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8/14/2019 Shaear Stresses
1/26
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
8/14/2019 Shaear Stresses
2/26
2002 The McGraw-Hill Companies, Inc. All rights reserved.
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
8/14/2019 Shaear Stresses
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2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
8/14/2019 Shaear Stresses
4/26
2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
8/14/2019 Shaear Stresses
5/26
2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
qI
QV
x
H
q
=
==+
=
=
=
axisneutralto
respecthmoment witfirst
0
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
6/26
2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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.
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
7/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
8/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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.
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
9/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
10/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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.
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
11/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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.
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
12/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
6 - 12
Sample Problem 6.2
SOLUTION:
Develop shear and bending momentdiagrams. Identify the maximums.
inkip90ftkip5.7
kips3
max
max
===
M
V
MECHANICS OF MATERIALSTE
8/14/2019 Shaear Stresses
13/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
8/14/2019 Shaear Stresses
14/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
8/14/2019 Shaear Stresses
15/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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.
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
16/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
17/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALShird
Edition
Beer Johnston DeWolf
6 - 17
Shearing Stresses in Thin-Walled Members
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 =
MECHANICS OF MATERIALSTh
Ed
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MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
6 - 18
Shearing Stresses in Thin-Walled Members
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.
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
19/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
6 - 19
Shearing Stresses in Thin-Walled Members
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.
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
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MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
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
Plastic Deformations
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
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
21/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
6 - 21
Plastic Deformations
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
MECHANICS OF MATERIALSTh
Ed
8/14/2019 Shaear Stresses
22/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
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
=
MECHANICS OF MATERIALSTh
Ed
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MECHANICS OF MATERIALShird
dition
Beer Johnston DeWolf
6 - 23
Unsymmetric Loading of Thin-Walled Members
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 =
MECHANICS OF MATERIALSTh
Ed
f
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24/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSirddition
Beer Johnston DeWolf
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.
Unsymmetric Loading of Thin-Walled Members
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 =
MECHANICS OF MATERIALSTh
Ed
B J h t D W lf
8/14/2019 Shaear Stresses
25/26 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSirdition
Beer Johnston DeWolf
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
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MECHANICS OF MATERIALSirdition
Beer Johnston DeWolf
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