shaear stresses

8/14/2019 Shaear Stresses

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MECHANICS OF

MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2002 The McGraw-Hill Companies, Inc. All rights reserved.

6Shearing Stresses in

Beams and Thin-

Walled Members


2/26


MECHANICS OF MATERIALSThird

Edition

Beer Johnston DeWolf

6 - 2

Shearing Stresses in Beams and

Thin-Walled Members

IntroductionShear on the Horizontal Face of a Beam Element

Example 6.01

Determination of the Shearing Stress in a Beam

Shearing Stresses

Further Discussion of the Distribution of Stresses in a ...

Sample Problem 6.2

Longitudinal Shear on a Beam Element of Arbitrary Shape

Example 6.04

Shearing Stresses in Thin-Walled Members

Plastic Deformations

Sample Problem 6.3

Unsymmetric Loading of Thin-Walled Members

Example 6.05

Example 6.06


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Edition


6 - 3

Introduction

( ) 00

0

00

= ===

===== ===

xzxzz

xyxyy

xyxzxxx

yMdAF

dAzMVdAF

dAzyMdAF

Distribution of normal and shearing

stresses satisfies

Transverse loading applied to a beam

results in normal and shearing stresses intransverse sections.

When shearing stresses are exerted on the

vertical faces of an element, equal stressesmust be exerted on the horizontal faces

Longitudinal shearing stresses must exist

in any member subjected to transverse

loading.

T


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Edition


6 - 4

Shear on the Horizontal Face of a Beam Element

Consider prismatic beam

For equilibrium of beam element

( )

=

+==

A

CD

ADDx

dAyI

MMH

dAHF 0

xVxdx

dMMM

dAyQ

CD

A

==

= Note,

flowshearI

VQ

x

Hq

xI

VQH

==

=

= Substituting,

MECHANICS OF MATERIALSTE


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Edition


6 - 5

Shear on the Horizontal Face of a Beam Element

flowshearI

VQ

x

H

q ==

=

Shear flow,

where

sectioncrossfullofmomentsecond

aboveareaofmomentfirst

'

2

1

=

=

=

=

+AA

A

dAyI

y

dAyQ

Same result found for lower area

HH

QQ

qI

QV

x

H

q

=

==+

=

=

=

axisneutralto

respecthmoment witfirst

0



6/26



Edition


6 - 6

Example 6.01

A beam is made of three planks,

nailed together. Knowing that the

spacing between nails is 25 mm and

that the vertical shear in the beam isV= 500 N, determine the shear force

in each nail.

SOLUTION:

Determine the horizontal force per

unit length or shear flow q on the

lower surface of the upper plank.

Calculate the corresponding shear

force in each nail.



7/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.


Edition


6 - 7

Example 6.01

( ) ( )

( ) ( )

( ) ( )

( ) ( )

46

2

3121

3

12

1

36

m1020.16

]m060.0m100.0m020.0

m020.0m100.0[2

m100.0m020.0

m10120

m060.0m100.0m020.0

=

+

+

=

=

==

I

yAQ

SOLUTION:

Determine the horizontal force per

unit length or shear flow q on the

lower surface of the upper plank.

mN3704

m1016.20

)m10120)(N500(

46-

36

=

==

I

VQq

Calculate the corresponding shear

force in each nail for a nail spacingof 25 mm.

mNqF 3704)(m025.0()m025.0( ==

N6.92=F





Edition


6 - 8

Determination of the Shearing Stress in a Beam

The average shearing stress on the horizontal

face of the element is obtained by dividing theshearing force on the element by the area of

the face.

ItVQ

xt

x

I

VQ

A

xq

A

Have

=

=

=

=

On the upper and lower surfaces of the beam, yx= 0. It follows that xy= 0 on the upper and

lower edges of the transverse sections.

If the width of the beam is comparable or large

relative to its depth, the shearing stresses atD1

andD2 are significantly higher than atD.





