CMats Lect2-Normal Bending Stress and Strain [Compatibility Mode]

8/3/2019 CMats Lect2-Normal Bending Stress and Strain [Compatibility Mode]

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Civil Engineering Materials 267Stresses in Materials


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Civil Engineering Materials 267 (Stresses)

Mechanics of Materialss u y o e re a ons p e ween e ex erna oa s on a

body and the intensity of the internal loads within the body.

Structural Analysis

moments, shear forces, axial forces, torsion moments) on.

Stresses in Material

Uses the member actions determined by

stresses set up within a body to resist the applied

Lecture 2 2

oa ng an o e o y oge er.


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Normal Stresses Due to Axial Forces

axial = a hi l x min E f f

en ng oments

en ng , E, fy, fu still valid

Lecture 2 3


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Normal Stresses due to Bending

Need to have an understanding of

Properties of Areas to be able to determine thestresses due to bendin .

Centre of Gravity

Second Moment of Area (I value)

Elastic Section Modulus;(Z value)

Lecture 2 4


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Centre of Gravity, Centroid What are they?

Where are they? Geometric centre for the area?

The centre of gravity of abody is the point where the

entire wei ht of the bod

appears to be concentrated.

the Centroid.

Lecture 2 5


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The Centroid of a body is the point where the

entire weight of the body appears to be

concentrated.

Does the Centroidal axis

40mm

mm

o or anUnsymmetrical axis

3

0

So, where is it?40mm

300mm

How do we find it?

Lecture 2 6


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en ro how do we find it?

F=0 a ance o orces to prevent t e o y rom mov ng. eg1: Axial stress * Cross sectional area of section = Applied axial force =

M=0 Balance of moments to prevent the body from rotating. eg1: Total applied moment action of a beam = internal stresses*area*lever arm

Shortcut equ r um sa s e a ong axes o symme ry equal forces and moments on either side of a symmetrical axis

Centroid sits somewhere alon the axis of s mmetr

Lecture 2 7

If there are two perpendicular axes of symmetry, the centroid is located at theintersection of the two axes


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ass xerc se:

Find location of centroidal axes for this t-beam

y

(ie: find centroid)yA

m

balancing t-shaped section on finger:

x

40mm

300 Need to find the point where (weight of each

element multiplied by its distance from thecentroid = zero

40mm300mm

ie Mcentroid = 0 (section in equilibrium)

Difficult to do when centroid location isnt yetknown.

Better method: Take moments about anotherarbitrary point (M=0 at every point).Usually use end of section

-

Weight of

section

Weight of

section

Lecture 2 8


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Class Exercise: Find location of centroid for this T-beam


y-axis is along axis of symmetry

x-axis: consider equilibrium to determiney y

x

A

wt(1)componentofWeight

;0M

1

AA

=

=

40mm00mm

area*densitMaterialw

wtwtactionRewt(2)componentofWeight

21

2

=

+== F = 0

300mm A)2(componentofarea

A)1(componentofarea

2

1

=

=

A

2com onentofcentroidtolanefromcetandis

y)1(componentofcentroidtoplanefromcetandis

ysectionholewofcentroidtoplanefromcetandis

1AA

AA

=

=

=

Weight of Weight of ]y*wty*wt[0]y*wty*wt[y*actionRe

2211

2211

+=

=+ M = 0

Reactionsection section

]AA[DensityMaterial

]y*Ay*A[DensityMaterialy

actionRe

21

2211

+

+=

Lecture 2 9

Refer to Engineering Mechanics 100 notes (Lecture 6B & C) (2004), located on Blackboard,for proper calculus derivation of centroid location (pp 10-21).


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Class Exercise: Find location of centroidal axis for this t-beam


y

x

A Area of eachcomponent of the

y

( )= yAy . dist. of each00mm

dist. to section

to arbitrary plane

300mm

Acentroid

Sum of area of components

Weight of Weight of

Reactionsection section

Refer to Engineering Mechanics 100 notes (Lecture 6B & C) for

Lecture 2 10

proper ca cu us er vat on o centro ocat on.


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m ar y, resu an o norma s resses rom an

axial load acts through the centroid (in orderto satisfy M=0)

= A.40mm

00mm

Stress in sectiondue to axial loads

x

40mmy

(acting through centroid)Stress in section

300mmA A due to axial loads

Lecture 2 11

Ci il E i i M i l 267 (S )


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,

about the x-axisy

x

1b1

d

componentAyy =1

2y1component

x x

b2 2

3y2

y3 d3

y3

y

***

)y*d*b()y*d*b()y*d*b(y 333222111

++=

Lecture 2 12

311

Ci il E i i M t i l 267 (St )


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,

about the y-axisy x1

x

1b1

d

componentAxx =1

2xcomponent

x x

b2 2x2

3x3

3 d3

y

***

)x*d*b()x*d*b()x*d*b(x 333222111

++=

Lecture 2 13

3211

Ci il E i i M t i l 267 (St )


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,What is an I-value?What does it mean?Why do we need to use it?

