Beams and Thin- Walled Members

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Third
Edition
5/3/2010
Third Edition
MECHANICS OF
CHAPTER
MATERIALS
Ferdinand P. Beer
6
E. Russell Johnston, Jr.
Shearing Stresses in
John T. DeWolf
Beams and Thin-
Lecture Notes:
Walled Members
J. Walt Oler
Texas Tech University
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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 τ in Common Types of Beams
xy
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
Sample Problem 6.3
Unsymmetric Loading of Thin-Walled Members
Example 6.05
Example 6.06
6 - 2
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
1Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Introduction
• Transverse loading applied to a beam
results in normal and shearing stresses in
transverse sections.
• Distribution of normal and shearing
stresses satisfies
F = σ dA=0 M = (yτ −zτ )dA=0
x x x xz xy
∫ ∫
F = τ dA=−V M = zσ dA=0
y ∫ xy y ∫ x
F = τ dA=0 M = (−yσ )=M
z ∫ xz z ∫ x
• 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.
6 - 3
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Shear on the Horizontal Face of a Beam Element
• Consider prismatic beam
• For equilibrium of beam element
F = 0=ΔH + (σ −σ )dA
∑ x C D

A
M − M
D C
ΔH = y dA

I
A
• Note,
Q = y dA

A
dM
M − M = Δx = V Δx
D C
dx
• Substituting,
VQ
ΔH = Δx
I
ΔH VQ
q= = = shear flow
Δx I
6 - 4
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
2Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Shear on the Horizontal Face of a Beam Element
• Shear flow,
ΔH VQ
q= = = shear flow
Δx I
• where
Q = y dA

A
= first moment of area above y
1
2
I = y dA

A+ A'
= second moment of full cross section
• Same result found for lower area
′ ′
ΔH VQ
q′= = =−q′
Δx I

Q+ Q = 0
= first moment with respect
to neutral axis
ΔH′= −ΔH
6 - 5
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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.
6 - 6
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
3Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Example 6.01
SOLUTION:
• Determine the horizontal force per
unit length or shear flow q on the
lower surface of the upper plank.
−6 3
VQ ( 500N )(120×10 m )
q= =
-6 4
I
16,20×10 m
N
= 3704
Q= Ay
m
= (0,020m× 0,100m)(0,060m)
−6 3
=120×10 m
• Calculate the corresponding shear
3
1
I = 0,020m 0,100m force in each nail for a nail spacing of
( )( )
12
3
25 mm.
1
+2[ 0,100m 0,020m
( )( )
12
2 F = ( 0,025m )q= ( 0,025m )( 3704 N m
+ 0,020m× 0,100m 0,060m ]
( )( )
F = 92,6 N
−6 4
=16,20×10 m
6 - 7
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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.
ΔH qΔx VQ Δx
τ = = =
ave
ΔA ΔA I tΔx
VQ
=
It
• On the upper and lower surfaces of the beam,
τ = 0. It follows that τ = 0 on the upper and
yx xy
lower edges of the transverse sections.
• If the width of the beam is comparable or large
relative to its depth, the shearing stresses at D
1
and D are significantly higher than at D.
2
6 - 8
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
4Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Shearing Stresses τ in Common Types of Beams
xy
• For a narrow rectangular beam,
2
 
VQ 3V y
 
τ = = 1−
xy
 2 
Ib 2 A
c
 
3V
τ =
max
2 A
• For American Standard (S-beam)
and wide-flange (W-beam) beams
VQ
τ =
ave
It
V
τ =
max
A
web
6 - 9
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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:
2
 
3 P y Pxy
 
τ = 1− σ =+
x
xy
 2 
2 A I
c
 
• 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.
6 - 10
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
5Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
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
• Determine the beam depth based on
concentrated loads shown. Knowing
allowable shear stress.
that for the grade of timber used,
• Required beam depth is equal to the
σ =12MPa τ = 0,83MPa
all all
larger of the two depths found.
determine the minimum required depth
d of the beam.
6 - 11
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Sample Problem 6.2
SOLUTION:
Develop shear and bending moment
diagrams. Identify the maximums.
V =11,25kN.m
max
M =15kN
max
6 - 12
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
6Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Sample Problem 6.2
• Determine the beam depth based on allowable
normal stress.
M
max
W =
σ
adm.
11,25 kN.m
2
0,015d =
3
12×10 kN
2
d = 0,0625 ⇔ d= 0,25m
1 3
I = b d
12 • Determine the beam depth based on allowable
I
2 shear stress.
1
W = = bd
6
c
V
3
max
2 τ =τ =
1
m all
= 0,090m d
( )
2 A
6
2
3 15kN
3 2
= (0,015m)d
0,82×10 kN m =
2 0,090m d
( )
d = 0,30m
• Required beam depth is equal to the larger of the two.
d = 0,30m
6 - 13
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Longitudinal Shear on a Beam Element
of Arbitrary Shape
• We have examined the distribution of
the vertical components τ on a
xy
transverse section of a beam. We now
wish to consider the horizontal
components τ of the stresses.
xz
• Consider prismatic beam with an
element defined by the curved surface
CDD’C’.
( )
∑ F = 0=ΔH + σ −σ dA

