CE 321 - Lecture 1 - Introduction

I. Introduce Self, Review Class Schedule, Comment on Textbook - Ask for questions

II. Chapter 1 Types of Structures and Loads

Structural Engineers Responsibilities:

Planning > Analysis > Design > Construction

Use connections member size Insure design is followed

Structural Form member forces

(Arch, Truss ..)

Loadings

materials

CE 321 focuses on the Analysis

Physical Structure>Idealized Structure>Mathematical model>Solution of equations>Interpit Results

Structural Members

Tie Rods -- Slender rods that can only resist tension -- Show Fig 1-1 Cross Sections

Beams -- Horizontal members that carry vertical loads

Supports-- Show Fig 1-2 Supports for beams

Beam Types-- Figs 1-3, 1-4, 1-5, 1-6, 1-7

Columns - Vertical members that resist compression Columns -- Fig 1-8

Members are assembled into Structures

Types of Structures Show Composite Figure 1-9 - 1-12

Have Class try to fill in matrix below

Type Uses Advantages Disadvantages

Trusses Bridges Light Weight Many connections

Roof support efficient Deep

towers

Cable Long Spans Light Weight Flexible (aero loads)

No Buckling

Arch Bridges Heavy Difficult to construct

Dome roofs Compression only

Openings

Frames Buildings Use standard sections Many connections

Surface Roof Structures lightweight Difficult to Design

Shells Strong Difficult to construct

LOADS

To be determined by engineer -- often one of the most important jobs !!!

Consider if you underdesign or overdesign

Requires experience

General Building Codes -- Specify requirements of governmental bodies for minimum design loads and minimum construction standards

Show Table 1-1

Dead Loads -- Things permanently a part of the structure

columns, beams, girders, floor slab, roofing, walls, etc.

See Table 1-2 for typical values

Live Loads -- Loading from things that the structure is designed for, but may change in magnitude and location

Items placed on structure, moving vehicles, people, etc.

Building loads See Table 1-3

Can use a reduction factor

L = Lo ( 0.25 + 15/ AI1/2)

Restricted to not less than 50% for members supporting one floor and

not less than 40% for members supporting more than one floor

NO REDUCTION FOR PUBLIC ASSEMBLY, GARAGES, ROOFS

Bridge Loads -- see AASHTO

Wind Loads -- q = .00256 v2

Plus various shape factors (See Text

Snow Loads

Earthquake Loads -- Beyond the scope of this class

Soil Pressure

Work Example Problem 1-8

40 in

Given:

This beam supports the floor of a light storage warehouse, beam is concrete with density of 150 lb/cu. ft.,

Wanted: Dead and live load per foot length of beam

Solution:

Dead Load = weight of beam

= area x density

= (10x18+8x40) in2 1 ft2 (150 lb/ ft3 )

144 in2

= 521 lb/ ft

Live Load = 125 lb/ ft2 x 40 in

12 in/ft

= 417 lb/ft

Problem 1-10 12 ft

Given:

7 ft

Wind velocity of 75 mph impacts sign, Shape factor = 0.8

Wanted: Resultant force of the wind

Solution: q = 0.00256 (75) 2 = 14.4 lb/ft2

p = 0.8 (14.4) = 11.52 lb/ft2

F = 11.52 x 12 x 7 = 968 lb

Acts at the centroid of the rectangle

Lecture 2 - Review Of Statics

1. Idealized Structure -- Show Overhead Fig 2-1,2-2 Point out Authors Sketch of Pinn and Fixed Support

Show Overhead Fig 2-4 A,B Acutal "physical" Beam

Idealized beam

Mathematical Model

Solutions to Equations - Shear/ Bending Diag,

Interpert Results - Stress, Displacement Criteria

2. Talk about Boundary Condition Idealization - Overhead, Models

3. Idealized Loadings -- Show Fig 2-11 One Way System

Show Fig 2-13 Two Way System

Criteria for one way, two way :See Fig 2-12

if ( L2/ L1 ) > 2 one way slab

4. Principle of Superposition:

1. Structural Material is Linear Elastic (What does this mean?)

2. Small Displacement (What does this mean?)

Do you remember Why? C.E. 203

5. Equations of Equilibrium

Sum forces x,y,z = 0 (three equations for 3D, 2 equations for 2D)

Sum of Moments x,y,z, = 0 (three equations for 3D, 1 equation for 2D)

New Stuff -- Determinacy and Stability!!!

Statically Determinant - Definition ?

Number of unknown forces = number of equations of Equilibrium

Statically Indeterminate - Definition?

Number of unknown forces > number of equations of equilibrium

Missing Condition - What if Number of unknown forces < number of equations of equilibrium?

UNSTABLE ( - consider roller-skate ) !!

Hibblers Formula - good n = "parts" of a structure

r = force and moment reaction components

Equation for determinacy:

r = 3n Statically determinate

r > 3n Satirically indeterminate

What is a "part" -- see example 2-1 - rigid members between pins

overheads for Example 2-1, 2-2, 2-3

Equation for Stability

r < 3n unstable

r > 3n unstable if member reactions are:

1. concurrent or

2. parallel or

3. some of the components form a collapsible mechanism

overheads for Example 2-4

Example Problem - 2-36

4 kip 3 kip

4 ft 6 ft 4 ft

Given: Wanted : Horizontal , vertical forces at joints

12 ft (400 lb/ft)

All Joints pinned

LECTURE 3 - Review of TRUSSES!!!

T.A's Introduction --Clark Seamans, Audrey Newell

Grading all papers

Available for Help during hours listed (email contact encouraged)

TYPES of Trusses -- See Textbook for several names

Classification of Coplanar -- Simple, Compound (combination of two or more simple, Complex)

Determinacy and Stability -- Important !!!

b = number of members (1 unknown per member)

r = number of reactions

j = number of joints

d = 2 for 2d, 3 for 3d

b + r < d * j Unstable

b + r = d * j Statically determinate, check stability

unstable if

1) truss support reactions are concurrent or parallel

2) some of the components of the truss form a collapsible mechanism

b + r > d * j Statically indeterminate, check stability

unstable if

External stability - requires

1. that the lines of action of the reactions do not intersect a common axis

( Reactions are concurrent)

2. that the lines of action of the reactions all be parallel to one another

(Reactions are parallel)

Internal stability - careful inspection of member arrangement - insure that each joint is held fixed by its supports or connecting members so that it cannot move with respect to the other joints -- i.e. Rigid Body Motion !!

Cramers rule - if determinate of defining equations is zero.

Show with Visual Analysis - Truss Unstable

Put member in to make it stable.

