Third 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Third
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-2
Third
Edition
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   0
• 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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-3
Third
Edition
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   D dA
A
H 
M D  MC
 y dA
I
A
• Note,
Q   y dA
A
M D  MC 
dM
x  V x
dx
• Substituting,
VQ
x
I
H VQ
q

 shear flow
x
I
H 
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-4
Third
Edition
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 with respect
to neutral axis
H   H
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-5
Third
Edition
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-6
Third
Edition
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
I
1 0.020 m 0.100 m 3
 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
F  (0.025 m)q  (0.025 m)(3704 N m
F  92.6 N
6-7
Third
Edition
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.
t ave 

H q x VQ x


A A
I t x
VQ
It
• 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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-8
Third
Edition
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 
t max 
y 2 
c 2 
3V
2A
• For American Standard (S-beam)
and wide-flange (W-beam) beams
VQ
It
V
t max 
Aweb
t ave 
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6-9
Third
Edition
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 10
Third
Edition
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 11
Third
Edition
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 12
Third
Edition
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.
0.5833 in.  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 3.5in.  d
t all 
d  10.71in.
• Required beam depth is equal to the larger of the two.
d  10.71in.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 13
Third
Edition
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 
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
VQ
x
I
q
H VQ

x
I
6 - 14
Third
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 15
Third
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.


VQ 600 lb  4.22 in 3
lb
q


92
.
3
I
in
27.42 in 4
q
lb
 46.15
2
in
 edge force per unit length
f 
For the upper plank,
Q  Ay  0.75in. 3 in .1.875 in .
 4.22 in 3
For the overall beam cross-section,
1 4.5 in   1 3 in 
I  12
12
3
3
 27.42 in 4
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
• Based on the spacing between nails,
determine the shear force in each
nail.
lb 

F  f    46.15 1.75 in 
in 

F  80.8 lb
6 - 16
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• Consider a segment of a wide-flange
beam subjected to the vertical shear V.
• The longitudinal shear force on the
element is
H 
VQ
x
I
• The corresponding shear stress is
t zx  t xz 
H VQ

t x It
• Previously found a similar expression
for the shearing stress in the web
t xy 
VQ
It
• NOTE: t xy  0
t xz  0
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
in the flanges
in the web
6 - 17
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• The variation of shear flow across the
section depends only on the variation of
the first moment.
q tt 
VQ
I
• For a box beam, q grows smoothly from
zero at A to a maximum at C and C’ and
then decreases back to zero at E.
• The sense of q in the horizontal portions
of the section may be deduced from the
sense in the vertical portions or the
sense of the shear V.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 18
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• For a wide-flange beam, the shear flow
increases symmetrically from zero at A
and A’, reaches a maximum at C and the
decreases to zero at E and E’.
• The continuity of the variation in q and
the merging of q from section branches
suggests an analogy to fluid flow.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 19
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
I
M

 Y  maximum elastic moment
• Recall: Y
c
• For M = PL < MY , the normal stress does
not exceed the yield stress anywhere along
the beam.
• For PL > MY , yield is initiated at B and B’.
For an elastoplastic material, the half-thickness
of the elastic core is found from
 1 yY2 
3
Px  M Y 1  2 
 3c 
2


• The section becomes fully plastic (yY = 0) at
the wall when
3
PL  M Y  M p
2
• Maximum load which the beam can support is
Pmax 
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Mp
L
6 - 20
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• Preceding discussion was based on
normal stresses only
• Consider horizontal shear force on an
element within the plastic zone,
H   C   D dA   Y   Y dA  0
Therefore, the shear stress is zero in the
plastic zone.
• Shear load is carried by the elastic core,
3 P 
t xy 
1


2A
t max 
y 2 
where A  2byY
2
yY 
3P
2 A
• As A’ decreases, tmax increases and
may exceed tY
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 21
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 6.3
SOLUTION:
• For the shaded area,
Q  4.31in 0.770 in 4.815 in 
 15.98 in 3
• The shear stress at a,
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.



VQ 50 kips  15.98 in 3
t

It
394 in 4 0.770 in 

t  2.63 ksi
6 - 22
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Unsymmetric Loading of Thin-Walled Members
• Beam loaded in a vertical plane
of symmetry deforms in the
symmetry plane without
twisting.
x  
My
I
t ave 
VQ
It
• Beam without a vertical plane
of symmetry bends and twists
under loading.
x  
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
My
I
t ave 
VQ
It
6 - 23
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Unsymmetric Loading of Thin-Walled Members
• If the shear load is applied such that the beam
does not twist, then the shear stress distribution
satisfies
D
B
E
VQ
t ave 
V   q ds F   q ds    q ds   F 
It
B
A
D
• F and F’ indicate a couple Fh and the need for
the application of a torque as well as the shear
load.
F h  Ve
• When the force P is applied at a distance e to the
left of the web centerline, the member bends in a
vertical plane without twisting.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6 - 24
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
e
Fh
I
• where
b
b VQ
Vb h
F   q ds  
ds   st ds
I0 2
0
0 I
Vthb 2

4I
2
1 3
1 3
 h 
I  I web  2 I flange  th  2 bt  bt   
12
 2  
12
1 th 2 6b  h 
 12
• Combining,
e
b
h
2
3b

4 in.
6 in .
2
34 in .
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
e  1.6 in .
6 - 25
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.06
• Determine the shear stress distribution for
V = 2.5 kips.
t
q VQ

t
It
• Shearing stresses in the flanges,
VQ V
h Vh
 st  
s
It
It
2 2I
Vhb
6Vb
tB 

2 1 th 2 6b  h  th6b  h 
t
12 

62.5 kips 4 in 
 2.22 ksi
0.15 in 6 in 6  4 in  6 in 
• Shearing stress in the web,
 
1
VQ V 8 ht 4b  h  3V 4b  h 
t max 


2
1
It
th 6b  h t 2th6b  h 
12

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
32.5 kips 4  4 in  6 in 
 3.06 ksi
20.15 in 6 in 6  6 in  6 in 
6 - 26