Edition


6 - 9

Shearing Stresses xyin Common Types of Beams

For a narrow rectangular beam,

A

V

c

y

A

V

Ib

VQxy

2

3

12

3

max

2

2

=

==

For American Standard (S-beam)

and wide-flange (W-beam)

beams

web

ave

A

VIt

VQ

=

=

max





Edition


6 - 10

Further Discussion of the Distribution of

Stresses in a Narrow Rectangular Beam

=2

2

12

3

c

y

A

Pxy

I

Pxyx +=

Consider a narrow rectangular cantilever beam

subjected to loadPat its free end:

Shearing stresses are independent of the distance

from the point of application of the load.

Normal strains and normal stresses are unaffected

by the shearing stresses.

From Saint-Venants principle, effects of the load

application mode are negligible except in immediate

vicinity of load application points.

Stress/strain deviations for distributed loads are

negligible for typical beam sections of interest.





Edition


6 - 11

Sample Problem 6.2

A timber beam is to support the three

concentrated loads shown. Knowing

that for the grade of timber used,

psi120psi1800 == allall determine the minimum required depth

dof the beam.

SOLUTION:

Develop shear and bending moment

diagrams. Identify the maximums.

Determine the beam depth based on

allowable normal stress.

Determine the beam depth based on

allowable shear stress.

Required beam depth is equal to the

larger of the two depths found.





Edition


6 - 12

Sample Problem 6.2

SOLUTION:

Develop shear and bending momentdiagrams. Identify the maximums.

inkip90ftkip5.7

kips3

max

max

===

M

V





Edition


6 - 13

Sample Problem 6.2

( )

( ) 2

261

261

3121

in.5833.0

in.5.3

d

d

dbc

IS

dbI

=

=

==

=

Determine the beam depth based on allowable

normal stress.

( )

in.26.9

in.5833.0

in.lb1090psi1800

2

3

max

=

=

=

d

d

S

Mall

Determine the beam depth based on allowable

shear stress.

( )

in.71.10

in.3.5

lb3000

2

3psi120

2

3 max

=

=

=

d

d

A

Vall

Required beam depth is equal to the larger of the two.

in.71.10=d

MECHANICS OF MATERIALST

E




Edition


6 - 14

Longitudinal Shear on a Beam Element

of Arbitrary Shape

We have examined the distribution of

the vertical components xy on a

transverse section of a beam. We

now wish to consider the horizontal

components xzof the stresses.

Consider prismatic beam with anelement defined by the curved surface

CDDC.

( ) +==a

dAHF CDx 0

Except for the differences inintegration areas, this is the same

result obtained before which led to

I

VQ

x

Hqx

I

VQH =

==

MECHANICS OF MATERIALSTh

E




Edition


6 - 15

Example 6.04

A square box beam is constructed from

four planks as shown. Knowing that the

spacing between nails is 1.5 in. and the

beam is subjected to a vertical shear of

magnitude V= 600 lb, determine the

shearing force in each nail.

SOLUTION:

Determine the shear force per unit

length along each edge of the upper

plank.

Based on the spacing between nails,determine the shear force in each

nail.


Ed




Edition


6 - 16

Example 6.04

For the upper plank,

( ) ( ) ( )

3in22.4

.in875.1.in3in.75.0

=

== yAQ

For the overall beam cross-section,

( ) ( )

4

31213

121

in42.27

in3in5.4

=

=I

SOLUTION:

Determine the shear force per unitlength along each edge of the upper

plank.

( )

lengthunitperforceedge

in

lb15.46

2

in

lb3.92

in27.42

in22.4lb6004

3

=

==

===

qf

I

VQq

Based on the spacing between nails,

determine the shear force in eachnail.

( )in75.1in

lb15.46

== fF

lb8.80=F


Ed



MECHANICS OF MATERIALShird

Edition


6 - 17


Consider a segment of a wide-flange

beam subjected to the vertical shearV.

The longitudinal shear force on the

element is

xI

VQH =

It

VQ

xt

Hxzzx =

=

The corresponding shear stress is

NOTE: 0xy0xz

in the flanges

in the web

Previously found a similar expression

for the shearing stress in the web

It

VQxy =


Ed




dition


6 - 18


The variation of shear flow across the

section depends only on the variation ofthe first moment.

