FUNdaMENTALS of Design - psnacet.edu.in · FUNdaMENTALS of Design Topic 8 Structures © 2000 Alexander Slocum 8-1 1/26/2005 Topic 8 Structures Topics • Beginnings

1/26/2005© 2000 Alexander Slocum 8-0

FUNdaMENTALS of DesignTopic 8

Structures


Topic 8Structures

Topics• Beginnings• FUNdaMENTAL Principles• Materials• Visualization• Layout• Stability• Loadings• Stiffness• Strength• Trusses • Laminates & Composites

Bryan R

uddy’smost am

azing Lazy-Tongs structure (2001)


Beginnings

• People have always sought to create ever larger, more complex structures– A structure might be able to hold its own weight, but then how much of a load could it carry?

• Bridges represent the greatest structural challenges:– Whenever a longer bridge was needed, adding more material also increased the weight…

• http://www.bizave.com/portland/bridges/Bridge-Gallery1.htmlhttp://www.discovery.com/stories/technology/buildings/brdg_explore.html#photo

• The great mathematicians of the 18th century set their minds to the task of developing mathematical formulas for predicting the strength of structures

– See Stephen Timoshenko, History of the Strength of Materials• History repeats itself (Patterns!)

– Your machine has limited size & weight, yet you want your machine to reach out to the worldJohn McBean’s longest 2.007 truss ever!

Rony Kubat’s most professional 2.007 machine EVER!

Barré de Saint-Venant1797-1886

Stephen Prokofyevich Timoshenko(1878-1972)

http://smitu.cef.spbstu.ru/timoshenko_en.htmLeonhard Euler

1707-1783


FUNdaMENTAL Principles• When the first sketch of the structure is made:

– Arrows indicating forces and moments should also be sketched– The path of how these forces and moments flow from the point of action to the

point of reaction, shows the structural loop• A sketch of the structural loop is a great visual design aid

– A closed structural loop indicates high stability and the likelyuse of symmetry to achieve a robust design

– An open structural loop is not bad, it means “proceed carefully”– Remember Aesop’s fables & “The Oak Tree and the Reeds”

• Example: automobiles to disk drives to semiconductors, exist because of double-disk grinders’ ability to create parallel flat surfaces:

Structural Loop


Materials• Materials make the machine just as sure as any creative design, and are often selected based

on strength, stiffness, manufacturability, and wear and corrosion resistance– Metals have very high strength-to-weight ratios and are easily machined formed, and joined– Woods high directional strength/weight and are easily joined– Plastics can have structural and low friction & wear-resistant properties and are easily molded,

formed, machined

Mild steel shapes in the 2.007 kit

Alum

inum shapes in the 2.007 kit

Plastic shapes in the 2.007 kit

It’s a bird! It’s a plane! NO, look, it’s a sheet of rubber from the 2.007 kit


Materials: Wear, Strength & Stiffness• Axial, torsion, and bending loads can be applied to structures and components

– An equivalent stress needs to be determined and compared to the material’s yield stress• Yielding in a component’s material can mean the component has failed, as will the machine,

or it can be used to form a component during manufacturing– Elastic instability (buckling) can affect shafts and columns in compression or torsion– Know your limits! (See pages 5-20 to 5-23!)

σyield

σultimate

Stre

ss(f

orce

/are

a)

Strain(elongation/length)

εyieldεplastic

deformation

εdeformationεelastic

springback

Mild steel (1018)

6061-T6 Aluminum

E = slope

Silicon MEMS relay structure from Jian Li’s Ph.D. thesis


Visualization

www.omax.com

F

FTrigger

F

FTrigger

Jammed!

Ouch!

• Develop your ability to imagine a structure deforming as loads are applied– Sequentially imagine that each element of the design is a piece

of rubber, while other elements are steel• Apply forces to the system and see how it deforms

– Does the deformation cause problems?– How can structure be changed to minimize deformations?

• Play the movie in your mind• Bracing elements with triangles (plate-type gussets or beam-type trusses) is the most efficient

method for strengthening a structure!• Creating CAD or paper and pushpin models is an effective way to visualize a structure

– Even if you are planning on using finite element analysis, a simple model can help you later determine if the results are meaningful!


Layout

Will Delhagen & Alex Jacobs in MIT’s 2001 2.007 contest Tiltilator!

• Layout is where the designer starts to define relative placement of elements and the supporting structure– The Layout is the first embodiment of the design intent and defines boundaries on the structure– You can create several simple stick-figure layout sketches of different concepts– Use appropriate analysis (e.g., 1st order error budget) guided by the layouts to help select the best!