Show 0 force members

Present example problems

LECTURE 4 - Review of BEAMS and Introduction to FRAMES!!!

1. Establish The accepted positive sign conventions:

Internal Forces at a Point - From Equations of Equilibrium

THE BIG PICTURE - The Forest

Shear and Moment Functions - From Equations of Equilibrium written in General

From these Functions - Equations for Displacement are determined

From these equations distorted plots can be generated

Shear and Moment Diagrams - Plots of Shear and Moment Functions

From Diagrams we can find points of Maximum Internal Load.

From these points we can get stress state (C.E. 203) s = My/I

t = VQ/Ib

From stress state we can determine if material will fail (C.E. 205) s < sf

Let's do some examples of this !! - BEAMS / Frames

Moment Diagram for a Frame --

Draw Moment on the Compression or tension side of Beam ?

Tension side common for concrete

Compression side - common convention used in Hibbler

This means

Compression Compression

CompressionLECTURE 5 - Review of CABLES

CABLES

Assumptions:

1. Perfectly flexible -- i.e. offers no resistance to shear, bending, or tension

2. Inextensible -- cable does not change length upon loading

When are these valid assumptions? When are they not ?

1. Concentrated Loads --

Unknowns - tensile force in each segment, 2 reaction components at each end

Equations - Sum of forces in x, Sum of forces in y at each "node"

+ Length of cable (Inextensible assumption helps here)

1. Uniform distributed Load

Difficult equations - the keys are that

A. The cable deforms in the shape of a parabola

y = (h/L2) x2

B. horizontal component of force is constant along the cable

FH = (wo L2)/2h

where h and L are shown on the figure

C. The maximum force in the cable is

Tmax = wo L ( 1 + (L/2h)2)1/2

(we will use this to solve the homework problems!!)

LECTURE 6 -- ARCHES -- Focus on Determinate Three Hinged

Analyze like a frame -- Extra moment = 0 equation

Lecture 7 -- Introduction to Influence lines

NEW SUBJECT!!

Bending Moment or Shear Diagram - The Function (moment or shear) at the location (x), due to the given static loading condition - this gives the effect of fixed loads at all points on the beam

Influence Diagram -- The Function (moment, shear, or reaction) at the given location, due to a unit load at the location (x). --This gives the effect of a moving load at a specified point on a beam

F(x)

PROCEDURE

1. Tabulate Values - Just locate the unit load at various positions (x) on the structure - build a table of the value of the function vs. x

x F(x)

Connect points

2. Influence Line Equations

Equations for the Reactions are usually obtained from a freebody of the total structure

Draw Free Bodies with the member "cut" at the location of interest

Place a unit load on each side of the "cut" member and determine the equation for the function to be determined using the equations of equilibrium.

i.e.

* C

Vc Mc

Mc Vc

MORE EXAMPLES Lecture 8 - Some fun stuff - Muller-Breslau Principle

The influence line for a function is (to the same scale) as the deflected shape of the beam when the beam is acted upon by the function

To do this

1. locate the position for the influence line

2. Apply the function

for a reaction apply a positive force - modify the beam at the load to not resist the applied force

for a shear - apply a positive shear - modify the beam so that it cannot resist shear at this point

For a moment - apply a positive moment - modify the beam so that it cannot resist moment at this point (a pin)

SHOW OVERHEAD OF CUTS Fig 6-12 through 6-14

3. The shape the rigid beam assumes is to scale as the influence function

Note - we do not have the magnitudes of the influence line - only its shapes

Why is this important ? -- You can now determine where to place loads for maximum influence

Can you determine these magnitudes of the peaks on the influence line developed using the Muller-Bresslau principle --? Yes - see Example 6-3 - place a unit load at the "peak" location and compute the function. You can usually use similar triangles to determine the other peaks.

Work Example 6-12, have class determine h. - see your solutionsLecture 9 - Influence lines for Floor Girders and Trusses

Show Slab, floor beam, girder column figure - girder takes concentrated loads

Point out what a panel

Show Truss Figure -- Explain

Show examples of influence lines - have students explain what is shown

Do Example 6-36

Break class into groups to determine a particular point in the table

let them work together

Do Example 6-53

break class into groups to determine a particular point in the table

let them work together

Lecture 10 - Placement of Live Loads for Maximum Internal Forces

Answer last period's question - Can you use Muller-Bresslau for the floor systems?

NO - Because (see development of Muller-Bresslau in Hibbler, the principle is based on one load - with floor systems we have the one load distributed to the girder to (at least) two loads

Live loads on Bridges

Show Loading diagrams for Trucks, Trains

To determine Maximum Function for "traveling loads" using Influence lines

1. Step by Step approach

Consider the loading shown in Fig 6-29

Place each of the three loads at "max position" and compute function

This is a straight forward approach -- no trouble

Show Figure

2. Slope change approach

Note that for loads moving along a sloping line DV = Ps(x2 - x1 )

For loads passing through jump DV = P(y2 - y1 )

Show Figure

For Moments - (no jumps so) DM = Ps(x2 - x1 )

Show Figure with Calculations

(It may be easier to use procedure 1)

Absolute Maximum Shear and Moment - this is neat stuff

We will only consider 2 beam configurations at this point

Cantilever Beam

Max Shear

Max Moment

Simply Supported Beam

Max shear

The maximum moment will be under a load, but not necessarily at the center of the beam, it depends upon the location of the total set of loads!!!

See pg. 283 of Hibbler

The absolute maximum moment in a simply supported beam occurs under one of the concentrated forces, such that this force is positioned on the beam so that it and the resultant force of the system are equidistant from the beams center line

Procedure:

1. Determine the magnitude of the resultant of the concentrated forces

S F = R

2. Determine the position, x' of the resultant

S M = > x'R = S x1F1 + x2F2 + x3F3 + x4F4 + ...

3. Show the Resultant in it's correct position

4. Position the loads such that the 1st load and the resultant are equidistant from the center - compute moment under the 1st load

Position the loads such that the 2nd load and the resultant are equidistant from the center - compute moment under the 2nd load

Choose position that give the Maximum Moment

Show Example

Lecture 11 - Review - (Note all notes are available on h:/pcapps/dept/ce/ce321 lectures.doc)

OLD STUFF ..