Imagine a rectangular piece of timber.1

Which is the best orientation if using it as a beam?2

Why would you orientate it this way? Less deflection Lower stresses induced thus is stronger in this orientation

I-value is a measure of the geometric stiffness of a shape.

I-value allows us to quantify deflections and stresses

Lecture 2 14

depending upon the orientation of the beam.



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,

,refer to the bending axes.

y

,shown here, it is referred to as bending about itsstrong axis (usually referred to as the x-axis)

Picture grabbing the x-axis with your hand and rolling itforward (bending about the axis). The beam will bend,

xx

,bottom, the same as it you applied a load to the top of asimply supported beam.y

We calculate Ixx (second moment of area about the x-axis) in order to quantifybehaviour deflections and stresses when bendin about the x-axis.

Similarly, we calculate Iyy (second moment of area about the y-axis) in orderto quantify behaviour when bending about the y-axis.

Lecture 2 15



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,

n r l f rm l f r

+= AhI

calculating I value:made u of a series of

rectangular elements)y Where:

x b

,an individual rectangular element

d = dimension perpendicular to axis being

xxd

cons ere , o an n v ua rec angu ar e emen

A = area of the individual rectangular element

h = the dimension between the centroid of the

individual rectangular element and the centroid ofthe whole section (must know location of centroidin order to calculate h and thus I value

Proper calculus proof given in

En ineerin Mechanics 100

Note the significance of the depth in the formula(being cubed). This indicates that a minor

Lecture 2 16

Lecture 6B&C notes (pp 22-32).It shows why the depth is cubed.

increase in depth provides the section withsignificantly greater stiffness and strength.



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,

3y

+= 2Ah12

Ix

b

3 0

xxd

+

= Ah12

I

3

gletanrec(about its centroid)

=12

y

Lecture 2 17



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,

General equation 23bd

about the x-axis

3*

12

yx

1b1 +=

11111x yy*)d*b(12I

12

1

y1

++2

222

22 yy*)d*b(12

2b2

d2

++2

333

33 yy*)d*b(12

d*b

x xy2

3 d3y3 Must know location of centroid first

Lecture 2 18

yb3



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,

General equation 23bd

about the y-axis

3

12

yx

1b1

1

+=2

11111

y xx*)d*b(12

I

12

1

x

++2

222

22 xx*)d*b(

12

b*d

2b2

d2x2

++

2

333

3

33 xx*)d*b(12

b*d

xx3

3 d3 Must know location of centroid first

Lecture 2 19

yb3



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,

Tricks for doubly symmetric sections about the x-axis

y 23

11 **d*b

+= 2Ah12

bdI

x b1

++

2

3

33

111x

*d*bd*b

12 Ix,flanges

1 d1

++

2

3

22 0*)d*b(d*b

12

b2 d2

x,we

b1=b3, d1=d3, y1=y3y3

3 d3

+

+=

12

d*by*)d*b(

12

d*b*2I

3

222

111

3

11x

Lecture 2 20

3



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,

Tricks for doubly symmetric sections about the y-axis

+= 2Ah12

bdI

y

For the I-value about the y axis, all of therectangle elements are centred about the

x b1

,

rectangle can be added together:

**33

Iv,flangesd1 +

=

*

1212I

3

3311y

b2 d2

+12 y,web

b1=b3, d1=d3

+

=

12

b*d

12

b*d*2I

3

22

3

11yd3

Lecture 2 21

3



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Elastic Modulus Z When external axial loads When external bending moments are considered,

,

internal resisting stressesare determined based on

couple. It therefore follows that we need a firstmoment of area to help determine the stresses in

e cross sec ona area othe section.

a eam su ec e o a momen .

Mbendin =

(Nmm)

Aaxial =

(MPa)(mm2)

(MPa)(mm3)

Compressive stress c

M

everarm

Lecture 2 22Tensile stress t



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x

xZ=

Elastic Section Modulus Zy

x

yProof given inEngineering Mechanics 100

Lecture 6B&C notes (pp 33-35)For a rectangle:

dbd3

3

xxd

2yan

12maxx ==

bdd

12Z2

gletanrec ==

2y

But for all other sections, must calculate I first, then determine Z

Lecture 2 23



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orma tresses ue to en ng

Consider a beam subject

o en ng

c

Bottom of beam fibres stretch tensile strainTop of beam fibres compress compressive strain

tomewhere between these two regions there must be a plane

in which the longitudinal fibres will not undergo a change in

Lecture 2 24

eng . s p ane s re erre o as e neu ra ax s .