• A Layout Drawing is the graphical interpretation of the FRDPARRC table’s Design Parameters– As a design progresses from Strategy, to Concept, to Modules, to Components, the layout is the

first step towards creating the details• A hand-drawn sketch & notes suffices for an initial layout, & it is a road map for creating a solid model• A solid model can serve as a layout, as long as one takes care to not add a lot of detail

– Use datum planes and curves referenced to a global coordinate system– Beware of referencing features to other features which may leater be moved or deleted


Layout: Sketches1) Early concept development (used in PREP)

2) Layout sketch (used to create simple 2D CAD drawing or solid model)

3) Solid model (and generated end view)

Alex Sprunt’s Wall-Crawler

• Use Motion diagrams and stick figures to help define and select your concept as initial starting points for your layout

– Design is like a flexible anagram• You are allowed not only to rearrange things

– You are allowed to add or subtract things!• Use your knowledge of the FRs and DPs and of fundamental

principles to catalyze the creation of the layout– E.g., the red-line-strategy machine grabs the pendulum with

a flexible arm and then goes to the end of the beam• What kinds of structures can enable a machine to do this?• In order to define the structure, you will also have to sketch

some basic ideas for the mechanism


Layout: FRDPARRC• Use a FRDPARRC table to guide creation of initial layout sketches

• Example: For the MIT & the Pendulum contest, create layouts for Concepts for Start pendulum swinging and collect balls and pucks Strategy

????1)Botherbot2)Pendulum clamp3)Cover goal

Block opponent

????1)Vehicle knocks pendulum as it drives by2)Fixed-to-ground spinning actuator

Actuate pendulum from ground

1)Gather 2 or 3 objects2)Gather 2 or 3 objects

1)Not enough time to make multiple trips2)Gather bin is too large

Physics text and past contests. Farm equipment websites

1)Time/Motion study, Friction/slip, Linkage design2)Friction, slip, linkage design

1)Pick up and score one at a time2)Harvest lots and dump loads

Gather pucks and balls and deposit in goal

Counter-measures

RiskReferencesAnalysisPossible Design Parameters

(Modules FR’s)

Functional Requirements


Layout: “GeekPlow” Example• Appropriate detail for layout sketches and Peer Review Evaluation Process (PREP) of a

machine (sketched with a Tablet PC) created according to the previous FRDPARRC table:

12

4 3


Layout: Analysis & Bench Level Experiments• Since layout involves creating the overall skeleton or supporting structure, it also is a

first chance to define the overall structural performance– The key is to understand how the structural loop behaves

• Sequentially imagine each component is made of a soft material…visualize…• First order calculations are in order• If the machine is complex, Finite Element Analysis (FEA) may also be used• If analysis is too costly (e.g., time to do), consider a Bench Level Experiment• Deflections, stresses, and vibration modes can all be initially estimated

?


• There are often two or more possible design paths– Use analysis, manufacturability, & robustness design reviews

• The overall structure must be defined before module development can commence in earnest

– Make the design amenable to evolution as detail later emerges• Use weighted design comparison charts

Comparison between one large or two small bearingsLWH55 (T2 preload)Moment (kgf-m) 42Deflection (rad, minutes) 0.000582 2Moment stiffness (kgf-m/rad) 71,463 distance to load (m) 0.12Load (kgf) 100resulting moment (kgf-m) 12resulting deflection (rad) 0.000168resulting deflection (microns) 20.2equivelent stiffness (N/micron, lbs/inch) 49 286,086 rated moment load capacity (kgf-m) 431equivelent load capacity at outside edge (kgf) 3,592

Lwhd15 (T2 preload)Force (kgf) 200deflection (microns) 10Lateral stiffness 20distance between bearing rails 0.2stiffness (kgf-m/rad) 800,000 rated load at edge of table (2 trucks) (kgf) 1900

Layout: Evolution & Comparison


Layout: Solid Models

• In order to create an appropriate level-of-detail solid model layout drawing:

– Use the FRDPARRC tables from Strategy, toConcept, to Modules, to Components to understand what functions the structure must perform

• The chicken-and-egg issue is that no detail yet exists, only sketches and spreadsheets

– Most machines must have an overall structure, a skeleton or frame, to which modules are attached

» Creating the skeleton or frame is the critical first step in catalyzing the generation of detail for all the modules

» Details are added as the design progresses

Rony Kubat’s solid model and machine. Rony’s solid model fit in the solid model size-constraint box as did his real machine in the real box. It performed as modeled and very nearly won the 2.007 MechEverest contest as it scored a perfect 50 points EVERY time…


Stability• Static Stability:– For robot contest machines, tipping-over stability is often a

prime Functional Requirement that drives the shape of the overall structure and where the modules will be located with respect to each other

• Dynamic stability and Buckling (see page 5-23!):– Are structural resonances excited that can lead to instability

and degradation of components or the process?– Do axial compression forces cause the component to buckle?