DETERMINATE TRUSS ANALYSIS : -- Method of joints, Method of Sections

DETERMINATE BEAM ANALYSIS: Development of Shear and Bending moment diagrams

CABLES: Assumptions: Perfectly flexible and Inextensible --

NEW STUFF ... a problem(s) likely from the (*) sections

LOADS: Dead Loads -- Things permanently a part of the structure

Live Loads -- Loading from things that the structure is designed for, but may change in magnitude and location. Items placed on structure, moving vehicles, people, etc. Load reduction factors, Idealized Loading (One Way, Two Way Systems)

(1) DETERMINACY, STABILITY:

Beams and Frames Equation for determinacy:

r = 3n satirically determinate, r > 3n satirically indeterminate r = force or moment reaction components, n = parts a "part" is a rigid members between pins

Beams and Frames Equation for Stability

r < 3n unstable, or if r > 3n unstable if member reactions are: 1. concurrent or 2. parallel or 3. some of the components form a collapsible mechanism

Trusses Determinacy and Stability -- trusses

b = number of members (1 unknown per member), r = number of reactions

j = number of joints, d = 2 for 2d, 3 for 3d

b + r < d * j Unstable

b + r = d * j Statically determinate, check stability unstable if

1) truss support reactions are concurrent or parallel

2) some of the components of the truss form a collapsible mechanism

b + r > d * j Statically indeterminate, check stability

External stability - requires 1. that the lines of action of the reactions do not intersect a common axis or 2. that the lines of action of the reactions all be parallel to one another

Cramers rule - if determinate of defining equations is zero.

MOMENT DIAGRAM FOR A FRAME :

Draw Moment on the Compression side of Beam - convention in Hibbler and for this class

(2) THREE HINGED ARCH : Analyze like a frame -- Extra moment = 0 equation at the pin

(3) INFLUENCE DIAGRAM: The Function (moment, shear, or reaction) at the given location, due to a unit load at the location (x). --This gives the effect of a moving load at a specified point.

Determined by: 1. Tabulated Values

2. Developed Equations

3. Muller:Bresslau Principle

Influence lines for Floor Girders and Trusses

To determine Maximum Function for "traveling loads" using Influence lines

1. Step by Step approach

Consider the loading shown in Fig 6-29

Place each of the three loads at "max position" and compute function

This is a straight forward approach -- no trouble

2. Slope change approach

Note that for loads moving along a sloping line DV = Ps(x2 - x1 )

For loads passing through jump DV = P(y2 - y1 )

For Moments - (no jumps so) DM = Ps(x2 - x1 )

(4) THE ABSOLUTE MAXIMUM MOMENT IN A SIMPLY SUPPORTED BEAM : This value occurs under one of the concentrated forces, such that this force is positioned on the beam so that it and the resultant force of the system are equidistant from the beams center line

Procedure:

1. Determine the magnitude of the resultant of the concentrated forces

S F = R

2. Determine the position, x' of the resultant

S M = > x'R = S x1F1 + x2F2 + x3F3 + x4F4 + ...

3. Show the Resultant in it's correct position

4. Position the loads such that the 1st load and the resultant are equidistant from the center - compute moment under the 1st load

5. Position the loads such that the 2nd load and the resultant are equidistant from the center - compute moment under the 2nd load

n. Position the loads such that the nth load and the resultant are

equidistant from the center - compute moment under the nth load

n+1. Choose position that give the Maximum Moment

Lecture 12- Deflections of Beams and Frames by INTEGRATION

The "Elastic Curve"

Error in text (safe) - unsafe ppg 341

Calculation of Displacements - Calculus -- Differential Equations

It is a straight forward process for beams!!

Show Boundary Conditions overhead

Show Deformed member overheads

Intuitive feel for deformed shape

Help can come from the moment diagram

Moments tell us about "curvature"

Show moment - deflected shape figure

Elastic Beam Theory

Assumptions- Impt !!

1. Beam is initially straight

2. Loads applied perpendicular to Beams axis

3. plane cross sections of the beam remain plane before and after bending !!

Bernoulli-Euler Beam - a very good approximation (Stocky Structure stuff )

4. Deflections and rotations are small

5. Material remains elastic

v" = M/EI

Go Through example

Lecture 13- Deflections of Beams and Frames by Moment Area

Review the Moment area theorems Show Diagrams

Important 1st theorem - change in rotation

2nd theorem - care in what it says!!!

Helps when doing Moment area:

1. Build your moment diagram by parts

2. Find a location on the beam where you know the displacement

or rotation --- like at a fixed support d = 0, q = 0

or at a pin d = 0.

Review the Example Problems Show Diagrams

Work Example problems in groupsLecture 14- Deflections of Beams and Frames Conjugate Beam

Review the Conjugate Beam theorems Show Diagrams

Important 1st theorem - rotation

2nd theorem - displacement

Easier concept that Moment Area I Believe

The Key is modifying the boundary conditions - Look at Table 8-2

Show the examples Fig 8-20

Procedure:

1. Develop the moment diagram for the real beam

2. Modify the real beam's boundary conditions to a "conjugate" beam

3. Place the moment diagram from the real beam on the conjugate beam for the loads

4. compute the "reactions" for the conjugate beam

4. Use Theorems 1 and 2 to compute rotations and displacements of the real beam

Do examples

Lecture 15 - Review Exam

Lecture 16 Work Energy Methods

External work and Strain Energy

Simple Concept but sometimes difficult to understand in its simplicity

Show realtionship between Load Displacement and Stress Strain

i.e. the difference between strain energy and strain energy density

D U = 1/2 PD P

Consider two situations -

1) loading slowing from an intial unstress state

2) loading already applied -- then another so that the first load rides through the displacement

produced by the second action

Case 1 Case 2

External Work - Force U = 1/2 PD U = PD

External Work - Moment U = 1/2 Mq U = Mq

Strain Energy -Axial Force U = P2L/(2AE) U = P2L/(AE)

Strain Energy - Bending U = I M2/2EI dx U = I M2/EI dx

Principle of Work and Energy

External work = internal strain energy (usually use Case 1)

(Show calcualtion for Fig 8-29)

Principle of Virual Work

Assumptions:

deformations do not have to be elastic ( New freedom)

Displacements must be compatible

Work of External loads = Work of internal Loads

Proceedure:

1. Place a unit load on a structure in the direction you desire to determine the displacement

This is an imaginary or "virtual" load. It doesn't exist in the real structure and is just used as a conveniece to determine the dispalcement at this point

2. This virtual external load induces virtual internal forces (u) and virtual dispalcements ( in the structure (u)

Lecture 17 Work Energy Methods - Trusses

For a truss member, the internal virtual work is

the virtual loads moving through the (real displacements associated with applied Real loads)

= n (NL/AE)

plus

the virtual loads moving through the real displacements associated with applied temperature changes

= n (a D T L)

plus

the virtual loads moving through the real displacements associated with Fabrication errors

= n ( D L)

Consider:

2P also let member BD be heated by DT, member CB Fab error of d

P A B

C D

Assume: "axial rigidity" = AE (New Word)

Compute the horizontal and vertical displacements of B.

