BENDING STRESSES IN BEAMS UNIVERSAL COLLEGE OF ENGINEERING AND TECHNOLOGY.
Files-2-Lectures LEC 06 Shear Stresses for Beams in Bending
Transcript of Files-2-Lectures LEC 06 Shear Stresses for Beams in Bending
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Chapter 4
Load and Stress AnalysisLoad and Stress Analysis
A. Bazoune
Shear Stresses for BeamsShear Stresses for Beams in Bendingin Bendingecec..
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4.84.8Shear Stresses forShear Stresses forBeams in BendingBeams in Bending
Most beams have both shear forces and bending momentpresent.
It is only occasionally that we encounter beams subjected, .
The flexural formula was developed on the assumption ofpure bending in order to eliminate complicating effects ofshear force.
For engineering purposes, the flexure formula is valid nomatter whether the shear force is present or not.
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Fig. 4.20 showsa beam ofconstant crosssection
subjected to ashear force Vand a bendingmoment M
Figure 4-20 Beam section isolation.Note: Only forces in x-direction are
shown on the dx element.
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Associate the hollowAssociate the hollowvector with yourvector with yourright hand.right hand.
If you place theIf you place thethumb of our ri htthumb of our ri ht
Figure 4-20 Beam section isolation.Note: Only forces in x-direction are
shown on the dx element.
hand in thehand in the negativenegativezz-- directiondirection, then, thenyour fingers, whenyour fingers, whenbent, will indicatebent, will indicatethethe direction ofdirection of
moment Mmoment M..
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Shear force andShear force andbending moment arebending moment arerelated by:related by:
dMV =
Figure 4-20 Beam section isolation.Note: Only forces in x-direction are
shown on the dx element.
dx
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At some pointalong the beam,we cut transverse
section dx at a
distancey1 abovethe neutral axis.
We remove thissection to studythe forces that acton it.
Figure 4-20 Beam section isolation.Note: Only forces in x-direction are
shown on the dx element.
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Because shear
force is present,the B.M. ischanging as wemove along the x-
.
We can designatethe B.M. as M onthe Near Side of
the section and as(M + dM) on theFar Side.
Figure 4-20 Beam section isolation.Note: Only forces in x-direction are
shown on the dx element.
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The moment M
produces a normalstress , and themoment (M + dM),a normal stress
+ d).
The compressive force on the far side being
greater than on the near side.
The Resultant of these two would cause thesection to tend to slide in the negative x-direction.
These normalstresses producenormal forces onthe vertical faces of
the element.
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This resultant mustbe balanced by a
shear force actingin the positive x-direction on thebottom of the
For the near face
(b)
where the limits indicate that we integrate from y = y1 to y = c.
sec on.
This shear forceresults in a shearstress.
1
c
Ny
F dA=
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For ,
Eq. (b) becomes
(b) (c)
y I=
c
N
MF y dA=
For the Far side
1y
( )1 1
c c
Fy y
M dMF d dA y dA
I
+= + = (d)
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The force on thebottom face is theshear stress timesthe area of thebottom face
(e)BF bdx=
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In this equation, the integralis the first moment of thearea of the isolated verticalface about the neutral axis.
1
c
yy dA Q=
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The moment is usually designated Q
==c
yAydAyQ
1
''(4-30)
V Q
I b = (4-31)
where, for the isolated area fromy1 to c, is the distance from they-axis to the area centroid andA is the area.
Finally, we write
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Shear Stresses in StandardShear Stresses in Standard--SectionSectionBeamsBeams
The shear stress distribution in a beam depends on how Q/b
varies as a function of y1.
g. - s ows a por on o a eam w a rec angu ar cross-
section.
It is subjected to shear force V and a bending Moment M.
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Figure 4-22Shear Stress in a rectangular beam
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A normal stress is developed on a cross-section such as A-A due toM.
Section A-A is in compression above the N.A. and in tension below.
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Consider an element of area dA at a distancey above the N.A. such that
dA bdy=
Eq. (4-30) becomes
( )1 1
2
2 2
1
12 2
cc c
y yy
by bQ y dA b y dy c y
= = = =
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Substitute this value into Eq.(4-31)
From Table A-18 and for a rectangular cross-section, we haveSubstitute and gives
( )2 212
Vc y
I = (4-32)
2
3
AcI=
3
12
bh
I=A bh= 2h c=
(b)
s ng s va ue or n q. - g ves
The maximum shear stress exists when y=0, which is at the bendingneutral axis. Thus
2
1
2
31
2
V y
c
=
(4-33)
(4-34)max
3
2
V =
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As we move away from the neutral axis, the shear stress
decreases parabolically until it is zero at the outer surfacewhere y = c .
Notice that the shears ress s max mum a
the bending neutralaxis, where thenormal stress due tobending is zero andthat the shear stress
is zero at the outersurfaces, where thebending stress is amaximum.
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Particular Case:Particular Case:Shear Stresses in Standard Section BeamsShear Stresses in Standard Section Beams
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Example 4Example 4--7 Textbook7 Textbook
AA beambeam 1212 inin longlong isis toto supportsupport aa loadload ofof 488488 lblb..ff actingacting 33 inin fromfrom thethe leftleftsupportsupport asas shownshown inin FigFig-- 44--2121 aa.. BasingBasing thethe designdesign onlyonly onon bendingbending stress,stress, aa
designerdesigner hashas selectedselected aa 33--inin columncolumn channelchannel withwith thethe crosscross--sectionalsectionaldimensionsdimensions asas shownshown.. IfIf thethe directdirect shearshear isis neglected,neglected, thethe stressstress beambeam maymay bebeactuallyactually higherhigher thanthan thethe designerdesigner thinksthinks.. DETERMINEDETERMINE thethe principalprincipal stressesstressesconsideringconsidering bendingbending andand directdirect shearshear andand comparecompare thenthen withwith thatthat consideringconsideringbendingbending onlyonly..
Figure 4Figure 4--2121..
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Example 4Example 4--7 Contd7 Contd
TheThe loading,loading, shearshear forceforce andandbendingbending momentmoment diagramdiagram areare
shownshown inin FigureFigure44--2121(b)(b)..
IfIf thethe directdirect shearshear forceforce isisincludedincluded inin thethe analysis,analysis, thethe
andand bottombottom ofof thethe beambeam willwillbebe thethe samesame asas if if onlyonlybendingbending werewere consideredconsidered..TheThe maximummaximum stressesstresses areare::
Figure 4Figure 4--21 (b)21 (b)..
( )1098 1.5992
1.66psi
Mc
I
=
= =
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Example 4Example 4--7 Contd7 Contd
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Example 4Example 4--7 Contd7 Contd
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yy QQ ss tt ss11,s,s
22
inin inin33 psipsi psipsi psipsi
00 0.6530.653 00 847847 847,847, --847847
.. .. ,, --
0.50.5 0.6310.631 331331 818818 670,670, --10001000
0.750.75 0.6050.605 496496 785785 575,575, --10711071
11 0.5680.568 661661 737737 477,477, --11381138
1.2271.227 0.5250.525 812812 681681 387,387, --12001200
1.51.5 00 992992 00 0,0, --992992