• Positive uses (apply reciprocity!)– Pile drivers, ultrasonic machining, triggers…

http://www.eng.iastate.edu/explorer/Bridge/collapse.htm

0.001

0.01

0.1

1.050 100 150 200

Drive point displacement (micrometers/Newton)Hz

1

234

5 67

8

Mode 7: Table bending

Mode 8: Column bending

Experimental modal analysis of a small surface grinder (performed by Eric Marsh when he was Prof. Slocum’s Ph.D. student)

F

F

Jammed!


Stability: Obstacles• Two wheel drive vehicles: The rear wheels have to push hard

enough to make the front wheels climb the obstacle• Four wheel drive vehicles: The rear wheels also provide the

normal force needed for the front wheels to apply a tractiveeffort to help climb over the obstacle

• What do the free-body diagrams show about “pushing”verses “pulling” the wheels over a bump?

– Is it better to try and climb a bump straight-on (both front wheels engage it at the same time) or one wheel at a time?

• Experiment with the spreadsheet Driving_over_step.xls

h

D

Lw

Lcg

hcgθ

FTr FTfFNr

FNf

ΓFwΓRw

mg

Option 1: Nice and easy slow drive over the step2 wheels contact

1 wheel contact

Normal force between a rear wheel and the ground, FNr/2 (N) 13.9 13.9Minimum required coefficient of friction, mumin2, mumin1 0.43 0.24Normal force between a front wheel and the step FNf/2 (N) 11.2 13.3

Total motor tractive effort, gm (N) 16.0 16Total motor limited tractive force from from both rear wheels, Frwmax (N) 8.0 8Total motor limited tractive force from from both front wheels, Ffwmax (N) 8.0 8

Total friction limited tractive effort, gmu (N) 15.1 16.3Total friction limited tractive force from from both rear wheels before slip, Frwmumax (N) 8.4 8.4Total friction limited tractive force from from both rear wheels before slip, Ffwmumax (N) 6.7 8.0

Total minimum tractive effort required, gmin (N) 21.7 13.22Total tractive effort required by both rear wheels, FTr (N) 12.06 6.77Total tractive effort required by both front wheels, FTf (N) 9.66 6.45

Step Climable?Can the machine climb over the step? no yes

Option 2: Ramming speed!Ideal forward velocity required to get over the step, v (mm/s) 626∆ potential energy on top of step, DPE (N-m) 0.98


Stability: Slopes & Balance • Can the machine drive up the hill?

– What is better for climbing hills, FWD or RWD?– Do you really need AWD?

• When will the machine tip over?– What happens when the force vector due to gravity just passes

through the rear wheels ground contact points?

θ

Lw

Lcg

X

Y

mg

FTr

FTf

FNr

FNf

hmg

Funny image found on w

ww

, photographer not credited, w

ould like to, email slocum

@m

it.edu

SystemRear wheels' diameter, Drw (mm) 100Front wheels' diameter, Dfw (mm) 100Distance between wheels, Lw (mm) 250Distance center of front wheel to center of gravity, Lcg (mm) 125Height of center of mass about plane, hmg (mm) 50Slope angle, theta (deg, rad) 20.00 0.35Machine mass, m (kg) 4Machine weight, mg (N) 39.2Maximum drive torque applied to both rear wheels, grw (N-mm) 400Maximum drive torque applied to both front wheels, gfw (N-mm) 400Coefficient of friction, mu 0.5

Enter 1 for rear wheel drive or AWD, γr 1Enter 1 for front wheel drive or AWD, γf 1

Nice and easy slow drive up the ramp (no acceleration)Normal force between both rear wheels and ground, FNr (N) 21.1Normal force between both front wheels and step FNf (N) 15.7Maximum rear wheel tractive force from drive torque, Frwmax (N) 8Maximum rear wheel force from drive torque, Ffwmax (N) 8Maximum rear wheel tractive force before slip, Frwmumax (N) 10.5Maximum front wheel tractive force before slip, Ffwmumax (N) 7.9Total tractive force generated by both rear wheels, FTr (N) 8.00Total tractive effort required by both front wheels, FTf (N) 7.87Total tractive effort of the machine, FTm (N) 15.87Force from gravity acting along the incline, Fg (N) 13.41Can the machine climb up the ramp? yes

Tip-over angle (deg, rad) 68.2 1.19Required minimum friction coefficients:

Front wheel drive only required coefficient of friction 0.85Rear wheel drive only required coefficient of friction 0.64All wheel drive required coefficient of friction 0.36

for FWD γf = 1 and γr = 0, for RWD γf = 0 and γr = 1, and for AWD γf = 1 and γr = 1

Driving_up_slope.xlsValues


• There are four basic types of loads:– Axial tension or compression: the applied force directly acts on the

material to cause tension or compression– Bending: an applied force acts via a lever to bend a beam, causing

tension on one side and compression on the other side of the structure– Torsion: a torque (e.g., twisting or two equal and opposite forces applied

about a point) causes twist of the structure– Shear: two equal and opposite essentially collinear forces act

perpendicular to a structure• Structures that fail in torsion are actually also failing in shear• Glue joints in laminates, subjected to bending, actually fail in shear

• Boundary conditions are critical!

Loadings: Axial, Bending, Torsion, & Shear

ends of beams joined

one beam resting on the other

3 4For equal diameter and length beams: 4νδ

νδ+

=

Ends free Ends rigidly connected


b

d

r

Ø D

d+r-D/2

b+r-D/2

D/2+rD/2

Loadings: Structural Cross-Sections torsional stiffnessPGI

K L=

( ) ( )4 4 4 4

max 4

162 64 32

o I o Iobending torsionNA

o

D D D DDy I ID

π πτ

π− − Γ

= = = =

3 43

max4 3

4torsion with a = b

16 4.83.36 12 12 3 12

2.25

bending torsionNAb a bb bay bI I a a a

bI

τ⎡ ⎤⎛ ⎞ Γ

= = = − − =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

=

( )( ) ( ) ( )( )( )

( ) ( )( )

3 2 23

max short side2 2

3torsion with a = b and uniform thickness t max long side

2 2 22 12 2

2

b a a b b abending torsionNA

b a a b b b a

a b a

a a b a bb b t t t t t ty I I

a b a bt t t t t t t

t b tIa bt t t

τ

τ

− − − − − Γ= = = =

+ − − − −

Γ= − =

− −

2mean of areas enclosed by boaundaries thickness

length of median boundary

average shear stress

4

2

ptA

IP

tAτ

=

Γ=

( )

3length of median boundary thickness

average shear stress 22

33 1.8

ptP

I

P t

tPτ

=

Γ +=

( )( )( ) ( )

( )( )

2 22 2 2 2

4 433 4

44

12 2 12 22

1 10.21 1 0.105 1 0.07 0.0763 12 3 192

bendingNA NA NA

torsion

b c ba d a da dab d c by y yIab d c b

b d d rb ba c b dbI Da c b b ba c b

⎡ ⎤ ⎡ ⎤+ − ⎛ ⎞ ⎛ ⎞= = + − + − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟+ − ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= − − + − − − + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟−⎢ ⎥ ⎝ ⎠−⎝ ⎠⎣ ⎦ ⎝ ⎠⎣ ⎦c

a

Do

ta

tb

b

Di

Neutral axis

bd

c

a

a

c

yNA

yNA

yNA

yNA

d b

b

a

yNA

a dbØ D

r

Ø Dr

r

flange

web

web

flangeØ D

leg

leg

( )( )( ) ( )

( )

2 22 2 2 2

34 43 4

44

22 2 212 2 12 22 2 2

1 12 0.21 1 0.105 13 12 2 3 1922 2

bendingNA NA NA

torsion

b c ba d a da dab d c by y yIab d c b

b c d db ba b dbI Dca ba b c b

⎡ ⎤ ⎡ ⎤+ − ⎛ ⎞ ⎛ ⎞= = + − + − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟+ − ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤⎛ ⎞⎡ ⎤ ⎢ ⎥⎛ ⎞ ⎜ ⎟⎛ ⎞= − − + − − − +⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ −⎝ ⎠⎣ ⎦ ⎢ ⎥−⎜ ⎟⎝ ⎠⎣ ⎦

0.07 0.076 rb

⎧ ⎫⎪ ⎪⎛ ⎞⎪ ⎪+⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎪ ⎪⎩ ⎭

Ignore r in Ibendingcalculations)

Ignore r in Ibendingcalculations)

( )( )

( )

33

34 43 4

44

2 2 22 12

1 12 0.21 1 0.105 1 0.15 0.103 12 2 3 1922 2

bendingNA

torsion

a a d c bc cy I

b c d d rb ba b dbI Dca b ba b c b

− − −= =

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪⎡ ⎤ ⎢ ⎥⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎪ ⎪= − − + − − − + +⎨ ⎬⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠−⎝ ⎠⎣ ⎦⎪ ⎪⎢ ⎥−⎜ ⎟⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

Be thankful for spreadsheets!