Solution:

1. Determine the internal forces due to the real loads:

(P)

(-2P) (-1.41P) (P)

(0)

2. Determine the internal forces due to the virtual loads located where we want to find the displacements

a) Horizontal (0)

( 0) (1.41) (-1)

(0)

b) Vertical (0)

( 0) (0) (1)

(0)

a) Temperature (0)

( 0) (0) (a(DT) L')

d (0)

Now, we build a table with these values

Mem L N T Fab nx ny Nnx L NnyL Tnx Tny Fabnx Fabny

AB L' P 0 0 0 0 0 0 0 0 0 0

AC L' -2P 0 0 0 0 0 0 0 0 0 0

BD L' P a D T L' 0 -1 1 -PL' PL' -a D T L' a D T L' 0 0

CD L' 0 0 0 0 0 0 0 0 0 0 0

CB 1.414L' -1.414P 0 d 1.414 0 -2.828PL' 0 0 0 1.414d 0

-3.828PL' PL' -a D T L' a D T L' 1.414d 0

For loads dx = S nx (NL/AE) = -3.828PL'/AE

For loads dy = S ny (NL/AE) = PL'/AE

For Temp Change dx = S nx (a D T L) = -a D T L'

For Temp Change dy = S ny (a D T L) = a D T L'

For Fab error dx = S nx (D L) = 1.414d

For Fab error dy = S ny (D L) = 0

Total dx = -3.828PL'/AE - a D T L' + 1.414d

Total dy = PL'/AE + a D T L' + 0

Have class work Example Problem

Lecture 18 Work Energy Methods - Beams

1 D = m mM/EI dx -> to calculate displacement

1 q = m m q M/EI dx -> to calculate rotation

The keys here are

1. a unit load is place at the location, in the direction you desire to determine the

displacement, the equations for the moment produced in the beam are then developed

2. a unit moment is placed at the location you desire to determine the rotation, the equations for the moment produced in the beam are then developed

Show Figure with descriptions

3. You have some felxibility in determining your coordiante systems to use to do the integration. HOWEVER YOU MUST USE THE SAME SET FOR M and m !!!

Show Figure of Example 8-18: Note the choice of Coordinate Systems !!

There are some helps in doing the integrations:

Show Integration helps Figure from inside front cover (Explain

Work example 8-18 using the figures

Unit Loading

Moment diagram for unit loading

Real Loading

Moment diagram for real loading

Integral evaluation using tables

1 D = m mM/EI dx

=1/EI (integral for x3 + integral for x2 + integral for x1 )

= 1/EI[1/6 (-7.5(80+2x70)(10) + 1/6 (70)( 2x-7.5 -15.0)(10) + 1/2 (15)(0)(15)]

=1/(29(103)(800) [-2750 - 3500 + 0 ]

= - 0.466 in.

Show Example 8-20

This is probably easier with the tables, note the parabolic portion

It is probably easier because you don't have to keep track of the different segments - it is done visually

Virtual Strain Caused by Axial Load, Shear, Torsion

(Very small, usually neglected)

and Temperature (Could be important)

1 D = S nNL/AE -> to calculate displacement from axial loads

1 D = m KvV/GA dx -> to calculate displacement from shear

1 D = m tT/GJ dx -> to calculate displacement from torsion

1 D = m maDTm /c dx -> to calculate displacement from temperature change

Show Example 8-21 to show how little these impact

Lecture 19 Castigliano's Theorem - Trusses

For this class we will not spend time deriving the theorem, rather just in its application

Simply stated

D = dU/dP

or, the displacement D , under a load P, is equal to the partial derivative of the strain energy in the structure with respect to the applied load P.

Apply this to Trusses

U = 1/2 S N2 L / (AE)

Now, let's take the partial derivative with respect to P understanding that the only functions of P are the internal forces, N

dU/dP = 1/2 S 2N(dN/dP) L / (AE) = S N(dN/dP) L / (AE) = D

Note we must treat P as a variable and thus each truss member force must be an expresses as a function of P.

Procedure:

1. Place a force P in the direction where the displacement is to be determined, if there is already a force there, assume it is P.

2. Compute the force in each member as a function of P.

3. Prepare a table for the pertinent values (see example)

4. In a separate column of the table, list the internal forces with P having it's numerical value, (if no P load is at the node of interest, P=0)

5. Apply Castigliano's theorem to determine the displacement.

Show Example 8-23 (Example where P=0 at node of interest)

Show Example 8-25 (Example where P= 4 k at node of interest)

Have Class Work Problem 8-77 (Find vertical deflection at C) given the initial start

H G F

10'

A B C D

10' 10' 10' 10'

500 lb 500 lb 500 lb

Since the truss is symmetric, consider only 1/2 and double the energy Give class Member and N columns - have them calculate the remainder

Note: The N column in this case has the load at C equal to 500 lb. as well as an additional load of the variable P .

Member N dN/dP N (P=0) L N(dN/dP) L

AH,EF -1060.66 -P/(2)1/2 -0.707 -1060.66 (200)1/2 (-0.707)(-1060.66)(200)1/2

AB,ED 750 + P/2 0.5 750 10 (0.5)(750)(10)

HB,FD 500 0 500 10 (0)(500)(10)

HG,FG -1000 -P -1.0 -1000. 10 ( -1.0)(-1000.)(10)

HC,FC 353.55+ P/(2)1/2 0.707 353.55 (200)1/2 (0.707)(353.55)(200)1/2

BC,DC 750 + P/2 0.5 750 10 (0.5)( 750)( 10)

GC 0 0.0 0 10 (0.0)( 0)(10)

D = 1/AE [2*(-0.707)(-1060.66)(200)1/2 + 4*(0.5)(750)(10) + 2*( -1.0)(-1000.)(10) + 2*(0.707)(353.55)(200)1/2]

Lecture 20 Castigliano's Theorem - Beams and Frames

as before, simply stated

D = dU/dP

Now for beam action we also have:

q = dU/dM

or, the displacement D , under a load P, is equal to the partial derivative of the strain energy in the structure with respect to the applied load P, and the rotation q, under a moment M, is equal to the partial derivative of the strain energy in the structure with respect to the applied moment M.

