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

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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

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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

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

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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

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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.

6 - 6

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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

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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.

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

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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

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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:

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

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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

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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

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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

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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 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

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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

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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.

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

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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

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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

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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

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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

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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

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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

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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

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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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