Stiffness (1/Compliance)

OUCH!EEOW!Ahhhh!

• All structures deform under load– Will the deformations create translational and angular displacements that

will cause other elements to become overloaded or interfere and then fail?• Make the deflection 3-5x LESS than critical clearances (Saint-Venant)

• Where are forces transmitted between members with respect to their neutral axes”?

– Position interface contact points at neutral axis planes!• System compliance = sum of structural and element compliances

– Machine elements (e.g., bearings) and joints should have a stiffness on the order of the structure itself

– During early design stage, before bearings and joints are designed, assume net stiffness will thus be structural stiffness/3

1 mm

long silicon microspring

electrical contacts. See U

S Patent 6,497,581

Differential Differential

Width-tapered silicon springs


Stress

-12.0-10.0

-8.0-6.0-4.0

-2.00.02.0

4.06.0

0 10 20 30 40 50 60 70 80 90 100

Distance from left end of beam (mm)

Stre

ss (P

a)

Slope

-0.150

-0.100

-0.050

0.000

0.050

0.100

0.150

0.200

0 10 20 30 40 50 60 70 80 90 100


Slop

e (m

illi r

adia

ns)

Stiffness: First-Order Analysis• Complex systems can often be modeled by superimposing simple models

Abbe errors exist, but reciprocity says they might be usable… Do you see a pattern?

3 2 2

3 2 2F F M ML L L L

EI EI EI EIδ α δ α= = = =

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( )

2

2 3

33 23 2 2

2 2 3

2 6 4

2 3 43

6 24 5 2

L a L aB B a

L aA a

L aA a

L a L aw w w wF F L a MwR M

F L a L a M L aw ww

EI EI EI

F a ML a L aa aL L Lw wL awEI EI EI

θ

δ

⎡ ⎤+ − − −= + = − − − + +⎢ ⎥

⎢ ⎥⎣ ⎦

⎡ ⎤− − − −= + + −⎢ ⎥

⎢ ⎥⎣ ⎦

− − + −⎡ ⎤− − += − + + +⎢ ⎥

⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( ) ( )( )

( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )

2

2 2 2 2222

22 2 22222

2 26 2 3

6 6 32 7 3 22

6 24 15 6

98 316 24 15

L a L aA B L a

L aA a

L aB a

F L a L a L a L aw w M Fa w w Mw wR RL L L L L L

aL M aLFa L a L a L a w w L a aLaLw aLEIL EIL EIL

Fa L a aLa w w L aLw aLEIL EIL

θ

θ

⎡ ⎤− − + − − += + − = + + − +⎢ ⎥

⎢ ⎥⎣ ⎦⎡ ⎤− + − − +− − − −

= − + − + −⎢ ⎥⎢ ⎥⎣ ⎦

⎡− − − + +⎢= + +−⎣

( )22 36

M aLEIL

⎤ −⎥ +

⎢ ⎥⎦

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( )

2 3 32 2 2 22 2

3 3 3 3 3 3

3 22 22

2 2

3 32 4 433 3

2 8 5 2 2 2 8 5 2

43

2 8 5 2 3

L a L a L aA Ba a

L a L aB a a

M Fa MF L a L a L a L a L a L a L aa L a aL Lw w w w w wL a L aw wR RL L L L L L

Fa ML a L a L aaL w w w wL aw wML L

− − −⎡ ⎤ ⎡ ⎤− + − − + + − − − += + + + − = + − + + +⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

− − ⎡ ⎤ ⎡ ⎤− − + − −= + + + − + +⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( )

( ) ( ) ( ) ( )( ) ( ) ( )

2

2

2 3

32

2 3 33

4 48 5 4L a

A a

a LL

Fa L a L a L a M L a a Lw wL awEIL EIL EIL

θ

−

⎡ ⎤− − − − + − −= − + + +⎢ ⎥

⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

2 3 32

3 3 3 3 3 3

2 3 22

2 2 2

22

2

2 3 2 6 3 2 3 2 62 10 2 2 10

4 32 33

12 5

L a L a L aA Ba a

L aA a

B

F L a L a L a L a Ma L a F L a L a L a L a Ma L aw w a w w w wL a L aw wR RL L L L L L

M aLFa L a L a L a aLw wL awML L L

F L aa LML

⎡ ⎤ ⎡ ⎤− + − − + − − + − − − + −= + + + − = + − + + +⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

− +⎡ ⎤− − − − += − + + −⎢ ⎥

⎢ ⎥⎣ ⎦

− −= + −( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )

2 2

3 2 3 2 3 3 22 10 6 30 2 6

L a a L a L aaa

L a L a L a L a Ma a Lw w w w w w wwa L awL L

⎧ ⎫⎡ ⎤− − + + − + − −⎪ ⎪+ + − − − − +⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

( )( )( )

( )( )

( )

20

2 30

2 3 42

2 3

Transverse shear2

Moment 2 6

Slope2 2 6 24

Displacement2 6

L aA a

a L aA A

a L aA AA

A AA A

x aw wV F x a x awR L a

x a x aw w wM x F x a M x aM R L a

F x a x a x a M x aw w wx xM REI EI EI EI EI L a EI

Fx xM RxEI EI

θ θ

δ θδ

− −= − − − − −

−

− − −= + − − − − + −

−

− − − − −= + + − − − +

−

= + + −+( )

( )

3 4 5 2

6 24 120 2a L ax a x a x a M x aw w w

EI EI EI L a EI− − − − −

− − +−

Deflection

-4.0

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.00 10 20 30 40 50 60 70 80 90 100


Def

lect

ion

(mcr

ons)

+

=

am

aw

Lx

A B

afwa

M

F wL


Stiffness: Advanced Analysis Methods• Singularity functions enable loads to be “turned on”

– Create expressions for the moment M(x) as a function of position – M(x) is used in moment-curvature & energy method calculations

x0

q = M<x - a>-2

x = a

M (N-m)

x0

q = F<x - a>-1

x = a

F (N)

x0

q = w<x - a>0

x = a

w (N/m)

x0

q = w<x - a>1

x = a

w (N/m/m)Q

dQ

∆ d∆

dU'

dUcomplementaryelastic energy U'

elastic energy U

A most useful website is http://www.integrals.com/index.en.cgi

dsis the distance along the beam (e.g., ds=

dxfor

straight beams, ds

= Rdq

for curved beam). For

determining the slope, differentiate not w

ith respect to a force Q

, but an applied mom

ent M(w

hich may also be a

virtual applied mom

ent at the point of interest)

Loading Variables Energy Deflection Axial P, E , A 2

0 2L P ds

EA∫ ( )0

L P QF dsEA∂ ∂

∫

Torsion Γ, G, Ipolar 2

0 2 polar

L

Ids

GΓ∫ ( )

0 polar

L QI

dsG∂Γ ∂Γ

∫

Transverse Shear(approximate for non-rectangular sections, else ½ => 3/5)

V, G, A 2

0 2L V ds

GA∫ ( )0

L V QV dsGA∂ ∂

∫

Bending M, E, I 2

0 2L M ds

EI∫ ( )0

L M M Q dsEI∂ ∂

∫

L

Q (virtual force at point of interest)

a


Cantilever Beam Bending: FEA vs Theory

0.86

0.88

0.90

0.92

0.94

0.96

0.98

1.00

1.02

1.04

1.06

0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0

Beam length/height

defle

ctio

n ra

tio

Theory total/FEATheory bending/FEA

Stiffness: Energy Methods Examples• Energy methods can be used to determine the deflections in a curved beam

– They can also be used to calculate the relative contributions of axial, shear, and bending; E.g., for a curved beam:

( )

( )

2

0 axial she

cos sin sin cos 1 cos sin

cos sin 0 sin cos 0 1 cos sin 1

sin coscos

x y x y x y o

x y o x y o x y o

xx

P V M R R ds RdF F F F F F MP P P V V V M M MR RF F M F F M F F M

U F Fx y RdF EA

φ

θ θ θ θ θ θ θ

θ θ θ θ θ θ

θ θ θδ θ

= − = + = − + + =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= = − = = = = = − = =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ −= = +∂ ∫

( ) ( ) ( )22 2 2

0 0ar bending

2 23

axial shear bending

cos sin 1 cos sin 1 cos 1 cossin

2 sin 2 1 2 sin 2 1 1 6 8scos cos4 2 4 2

o

x x x x

RF F F R F R Mx y x yRd RdGA EI

R RFF F F F Ry yEA GA EI

φ φθ θ θ θ θ θ θθ θ

φ φ φ φ φ φ φδ

+ − + − + −+

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+ − − − −⎛ ⎞ ⎛ ⎞= + +− +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

∫ ∫

( )

( )

223

2 2 22 2

0 0 0axial shear bending

2

axial

in sin 2 1 2cos cos sin4 2

cos sin sin 1 cos sinsin cos sincos sin

1 cos

oy

oy

y

y

F M RR

U RF F F F F R F R Mx y x y x yRd Rd RdF EA GA EI

RF xEA

φ φ φ

φ φ φ θ φ φ

θ θ θ θ θ θ θ θθ θδ θ θ θ

φδ

⎡ ⎤⎛ ⎞+ − +⎛ ⎞ + + −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∂ + + − + +−= = + +∂