Apply this to Beams and Frames

For Displacement:

Bending

Ub = m M2/2EI dx d Ub/ dP = m M(d M/ dP ) /EI dx

Shear

Us = K m V2/2GA dx d Us/ dP = K m V(d V/ dP ) /GA dx

Torsion

Ut = m T2/2GJ dx d Ut/ dP = m T(d T/ dP ) /GJ dx

For Rotation:

Bending

Ub = m M2/2EI dx d Ub/ dP = m M(d M/ dM' ) /EI dx

Shear

Us = K m V2/2GA dx d Us/ dP = K m V(d V/ dM' ) /GA dx

Note we must treat P and M' as variables and thus each Beam member force must be an expresses as a function of P or M'.

Procedure:

1. Place a force P (or M') in the direction where the displacement (or rotation) is to be determined, if there is already a force (or moment) there, assume it is P (or M').

2. Compute the internal forces in as a functions of P (or M') considering individual "x" segments between concentrated loads.

3. Determine the partial derivatives of the moment with respect to the applied loads in each segment

4. Apply Castigliano's theorem to determine the displacement.

Show Example 8-26 -

Show Example 8-29 - Don't let the sloped member cause concern

Have Class Work Problem 8-48

Lecture 21 Review - Deflections in Beams and Frames - Chapter 8

Elastic Beam Theory

Assumptions- Important !!

1. Beam is initially straight

2. Loads applied perpendicular to Beams axis

3. plane cross sections of the beam remain plane before and after bending !!

Bernoulli-Euler Beam - a very good approximation (Stocky Structure stuff )

4. Deflections and rotations are small

5. Material remains elastic

v" = M/EI

This is a very important equation - Second order differential equation relating the displacement (v) to the moment (M), the material (E) and the geometrical feature of the beam (I)

Double Integration Method -

Use the above equation - rearrange

EI v" = M(x)

Now use the principles of ordinary differential equations to solve

i.e. integrate twice and match boundary conditions

For example where the displacement and slope are 0 - fixed end

where the displacement is 0 - simple support

or match where two segments of a beam must be compatible

Moment Area

Review the Moment area theorems

Theorem 1: The change in slope between any two points on the elastic curve equals the area of the M/EI diagram between these two points

Theorem 2: The deviation of the tangent at point B on the elastic curve with respect to the tangent at point A equals the "moment" of the M/EI diagram between the two points A and B computed about point A( the point on the elastic curve) where the deviation tA/B is to be determined.

Important 1st theorem - change in slope

2nd theorem - care in what it says!!!

Both deal with the M/EI diagram !!!

i.e. Loaded Structure M Diagram M/EI Diagram

Sometimes it takes some ingenuity to set up - See Example 8-7

Conjugate Beam

This is really fun !!

Procedure:

1. Develop the M/EI diagram for the real beam

2. Modify the real beam's boundary conditions to a "conjugate" beam

3. Place the M/EI diagram from the real beam on the conjugate beam for the loads

4. compute the "reactions" for the conjugate beam

5. Use Theorems 1 and 2 to compute rotations and displacements of the real beam

Theorem 1. The slope at a point on the real beam is equal to the shear at the corresponding point in the conjugate beam

Theorem 2. The displacement of a point on the real beam is equal to the moment at the corresponding point on the conjugate beam

Virtual Work

Assumptions:

deformations do not have to be elastic ( New freedom)

Displacements must be compatible

Work of External loads = Work of internal Loads

Procedure:

1. Place a unit load on a structure in the direction you desire to determine the displacement

This is an imaginary or "virtual" load. It doesn't exist in the real structure and is just used as a convenience to determine the displacement at this point

2. This virtual external load induces virtual internal forces (u) and virtual displacements ( in the structure (u)

Virtual work - trusses

For a truss , the external virtual work is

1 x D =

For a truss member, the internal virtual work is

the virtual loads moving through the (real displacements associated with applied Real loads)

= n (NL/AE)

plus

the virtual loads moving through the real displacements associated with applied temperature changes

= n (a D T L)

plus

the virtual loads moving through the real displacements associated with Fabrication errors

= n (D L)

Create a table to sum these contributions for the whole truss!!

Virtual Work - Beams and Frames

1 D = m mM/EI dx -> to calculate displacement

1 q = m m q M/EI dx -> to calculate rotation

The keys here are

1. a unit load is place at the location, in the direction you desire to determine the

displacement, the equations for the moment produced in the beam are then developed

2. a unit moment is placed at the location you desire to determine the rotation, the equations for the moment produced in the beam are then developed

3. You have some flexibility in determining your coordinate systems to use to do the integration. HOWEVER YOU MUST USE THE SAME SET FOR M and m !!!

Castigliano's Theorem

Simply stated

D = dU/dP

or, the displacement D , under a load P, is equal to the partial derivative of the strain energy in the structure with respect to the applied load P.

Castigliano's Theorem - Trusses

U = 1/2 S N2 L / (AE)

Now, let's take the partial derivative with respect to P understanding that the only functions of P are the internal forces, N

dU/dP = 1/2 S 2N(dN/dP) L / (AE) = S N(dN/dP) L / (AE) = D

Note we must treat P as a variable and thus each truss member force must be an expresses as a function of P.

Procedure:

1. Place a force P in the direction where the displacement is to be determined, if there is already a force there, assume it is P.

2. Compute the force in each member as a function of P.

3. Prepare a table for the pertinent values (see example)

4. In a separate column of the table, list the internal forces with P having it's numerical value, (if no P load is at the node of interest, P=0)

5. Apply Castigliano's theorem to determine the displacement.

Castiglaino's Theorem - Beams and Frames

as before, simply stated

D = dU/dP

Now for beam action we also have:

q = dU/dM

Apply this to Beams and Frames

For Displacement:

Bending

Ub = m M2/2EI dx d Ub/ dP = m M(d M/ dP ) /EI dx

Shear

Us = K m V2/2GA dx d Us/ dP = K m V(d V/ dP ) /GA dx

Torsion

Ut = m T2/2GJ dx d Ut/ dP = m T(d T/ dP ) /GJ dx

For Rotation:

Bending

Ub = m M2/2EI dx d Ub/ dP = m M(d M/ dM' ) /EI dx

Shear

Us = K m V2/2GA dx d Us/ dP = K m V(d V/ dM' ) /GA dx

Note we must treat P and M' as variables and thus each Beam member force must be an expresses as a function of P or M'.

Procedure:

1. Place a force P (or M') in the direction where the displacement (or rotation) is to be determined, if there is already a force (or moment) there, assume it is P (or M').

2. Compute the internal forces in as a functions of P (or M') considering individual "x" segments between concentrated loads.

