Structural Design Lab.(Prof. Ho-Kyung Kim) Dept. of Civil & Environmental Eng.
Seoul National University
457.649 Advanced Structural Analysis
Part II:
Displacement method
§ Reverse of force method
§ Also called “stiffness method” or “equilibrium method”
▶ See “Step in the displacement method”
These are “equilibrium equations”. Solve for {r}.
If linear, can use superposition. → greatly simplifies the calculation
Displacement Method
Nodes Elements
{R} {S} Forces
{r} Geometry {v} Displacements
(outside-in) Equilibrium (inside-out) known loads
(function of {r}) Action-Deformation (stiffness form)
(functions of {r}) (functions of {r})
▶ Degree of Freedom
§ Primary unknown in the displacement method
§ Kinematic constraints(support conditions, axially rigid members, etc.) could reduce the number of DOF
▶ Action-Deformation Relations
𝑆 = 𝑘 {𝑣} 𝑣 = 𝑓 {𝑆}
𝑘 = !"# 𝑓 = !"#
𝑘 = $!%# 2 1
1 2 𝑓 = &!%# 2 −1
−1 2 𝑘 = '!%# 𝑓 = '!%#
{r}, and {S}=[k]{v}
Truss Frame 2D(X-Y plane) Δx, Δy Δx, Δy, θz
3D Δx, Δy, Δz Δx, Δy, Δz, θx, θy, θz
▶ Geometry of Deformation
{v}=[a]{r}
▶ Modeling options
can model this several ways.
(1) Assume EA= finite (i.e. consider axial deformations)
Essentially, if we allow axial deformations, each node has →, ↑and dofs (unless restrained by displacement B.C.)
{R}=[K]{r}
10
(2) Assume EA = finite for columns, infinite for girders.
(a common assumption for tall building frames)
(3) Assume EA = ∞ for both girders and columns.
Reduces number of unknown displacements.
{R}=[K]{r}
(4) Assume girder is also rigid in bending (and EI = ∞ throughout).
All rotations are zero (“shear building”).
▶ Compare
{R}=[K]{r}
▶ Stiffness Coefficients
▶ Flexibility Coefficients
Definition of Stiffness & Flexibility Coefficients
𝑘(( = 4𝐸𝐼
𝐿 + 4𝐸𝐼
𝐿 = 8𝐸𝐼 𝐿 𝑘$( = 2𝐸𝐼
𝐿 𝑘'( = −6𝐸𝐼
𝐿$
Determination of Stiffness Coefficients by Equilibrium
▶ Work equivalence for calculating {R} acting on non-DOFs
§ But what is 𝑅 = ?
?
§ For cases as this, we can not get {R} simply by inspection, but must use a work approach.
{R} acting on Non-DOFs
§ Procedure: Impose r1 with all other r=0.
{R} acting on Non-DOFs
▶ E.g. 2-DOF frames with a hinge support
§ Case 1) r1=1, r2=0
{R} acting on Non-DOFs
a
b c
r1 r2
applied
M
§ Case 2) r1=0, r2=1
1 -1 2
R2
R1 R2
R1
( )
equiv 1
actual applied applied
1
1 2 W R
W M
R M
= ´
= ´ -
\ =-
equiv 2
actual applied 2
1 0 0
W R
W M
R
= ´
= ´
\ =
1
actual applied equiv
W M c
W
q
=
=
▶ E.g. Distributed load for a 3-DOF frame
{R} for Distributed Loads
• Fixed-End-Forces
• Equivalent concentrated loads on nodes
( ) ( )( )2
2 1
2.5 12 96
1 96 kip-in
20 20
R = wL = =
( ) ( )( )2
2 2
2.5 12 96
1 64 kip-in
30 30
R wL -
=- = = -
r1 r3
r2
▶ Notations
§ {r} = Unknown node displacements
§ {R} = Known forces corresponding to {r}
§ {S} = Element actions
§ {v} = Element deformations
§ [a] = Displacement transformation matrix
§ [k] = Element stiffness matrix
§ [K] = Structure stiffness matrix
▶ Geometry
▶ Action-deformation
Matrix Formulation of the Displacement Method
▶ Equilibrium using the Principle of Virtual Displacements
Matrix Formulation of the Displacement Method
▶ Steps in the displacement method
1. Define the joints, members and boundary conditions of the structure.
2. Identify the displacement degrees of freedom (ri, i=1,…,N) of the structure (the unknowns of the problem). Take account of all kinematic constraints (support conditions, axially rigid
members, etc.).
3. Define the measures of member deformation which are to be used.
4. Consider N separate displaced states (cases). For “Case i”, impose ri=1 with all other r=0. For each Case, determine the member deformations. This is a geometry problem.
5. For each Case multiply the member deformations by the member stiffnesses to get the member actions.
6. For each Case calculate the joint loads (Ri, i=1,…,N, corresponding to ri) required to hold the structure in equilibrium. This is an “inside-out” static problem. It will usually be convenient to use the Virtual Displacement Principle. For Case j, let the load corresponding to
displacement rj be kij. The kij values are stiffness coefficients for the structure.
7. The actual value, Ri, of the load corresponding to ri will always be known. Hence, by superposition .
Ri=∑j kijrj These are the equilibrium equations.
8. Solve for the r values.
Matrix Formulation of the Displacement Method
▶ Example: Truss by displacement method
Matrix Formulation of the Displacement Method
▶ Action-deformation:
▶ Statics: Virtual displacements procedure
▶ Equilibrium equations:
Matrix Formulation of the Displacement Method
To get Real act/force Imag displ./def WE WI
k11 1 1 k11 Σ!v1EAv1/L
k21 1 2 k21 Σ!v2EAv1/L
k12 2 1 k12 Σ!v1EAv2/L
k22 2 2 k22 Σ!v2EAv2/L
Member EA/L (·10-4) V1 V2 V1·EA/L·V1 V2·EA/L·V2 V1·EA/L·V2
AC .2 .8 0 .128 0 0
BC .143 1.0 0 .143 0 0
BD .2 0 .6 0 .072 0
CD .333 -.6 .8 .120 .213 -.160
.391·104 .285·104 -.160·104
k11 k22 k12or k21
Stiffness Coefficients by Virtual Displacement Procedure
R 1 1
T
E ij
W = r = ´R = ´k
( )
4 2 2 4
T T
I
i i j j
i i j j i j
W v S v kv
P M M
EA EI EI EI EI
L L L L L
q q
q q q q q q
= å =å
= å D + +
é æ ö æ öù
= åêëD × D+ çè + ÷ø+ çè + ÷øúû
▶ Example: Frame by displacement method
Matrix Formulation of the Displacement Method
▶ Action-deformation:
▶ Statics: Given member moments in each case, can find k11, k21, etc. by equilibrium equations. Alternatively, can use virtual displacements. For example:
▶ Equilibrium equations:
Matrix Formulation of the Displacement Method
To get Real act/force Imag displ./def WE WI
k11 1 1 k11 Σ !v1 T k v1
k13 3 1 k13 Σ !v1 T k v3