Pure Bending – Multiple Materials & Plastic Bending

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LECTURE LECTURE

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Bending

of

Composite Sections

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be

be written…written…

0 0 L L ! ! " " ==

•• AA clclososerer lolookok atat ththee dedefoformrmatatioionn o

off tthhee bbeeaamm aass aa ffuunnccttiioonn ooff R R provides… provides… •• WWee c ac ann s es eee t ht ha ta t…… R R L L y y 0 0 = = ! ! R R y y L L00 == ! ! •• SSoo…… R R y y = = ! !

•• ThThee foforcrcee onon ananyy elelememenentalstrtalstripip ofof mamateteririalal isis…… P P ii==! ! iiAAii

•• ItItss momomenmentt ababouttheoutthe neneututrarall axiaxis…s… M M ii== P P ii y yii==! ! ii A Aiiyyii ii ii ii ii total

total M M A Ayy

M

M =="" ==_"_"! !

•• SothetotSothetotalmomalmomenentt isis……

•• FoForr momomenmentsts inin ththee elelasastiticc rarangnge…e…" " ii==E E ii ! ! ii

•• CoCombmbinininingg wiwithth ourstrourstraiain…n…

R R y y E E ii ii ii== ! ! • • InInsesertrtiningginintotoouourrtototatallmomomementnt&&rereararrarangngining…g… _!!!! "" ## $$ $$ %% && ' ' = = R R y y A A E E M M ii ii ii total total 2 2 • • N oN ot it in gn gI=I=!!_A_A i iyyii22isisththeemomomementntofofininerertitia,a,ififEEisisconconststanant…t… _R_R E E I I M M total total = =

•• CoCombmbinininingg wiwith…th… ii yyii

R R E E = = !

! …&…& rearrarearranging…nging…

I I y y M M total total ii ii== ! ! W Wee find…find…

I I y y M M total total maxmax max max== ! ! max max y y I I M M all all all all ! ! = = $ $

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I I y y M M total total maxmax max max== ! ! max max y y I I M M all all all all ! ! = = 6 6

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momementnt onon ththee sesectctioionn gigivevenn by…by…

•• ReRecacalllliningg frfromom bebefoforere……

R R y y E E ii ii ii== ! !

•• WWee cacann ththenwrienwritete……

!! !! "" ## $$ $$ %% && ' ' = = R R y y A A E E M M ii ii ii total total 2 2

•• WWee nnootteedd II==!!_A_A

iiyyii22wawass ththee momomementnt ofof ininerertitia,a, &&

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ii E E ii total total ii _I_I E E

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•• LeLet’t’ss ininststeaeadd dedefifinene IIEE==!!EEiiAAiiyyii22..

R R I I M M E E total total 11 = = …or… …or… R R y y E E ii ii ii 11 = = ! !

•• WWee cacann ththenwrienwritete……

ii ii ii E E total total y y E E I I M M ! ! = =

•• WhWhicichh memeanans…s…

•• I tI t i si s o fo ft et enn c oc on vn ve ne ni ei en tn t t ot o n on or mr ma la li zi zee a la lll EEiivavalulueses toto ononee ofof ththee vavalulueses

(Sa

(Say,Ey,Eref ref =E=Eminmin)) ……

Ref Ref E E E E n n ii ii = =

•• WWee ththendefendefininee IInn==!!nniiAAiiyyii22..

ii n n ii total total ii nn I I y y M M = = ! !

•• AnAndd ouourr ststreresssseses bebecomcome…e…

•• OrmorOrmoree gegeneneraralllly…y…

( ( )) ii n n ii total total ii n n ii nn I I y y Y Y M M n n A A P P !! + + = = " "

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Ref: Hibbeler.

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

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Basic Tabular Method for Section Properties& Bending Analysis w/ Multiple Materials

Step 2: Compute Section Properties.

Step 1: Idealize & Characterize Section (Break it into slices)

Step 3: Determine loading.

ULooks like top & bottom of section in this case.

UWherever the y is greatest on area..

Step 4: Determine locations of potentially critical stress levels.

Step 5: Determine locations of potentially critical stress levels.

UCalculate Stresses.. Example: ( ) i n i i n i _I n y Y M n A P f ! + =

UIdealize segments for material and geometry..

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SOLUTION:

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

• 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.

• 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.

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.

Fig. 4.22a Composite, sandwich structure cross section.

