instant Structural Analysis (iSA) Workshop 2015 Structural Analysis (iSA... · Getting started: Define grid size •Model components, such as element and support, must be added AFTER

INSTANT STRUCTURAL

ANALYSIS (ISA)

WORKSHOP 2015

Paul Lam

Department of Architecture and Civil Engineering

City University of Hong Kong

7 Dec 2015

Workshop outline

• Introduction

• Getting started:

• Beam examples

• A portal frame example

• A truss example

• A building frame example

• Semi-rigid support condition

• Various types of hinges

• Dynamic modal analysis

• Second order analysis

• Moving load analysis

• Concluding remarks

iSA Workshop 2015, University of Queensland, Brisbane, Australia 2

Introduction: What is iSA?

• iSA (instant Structural Analysis) is a user-friendly 2-

dimensional structural analysis program.

• It can be used to do:

• First and second order static analysis;

• Dynamic modal analysis;

• Stability analysis;

• Moving load analysis;

• Lack-of-fit; Support settlement; Temperature loading; etc.

• It can be used for the analysis of:

• Truss

• Continuous beam

• Frame (with different types of pinned joints)


Introduction: Assumptions

• Member axis is straight.

• Material is homogeneous, isotropic and linear elastic.

• Length of a member is large when compared to the

section dimensions, and shear deformation is neglected.

• Plane sections remain plane after deformation.

• Hook’s law is valid without any restrictions.

• All loads are assumed to act statically and do not change

the direction due to deformation.

• * Deformations are small so that equilibrium on the

undeformed structure can be assumed.


Introduction: Element stiffness matrix

41

2

3

5

6

' ',ix ixq d

' ',iy iyq d

' ',iz izq d

' ',jx jxq d

' ',jz jzq d

' ',jy jyq d

3 2 3 2' '

'

2 2'

'

'

'3 2 3 2

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

ix ix

iy

iz

jx

jy

jz

EA EA

L L

EI EI EI EIq d

L L L Lq dEI EI EI EIq L LL L

q EA EA

L Lq

EI EI EI EIq

L L L L

EI EI EI EI

L LL L

'

'

'

'

'

iy

iz

jx

jy

jz

d

d

d

d


Introduction: System stiffness matrix


m = 0

m ≤ number of

members?

m ≤ number of

members?

Calculate the local element

stiffness matrix of the mth

member [km]l

Transform the [km]l to the

global element stiffness

matrix [km]g

Assemble the system stiffness

matrix [K]

m = m + 1

Obtain [K]YES NO

Introduction: Solution method

• After obtaining K and Q, iSA will calculate the global

displacement vector and support reactions.

• Then obtain the local displacement vector, and calculate

the member end forces.

• Finally, calculate the axial, shear force, and bending

moment diagrams. *** All within 0.1 sec!!


s

f

s

f

D

D

KK

KK

Q

Q

2221

1211

sfs DKDKQ 2221

sff DKQKD 12

1

11

Introduction: iSA graphical user interface


Menu bar

Tool bar

Load tree

Drawing area

Status bar

Coordinate system

x

z

Introduction: Toolbars


Results: axial force, shear force, and bending moment diagrams, etc.

Load: point, distributed and temperature loads, lack of fit etc.

Structure information: node and element numbers

Zoom

System (Model components): node, support, member, and hinge

General

Getting started: A beam example

• Calculate the support reactions, the shear force and

bending moment diagrams of the cantilever beam.


4 m4 m

10 kN/m

A

B Cx

y

Getting started: Define nodes


Getting started: Define elements


Message to the users!!

Getting started: Define support


Warning message to the users!!

Default: pin support! The fail mechanism!

Getting started: Define support condition


Message to the users!!

Right mouse click at the

support to open the Support

properties window.

Getting started: Apply loading


The analysis is finished

instantly!!!

Defection shape!

Default: 10 kN/m

It is automatically

assigned to the

‘imposed loads’

case.

Getting started: Result presentation


Support reactions

Getting started: Detail member report


The detail report of the

selected member.

Getting started: Another beam example

• Calculate the support reactions, shear force and

bending moment diagrams of the beam. Determine also

the deflection shape.

• Member 1: E = 200 GPa, A = 0.0150 m2, I = 2.8 104 m4

• Member 2: E = 200 GPa, A = 0.0080 m2, I = 2.2 104 m4


4 m4 m

10 kN/m

A B Cx

y

1 2

Getting started: Additional supports


Release the horizontal

displacement to model

a roller support.

The analysis is finished instantly!!

But, this is not the final answer.

Getting started: Modify member properties


Modify the properties of

both members!

