Transcript 6 - unisa.it

Fourth Edition
CHAPTER
6
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
Shearing Stresses in
Beams and ThinWalled Members
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Beams and
Thin-Walled Members
Introduction
Shear on the Horizontal Face of a Beam Element
Example 6.01
Determination of the Shearing Stress in a Beam
Shearing Stresses txy in Common Types of Beams
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-2
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Introduction
• Transverse loading applied to a beam
results in normal and shearing stresses in
transverse sections.
• Distribution of normal and shearing
stresses satisfies
Fx    x dA  0
Fy   t xy dA  V
Fz   t xz dA  0


M x   y t xz  z t xy dA  0
M y   z  x dA  0
M z    y  x   M
• 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
loading.
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6-3
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shear on the Horizontal Face of a Beam Element
• Consider prismatic beam
• For equilibrium of beam element
 Fx  0  H    D   C dA
A
H 
• Note,
M D  MC
y dA

I
A
Q   y dA
A
M D  MC 
dM
x  V x
dx
• Substituting,
VQ
x
I
H VQ
q

 shear flow
x
I
H 
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-4
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shear on the Horizontal Face of a Beam Element
• Shear flow,
q
H VQ

 shear flow
x
I
• where
Q   y dA
A
 first moment of area above y1
I
2
 y dA
A A'
 second moment of full cross section
• Same result found for lower area
H  VQ

 q
x
I
Q  Q  0
q 
 first moment wit h respect
to neutral axis
H   H
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-5
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.01
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.
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-6
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.01
SOLUTION:
• Determine the horizontal force per
unit length or shear flow q on the
lower surface of the upper plank.
VQ (500 N )(120  10 6 m3 )
q

I
16.20  10-6 m 4
 3704 N
m
Q  Ay
 0.020 m  0.100 m 0.060 m 
 120  106 m3
1 0.020 m 0.100 m 3
I  12
1 0.100 m 0.020 m 3
 2[12
• Calculate the corresponding shear
force in each nail for a nail spacing of
25 mm.
 0.020 m  0.100 m 0.060 m 2 ]
 16.20  106 m 4
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F  (0.025 m)q  (0.025 m)(3704 N m
F  92.6 N
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Determination of the Shearing Stress in a Beam
• The average shearing stress on the horizontal
face of the element is obtained by dividing the
shearing force on the element by the area of
the face.
H q x VQ x


A A
I t x
VQ

It
t ave 
• On the upper and lower surfaces of the beam,
tyx= 0. It follows that txy= 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 D1
and D2 are significantly higher than at D.
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6-8
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses txy in Common Types of Beams
• For a narrow rectangular beam,
VQ 3 V 
t xy 

1

Ib 2 A 
3V
t max 
2A
y 2 
c 2 
• For American Standard (S-beam)
and wide-flange (W-beam) beams
VQ
It
V
t max 
Aweb
t ave 
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6-9
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Further Discussion of the Distribution of
Stresses in a Narrow Rectangular Beam
• Consider a narrow rectangular cantilever beam
subjected to load P at its free end:
3 P 
y 2 
t xy 
1 2

2 A  c 
x  
Pxy
I
• 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-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
negligible for typical beam sections of interest.
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6 - 10
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 6.2
SOLUTION:
• Develop shear and bending moment
diagrams. Identify the maximums.
• 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,
 all  1800 psi
t all  120 psi
• Determine the beam depth based on
allowable shear stress.
• Required beam depth is equal to the
larger of the two depths found.
determine the minimum required depth
d of the beam.
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6 - 11
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 6.2
SOLUTION:
Develop shear and bending moment
diagrams. Identify the maximums.
Vmax  3 kips
M max  7.5 kip  ft  90 kip  in
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 6.2
• Determine the beam depth based on allowable
normal stress.
 all 
M max
S
1800 psi 
90  103 lb  in.
0.5833in.  d 2
d  9.26 in.
1 bd3
I  12
I
S   16 b d 2
c
 16 3.5 in. d 2
 0.5833 in. d 2
• Determine the beam depth based on allowable
shear stress.
3 Vmax
2 A
3 3000 lb
120 psi 
2 3.5 in.  d
d  10.71in.
t all 
• Required beam depth is equal to the larger of the two.
d  10.71in.
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Longitudinal Shear on a Beam Element
of Arbitrary Shape
• We have examined the distribution of
the vertical components txy on a
transverse section of a beam. We now
wish to consider the horizontal
components txz of the stresses.
• Consider prismatic beam with an
element defined by the curved surface
CDD’C’.
 Fx  0  H    D   C dA
a
• Except for the differences in
integration areas, this is the same
result obtained before which led to
H 
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
VQ
x
I
q
H VQ

x
I
6 - 14
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.04
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.
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.
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