−= −

∫ ∫ ∫

( )

( )

2 223 3

shear bending

0 bending

ben

2 sin 2 1 2 sin 2 1 1 2cos 2 sin 2cos cos 1 cos2 4 2 4 2 4

1 cos sin

y y ox y

x y o

o

RF FF F F M RR RxGA EI

U R RF F M RdM EI

φ

φ φ φ φ φ φ θ φ φ φ

θ θθ θ

θ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − + − + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ++ + + + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∂ − + += =∂

=

∫( ) ( )2 2

ding

1 sin 1 cos ox y RF F MR REIφ φ φ φ⎡ ⎤− + − +⎣ ⎦

3 63 5F FLL

EI GAδ = +

See Beam_curved.xls

Fx

Fy

Mo

R

Rsin R(1-cos )


Stiffness: Plates• A few simple loading cases give

insight into the nature of stresses & deformations in plates

• Check out the many different spreadsheets!

( )3

212 1EtD

ν=

−

( ) ( ) ( )

( ) ( ) ( )

2' 2

2 2 '2

max @ r = 02 2 '

'2

max @ r = 02 '

1.6 0.675 if 0.5

4 1 ln 1 4 1 ln 116 4

4 1 ln 1 4 4 1 ln 116 4

1 ln4 1

o o o

or r

o

ot t

o

t ttr r r

W a W aa r rM Mr a r r

W a W arM Mr r r

WrD

ν ν νπ π

ν ν νπ π

θπ ν

= + − <

⎡ ⎤⎛ ⎞ ⎡ ⎤−= + + − = + +⎢ ⎥⎜ ⎟ ⎢ ⎥

⎣ ⎦⎝ ⎠⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞

= + + − − = + +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦

= ++ ( )

( ) ( )( )

max @ r = a

22 2 2

max @ r = 0

4 1

33 2 ln16 1 16 1

a War D

WW a aa r r

D r D

θπ ν

ννδ δπ ν π ν

⎛ ⎞ =⎜ ⎟ +⎝ ⎠

− +− +⎛ ⎞= − − =⎜ ⎟+ +⎝ ⎠

( ) ( ) ( )

( ) ( ) ( )

2' 2

'2

max @ r = 0 min @ r = a2 '

'2

max @ r = 0 min @ r = a2 '

1.6 0.675 if 0.5

11 ln 1 1 ln

4 4 4 4

11 ln 1 ln

4 4 4 4

4

o o o

or r r

o

ot t t

o

t ttr r r

W a W a WrM M Mr r r

W a W a WrM M Mr r r

WrD

νν ν

π π π

ν ν νν ν νπ π π

θπ

= + − <

⎡ ⎤− −= + − + = + =⎢ ⎥

⎢ ⎥⎣ ⎦⎡ ⎤− −

= + − + = + =⎢ ⎥⎢ ⎥⎣ ⎦

= max @ r = 0.368a

22 2

max @ r = 0

ln 0.0293

1 2ln16 16

a War D

W a W aa rD r D

θ

δ δπ π

=

⎛ ⎞− −⎛ ⎞= − + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Circular plates with support aroundentire outer boundary

aro

W=qπro2

ro << ar

aro

W=qπro2

ro << ar

Uniformly_loaded_circular_plate.xls

-3.500

-3.000

-2.500

-2.000

-1.500

-1.000

-0.500

0.0000 10 20 30 40 50 60 70 80

Radius (mm) (0 = center)

Def

lect

ion

(mm

)

d (mm)d (mm)

Circular plates with support(simply supported or fixed)

around entire outer boundaryar

rot

a

b

simply supported = S

S

SS

Platelength a (mm) 300width, b (mm) 200thickness, t (mm) 1.5Modulus of elasticity, E (N/mm^2) 68947.6Poisson ratio, v 0.3Total area 60000a/b, aob 1.50

Loadinguniform loading pressure over entire plate, q (N/mm^2) 0.00017total load applied to the plate, W (N) 10uniform over small concentric circle of ro, qo (N/mm^2) 10load application radius, ro (mm) 2

total load applied to center of plate, Wo (N) 126Simply supported outer boundary:

Distributed load:Center displacement, dc (mm) -0.096Bending stress at center (N/mm^2) 1Reaction load at center of long side (N/mm) 0

Centrally applied concentrated load:Center displacement, fdc (mm) -3.621Bending stress at center (N/mm^2) 190