I

VQtq ==

For a box beam, q grows smoothly fromzero at A to a maximum at Cand Cand

then decreases back to zero atE.

The sense ofq in the horizontal

portions of the section may be deducedfrom the sense in the vertical portions

or the sense of the shearV.


Ed




dition


6 - 19


For a wide-flange beam, the shear flow

increases symmetrically from zero atA

andA, reaches a maximum at Cand the

decreases to zero atEandE.

The continuity of the variation in q andthe merging ofq from section branches

suggests an analogy to fluid flow.


Ed




dition


6 - 20

The section becomes fully plastic (yY = 0) at

the wall when

pY MMPL ==2

3

ForPL > MY, yield is initiated atB andB.

For an elastoplastic material, the half-thickness

of the elastic core is found from

=2

2

3

11

2

3

c

yMPx YY


momentelasticmaximum== YYc

IM Recall:

For M = PL < MY, the normal stress does

not exceed the yield stress anywhere along

the beam.

Maximum load which the beam can support is

L

MP

p=max


Ed




dition


6 - 21


Preceding discussion was based on

normal stresses only

Consider horizontal shear force on an

element within the plastic zone,

( ) ( ) 0=== dAdAH YYDC

Therefore, the shear stress is zero in theplastic zone.

Shear load is carried by the elastic

core,

A

P

byA

y

y

A

PY

Y

xy

=

=

=

2

3

2where1

2

3

max

2

2

AsAdecreases, max increases and

may exceed Y


Ed




dition


6 - 22

Sample Problem 6.3

Knowing that the vertical shear is 50

kips in a W10x68 rolled-steel beam,

determine the horizontal shearing

stress in the top flange at the point a.

SOLUTION:

For the shaded area,

( ) ( ) ( )

3in98.15

in815.4in770.0in31.4

=

=Q

The shear stress at a,

( )

( )( )in770.0in394in98.15kips50

4

3

==It

VQ

ksi63.2

=


Ed




dition


6 - 23


Beam loaded in a vertical plane

of symmetry deforms in thesymmetry plane without

twisting.

It

VQ

I

Myavex ==

Beam without a vertical plane

of symmetry bends and twists

under loading.

It

VQ

I

Myavex =


Ed

f



MECHANICS OF MATERIALSirddition


6 - 24

When the force P is applied at a distance e to theleft of the web centerline, the member bends in

a vertical plane without twisting.


If the shear load is applied such that the beam

does not twist, then the shear stress distributionsatisfies

FdsqdsqFdsqVIt

VQ E

D

B

A

D

Bave =====

FandFindicate a coupleFh and the need forthe application of a torque as well as the shear

load.VehF =


Ed

B J h t D W lf



MECHANICS OF MATERIALSirdition


6 - 25

Example 6.05

Determine the location for the shear center of the

channel section with b = 4 in., h = 6 in., and t= 0.15 in.

I

hFe =

where

I

Vthb

dsh

st

I

Vds

I

VQdsqF

b bb

4

22

0 00

=

===

( )hbth

hbtbtthIII flangeweb

+

++=+=

6

212

12

12

12

2121

233

Combining,

( ).in43

.in62

in.4

3

2 +=

+=

b

h

be .in6.1=e

MECHANICS OF MATERIALSThi

Ed

B J h t D W lf


26/26

MECHANICS OF MATERIALSirdition


Example 6.06

Determine the shear stress distribution for

V= 2.5 kips.

ItVQ

tq ==

Shearing stresses in the flanges,

( )

( )( ) ( )( ) ( )

( ) ( ) ( )ksi22.2

in6in46in6in15.0

in4kips5.26

66

62

22

2121

=+

=

+=

+=

===

hbthVb

hbth

Vhb

sI

Vhhst

It

V

It

VQ

B

Shearing stress in the web,( ( )

( )

( )

( )

( ) ( )

( ) ( ) ( )ksi06.3

in6in66in6in15.02

in6in44kips5.23

62

43

6

4

212181

max

=

+

+=

++

=+

+==

hbth

hbV

thbth

hbhtV

It

VQ

shaear stresses

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