3. Determine the partial derivatives of the moment with respect to the applied loads in each segment

4. Apply Castigliano's theorem to determine the displacement.

Lecture 22 Approximate Methods - Trusses

Begin our study of Indeterminate structures - Approximate Methods

Indeterminate structures have more unknowns that equations of equilibrium available

We will find shortly that to solve them, we must know E,I,A of the members>

However. To get an approximate solution we can make some assumptions to get the additional equations:

One method of doing this is by making assumptions about cross members on trusses

1. If members are slender they cannot carry compression force - i.e. they buckle

Therefore assume tension only members

2. If members are stout - assume that half of the shear is carried by each member

Show Examples

Have class work 7-1, 7-2 -- Compare answers and highlight the differences -- explain why the differences

Lecture 23 Approximate Methods - Beans and Frames

Approximate methods for multitstory frames - vertical loads

Assume a pin at the location of 0 moment -- a very good assumption

Approximate this location to be .1 l

Show Figure 7-5

Review Example 7-3

Portal Frames and trusses

For portals - again assume a hinge is located at 0 moment position

This location will vary due to boundary conditions, stiffenss of beams and columns etc, , but approximate location can set

Show Figure 7-7 7-8 and 7-9

Review Example 7-4

Lecture 24 Approximate Methods - Portal Method

Using the "lessons" from previous approximations on Portals - we can extend the method now from a single bay, to multiple bays and a few stories

This approximate method is most suitable when used for buildings with low elevations:

It is based on a "shear dominant" argument

Assumptions to get enough equations (or releases to make it a determinate condition)

1. A hinge is placed at the center of each girder since this is assumed to be a point of 0 moment

2. A hinge is placed at the center of each column since this is assumed to be a point of 0 moment

3. At a given floor level, the shear at the interior column hinges is assumed to be twice that at the exterior column hinges since the frame is considered to be a superposition of portals

Enough assumptions (releases) have now been made to allow solution solely with equations of equilibrium. Proceed by cutting the structure into pieces between the assumed hinges. Applying the known forces and assumed shear distribution.

Work Example 7-29

Lecture 25 Approximate Methods - Cantilever Method

Recall that the portal method is most suitable when used for buildings with low elevations:

It is based on a "shear dominant" argument

For taller structures, a approach based on a cantilever beam (i.e. the cantilever method) is more suitable

It is based on a normal stress assumption

Assumptions to get enough equations (or releases to make it a determinate condition)

1. A hinge is placed at the center of each girder since this is assumed to be a point of 0 moment

2. A hinge is placed at the center of each column since this is assumed to be a point of 0 moment

(The above two assumptions are the same as for the portal method)

3. The axial stress in a column is proportional to its distance from the centroid of the cross- sectional areas of the columns at a given floor level. Since stress equals force per area, then in the special case of the columns having equal cross-sectional areas, the force in a column is also proportional to its distance from the centroid of the column areas.

Work Example 7-30

Lecture 26 Review Exam II

Review Exam - Have Students work problems

Lecture 27 Statically Indeterminate Beams

Review what it means to be statically indeterminate - More unknowns than allowable equations of equilibrium

Comparison of Determinate and Indeterminate Structures:

Determinate Indeterminate

Stress (Assuming same E,I) High Low

therefore can use smaller sections

Load Redistribution Better

(Consider Plastic Hinge)

Support Settlement Introduces Stress No stress

Construction More expensive

Three basic sets of equations:

1. Equilibrium - We already know about these

2. Compatibility - This is the focus for the next few lectures, They are the equations that

specify that the structure must "fit together" or have some specified displacement.

3. Force-Displacement (Consitiutive Relation) - Stress-Strain behavior - C.E. 205

We will assume this to be linear elastic for now.

2 basic methods

Unknowns Equations

Force (or Flexibility) Forces Compatibility, Force/Displacement

Displacement (or Stiffness) Displacement Equilibirum, Force/Displacement

this is the method that is usually computerized

Force Method - General Procedure:

Consider first the Beam deflection and Slope tables in the front of the book (Show Overhead)

Ask - How did we get these ? (Virtual Work, Moment Area, Conjugate Beam)

in other words, you can derive these for determinate situations - REMEMBER THIS !!!!

NOTE -. IF WE USE P (OR M) AS A VARIABLE, WE CAN FORCE THE DETERMINATE BEAM TO A SPECIFIC DISPLACEMENT AT ANY LOCATION WE WANT --- WE WILL USE THIS CAPABILITY TO DEVELOPE THE FORCE METHOD, IN FACT, THE VALUE OF THIS FORCE TO PRODUCE A SPECIFIED DISPLACEMENT IS THE "COMPATIBILITY CONDITION" WE DESIRE.

General Procedure:

1. Given an Indeterminate Beam, Remove enough boundary conditions to make it determinate

These remove loads are called "redundants"

2. The determinate beam generated is called the "primary structure"

3. Displacements of the Primary structure can be computed by Moment Area, Conjugate Beam, Virtual Work - or by simply using the tables at the front of the book.

4. Again, using Moment Area, Conjugate Beam, Virtual Work - or by simply using the tables at the front of the book displacements (or rotations) at the location of the Redundants can be computed as linear functions of the Redundant ( i.e. the flexibility times the Redundant)

5. A compatibility equation (or equations) can be written to set the actual displacement at a point

6. Solving these compatibility equations, where the unknowns are the Redundant Forces provides the extra equations we need to analyze the indeterminate beam.

Show Figures 9-3, 9-4, and 9-5 - Discuss

Describe the flexibililty coefficient notation (excuse the clumbsy notation to keep consistane with Hibblers definitions) --

fAB Displacement at A Due to a unit force at B

For example, From tables let:

x = A P=1

a=B b

v(x) = - Pbx/6LEI (L2 - b2 -x)

fAB = v(A) = -(1)(L-B)(A)/6LEI (L2 - (L-B)2 -A)

Review Example 9-2 and Example 9-5 Lecture 28 Force Method of Analysis for Frames

The force method for FRAMES is identical to the beam procedure.

Note - as Hibbler says - Problems involving multistory frames or those with a high degree of indeterminacy are best solved by other methods.

Show Example 9-6

Comment on the flexibility coefficient fBB - Note that the integral is simply

the square of the unit moment!!! - a simple way to determine flexibility coefficients

The force method for TRUSSES follows along the same lines

Show Example 9-8

Composite structures are made of both BEAMS and TRUSSES

The force method can be applied to these as well

Show Example 9-10

In general the flexibility method gives us a set of equations as;

f11 R1 + f12 R2 + f13 R3 + f14 R4 + f15 R5 + ...... = D1

f21 R1 + f22 R2 + f23 R3 + f24 R4 + f25 R5 + ...... = D2

......

