Pure Bending – Multiple Materials & Plastic Bending

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LECTURE

07 !"#$%& '()*+(*),- /0+1,23+4 5

Bending

!"#$ &$'()'*+

of

,"-.)/-$ ,0.$#)0-1 2 3'$-01.)4 5 &267 890/.$# :;: .< :;=

Composite Sections Todd Coburn Cal Poly Pomona Chart by Todd Coburn.

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>$?)$@+ A"'(0B$'.0-1
>$?)$@+ A"'(0B$'.0-1
• Cons Consid ider er a stra straig ight ht beam beam subj subjec ecte ted d to bendi bending ng… … • Th e st ra ra in in a t a ny ny p os osi ti ti on on y c an an be written… " !

!

max

=

M total y

=

L0

M all

• A clos closer er look look at the the defo deform rmat atio ion n of the beam as a function of R provides… !

• We c an an s ee ee t ha ha t… t… • So …

!

=

!

R

L0

y =

E; F; I;

R

=

R

• The The forc forcee on any any elem elemen entalstr talstrip ip of mate materi rial al is… is… P i • Its Its momen momentt aboutthe aboutthe neut neutra rall axis… axis… M i • Sothetotalmom Sothetotalmomen entt is… is…

M total

=

=

" M i

P i yi

=

=

=

! i

Ai

i Ai yi

:; L;

!

"! i Ai yi

! • For For momen moments ts in the the elas elasti ticc rang range… e… " i E i ! i y • Comb Combin inin ing g with with ourstrai ourstrain… n… i E i i R • In Inse sert rtin ing g in into to ou ourr to tota tall mo mome ment nt & re rear arra rang ngin ing… g… M total =

!

=;

=

=

& A y 2 # ' $ E i i i ! R "! %$

• N ot ot in in g I=!Aiyi2 is th thee mo mome ment nt of in iner erti tia, a, if E is con const stan ant… t… • Comb Combin inin ing g with… with…

!

i

E =

…& rearranging… rearranging…

yi

R

I

all

y

max

!""#$%&'()"* !""#$%&'()"*

L0

y

y

! =

max

I

! i

=

M total y i

M total

E =

I

!

max

M total y

M all

We find…

! =

max

I

R

I

Chart by Todd Coburn.

=

!-0'$ 1$4.)<'1 1$4.)<'1 #$B0)' #$B0)' /-0'$ G$'*.9
I

all

y

max

$


6

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D-01.)4 &$'()'*+ ,"-.)/-$ ,0.$#)0-1

D-01.)4 &$'()'*+ ,"-.)/-$ ,0.$#)0-1

Ref: Hibbeler. Mechanics of Materials. 9th Edition. Pearson, 2014., Sect. 6.6

• In our original derivation, we found the mome moment nt on the the sect sectio ion n give given n by… M total

=

& A y 2 # ' $ E i i i ! R "! %$

• We noted I=!Aiyi2 was was the the mome moment nt of iner inerti tia, a, & proceeded with a constant E. • Let’ Let’ss inst instea ead d defi define ne IE=!EiAiyi2. • We can can thenwri thenwrite te… …

M total

• Reca Recall llin ing g from from befo before re… … • We can can thenwri thenwrite te… … • Whic Which h mean means… s…

! i

=

=

I E !

i

M total

1

=

=

n

R

E i

yi

…or…

R

! i

!

i

E i yi

=

1

=

E i E Re f

R

Option 1

I E E i yi M total yi E i I E

Option 2

• I t i s o ft ft en en c on on ve ve ni ni en en t t o n or or ma ma li li ze ze a ll ll Ei valu values es to one one of the the valu values es (Say,E (Say,E ref =Emin) … n E i

=

i

E Ref

• Ormore Ormore gene genera rall lly… y…

• We thendef thendefin inee In=!niAiyi2. • And And our our stre stress sses es becom become… e… Chart by Todd Coburn.

! i

=

M total yi I n

ni

" i =

P An

ni

+

M total (Y ! yi ) I n

ni 7


Pictures Courtesy of Pearson (Hibbeler’s Mechanics of Materials, 9th Edition).

&

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,$.9<(+ &$'()'* P'0-T1)1
DH0B/-$ E+ &26 8<'4$/. P//-)40.)<' :;I SOLUTION:

Basic Tabular Method for Section Properties& Bending Analysis w/ Multiple Materials Step 1: Idealize & Characterize Section (Break it into slices)

• Transform the bar to an equivalent cross section made entirely of brass

U Idealize segments for material and geometry. . Step 2: Compute Section Properties.

• Evaluate the cross sectional properties of the transformed section • Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar.