Getting started: Member properties


Material

Database

Cross-section

Database

Material

Properties

Cross-sectional

Properties

Getting started: Distributed loads

• Three type of distributed loads are available in iSA. They

are summarized in the table below.


Getting started: A portal frame example


4 m

1 m

7 m 7 m

Nodes Supports Elements

Start a new project!!

Getting started: Loading


Different types of loads

The analysis will be completed right after

defining the load!

The green line shows the deflection shape.

Getting started: Results presentation


Getting started: Member report

• For each

member, a detail

report on the

internal member

forces and

displacements at

different locations

on the member is

available.

• Users can

change the

subdivision

points.


Getting started: Element & system K


Contribution of the each

element in the system

stiffness matrix

Local and global element

stiffness matrices

Getting started: A truss example

• Calculate the support reactions and axial force diagram

of the truss. Determine also the deflection shape.

• All member: E = 210 GPa, A = 0.0150 m2


10 kN 10 kN10 kN

3 m 3 m 3 m 3 m

3 m

2 m

Getting started: Structure generator


Getting started: Truss shape/configuration


Getting started: Define truss location


Modelling completed!!

Getting started: Loading


Getting started: Modify member properties


Right mouse click at one of

the selected members!!

Getting started: Axial force diagram


Getting started: A building frame example

• Construct the axial force, shear force, and bending moment diagrams of the building frame.

• Determine also the top draft.

• Given: • All columns are:

UC 356 368 202

• All beams are:

UB 406 178 74


6 s

tori

es @

3.5

m4

m

6 m 6 m

50 kN

15 kN/m

30 kN

30 kN

50 kN

30 kN

30 kN

30 kN

15 kN/m

15 kN/m

15 kN/m

15 kN/m

15 kN/m

15 kN/m

Getting started: Structure generator


Getting started: Delete members


Select the unwanted

members

Delete the selected

members

Getting started: Change first story height


Select all nodes at

the ground floor.

Right mouse click

to open the ‘Move

selected nodes’

window.

Getting started: Apply vertical load


Select all beams and

apply the load on one

of them.

Change the default

load magnitude from

10 to 15 kN/m.

Getting started: Apply horizontal load


Change the default load

from 10 kN vertical to the

desired magnitudes and

directions.

Getting started: Result presentation


Getting started: Define grid size

• Model components, such as element and support, must

be added AFTER the definition of nodes.

• In general, Nodes can ONLY be added to the grid points

in the drawing area.

• In order to put nodes at the desired positions, one may

need to re-define the grid spacing in the drawing area.


Getting started: Load on node or member


Getting started: Display information

• One can switch on/off the display of:

• Loads

• Node number

• Element number

• Material properties

• Geometrical properties (e.g., length of member)


Semi-rigid connection

• A semi-rigid connection can be modeled by a rotational

(torsional) spring with stiffness equal to that of the actual

connection.


m

x

y

bb ee

11,dq22 ,dq

33,dq

44 ,dq

L

Infinitesimal

Rotational springs

mbb ee

1q2q

3q

4q1q

2q

1q2q

3q

4q3q

4q

qq

The equilibrium

of the springs:

Semi-rigid connection


bk ek

2d

2d

11 dd

33 dd

4d

4dDisplaced position

Initial position

mbb ee

L

Infinitesimal

11 dd

33 dd

222 ddkq b

444 ddkq e

bk

qdd 2

22

ek

qdd 4

44

Support conditions


Allow users to model supports

as different types of springs.

Allow users to rotate

the support.

Hands-on: Semi-rigid support

• Model a 4m by 6 m steel frame (using default section) and

apply a horizontal load of 10kN at the top left connection

of the frame. Show the bending moment diagram.


• Change one of the

supports to semi-rigid by

inserting a rotational

spring of 100000kNm/rad,

and observe the changes.

• Change the rotational

stiffness to 1000kNm/rad,

and observe the changes.

Various types of hinges

• There are three types of hinges in iSA. They are:

• Full hinge: all members are connected to this joint as a

pin connection.

• Half hinge: this allows users to define pin joints at the

end of individual members.

• Pinjointed hinge: this allows users to model the

situation when two members are hinged at the middle.


Select

Node

Support

MemberMember polygon

Full hinge

Pinjointed hinge

Half hinge

Split member

Hands-on: Hinges in iSA

• Consider a 4 m by 4 m steel frame under a horizontal load

at the top left connection.

• Record the horizontal displacement.


Hands-on: Half hinge

• Install two brace members to the steel frame (using

default member).

• Record the horizontal displacement.


Hands-on: Pinjointed hinge

• Install a pinjointed hinge at the middle of the two braces to

further reduce the horizontal displacement of the system.