Enters numbers in BOLD, Results in RED

From Roark & Young, Formulas for Stress and Strain, 5th eddition, pagTo determine deflection of a rectangular platePlate_Rectangular.xls

By Alex Slocum, Last modified 1/1/04 by Alex Slocum


Stiffness: Plate Examples• Al plate 200 x 300 x 1.5mm

uniformly loaded with 10N

100 mm

.03 mm

L

d

D

δ

D' α

α

Fixed-corners (L brackets would have to be added)

Free-corners: 1 mradslope on flange ends results

Shaft misaligned in bearing from slope in plate


Strength• Stress=Moment*distance of farthest fiber from Neutral axis/Moment of Inertia:

– Stress ratio = Applied stress / Maximum allowable stress• The parallel axis theorem can be used to evaluate any cross section’s inertia:

– Most important: The parallel axis theorem tells us that a section stiffens with the square of the distance from the neutral axis!

• E.g., when designing a laminate (1/16” AL sheet separated by wood core), double the core thickness and quadruple the panel stiffness!

• Torsional shear stress=Torque*radius/polar moment of inertia• Stresses caused by multiple loads can be combined into an equivalent stress by

the von Mises equivalent stress formula:

IMc

=σ

Ten 80 lb bags of concrete

Funny image found on www, photographer not credited, would like to, email [email protected]

( ) ( ) ( )2 22

2 2 23 3 32

x y x z y zxy xz yzequivalent

σ σ σ σ σ στ τ τσ

− + − + −= + + +

c

a

Do

ta

tb

b

Di

Neutral axis

bd

c

a

a

c

yNA

yNA

yNA

yNA

d b

b

a

yNA

a dbØ D

r

Ø Dr

r

flange

web

web

flangeØ D

leg

leg

OK

, what’s the ratio of the

stress ratio in the bar to the stress ratio in the knees?


H

H

Strength: 1st Order Analysis• Square tube with point loads

(from wheels) on two opposite sides

Model plane of symmetry

Yes, yes, yes, for all you FEA avocados out there, only a 1/8 models with zero slope constraints at the planes of symmetry is actually needed


Strength: Life, Fatigue and Stress Concentration• Fatigue:

– To obtain “infinite” life (endurance limit) in a steel structure• Applied stress should be less than ½ yield stress• Aluminum has no endurance limit: Infinite life requires zero stress

– A structure can be subjected to simultaneous continuous and alternating stresses• Stress concentration:

– Stresses must flow around features, and the sharper the feature, the more severe the turn

• This causes a kink, and the stress rises

Alte

rnat

ing

stre

ss: S

a

Mean tensile stress: Sm

Goodman line

Ultimate tensilestrength: Su

Endurance limit stress: Sen

Sax

Smx

N: number of cycles to failure

S: st

ress

Sen Type 1

Type 2


Trusses• Trusses can carry huge loads while being very light weight

– Saint-Venant & trusses • Web member (braces) spacing is typically on the order of the truss height• Greater spacing leads to greater deflections and greater chance of chord buckling

– Fundamentally, the farther away from the neutral axis you can add area, the better squared you will be!

– Trusses are easily fabricated from spot-welded steel welding rod!

∑

∑

=

== N

iii

N

iiii

NA

AE

yAEy

1

1

NA H

AH/2

W

2NA zz z NAAyI I −= +

H

ALyNA

AU

W

NA


Laminates & Composites• Laminates are made of stress-carrying elements bonded to a core (see Laminate.xls)

– They have tensile/compressive carrying members on the outside (chords), and a shear carrying member (core) on the inside

• Metal chords with a wood core can give nearly the strength of an I beam• The shear stresses in the adhesive and core materials must be carefully considered

• Composites use a matrix, typically a polymer such as epoxy, to bond together structural fibers, cloth…

– Thin composite members are then often laminated to a core…

Simple Platet 0.0625 thicknessw 1 widthL 10 lengthE 1.00E+07 modulusmaxstress 20000I 2.03E-05A 6.25E-02Ioc 6.51E-04Fmax 1.30deflection 2.13

Sandwhich beam (2 plates with wood core)Assume I and E of wood core are ignored

tcore 0.5Icore 0.0099Ioccore 0.0318Fmaxcore 63.5deflection 0.21

NAA

A1b

NAh

h/2y

b

NAh

h/2t

bDrill holes in plywood to reduce weight of laminate core

Solid plywood

Documents

FUNdaMENTALS of Design - psnacet.edu.in · FUNdaMENTALS of Design Topic 8 Structures © 2000 Alexander Slocum 8-1 1/26/2005 Topic 8 Structures Topics • Beginnings