MECHANICS OF MATERIALS CHAPTER 6 SHEARING STRESSES IN BEAMS AND THIN-WALLED MEMBERS

CHAPTER MECHANICS OF

6 MATERIALS
Shearing Stresses in
Beams and Thin-
Walled Members

MECHANICS OF MATERIALS

Introduction

Transverse loading applied to beam results in
normal and shearing stresses in transverse
sections.

Distribution of normal and shearing stresses
satisfies ( from equilibrium)

Fx   xdA  0 M x   y xz  z xy dA  0
Fy   xydA  V
Fz   xzdA  0 M y   z xdA  0
M z    y x   M

When shearing stresses are exerted on
vertical faces of an element, equal stresses
exerted on horizontal faces

Longitudinal shearing stresses must exist in
any member subjected to transverse
loading.

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MECHANICS OF MATERIALS

Vertical and Horizontal Shear Stresses

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MECHANICS OF MATERIALS

Shear Stress in Beams

Two beams glued together
along horizontal surface

When loaded, horizontal
shear stress must develop
along glued surface in
order to prevent sliding
between the beams.

MECHANICS OF MATERIALS

Shear on Horizontal Face of Beam Element

Consider prismatic beam

Equilibrium of element CDC’D’

 Fx  0  H    D   C dA

A

H  M D  MC  y dA

IA

Let, Q   y dA
A

M D  M C  dM x  V x
dx

H  VQ x
I

q  H  VQ  shear flow
x I

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MECHANICS OF MATERIALS

Shear on Horizontal Face of Beam Element

Shear flow,

q  H  VQ  shear flow
x I

where

Q   y dA

A

 first moment of area above y1
I   y2dA

A A'

 second moment of full cross section

Same result found for lower area

Q  Q   y dA  0
A1  A2

(first moment of area wrt NA is zero)

q  H   VQ  V (Q)  q
x I I

H   H

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MECHANICS OF MATERIALS

Example 6.1

Beam made of three planks, nailed
together. Spacing between nails is 25
mm. Vertical shear in beam is
V = 500 N. Find shear force in each
nail.

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MECHANICS OF MATERIALS

Example 6.1

SOLUTION:
Find horizontal force per unit length or

shear flow q on lower surface of
upper plank.

VQ (500N)(120 106 m3)
q 
I 16.20 10-6 m4

Q  Ay  3704 N m

 0.020m  0.100m0.060m Calculate corresponding shear force in
each nail for nail spacing of 25 mm.
 120 106 m3
F  (0.025m)q  (0.025m)(3704 N m
I  112 0.020m0.100m3 F  92.6 N

 2[ 112 0.100m0.020m3
 0.020m  0.100m0.060m2]

 16.20 106 m4

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MECHANICS OF MATERIALS

Determination of Shearing Stress

Average shearing stress on horizontal face of
element is shearing force on horizontal face
divided by area of horzontal face.
 ave  H  q x  VQ x
A A I t x

   ave  VQ ; tq
It

Note averaging is across dimension t (width)
which is assumed much less than the depth, so
this averaging is allowed.

On upper and lower surfaces of beam, tyx= 0. It
follows that txy= 0 on upper and lower edges of
transverse sections.

If width of beam is comparable or large relative to
depth, the shearing stresses at D’1 and D’2 are

significantly higher than at D, i.e., the above
averaging is not good.

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MECHANICS OF MATERIALS

Shearing Stresses txy in Common Types of Beams

For a narrow rectangular beam,

VQ 3 V  y2 
 xy   1  2 
Ib 2 A  c   

 max  3 V
2A

For I beams

 ave  VQ
It

 max  V
Aweb

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MECHANICS OF MATERIALS

Example 6.2

Timber beam supports three
concentrated loads.

 all  12 MPa  all  0.8 MPa

Find minimum required depth d of
beam.

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MECHANICS OF MATERIALS

Example 6.2

Vmax  14.5 kN
M max  10.95 kNm

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MECHANICS OF MATERIALS

Example 6.2

Determine depth based on allowable normal stress.

 all  M max
S

61210 Pa  10.95103 Nm

0.015 m d 2

d  0.246 m  246mm

I  112 b d 3 Determine depth based on allowable shear stress.

S  I  61 b d 2  all  3 Vmax
c 2A

 1 0.09 md 2 0.8106 Pa  3 14500
6
2 0.09 md
 0.015 md 2
d  0.322 m  322 mm

Required depth d  322 mm

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MECHANICS OF MATERIALS

Longitudinal Shear Element of Arbitrary Shape

Have examined distribution of vertical

components txy on transverse section.
Now consider horizontal components

txz .

Consider element defined by curved
surface CDD’C’.

 Fx  0  H    D  C dA
A

So only the integration area is different,

hence result same as before, i.e.,

H  VQ x q  H  VQ
I x I

Will use this for thin walled members

also

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MECHANICS OF MATERIALS

Example 6.3

Square box beam constructed from four
planks. Spacing between nails is 44
mm. Vertical shear force V = 2.5 kN.
Find shearing force in each nail.

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MECHANICS OF MATERIALS

Example 6.3

SOLUTION:

Determine the shear force per unit
length along each edge of the upper
plank.

VQ 2500 N64296 mm3  N
q   15.6
I 10332 mm 4 mm

f  q  7.8 N
2 mm

For the upper plank,  edge force per unit length

Q  Ay  18mm76 mm47 mm Based on the spacing between nails,
determine the shear force in each
 64296 mm3 nail.

For the overall beam cross-section,  N

I  1 112 mm4  1 76 mm4 F  f   7.8 44 mm
12 12
 mm 

 10332 mm4 F  343.2 N

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MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

Shear stress assumed constant through
thickness t, i.e., due to thinnness our
averaging is now accurate/exact.

Consider I-beam with vertical shear V.

Longitudinal shear force on element is

H  VQ x
I

Corresponding shear stress is

 zx   xz  H  VQ
t x It

Previously had similar expression for
shearing stress web

 xy  VQ
It

NOTE:  xy  0 in the flanges
 xz  0 in the web

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MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

The variation of shear flow across the
section depends only on the variation of
the first moment.

q   t  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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MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

For wide-flange beam, shear flow q
increases symmetrically from zero at A
and A’, reaches a maximum at C and
then 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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MECHANICS OF MATERIALS

Example 6.4

Q  108 mm19.6 mm122.2 mm

 258700 mm3

Shear stress at a,

Vertical shear is 200 kN in a VQ 200103 N258.7 106 m3 
W250x101 rolled-steel beam. Find  
horizontal shearing stress at a. It 16410 m 0.0196 m
6 4

  16.1MPa

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