F R = D

or R = F-1 D

Such equations are well suited for computer application

Note that after you get the Reactions, you can draw a free body of the member and create the moment diagram -- this is an important step

Lecture 29 Three Moment Equation

Begin with the powerpoint file 3moment.ppt on H:/pcapps/dept/ce/ce321

Have class work the example problem at the end of the PPT file

(i.e. a fixed fixed beam with a disributed load)

Try it with a Concentrated Load at the center

Compare these solutions to the tables in the back!!

Review Example 9-12

Lecture 30 Indeterminate Influence Lines

Revisit Muller Bresslau Principle --

The influence line for a function is (to the same scale) as the deflected shape of the beam when the beam is acted upon by the function

To do this

1. locate the position for the influence line

2. Apply the function

for a reaction apply a positive force - modify the beam at the load to not resist the applied force

for a shear - apply a positive shear - modify the beam so that it cannot resist shear at this point

For a moment - apply a positive moment - modify the beam so that it cannot resist moment at this point (a pin)

3. The shape the elastic beam assumes is to scale as the influence function

Note - we do not have the magnitudes of the influence line - only its shapes

However, Hibbler derives

for reaction Ay Ay = (1/fAA) fDA (1)

Scale factor * displacement at general point "D" due to a unit load at A

Scale factor = displacement at A due to a unit load at A

for a Shear at E VE = (1/fEE) fDE (2)

Scale factor * displacement at general point "D" due to a unit load at E

Scale factor = displacement at E due to a unit load at E

for a Moment at E ME = (1/aEE) fDE (3)

Scale factor * rotation at general point "D" due to a unit load at E

Scale factor = displacement at E due to a unit load at E

Now, we have a general method to get the influence lines, they are the equations 1,2,and 3 above!

It requires the calculation of the displacements of the indeterminate structure, (or calculation of the flexibility coefficients anywhere along the beam)

It is best to determine influence lines of indeterminate structure "Qualitatively" first -- Do this by Mueller Bresslau

To compute the actual coordinate values of the influence line,

1. place a unit load (or moment) at the point on the beam you are evaluating.

2. compute the displacements at enough locations on the curve to be meaningful

3. compute the displacement at the point you placed the load (or moment)

4. divide the computed displacements in step 2 by the value of step 3 to obtain the

actual influence line.

Review Examples 9-14, 9-15, 9-16.

Lecture 31 Slope Deflection Equations

Important things from Influence lines -- The qualitative diagram -- shows where to place loads

(Most of the problems we are working require this)

Qualitative influence lines for frames - locate where to place loads for maximum "influence"

Show Figure 9-30.

Now lets focus on the Displacement Method - Chapters 10,11,12,13,14,15

Recall --

Unknowns Equations

Force (or Flexibility) Forces Compatibility, Force/Displacement

Displacement (or Stiffness) Displacement Equilibrium, Force/Displacement

this is the method that is usually computerized

Degrees of Freedom:

We want to focus on specific points on a structure - define these points as "nodes"

Nodes are usually located at joints, supports, ends of members, or where members change in cross-section

Nodes

Each node in the above 2D can displace in the x,y directions and rotate about the z

(3 Degrees of freedom per node)

If the problem is 3D, each node can displace in the x,y,z and rotations about x,y,z

(6 Degrees of freedom per node)

Show Fig 10-1 to demonstrate this (Note the trusses don't consider rotation for there members) ask WHY?

For the displacement method, we consider the degree of freedom (displacement or rotation) at each node to be the unknowns. (There is an associated Force or Moment with each degree of freedom)

Slope-Deflection Equations:

A very useful case is to consider a beam (between two nodes): Assume all displacements ( Degrees of Freedom) are fixed:

A MAB MBA B

For a 2D beam we will consider only vertical displacements (y direction) and rotations

Now, we can get a relationship between one of the degrees of freedom and ( to begin with) the moments at A and B (Call them MAB and MBA )

MAB MBA

We can use conjugate beam to get the relationship between the moments and the degree of freedom qA

MAB = 4EI/L (qA)

MBA = 2EI/L (qA)

Similarly: by imposing a relative displacement between the ends, D. Note qA

and qB are both 0.0

MAB MBA

Again by using conjugate beam we get

MAB = MBA = -6EI/L2 (D)

The moments are "induced moments" because they are induced by the forced displacement. The sign convention used now (hold your breath) is

+ for clockwise

- for counter clockwise

Likewise, if we were to apply loads, the moments (and they will be fixed end moments) For example ( and recall that all Degrees of freedom are 0)

MAB A B MBA

Induced moments are:

MAB = PL/8 MBA = -PL /8

Using superposition, the slope deflection equation (s) can then be written:

MN = 2E (I/L) [2 qN + qF - 3( D/L)] + (FEM)N

This actually represents 2 equations (right end can be the near or the left end can be the near)Lets work an example problem - these are neat because a lot of the hard work is already done.

Lecture 32 Moment Distribution: Beams

Moment distribution has been one of the most popular methods to use to analyze indeterminate structures. It is an iterative technique that (although may be prone to error because of the number of steps required) is relatively simple to use. It has some basis in the Slope deflection procedure just studied.

Sign Convention - Same as SDP - Clockwise +

Counterclockwise -

Fixed End Moments - Required to begin the procedure

Member Stiffness Factor - From our development for Slope Deflection:

MAB MBA

MBA = .5 MAB

Now Divert for a minute - consider "stiffness" via simple springs

1 2 1 2 Kd 9Kd

K 9K

10 Kd

From "spring Constants" we know that K*d = F

Thus, the forces in each spring (when they each have the same d) is F1 = K d, F2 = 9 K d

in other words, the internal force is in direct proportion to the stiffness of the member

In addition the total stiffness of the structure is the sum of the stiffness of the members

(9K + K) d = F

The same relationship holds true for beams in bending

K M, q 9K

What does Moment equilibrium on this joint say? Note that since the angle of rotation at the joint is going to be the same on both sides of the applied moment - the internal moments in the members will be in proportion to the stiffness of the beam!!

(9K + K) q= M

Thus, the moments in each beam (when they each have the same q) is M1 = K q, M2 = 9 K q

in other words, the internal moment is in direct proportion to the stiffness of the member

In addition the total stiffness of the structure is the sum of the stiffness of the members

Member Stiffness

K = 4EI/L

Joint Stiffness

S Ki

Distribution Factor

DFi = Ki / S Ki

Carry-over Factor

MAB MBA

MBA = .5 MAB

Procedure:

Initial Setup:

1. Select Nodes

2. Calculate Stiffness Factors for each Member

3. Compute the Distribution factors at each node

4. Compute fixed end moments

Iterative Procedure:

5. determine the moment that is needed to put each joint in equilibrium

6. Release or "unlock" the joints and distribute the moments into each connecting span

7. Carry the moments in each span over to its other end by multiplying by the carry-over factor

8. Return to 5

Follow through Example 11-6 (Both Ways)

Have class compute Stiffness Factors and Distribution Factors FEM's as a preliminaryLecture 33 Moment Distribution: Frames

Work Problem 11-4

1st using traditional approach

Introduce Modified stiffness 3EI/L with carry over factor of 0.0 for far end pinned condition

Show how quick it converges

Lecture 34 Moment Distribution: Symmetry and Sidesway

Questions about Moment Distirbution ???

