ 2 β A – φ3 ρ )
L
DB
M A = ------- ( φ 1 β A + φ 2 β B – φ3 ρ )
L
DB φ 3
- ( β B + β A – 2ρ ) – Pρ
V B = – -----------L2
V A = – VB
Write member equations as
DS ⁄ L
MB
0
0
–DS ⁄ L
0
2D B φ 3
--------------L3
DB φ3
– -----------L2
FB
VB =
0
DB φ3 uB
– -----------vB +
L2
DBφ1 β B
-----------L
k BB
FB
0
0
0
0
2D B φ 3
– --------------L3
DB φ3
-----------L2
DS er
DBφ3 uA
– -----------v A + – Pρ
L2
0
DB φ2 βA
------------L
k BA
uA
uB
FB( r )
( i)
If FB = Forces due to span loads
( r)
Likewise
–DS ⁄ L
FA
VA =
MA
FB
( i)
F B = k BB u B + k BA uA + FB + F B
0
0
0
2D B φ 3
– --------------L3
DB φ3
– -----------L2
DS ⁄ L
0
DB φ3 uB
------------vB +
L2
DB φ2 βB
-----------L
k TBA
uB
0
0
0
0
2D B φ 3
--------------L3
DB φ3
-----------L2
k AA
–DS er
DB φ3 uA
-----------v A + Pρ
L2
DB φ1 βA
0
-----------L
uA
F A( r )
(i)
T u + k u + F (r) + F
F A = k BA
A
A
B
AA A
Note: This formulation is wrt local frame and must be transformed to use global frame
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 11 of 17
5.5 Applications
1. A Frame
W
W
IB
H
H
IC
IC
Consider
IB = ∞
W
Boundary Conditions
vB
H
B
βA = βB = 0
vA = 0
vB
ρ = ----L
vB = v
P = W
VB = H
x
A
y
End actions at B
2D B φ 3
- v – Wρ =
V B = --------------L3
 2D B φ 3
 --------------3
L

W
– ----- v = H
L
DBφ3
-v
M B = – -----------L2
Write
V B = H = kv
12D B  φ 3 WL2 
-  ----- – -------------- 
k = ------------L 3  6 12D B 
For W = 0
φ3 = 6
If we apply W = P cr = π 2 D B ⁄ L 2
12D B
k = ------------L3
µL = π
φ3 = π 2 ⁄ 2
12D B  φ 3 W π 2 
-  ----- – ------- ------  = 0 for W = Pcr
k = ------------L 3  6 Pcr 12 
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 12 of 17
2 Cantilever
W
Boundary Conditions
H, vB
βA = v A = 0
B
MB = 0
VB = H
P = W
A
Solution
vB
M B = φ 1 β B – φ3 ----- = 0
L
DB φ3 
v 
v
-  β B – 2 ----B-  – P ----B- = k H v B
V B = H = – -----------L
L
L2 
DB φ3 
φ 
-  2 – ----3-  – P
--k H = – -----------φ1  L
L3 
Note, from elementary mechanics for a cantilevered column
Then
π 2 DB
π2D
P cr = -------------B2- = -----------( 2L )
4L 2
If W = 0
P = 0
φ1 = 4
φ2 = 2
φ3 = 6
3D B
- ← Linear case
k H = --------L3
If W ≠ 0
kH
DB 
φ 32