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

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• Evaluate the transformed cross sectional properties

(

)( )

4 3 12 1 3 12 1 in. 063 . 5 in. 3 in. 25 . 2 = = = b h I _T SOLUTION:

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

in 25 . 2 in 4 . 0 in 75 . 0 933 . 1 in 4 . 0 933 . 1 psi 10 15 psi 10 29 6 6 = + ! + = = ! ! = = T b s b E E n

• Calculate the maximum stresses

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)( )

ksi 85 . 11 in. 5.063 in. 5 . 1 in. kip 40 4 = ! = = I Mc m " ( ) ( )_max 1.933 11.85ksi max ! = = = m s m b n" " " " ( ) ( ) 22.9ksi ksi 85 . 11 max max = = s b ! !

Fig. 4.22b Bar length and height dimensions.

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Given

Solution Find

U Steel Core with E_St=29 Msi U Bronze Plating with EBz=15 Msi

U M = 40 in-kip U Max Stress in Steel U Max Stress in Bronze

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)

i n i total i n i n I y Y M n A P ! + = "

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

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

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Given

Solution Find

U Wood Block with EW=12 GPa

U Steel Plate with E_St=200 GPa U M = 2 kN-m

U Max Stress in Wood U Max Stress in Steel

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Given

Find

• Aluminum Section Shown.

– E_Al=10.0Msi

• Titanium Fail Safe Chord.

– E_Ti=16.0 Msi

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

• Mx=10,000 in-lb.

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1D Solution

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conc st E E n = Basic Procedure:

U Determine Transformation Factor for Steel U Determine Neutral Axis of Transformed Area

( ) ( )' ' 0 2 ' ' !nA d !h = h h b _st ‘

(solve using quadratic equation) 0 ' ' ' 2 1 2 = ! +nAh nAd bh _st _st

U Proceed Using Composite Beam Approach

• Some materials cannot take tension.

• This necessitates additional effort to find the neutral axis.

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Stress Concentrations

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

loads are applied

I Mc K

m=

!

• in the vicinity of abrupt changes in cross section

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.

Maximum stress:

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Plastic Bending

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

m x c y ! ! =_"

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

I My

x=!

"

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

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

Fig. 4.28 Material with nonlinear stress-strain diagram.

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

Recall from ARO326 (Lecture 7)…

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• 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 .

• R_Bmay be used to determine M _Uof any member made of the same material and with the same cross sectional shape but different dimensions.

• The modulus of rupture in bending, R B, is found

from an experimentally determined value of M _U and a fictitious linear stress distribution.

I c M F F R U Rb b B= = =

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

Fig. 4.30 Member stress distribution at ultimate momentM U .

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• Rectangular beam made of an elastoplastic material

moment elastic maximum = = = = ! Y Y Y m m Y x c I M I Mc " " " " " "

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

= _Y Y y_Y c y M M

• As the moment increase, the plastic zones expand, and at the limit, the deformation is fully plastic.

shape) section cross on only (depends factor shape moment plastic 2 3 = = = = Y p Y p M M k M M

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.

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• Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough.

• 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.

• 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).

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

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

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Elastic Bending: I c M e all all = ! 2 6 bh M e all = 3 12 1 2 bh h M e all ! " # $ % & = all all bh M e ! 2 6 1 = So… Plastic Bending: ( )

[

]

! " # $ % & = 2 h A F M p p ult ult 4 2 h b all ! = ! " # $ % & ' ( ) * + , ! " # $ % & = 2 2 h h b M ult p - all e p all ult M M 4 6 = =1.5 e p all ult M M

k = for rectangular sections

Plastic Bending Shape Factor:

all all bh h b ! ! 2 2 6 1 4 = 5 . 1 =

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

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

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

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

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,0.$#)0-Ref: Bruhn. Analysis & Design of Flight Vehicle Structures. 2nd Ed. 1973., Sect. C3.4

Given Solution Find 7075-T6 Al Die Forging U Ftu=75 ksi U F_ty= 65 ksi

U Max Elastic Moment

U Max Plastic Moment assuming Elasto-Plasto Material

Elastic Allowable: Elasto-Plasto Assumption:

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

Given Solution Find 7075-T6 Al Die Forging U F_tu=75 ksi U Fty= 65 ksi

U Max Elastic Moment

U Max Plastic Moment assuming Elasto-Plasto Material

Elastic Allowable: Elasto-Plasto Assumption:

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What is the maximum stress this section can

withstand?

F

_tu

What is the maximum elastic moment this

section can withstand?

M

max_el

= (I/c) F

tu

What is the maximum elasto-plastic moment a

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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?

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

2. Hibbeler. Mechanics of Materials. 9th_{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. 9th_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.

Pure Bending – Multiple Materials & Plastic Bending

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Cal Poly Pomona

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Bending

Bending

of

of

Composite Sections

Composite Sections

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