Chapter Six Shearing Stresses in Beams and Thin-Walled Members

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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 Presentation

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Chapter Six

Shearing Stresses in Beams and Thin-Walled Members

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]

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)

Materials weak in shear resistance shear failure could occur.

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)

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

6.3 Determination of the Shearing Stresses in a Beam

aveH VQ xA I t x

aveVQIt

VQH xI

(6.6)

= ave. shear stress

xy = 0 at top and bottom fibers

Variation of xy < 0.8% if b h/4

-- for narrow rectangular beams

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)

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)

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:

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)

6.6 Longitudinal Shear on a Beam Element of Arbitrary Shape

xz = ?

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)

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)

VQH xI

ave xzVQIt

( 6.4)

( 6.6)

From Sec. 6.2

aveH VQ xA I t x

We have:

ave xzVQIt

(6.6)

-- This equation can be applied to a variety of cross sections.

6.8 Plastic Deformation

2

2

3 1(1 )2 3

YY

yPx Mc

(6.14)

2

' 2

3 (1 )2

xyY

P yA y

max '

32

PA

(6.15)

(6.16)

6.9 Unsymmetric Loading Of Thin-Walled Members; Shear Center

xMyI

aveVQIt

(4.16)

(6.6)

B

AF qds

D

BV qds

FheV

(6.17)

(6.18)

(6.19)

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)

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)

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