chapter six shearing stresses in beams and thin-walled members
DESCRIPTION
Chapter Six Shearing Stresses in Beams and Thin-Walled Members. 6.1 Introduction. -- In a long beam, the dominating design factor: . -- Primary design factor. -- Minor design factor. [due to transverse loading]. -- In a short beam, the dominating design factor: . - PowerPoint PPT PresentationTRANSCRIPT
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Chapter Six
Shearing Stresses in Beams and Thin-Walled Members
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6.1 Introduction
-- In a long beam, the dominating design factor:
mMcI
( )xyVaverageA
-- Primary design factor
-- Minor design factor
-- In a short beam, the dominating design factor:
32maxVA
[due to transverse loading]
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y component: xydA V 0z component: xzdA
Equation of equilibrium:
-- Shear stress xy is induced by transverse loading.
-- In pure bending -- no shear stress
(6.1)
(6.2)
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Materials weak in shear resistance shear failure could occur.
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6.2 Shear on the Horizontal Face of a Beam Element
0 0: ( )x D CF H dA A
MyI
C DI I I Knowing and
D CM MH ydAI
AWe have (6.3)
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Since dMVdx
( / )D CM M M dM dx x V x
VQH xI
H VQqx I
Therefore,
and
(6.4)
(6.5)
Defining Q ydA
= shear flow
= horizontal shear/length
here Q = the first moment w.t.to the neutral axis
Q = max at y = 0
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6.3 Determination of the Shearing Stresses in a Beam
aveH VQ xA I t x
aveVQIt
VQH xI
(6.6)
= ave. shear stress
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xy = 0 at top and bottom fibers
Variation of xy < 0.8% if b h/4
-- for narrow rectangular beams
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6.4 Shearing Stresses xy in Common Types of Beams
-- for narrow rectangular beams
xyVQIt
2 21 12 2
( ) ( ) ( )Q Ay b c y c y b c y
1 ( )2
y c y
t = b (6.7)
Q Ay
Also,3
3212 3bhI bc
Hence,2 2
334xy
VQ c y VIb bc
(6.8)
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Knowing A = 2bc, it follows
2
23 12
( )xyV yA c
maxxy
0xy
(6.9)
This is a parabolic equation with
@ y = c
@ y = 0 -- i.e. the neutral axis
At y = 0, max32VA
(6.10)
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aveVQIt
maxweb
VA
Special cases:
American Standard beam (S-beam)
or a wide-flange beam (W-beam)
-- over section aa’ or bb’
-- Q = about cc’
(6.6)
(6.11)
For the web:
For the flange:
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6.5 Further Discussion of the Distribution of Stresses in a Narrow Rectangular Beam
2
23 12
( )xyP yA c
xPxyI
(6.12)
(6.13)
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6.6 Longitudinal Shear on a Beam Element of Arbitrary Shape
xz = ?
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H VQqx I
VQH xI
0 0: ( )x D CF H dA A
Using similar procedures in Sec. 6.2, we have
= shear flow (6.5)
(6.4)
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6.7 Shearing Stresses n Thin-Walled Members
VQH xI
H VQqx I
These two equations are valid for thin-walled members:
(6.4)
(6.5)
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VQH xI
ave xzVQIt
( 6.4)
( 6.6)
From Sec. 6.2
aveH VQ xA I t x
We have:
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ave xzVQIt
(6.6)
-- This equation can be applied to a variety of cross sections.
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6.8 Plastic Deformation
2
2
3 1(1 )2 3
YY
yPx Mc
(6.14)
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2
' 2
3 (1 )2
xyY
P yA y
max '
32
PA
(6.15)
(6.16)
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6.9 Unsymmetric Loading Of Thin-Walled Members; Shear Center
xMyI
aveVQIt
(4.16)
(6.6)
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B
AF qds
D
BV qds
FheV
(6.17)
(6.18)
(6.19)
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2 VQ VsthqI I
0 0 02 2
b b bVsth VthF qds ds sdsI I
2
4VthbFI
2 2 2
4 4 Fh Vthb h th beV I V I
3 3 21 12 2[ ( ( ) ]12 12 2web flang
hI I I th bt bt bt
(6.20)
(6.21)
(6.22)
(6.23)
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3 2 21 1 1 (6 )12 2 12
I th tbh th b h
2 q VQ Vh st It I
212( )(6 )12
BVhb
th b h
1 1 1 1( ) ( ) (4 )2 2 4 8
Q bt h ht h ht b h
max2
1( )(4 ) 3 (4 )81 2 (6 )(6 )12
V ht b hVQ V b hIt th b hth b h t
(6.23)
(6.25)
(6.26)
(6.27)