P
------
= 3  2φ 3 – ----- – --------------------2
φ1 ( DB ⁄ L ) 
L 
At the critical load
π
µL = --2
π
D = 2 – --2
π
Dφ 1 = --2
π 2
Dφ 3 =  ---
 2
kH = 0
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 13 of 17
1
At P = --- P cr
2
π 1
µL = ---  ------- ≅ 1.11
2  2
cos µ L ≅ 0.445
sin µ L ≅ 0.9
φ1 ≅ 4
φ3 ≅ 6
kH
DB 
φ 32 4P 
DB
-----= 3  2φ 3 – ----- – ------2-  ≅ 1.77 ------3φ1 π 
L 
L
So, about a 40% reduction in stiffness due to axial loading.
3 Simply Supported Beam-Column
v
MB *
P
A
B
Boundary Conditions
MA = 0
M B = MB *
Solution
vA = 0
vB = 0
MA = 0 = φ1β A + φ2β B
φ2
β A = – ----- β B
φ1
DB
φ 22
M B = -------  φ 1 – ----- β B = k B β B
L
φ1
If P = 0
DB
k B = 3 ------- (Linear case)
L
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 14 of 17
4 Multi Span Beam-Column
A
H, v C
B
L1
L2
Segment A-B (from example 3)
M B, β B
MB = k B β B
P
A
P
C
B
L1
Segment B-C
H, v C
βB
kB
B
vC
ρ = ----L2
βC
C
P
L2
Boundary Conditions
MC = 0
vB = 0
M B = – k B β B (be careful of sign convention)
VC = H
H
M B, β B
P
Solution
i)
DB
vC
M B = -------  φ 1 β B + φ 2 β C – φ 3 ----- = – k B β B
L2
L2
ii)
DB
vC
M C = -------  φ 1 β C + φ 2 β B – φ 3 ----- = 0
L2 
L 2
Use i) and ii) to solve β B and β C in terms of v C
DB vC
DB  DB

β B  k B + ------- φ 1  + ------- φ 2 β C = ------- φ 3 ----L 2  L2
L2 L 2

φ
φ v
----2- β B + β C = ----3- -----C
φ1
φ 1 L2
Then substitute in expression for H
v 
D B φ3 
v
-  – β C – β B + 2 -----C  – P -----C
H = -----------L2 
L2
L 22 
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 15 of 17
5.5 Incremental Formulation
5.5.1 Some simplifications
1.
1
e r = ------ ∫ v, x2 dx
2L
v –v
L
B
A
- = ρ
Take v,x ≅ ----------------
Then
ρ2
1
e r ≅ ------ ( ρ 2 L ) = ----2L
2
2. Assume φ ‘s are constant during incremental motion. This results in k BB , k AA , and
k BA constant. (ie dk = 0 )
3. Consider small increment, ie work with first order change.
(r )
(i)
dF B ≅ k BB ∆u B + k BA ∆u A + dF B + dFB
(r)
(i)
T ∆u + k ∆u + dF
dF A = k BA
A + dF A
B
AA
A
(r)
(r)
dF A = – dFB
(r)
dF B = dF
F
(r)
(r)
D
------S ρ 2
2
FB ρ =
FBρ
0
0
DS er
=
DS
F B = ------ ( u B – u A ) + D S e r
L
operating on dF ( r )
dF
(r)
D S ρdρ
= F dρ + dF ρ
B
B
0
∆v B – ∆v A
dρ = -----------------------L
D
∆v B – ∆v A
dF B = -----S- ( ∆u B – ∆u A ) + D S ρ  -----------------------

L
L
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 16 of 17
Then
dF
dF
(r)
(r)
DS ρ
--------- ( ∆v B – ∆v A )
L
= F
DS ρ
DS ρ 2
- ( ∆u B – ∆u A ) + ----------------B- ( ∆v B – ∆v A ) + --------- ( ∆v B – ∆v A )
L
L
L
0
= K R ( ∆u B – ∆u A )
0
DS ρ
--------L
0
KR = D ρ F
DS ρ 2
S
--------- -----B- + ------------ 0
L L
L
0
0
0
5.5.2 Summarizing
(i )
dFB ≅ ( k BB + K R )∆u B + ( k BA – K R )∆u A + dF B
(i)
dF A ≅ ( k BA + K R ) T ∆u B + ( k AA + K R )∆u A + dF A
These two expressions define the tangent stiffness for a member. One usually retains
the non-linear terms only when the axial force is compressive since the stiffness is
reduced. A tensile axial force increases the stiffness.
5.5.3 Linearized Stability Analysis
1. Take ρ = 0 in KR .
2. Find axial forces with linear analysis (ie ignore KR .) Evaluate φ ‘s.
3. Test determinant of tangent stiffness matrix.
1.571 Structural Analysis and Control
Prof Connor
Section 5
Page 17 of 17