Hands-on: Continuous beam analysis

• Use iSA to model a continuous beam with four spans (4m,

6m, 4m, 6m).

• Apply various design actions onto the beam to consider

the most unfavorable conditions.

• Calculate the design moment envelop.


Hands-on: Truss analysis

• Calculate the support reactions and internal member

forces of the truss.

• Given: E = 200 GPa and A = 0.0077 m2


6 m

5 m5 m 5 m 5 m

15kN15kN

15kN

Hands-on: Rigid frame analysis

• Given:

• E = 210 GPa for all members.

• Columns: UC 305 x 305 x 283

• Beams: UB 686 x 254 170

• Plot the axis force, shear force,

bending moment diagrams

under:

• Vertical load ONLY

• Horizontal load ONLY

• Combine vertical and horizontal load

• [Demonstrate temperature load,

support settlement and lack of fit]


3 m

4 m

6 m

20 kN

10 kN

3 kN/m

3 kN/m

Hands-on: Temperature load

• Two types of

temperature loads are

available in iSA.

• The constant

temperature across the

depth of a section, and

• Linear varying

temperature along the

depth of a section.


Hands-on: Lack of fit & support movement


Dynamic modal analysis

• Consider the undamped free vibration of an N-DOF

system, the equation of motion:

• where M and K are the system mass and stiffness

matrices of the structure. Assuming a solution form of:

• The natural frequencies and mode shapes of the

structural system can be calculated by solving the

eigenvalue problem.


0KXXM

tsinφX

0φMK 2

Dynamic modal analysis


Second order analysis

• is the nodal displacement due to sway induced by loads

in the frame and imperfection.


P

F

H

P

F

1

P

F

1 2

FH FH+P1 FH+P1+P2

(a) (b) (c)

…

(a) (b) (c)

P

F

H

P

F

1

P

F

1 2

FH FH+P1 FH+P1+P2

(a) (b) (c)

…

P

F

H

P

F

1

P

F

1 2

FH FH+P1 FH+P1+P2

(a) (b) (c)

…

Second order analysis


ww

P

8

2wL

ww

P

11

2

8P

wL

ww

P

221

2

8 PP

wL

(a)

(b)

(c)

Second order analysis: Formulation


x

x

w

yy + y

M

M + M

V + V

V

P

P

A



0 yF 0 xwVVV wx

V

0x wdx

dVWhen .......................(1)

0 AM

02

2

yPx

wxVVMMM

0 yPxVMNeglecting higher order terms

DM

Dx-P

Dy

Dx=V

When dM

dx-Pdy

dx=V ..................(2) 0x



EI

M

dx

yd

2

2

2

2

dx

ydEIM

Vdx

dyP

dx

ydEI

3

3

wdx

ydP

dx

ydEI

2

2

4

4

EI

w

dx

yd

EI

P

dx

yd

2

2

4

4

…………………..(3)

…………………..(3a) With (2)

With (1)

Second order analysis: iSA solution


Calculate the first order

displacement vector {u}I

Determine the axial forces

N(n) using {u}I

Assemble the second order

stiffness matrix [K]II

Calculate the second order

displacement vector {u}II

Determine the axial forces

N(n) using {u}II

n = 1

n = n + 1

nN

nNnN 1

nN

nNnN 1YES

NO

Calculate member forces

Hands-on: Second order analysis

• Given: • All columns are UC 15215230

• All beams are UB 17810219

• Use iSA to model the frame together with the two vertical loads.

• Apply a horizontal load of 1 kN at the top left corner of the frame and carry out second order analysis. What happened?

• Increase the vertical loads to 550kN. What happened?


4 m

3 m

5 m

400 kN400 kN

1 kN

Moving load analysis


Hands-on: Vehicle load on bridge

• Given:

• The I & A of the bridge desk are 2.13 107 cm4 and 4.00 104 cm2.

• The I & A of the columns are 1.80 107 cm4 and 2.25 104 cm2.

• Plot the bending moment and displacement envelopes for

a point load of 10 kN moving on the bridge deck.


16 m 22 m 22 m 16 m

10 m 10 m12 m

A B C D E

FG

H

Concluding remarks

• Various features of iSA in 2D truss, beam and

frame analysis were demonstrated.

• Through a series of hands-on exercises,

participants should be able to use iSA in:

• Structural analysis of 2D structures

• Modeling semi-rigid supports

• Dynamic modal analysis

• Second order analysis

• Moving load analysis



Documents

instant Structural Analysis (iSA) Workshop 2015 Structural Analysis (iSA... · Getting started: Define grid size •Model components, such as element and support, must be added AFTER