Step 3: Determine loading. Fig. 4.22a Composite, sandwich structure cross section.

Step 4: Determine locations of potentially critical stress levels.

U Wherever the y is greatest on area. . U Looks like top & bottom of section in this case. Step 5: Determine locations of potentially critical stress levels.

U Calculate Stresses. . Example:

f i

=

P An

ni

+

M (Y ! yi ) I n

ni

8


Bar is made from bonded pieces of steel ( E s = 29x106 psi) and brass ( E b = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied. Chart by Todd Coburn.

• Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity.

Chart developed from content provided by McGraw-Hill for [1].

69:

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DH0B/-$ E+ &26 8<'4$/. P//-)40.)<' :;I V8<'.O(W

DH0B/-$ J &26 8<'4$/. P//-)40.)<' :;I VP-.$#'0.$W Given

SOLUTION:

U U U

• Transform the bar to an equivalent cross section made entirely of brass. n

=

bT

E s E b

=

6

=

29 !10 psi 6

15 !10 psi

=

Find

1.933

0.4 in + 1.933! 0.75 in + 0.4 in

=

U U

2.25 in

• Evaluate the transformed cross sectional properties I Fig. 4.22b Bar length and height dimensions.

=

3 1 b h 12 T

=

1 12

Steel Core with ESt=29 Msi Bronze Plating with EBz=15 Msi M = 40 in-kip Max Stress in Steel Max Stress in Bronze

Solution

(2.25 in.)(3 in.)3

4

=

5.063 in.

• Calculate the maximum stresses Mc "

m

=

=

5.063 in.4

I

(" b )max (" s )max Chart by Todd Coburn.

(40 kip ! in.)(1.5 in.)

=

" m

=

n" m

=

" =

i

=

11.85 ksi

1.933 !11.85 ksi

(! b )max (! s )max

=

=

ni

+

M total (Y ! yi ) I n

ni

11.85 ksi 22.9 ksi


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P An

69;


<=

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DH0B/-$ F+ M)KK-$# DH0B/-$ =;EX

DH0B/-$ F+ M)KK-$# DH0B/-$ =;EX V8<'.O(W

P 4
_<-".)<' U `$ @)-- .#0'1C<#B .9$ 1$4.)<' )'.< <'$ B0($ $'.)#$-T
• Therefore,E Ref =Est, nst= 1 & nw=Ew/Est, &

=

=

E Ref

=

=

E st

U \9$ -<40.)<'
=

'1 '1 3 2$ 3 %&12 (0.15)(0.02) + ( 0.15)( 0.02)( 0.03638 ! 0.01) "# + %&12( 0.009)( 0.15) +( 0.009)( =

(

9.358 10

!6

)m

) $"

0.15 0.095 ! 00.03638

)(

2

#

4

U P//-T)'* .9$ C-$H"#$ C<#B"-07 .9$ '<#B0- 1.#$11 0. &O 0'( $ )1 2(0.17 ! 0.03638 ) " B'

" C

(

=

=

9.358 10!6 2(00.03638)

(

9.358 10!6

)

)

=

=

28.6 MPa

7.78 MPa (Ans)

U \9$ '<#B0- 1.#$11 )' .9$ @<<( 0. " )1 12 (28.56) 1.71 MPa (Ans) n! B ' ! B 200 =


Picture & Problem Courtesy of Pearson ( Hibbeler’s Mechanics of Materials, 9th Edition).

<<


=

=

<%

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DH0B/-$ F+ M)KK-$# DH0B/-$ =;EX VP-.$#'0.$W

DH0B/-$ I+ &$'()'*
Given U U U

Given

Wood Block with E W=12 GPa Steel Plate with ESt=200 GPa M = 2 kN-m

• Aluminum Section Shown. – EAl=10.0Msi

• Titanium Fail Safe Chord.

Find U U

– ETi=16.0 Msi

Max Stress in Wood Max Stress in Steel

• Mx=10,000 in-lb.

Solution Find • S tr ess es a t A , B , C , D , E, F.

" =

i

P ni An

+

M total (Y ! yi ) ni I n


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<6

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&$'()'* 5 >$)'C<#4$( 8<'4#$.$ &$0B1

DH0B/-$ I+ &$'()'*
Solution


• Some materials cannot take tension. • This necessitates additional effort to find the neutral axis.

‘

Basic Procedure: Determine Transformation Factor for Steel

U

Determine Neutral Axis of Transformed Area

( ) h ! nA st (d '!h' ) '

b h' 1

2

bh '

2

U Chart by Todd Coburn.