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g g ( )

u

a relationship needs to be developed between the appliedbendin moment and the lon itudinal stress distribution in a

beam developed to resist/support the external moment.

Beam theory assumes small deflections and strains

the material behaves in a linear-elastic manner

Hookes law applies ie: = E cbe a consequence of a linear

Lecture 2 25

t



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g g ( )b

ax. ompress ve s ress c

CF=0 = N.A. rarm

Lev

e

T

We have a linear stress distribution. (These internal stresses are induced

. t

The resultant of the internal tensile and compressive stresses are atens on an compress on orce w c orm a coup e to res st t eexternally applied moment (thus M=0 is satisfied)

Lecture 2 26

The tension and compression forces must be equal to satisfy F=0.(ie: T = C)



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g g ( )b ax. ompress ve s ress c

C e pos on o e neu ra

axis can be located bysatisf in the condition that N.A. r

arm

the tension and compressionforces are equal. L

ev

e

For a rectangular section it

can be easil seen that the

T

neutral axis will occur at themid-height of the beam.

. t

The location of the neutralaxis occurs at the centroidalaxis for all sections underelastic loading.

Lecture 2 27



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g g ( )b

ax. ompress ve s ress c

C =ie: C=T

*N.A.

rarm

=Top half of beam has compressive stressCompression force

Lev

e

=av. compr. Stress*area of top half of beam

T

. t

2*b*

2C c

=

Force acts closer to top of beam (more stress at the top of the beam)1 3

triangle from the base of the triangle).

Lecture 2 28



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bax. ompress ve s ress c

C

N.A.rarmd

*b*C c

=

Lev

e

22

T

M=0Internal moment (= C*lever arm)

. t

= Externally applied bending moment (=M)

Lever arm = =+

Lecture 2 29



30/37

b. c

CM=0N.A.

rarm

Internal moment= armlever*C

Lev

e

=3

d2*

d*b*

2

cT

=

= Moment)(ExternalMbd2

c

. t

= bd

Z2

letanrec M =

For a doubly symmetric section

Lecture 2 30

(such as this) t = c = max Beam Equation



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Normal Bending Stresses:From Structural Anal sis:

General Beam Equation: ==

My=

I

Imax

max =

IZ= M=We know:As determinedin previous

maxy Z examp e

Lecture 2 31

resses are re a e o va ue



32/37

Normal Bending Stresses:From Structural Anal sis:

General Beam Equation: ==

M1

EIR

)ttancons(=

Constant defined by load application pattern and beam length.

EI

wL

384

5:eg =

Lecture 2 32

Deflections are related to I value(and E).



33/37

y

x

topZ =about the x-axisx

fibretop

1max top

fibreottombbottom y

=2ytop fibrey

xyM

=

xx

ybottom fibrex

= dist. from neutral axis to a oint

3

on the section in direction of y-axis

y

x

tomax

= xottombmax =

Lecture 2 33

top bottom



34/37

y yI=

about the y-axis

x fibreHSL

I

x1

fibreHSRRHS x

=2x

yxM

=

x xxRHS fibrexLHS fibre

y

x= dist. from neutral axis to a oint

3on the section in direction of x-axisy

max RHS

y=max LHS

y

HSRmax=

Lecture 2 34

LHS RHS



35/37

xxZ =

Important Formulae:

maxymax= dist. from neutral axis to

componentAy

y

=

extreme fibre of section in directionof y-axiscomponent

My

=3

I += 12 = dist. from neutral axis to a ointon the section in direction of y-axis

M

=

bdZ

2

gletanrec =

Lecture 2 35

max



36/37

Stresses due to bending

(moments)

Independent of materials used

Proportional to stresses

Dependent on the material used E value Low E-value not a very stiff material

hi h strains hi h deflections

Lecture 2 36

Civil Engineering Materials 267 (Stresses)10


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10

400

40

0

y

Class exercise/tutorial

120

x x

y x x

10mm wallthickness

x x

20

Whi h i n will h v

80y

100y

all aroundy

10010

the:

Section 1 Section 2 Section 3x-

smallest Ix-value?

y 10 y0

y32 greatest Iy-value?

28

0

x x 260

x x

1032

x x

y-

y

350

10 y

350

20y

100

32

Lecture 2 37

Section 4 Section 5 Section 6

CMats Lect2-Normal Bending Stress and Strain [Compatibility Mode]

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