fourth edition mechanics of materialsyunus.hacettepe.edu.tr/~boray/6_1_shearing_stresses...

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

Shearing Stresses in John T. DeWolf

Lecture Notes:Beams and Thin-Walled Members

J. Walt OlerTexas Tech University

Walled Members

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Shearing Stresses in Beams and Thi W ll d M bThin-Walled MembersIntroductionShear on the Horizontal Face of a Beam ElementExample 6.01Determination of the Shearing Stress in a BeamDetermination of the Shearing Stress in a BeamShearing Stresses τxy in Common Types of BeamsFurther Discussion of the Distribution of Stresses in a ...S l bl 6 2Sample Problem 6.2Longitudinal Shear on a Beam Element of Arbitrary ShapeExample 6.04pShearing Stresses in Thin-Walled MembersPlastic DeformationsSample Problem 6 3Sample Problem 6.3Unsymmetric Loading of Thin-Walled MembersExample 6.05

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 6 - 2

Example 6.06


FourthEdition


Introduction• Transverse loading applied to a beam

results in normal and shearing stresses in i

• Distribution of normal and shearing

transverse sections.

( )dAMVdAF

dAzyMdAF xyxzxxx =−===

∫∫∫∫ ττσ

000

stresses satisfies

( ) MyMdAFdAzMVdAF

xzxzz

xyxyy=−=====−==

∫∫∫∫

στστ

00

• When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces

• Longitudinal shearing stresses must exist in any member subjected to transverse


y jloading.


FourthEdition


Shear on the Horizontal Face of a Beam Element

• Consider prismatic beam

• For equilibrium of beam element• For equilibrium of beam element( )∑ ∫ −+==

ACDx

MM

dAHF σσ∆0

∫−

=A

CD dAyI

MMH∆

dAQ ∫

• Note,

xVxdxdMMM

dAyQ

CD

A

∆=∆=−

∫=

dx

VQH ∆∆

• Substituting,

flowshearI

VQxHq

xI

VQH

==∆∆

=

∆=∆


Ix∆


FourthEdition


Shear on the Horizontal Face of a Beam Element

flowshearI

VQxHq ==∆∆

=

• Shear flow,

Ix∆

• where∫= dAyQ

above area ofmoment first 2

1

∫=

=A

dAyI

y

sectioncrossfullofmoment second'

=

∫+AA

y

S lt f d f l• Same result found for lower areaq

IQV

xHq

′

′−=′

=∆

′∆=′

QQ==′+

axisneutraltorespect h moment witfirst

0


HH ∆−=′∆


FourthEdition


Example 6.01

SOLUTION:

• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank.pp p

• Calculate the corresponding shear force in each nail.

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 is V = 500 N, determine the shear force ,in each nail.



FourthEdition


Example 6.01

SOLUTION:

• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank.

AQ

pp p

m1016.20)m10120)(N500(

46-

36

×

×==

−

IVQq

( )( )36m10120

m060.0m100.0m020.0−×=

×== yAQ

mN3704=

( )( )

( )( )31

3121

m0200m1000[2

m100.0m020.0

m10120

+

=

×

I• Calculate the corresponding shear

force in each nail for a nail spacing of 25 mm.( )( )

( )( )46

212

m102016

]m060.0m100.0m020.0

m020.0m100.0[2

−×

×+

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

N6.92=F


m1020.16 ×=


FourthEdition


Determination of the Shearing Stress in a Beam

• The average shearing stress on the horizontal face of the element is obtained by dividing the y gshearing force on the element by the area of the face.

xVQxqH ∆∆∆

VQxtx

IVQ

Axq

AH

ave

=

∆∆

=∆∆

=∆∆

=τ

It

• On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 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 at D1and D2 are significantly higher than at D.


and D2 are significantly higher than at D.


FourthEdition


Shearing Stresses τxy in Common Types of Beams

• For a narrow rectangular beam,

VVQ 3 2 ⎞⎛

Vcy

AV

IbVQ

xy

3

123

2

2

⎟⎠

⎞⎜⎜⎝

⎛−==τ

A2max =τ

• For American Standard (S-beam) and wide-flange (W-beam) beams

ave

VIt

VQ=τ

webAV

=maxτ



FourthEdition


Further Discussion of the Distribution of St i N R t l BStresses in a Narrow Rectangular Beam

• Consider a narrow rectangular cantilever beam bj t d t l d P t it f d

⎟⎠

⎞⎜⎜⎝

⎛−= 2

21

23

cy

AP

xyτI

Pxyx +=σ

subjected to load P at its free end:

⎠⎝2 cA I

• Shearing stresses are independent of the distance from the point of application of the load.o t e po t o app cat o o t e oad.

• Normal strains and normal stresses are unaffected by the shearing stresses.

• From Saint-Venant’s principle, effects of the load application mode are negligible except in immediate vicinity of load application points.

• Stress/strain deviations for distributed loads are li ibl f t i l b ti f i t t


negligible for typical beam sections of interest.


FourthEdition


Sample Problem 6.2

SOLUTION:SOLUTION:

• Develop shear and bending moment diagrams. Identify the maximums.g y

• Determine the beam depth based on allowable normal stress.

A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used

• Determine the beam depth based on allowable shear stress.

that for the grade of timber used,

psi120psi1800 == allall τσ

d t i th i i i d d th

• Required beam depth is equal to the larger of the two depths found.

determine the minimum required depth d of the beam.



FourthEdition


Sample Problem 6.2SOLUTION:

Develop shear and bending momentDevelop shear and bending moment diagrams. Identify the maximums.

kips3=Vinkip90ftkip5.7

kips3

max

max⋅=⋅=

=MV



FourthEdition


Sample Problem 6.2• Determine the beam depth based on allowable

normal stress.maxM

in.lb1090psi1800 2

3

max

⋅×=

=S

Mallσ

31 dbI

( )in.26.9

in.5833.0p 2

=dd

D i h b d h b d ll bl2

61

3121

dbcIS

dbI

==

= • Determine the beam depth based on allowable shear stress.

3 max=V

allτ( )

( ) 2

261

in.5833.0

in.5.3

d

d

=

=

( )in.3.5lb3000

23psi120

2

=d

Aallτ

( ) ( )in.71.10=d

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


in.71.10=d


FourthEdition


Longitudinal Shear on a Beam Element f A bit Shof Arbitrary Shape

• We have examined the distribution of th ti l tthe vertical components τxy on a transverse section of a beam. We now wish to consider the horizontal

hcomponents τxz of the stresses.

• Consider prismatic beam with an element defined by the curved surfaceelement defined by the curved surface CDD’C’.

( )∑ ∫ −+∆== dAHF CDx σσ0a

• Except for the differences in integration areas, this is the sameintegration areas, this is the same result obtained before which led to

VQHqxVQH =∆

=∆=∆


Ixq

I ∆


FourthEdition


Example 6.04

SOLUTION:

• Determine the shear force per unit length along each edge of the upper plank.

• Based on the spacing between nails,

A b b i d f

determine the shear force in each nail.

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.


g

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