Lecture Mechanics of materials (Third edition) - Chapter 6: Shearing stresses in beams and thinwalled members

Third Edition

CHAPTER

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

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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 τxy 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
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Introduction
• Transverse loading applied to a beam
results in normal and shearing stresses in
transverse sections.
• Distribution of normal and shearing
stresses satisfies

Fx = ∫ σ x dA = 0
Fy = ∫ τ xy dA = −V
Fz = ∫ τ xz dA = 0

(

)

M x = ∫ y τ xz − z τ xy dA = 0
M y = ∫ z σ x dA = 0
M z = ∫ (− y σ 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.
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Shear on the Horizontal Face of a Beam Element
• Consider prismatic beam
• For equilibrium of beam element
∑ Fx = 0 = ∆H + ∫ (σ D − σ 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 =

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MECHANICS OF MATERIALS

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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
= first moment with respect
to neutral axis
∆H ′ = − ∆H
q′ =

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6-5

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MECHANICS OF MATERIALS

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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.

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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
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
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F = (0.025 m)q = (0.025 m)(3704 N m

F = 92.6 N

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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

τ ave =

• 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 D1
and D2 are significantly higher than at D.
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Shearing Stresses τxy in Common Types of Beams
• For a narrow rectangular beam,
VQ 3 V ⎛⎜
1−
τ xy =
=
⎜
Ib 2 A ⎝

τ max =

y 2 ⎞⎟
c 2 ⎟⎠

3V
2A

• For American Standard (S-beam)
and wide-flange (W-beam) beams
VQ
It
V
τ max =
Aweb

τ ave =

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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 ⎛⎜
τ xy =
1−
2 A ⎝⎜

y 2 ⎞⎟
c 2 ⎟⎠

σx = +

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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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,
σ all = 1800 psi

τ 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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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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Sample Problem 6.2
• Determine the beam depth based on allowable
normal stress.
σ 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 (3.5 in.) d
d = 10.71in.

τ all =

• Required beam depth is equal to the larger of the two.
d = 10.71in.
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Longitudinal Shear on a Beam Element
of Arbitrary Shape
• We have examined the distribution of
the vertical components τxy on a
transverse section of a beam. We
now wish to consider the horizontal
components τxz of the stresses.
• Consider prismatic beam with an
element defined by the curved surface
CDD’C’.

∑ Fx = 0 = ∆H + ∫ (σ D − σ C )dA
a

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

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VQ
∆x
I

q=

∆H VQ
=
∆x
I
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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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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
in
2
= 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
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• Based on the spacing between nails,
determine the shear force in each
nail.
lb ⎞
⎛
F = f A = ⎜ 46.15 ⎟(1.75 in )
in ⎠
⎝

F = 80.8 lb
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MECHANICS OF MATERIALS

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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
τ zx = τ xz ≈

∆H VQ
=
t ∆x It

• Previously found a similar expression
for the shearing stress in the web
τ xy =

VQ
It

• NOTE: τ xy ≈ 0
τ xz ≈ 0
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in the flanges
in the web
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Shearing Stresses in Thin-Walled Members
• The variation of shear flow across the
section depends only on the variation of
the first moment.
q =τt =

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.

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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.

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Plastic Deformations
I
c

• Recall: M Y = σ Y = maximum elastic moment
• 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 =
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Mp
L
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Plastic Deformations
• Preceding discussion was based on
normal stresses only
• Consider horizontal shear force on an
element within the plastic zone,
∆H = −(σ C − σ D )dA = −(σ Y − σ Y )dA = 0

Therefore, the shear stress is zero in the
plastic zone.
• Shear load is carried by the elastic core,
3 P ⎛⎜
τ xy =
1−
2 A′ ⎜⎝

τ max =

y 2 ⎞⎟
where A′ = 2byY
2⎟
yY ⎠

3P
2 A′

• As A’ decreases, τmax increases and
may exceed τY
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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.

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(
)

)

VQ (50 kips ) 15.98 in 3
τ=
=
It
394 in 4 (0.770 in )

(

τ = 2.63 ksi

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Unsymmetric Loading of Thin-Walled Members
• Beam loaded in a vertical plane
of symmetry deforms in the
symmetry plane without
twisting.
σx = −

My
I

τ ave =

VQ
It

• Beam without a vertical plane
of symmetry bends and twists
under loading.
σx = −

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My
I

τ ave ≠

VQ
It

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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
VQ
τ ave =
V = ∫ q ds
It
B

B

E

A

D

F = ∫ q ds = − ∫ q ds = − F ′

• 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.

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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.)

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e = 1.6 in.

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