x D C
a
• Except for the differences in
integration areas, this is the same
result obtained before which led to
VQ ΔH VQ
ΔH = Δx q= =
I Δx I
6 - 14
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
7Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
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 45 mm. and the
beam is subjected to a vertical shear of
magnitude V = 2,7 kN, determine the
shearing force in each nail.
6 - 15
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Example 6.04
SOLUTION:
• Determine the shear force per unit
length along each edge of the upper
plank.
3
2700N 68590mm
( )
( )
VQ N
q= = =16,4
6 4
I mm
11,29×10 mm
q N
f = = 8,2
2 mm
For the upper plank,
= edge force per unit length
Q = A′y = 19mm 76mm 47,5mm
( )( )( )
• Based on the spacing between nails,
3
= 68,590mm
determine the shear force in each
nail.
For the overall beam cross-section,
N
 
F = f  = 8,2 45mm
4 4 ( )
 
1 1
I = 114mm − 76mm mm
( ) ( )  
12 12
4
=11,29×10mm
F = 369N
6 - 16
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
8Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
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
VQ
ΔH = Δx
I
• The corresponding shear stress is
ΔH VQ
τ =τ ≈ =
zx xz
tΔx It
• Previously found a similar expression
for the shearing stress in the web
VQ
τ =
xy
It
τ ≈ 0
• NOTE: in the flanges
xy
τ ≈ 0 in the web
xz
6 - 17
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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.
VQ
q=τ t =
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.
6 - 18
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
9Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
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.
6 - 19
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Sample Problem 6.3
SOLUTION:
• For the shaded area,
Q= 110mm 19,6mm 122,2mm
( )( )( )
4 3
= 26,35×10 mm
• The shear stress at a,
3 4 3
220×10 N 26,35×10 mm
( )( )
VQ
Knowing that the vertical shear is
τ = =
6 4
It
164×10 mm 19,6mm
( )( )
220 kN in a W250 x 101 rolled-steel
beam, determine the horizontal
τ =18,03MPa
shearing stress in the top flange at the
point a.
6 - 20
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
10Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Unsymmetric Loading of Thin-Walled Members
• Beam loaded in a vertical plane
of symmetry deforms in the
symmetry plane without
twisting.
My VQ
σ =− τ =
x ave
I It
• Beam without a vertical plane
of symmetry bends and twists
under loading.
My VQ
σ =− τ ≠
x ave
I It
6 - 21
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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
τ = V = q ds F = q ds = − q ds = −F′
∫ ∫ ∫
ave
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.
Fh= 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.
6 - 22
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
11Third Third
Edition Edition
5/3/2010
Beer • Johnston • DeWolf
MECHANICS OF MATERIALS
Example 6.05
• Determine the location for the shear center of the channel
section with b = 100 mm, h = 150 mm, and t = 3,8 mm.
F h
e=
V
• where
b b b
VQ V h
F = q ds = ds = st ds
∫ ∫ ∫
I I 2
0 0 0
2
Vthb
=
4I
2
 
1 1  h
3 3
I = I + 2I = th + 2 bt + bt
   
web flange
12 12 2
 
 
 
2
1
≅ th (6b+ h)
12
• Combining,
b 100mm
e= =
e= 40mm
h 150
2+ 2+
3b 3(100mm)
6 - 23
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
Example 6.06
• Determine the shear stress distribution for
V = 12 kN.
q VQ
τ = =
t It
• Shearing stresses in the flanges,
VQ V h Vh
τ = = (st) = s
It It 2 2I
Vhb 6Vb
τ = =
B
2
1
th 6b+ h
2 th 6b+ h ( )
( )
( )
12
6 12000N 100mm
( )( )
= =16,8MPa
3,8mm 150mm 6×10mm+150mm
( )( )( )
• Shearing stress in the web,
1
V ht (4b+ h) 3V 4b+ h
VQ ( ) ( )
8
τ = = =
max
2
1
It th 6b+ h t 2th 6b+ h
( ) ( )
12
3 12000N 4×100mm+150mm
( )( )
= = 29,2 MPa
2 3,8mm 150mm 6×10mm+150mm
( )( )( )
6 - 24
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