CMats Lect2-Normal Bending Stress and Strain [Compatibility Mode]
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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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Civil Engineering Materials 267 (Stresses)
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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Civil Engineering Materials 267 (Stresses)
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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Civil Engineering Materials 267 (Stresses)
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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Civil Engineering Materials 267 (Stresses)
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
Civil Engineering Materials 267 (Stresses)
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
Civil Engineering Materials 267 (Stresses)
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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Civil Engineering Materials 267 (Stresses)
,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
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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
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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).
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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
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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
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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
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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
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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
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