Introduce Symmetry conditions

Note: if structure and loads are symmetric -

Change stiffness on symmetric section to K = 2EI/L

(Don't carryover across the symmetry line)

Super convergence

if structure is symmetric and loads are antisymmetric

Change stiffness on symmetric section to K = 6/EI/L

(Don't carryover across the symmetry line)

Super convergence

Show Example 11-4

Use this to tell when to stop

1. if moments are balanced and no carryover

2. when carryover becomes "small"

Up to this point we have considered only rotation of a joint, What if it displaces as well?

This is the situation for "sidesway"

Show Fig 11-16

This situation is solved using a combination of moment distribution and "consistent applied Displacement"

Go through Example 11-7 (Has it all)

Point out

a) The superposition problem to solve

b) Modified stiffness for Joint C - Rapid convergence

c) Calculation of R'

d) Method to compute FEM for assumed D"

e) Computation of R'

f) Superposition of both solutionsLecture 35 Review Approximate Methods, Indeterminate Analysis, Moment Distribution

Approximate Methods - Trusses

One method of doing this is by making assumptions about cross members on trusses

1. If members are slender they cannot carry compression force - i.e. they buckle

Therefore assume tension only members

2. If members are stout - assume that half of the shear is carried by each member

Portal Method

Assumptions to get enough equations (or releases to make it a determinate condition)

1. A hinge is placed at the center of each girder since this is assumed to be a point of 0 moment

2. A hinge is placed at the center of each column since this is assumed to be a point of 0 moment

3. At a given floor level, the shear at the interior column hinges is assumed to be twice that at the exterior column hinges since the frame is considered to be a superposition of portals

Cantilever Method

Method of Consistent Displacements (Force Method):

General Procedure:

1. Given an Indeterminate Beam, Remove enough boundary conditions to make it determinate

These remove loads are called "redundants"

2. The determinate beam generated is called the "primary structure"

3. Displacements of the Primary structure can be computed by Moment Area, Conjugate Beam, Virtual Work - or by simply using the tables at the front of the book.

5. A compatibility equation (or equations) can be written to set the actual displacement at a point

6. Solving these compatibility equations, where the unknowns are the Redundant Forces provides the extra equations we need to analyze the indeterminate beam.

Show Figures 9-3, 9-4, and 9-5

In general the flexibility method gives us a set of equations as;

f11 R1 + f12 R2 + f13 R3 + f14 R4 + f15 R5 + ...... = D1

f21 R1 + f22 R2 + f23 R3 + f24 R4 + f25 R5 + ...... = D2

......

F R = D

or R = F-1 D

Indeterminate Influence Lines

Muller Bresslau Principle --

The influence line for a function is (to the same scale) as the deflected shape of the beam when the beam is acted upon by the function

To do this

1. locate the position for the influence line

2. Apply the function

for a reaction apply a positive force - modify the beam at the load to not resist the applied force

for a shear - apply a positive shear - modify the beam so that it cannot resist shear at this point

For a moment - apply a positive moment - modify the beam so that it cannot resist moment at this point (a pin)

3. The shape the elastic beam assumes is to scale as the influence function

Note - we do not have the magnitudes of the influence line - only its shapes Best Use - Determine where loads are applied.

Slope Deflection Equation (s):

MN = 2E (I/L) [2 qN + qF - 3( D/L)] + (FEM)N

This actually represents 2 equations (right end can be the near or the left end can be the near)

Moment Distribution:

It has some basis in the Slope deflection procedure just studied.

Sign Convention - Same as SDP - Clockwise + Counterclockwise -

Fixed End Moments - Required to begin the procedure

Member Stiffness Factor - 4EI/L

MAB MBA

Carry over Factor MBA = .5 MAB

Moment Distribution Procedure:

Initial Setup:

1. Select Nodes

2. Calculate Stiffness Factors for each Member

3. Compute the Distribution factors at each node

4. Compute fixed end moments

Iterative Procedure:

5. determine the moment that is needed to put each joint in equilibrium

6. Release or "unlock" the joints and distribute the moments into each connecting span

7. Carry the moments in each span over to its other end by multiplying by the carry-over factor

8. Return to 5

Potential Problems for Test:

Force Method: - one problem

Three-Moment Equations: - one problem

Slope Deflection: - one problem

Moment Distribution: (initial Setup)

(Influence Lines - Qualitiative only if at all)Lecture 36 Stiffness Method: Trusses

Go to PowerPoint File Stiff1

Overview of a stiffness matrix for a truss member

Lecture 37 Stiffness Method: Beams and Frames

Review element stiffness matrix for a truss member at angle

Show Fig 14-6, Equation 14-16

Discuss Row members where there are 2 (or more) degrees of freedom per node

Discuss Example 14-1

Now, for the element stiffness matrix of a beam Show Figs 15-7 and Equ 15-1

Some of these terms come directly from the slope-deflection equation!!

If the beam is at an angle, we must consider the direction cosines

Show Fig 15-9 and Equ 15-10

Note -- All we have to know to determine the stiffness matrix is

1. The cross-section characteristics of the Beam - A,I

2. The Material it is made from - E

3. The location of the nodes - L, l's

This information, coupled with the displacements where we know them, and the forces where we know them, is all that is required to set up a problem for the stiffness matrix

K U = F

Let the computer do the boring stuff - it is great at that.

Consider Example 15-2Lecture 38 Introduction to Computer Analysis

Visual Analysis is one of many computer programs that are available to do structural analysis

If you like, you can order yourself a personal copy (for educational use)

Lets work some examples

Truss - Statically indeterminate Example 9-8

Frame - Statically indeterminate Example 11-5

Statically indeterminate Example 10-9

Compare your results withe the example results - note that displacements are also computed

Reports can be printed directly

Note that the results will change for an indeterminate structure if the member sizes are changed

the results will not change for a determinate structure if the member sizes are changed.

Lecture 39 Computer Analysis of Structures

Lecture 40 Computer Analysis of Structures

Lecture 41 Review of Exam III

Lecture 42 REVIEW (Stiffness Method)