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2

+ nA st h '

=

E st =

E conc

0

! nA st d ' = 0

(solve using quadratic equation)

Proceed Using Composite Beam Approach

Pictures Courtesy of Pearson (Hibbeler’s Mechanics of Materials, 9th Edition). Chart by Todd Coburn.

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n

U

<&


&$'()'* 5 >$)'C<#4$( 8<'4#$.$ J DH0B/-$

&$'()'* 5 >$)'C<#4$( 8<'4#$.$ J DH0B/-$

\9$ #$)'C<#4$( 4<'4#$.$ K$0B 901 .9$ 4#<11J1$4.)<'0- 0#$0 01 19<@'; 3C ). )1 1"KZ$4.$( .< 0 K$'()'* B
_<-".)<' U \9$ .<.0- 0#$0
A st

=

(

2 ! 12.5

( ) (982) 25(10 )

)

2

2

=

982 mm7

.9"1

3

A'

=

nA st

200 10 =

3

2

=

7856 mm

U `$ #$a")#$ .9$ 4$'.#<)( .< -)$ <' .9$ '$".#0- 0H)1;

# y~ A

300 =

(h' )

300

0 400

h' 2

" 7856(400 " h') = 0

h'2 +52.37h'"20949.33 = 0 ! h' = 120.90 mm

7856mm2

U \9$ B
<8

=

2 '1 $ . 120.9 + 3 2 6 4 ) + 7856(400 ( 120.9 ) " = 788.67 !10 mm % (300)(120.9) + 300(120.9), - 2 * %&12 "#


Picture & Problem Courtesy of Pearson (Hibbeler’s Mechanics of Materials, 9th Edition).

<:

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&$'()'* 5 >$)'C<#4$( 8<'4#$.$ J DH0B/-$ _<-".)<' U P//-T)'* .9$ C-$H"#$ C<#B"-0 .< .9$ .#0'1C<#B$( 1$4.)<'7 .9$ B0H)B"B '<#B0- 1.#$11 )' .9$ 4<'4#$.$ )1 ' 1m $ " & 1000mm #

(

) conc

'

) conc

)max

=

=

=

mm

4

400

0.920 MPa (Ans)

1m 4 ) 1000mm ' 1m $ 60,000 Nm(400mm ( 120.9mm )% " & 1000mm # 1m 6 4 788.67 ! 10 mm( ) 1000mm 6

788.67 ! 10

Stress Concentrations

300

60,000 Nm(120.9mm )% (

7856mm2 =

21.23 MPa

U \9$ '<#B0- 1.#$11 )' $049
' st

=

n' 'conc

=

& 200(103 ) # $$ !21.23 169.84 MPa (Ans) 3 ! % 25(10 ) " =

Picture & Problem Courtesy of Pearson (Hibbeler’s Mechanics of Materials, 9th Edition).



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Plastic Bending Fig. 4.24 Stress-concentration factors for flat bars with fillets under pure bending.

Fig. 4.25 Stress-concentration factors for flat bars with grooves (notches) under pure bending.

Stress concentrations may occur: • in the vicinity of points where the loads are applied

Maximum stress:

• in the vicinity of abrupt changes in cross section Chart by Todd Coburn.

! m


=

K

Mc I


6 9 %<

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Recall from ARO326 (Lecture 7)…

!-01.)4 b$C<#B0.)<'1

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!-01.)4 b$C<#B0.)<'1

• For any member subjected to pure bending, the strain varies linearly across the section as follows… !

x

Fig. 4.27 Linear strain distribution in beam under pure bending.

y =

"

=

m

My

!

• When the maximum stress is equal to the ultimate strength of the material, failure occurs and the corresponding moment M U is referred to as the ultimate bending moment .

!

c

• If the member is made of a linearly elastic material , the neutral axis passes through the section centroid and we write… " x

Fig. 4.28 Material with nonlinear stress-strain diagram.


Fig. 4.29 Nonlinear stress distribution in member under pure bending.

I

• For a member with vertical & horizontal planes of symmetry & the same tensile & compressive stressstrain relationship, the neutral axis is located at the section centroid & the stress-strain relationship maps the strain distribution from the stress distribution.

• The modulus of rupture in bending, R B, is found from an experimentally determined value of M U and a fictitious linear stress distribution. R B

=

F b

=

F Rb

M U c =

I

• R B may be used to determine M U of any member made of the same material and with the same cross sectional shape but different dimensions. Fig. 4.30 Member stress distribution at ultimate moment M U .

Fig. 4.29 Nonlinear stress distribution in member under pure bending. Chart by Todd Coburn.


6 9 %$



6 9 %6

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,$BK$#1 ,0($
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>$1)("0- _.#$11$1 • Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough.

• Rectangular beam made of an elastoplastic material Mc

! " Y

" x

" m

=

" m

=

M Y

" Y

I I

=

c

" Y

=

maximum elastic moment

• Since the linear relation between normal stress and strain applies at all points during the unloading phase, it may be handled by assuming the member to be fully elastic.

• If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core. & y 2 # M 3 M Y $1 ' 1 Y ! yY elastic core half - thickness 2 $ 3 c2 ! % " =

• Residual stresses are obtained by applying the principle of superposition to combine the stresses due to loading with a moment M (elastoplastic deformation) and unloading with a moment -M (elastic deformation).

=

• As the moment increase, the plastic zones expand, and at the limit, the deformation is fully plastic. Fig. 4.33 Bending stress distribution in a beam for: (b) yield impending, M = My, (c) partially yielded, M > My, and (d) fully plastic, M = Mp. Chart by Todd Coburn.

M p k

=

3 M 2 Y

M p

=

=

M Y

Fig. 4.37 Elastoplastic material stress-strain diagram with load reversal.

plastic moment

=

shape factor (depends only on cross section shape)



6 9 %7

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&$'()'* 5 3'$-01.)4 c !-01.)4

! all

e

=

=

I

1

bh

So…

M all

=

e

e

=

bh

3

2

Find

2

bh

6

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6 9 %&

Ref: Bruhn. Analysis & Design of Flight Vehicle Structures. 2nd Ed. 1973., Sect. C3.4

7075-T6 Al Die Forging U Ftu=75 ksi U Fty = 65 ksi

6 M all

12

1


Given

Plastic Bending: M all c

• The final value of stress at a point will not, in general, be zero.

DH0B/-$ !#
Elastic Bending:

& h # M all $ ! e % 2 "


Ref. [7], Fig. 6-48 (c)

! all

U U

Max Elastic Moment Max Plastic Moment assuming Elasto-Plasto Material

Ref. [7], Fig. 6-48 (d)

Solution

Ref. [2], Fig. 6-48 (a)

Plastic Bending Shape Factor: M ult p

all b

M all e

k

=

Ref. [2], Fig. 6-48 (b)

h

! =

1

bh

!

=

=

1.5

4

M ult

p

p =

=

all

M ult p

M ult

M all

6

4

2

6

Ref. [7], Fig. 6-48 (e)

Ref. [7], Fig. 6-48 (f)

2

1.5

=

[ F ult ( A)]& $% h # "! p

Elastic Allowable:

2

, & h #)& h # *- all $ b !'$ ! % 2 "(% 2 " +

=

all b

h

2

Elasto-Plasto Assumption:

!

4

for rectangular sections

e


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%8

Ref: Bruhn. Analysis & Design of Flight Vehicle Structures. 2nd Ed. 1973., Sect. C3.4

DH0B/-$ !#
!"#$%& What is the maximum stress this section can withstand? Ftu What is the maximum elastic moment this section can withstand? Mmax_el = (I/c) Ftu

7075-T6 Al Die Forging U Ftu=75 ksi U Fty = 65 ksi U U

%:

8<'4$/."0- d"$1.)<'1

Given

Find


Max Elastic Moment Max Plastic Moment assuming Elasto-Plasto Material

Solution

What is the maximum elasto-plastic moment a square section can withstand? Mmax_EP = 1.5 (I/c) F tu Elastic Allowable: Elasto-Plasto Assumption:


%;


% 9 $=

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8<'4$/."0- d"$1.)<'1

D'Z
If you wish to determine the maximum elasto-plastic moment this section can withstand, what is the best way to idealize the thing for computing properties?

Why?


% 9 $<

!"#$%&

>$C$#$'4$1 2 &)K-)<*#0/9T 1.

Beer , Johnson, DeWolf, & Mazurek. Mechanics of Materials. 7th Edition. McGraw Hill. 2015.

2.

Hibbeler. Mechanics of Materials. 9 th Edition. Pearson, 2014.

3.

Shanley. Strength of Materials. McGraw-Hill. 1957.

4.

Bruhn. Analysis & Design of Flight Vehicle Structures. S.R. Jacobs & Associates. 1973.

5.

Peery & Azar. Aircraft Structures. 2nd Edition. McGraw-Hill . 1982.

6.

Budynas & Nisbett. Shigley’s Mechanical Engineering Design. 9 th Edition. McGraw Hill. 2011.

7.

Roark, Young, Budynas, & Sadegh. Roark’s Formula’s for Stress & Strain, 8th Edition. McGraw Hill. 2012.

8.

Ugural & Fenster. Advanced Strength & Applied Elasticity. 4th Edition. Prentice Hall, 2003.


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Pure Bending – Multiple Materials & Plastic Bending

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