SHEAR STRESS

a) Average shear stress

The average shear stress acting over a surface on which a shearing force is acting is given by

Average shear stress (  ) =

b) Shear strain
The action of shear stress on a material gives rise to shear strain. Shear strain (  ) is measured as
the angular distortion of a small rectangular element of the material. Shear strain is measured in
radians and is dimensionless

c) Shear modulus
For a linearly elastic material shear stress and shear strain are related by the shear modulus ( G ),
sometimes referred to as the modulus of rigidity. The shear modulus is given by the formula

d) Complementary shear stress

A shear stress acting on either of two planes which intersect at right angles is always
accompanied by a complementary shear stress of equal magnitude but opposite sign acting on the
other plane

e) Shear stresses in Beam Sections

The general formula for calculating the shear stress in a beam section which is subjected to a
shearing force is given by

QA y

Ib
Where  = the shear stress;
Q = the transverse shear force acting at the section;
A = the area to the outside of the level where shear stress is being calculated;
_
y = the distance to the centroid of the area A measured from the neutral axis;
I = the second moment of area of the whole cross-section about the neutral axis; and
b = the width of the cross section at the level where shear stress is being calculated.

The terms used in this equation are illustrated

In commonly occurring structural sections the maximum shear stress calculated by use of the
above formula usually occurs at the level of the neutral axis. In rectangular, T-shaped and I-
shaped beams and other commonly occurring sections the shear stress varies parabolically
throughout the depth of the section, with abrupt changes of stress where the geometry of the
section changes suddenly, such as where the web and flanges of an I-section meet.

f) Shear stresses in rectangular beam sections

For a rectangular beam section the maximum shear stress is 50% greater than the average shear
stress.

EXAMPLES

Q1. Show that in a beam of rectangular cross-section subjected to a transverse shearing force, the
value of the maximum shear stress is 50% greater than the average shear stress.

S1.

Let the dimensions

d/2 of the beam be as
d/4 shown beside.
The maximum shear
d stress will occur at
the level of N.A.,
which is at mid depth
of the section in this
case.
b

2
 _
QA y
The shear stress  , is given by  
Ib
For a rectangular cross section, I  bd 3 12
_
And for shear stress at mid-depth A y  b * d 2 * d 4  bd 2 8

Hence  max 

Q * bd 2 8 
 1.5 Q bd

bd 3 12 * b 
The average shear stress is given by the total shearing force Q per unit gross beam
section ( b* d ). Hence
 max  1.5 * average shear stress = 50% greater than average shear stress

Q2. The I-section beam shown in Fig. Q2 is subjected to a transverse shearing force of 450kN.
(i) Compute the value of the maximum shear stress and sketch the distribution of shear in
the web.
(ii) Calculate the percentage difference between the maximum and the average shear stress in
the web of this section.
170
40

15

400 X X

40

S2.

b) (i) Calculate I XX
170
40

I XX  I CC  Ah 2 but Ah 2  0
15
 I xx  I CC  bd 3 12
400 X X
170 * 4803 155 * 4003
I xx    740.053*106 mm 4
12 12
I xx  740.053*10 mm 4
6

40

3
 _
Calculate first moment of area A y (taking upper section above N.A.)
_
A y  Ay flange  Ay web  (170 * 40) * 220  (15 * 200) *100
_
A y  1.796 * 10 6 mm 4
_
QA y 450 * 103 * 1.796 * 106
Hence  max   6
 72.81 N mm 2
Ib 740.053* 10 * 15
Sketch

X X

72.81N/mm2

Q 450 *103
(i) Average shear stress,  av    62.5 N / mm 2
bd 15 * 480
72.81  62.5
Difference =  16.50%
62.5

Q3. The I-section beam shown in Fig. Q3 is subjected to a transverse shearing force of 500 kN
(i) Obtain the value of the maximum shear stress and sketch the distribution of shear stress
in the web. [Answer: max shear stress (at Neutral axis) = 73.01 N/mm2]
(ii) Calculate the percentage difference between the maximum and the average shear stress in
the web of this section.[Answer: 12.15 %]

300
20

12

600 X X

20

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Q4. Consider the beam section shown hereunder. Plot the distribution of shear stress along the
vertical axis and determine the maximum shear force which can be carried by the section if the
permissible maximum shear stress must not exceed 90 N/mm2.

S4:
Position of N.A from bottom = 32.5 mm
IXX = 2,354.16 * 103 mm4
Q = 70.55 kN

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Q5: A universal beam section is strengthened by welding a 200 mm * 20 mm plate to the top
flange as shown hereunder.
a) For the compound section determine:
i) The depth to the neutral axis
ii) The second moment of area about the neutral axis
b) If this compound section is then used as a simply supported beam and the permissible
bending stresses are limited to 140 N/mm2 and 80 N/mm2 in tension and compression,
respectively, and the permissible shear stress is 70 N/mm2, then determine:
i) The maximum bending moment
ii) The maximum shearing force which the beam can sustain

S5
Note that the section properties of the universal beam are given and the calculation of the section
properties of the composite section will defer slightly from the previously worked examples.
In the following calculations moments of area are taken about the bottom face of the beam

Part Area (A) y Ay ICC h Ah2

(mm2) (mm) (mm3*103) (mm4*106) (mm) (mm4*106)
Plate 4000 270 1080.0 0.13 80.8 26.11
I-beam 5460 130 709.8 65.82 59.2 19.14
9460 1789.8 65.95 45.25

The bending moment that will cause limiting tensile stress on the bottom face of the beam is
given by

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Q6: The Figure hereunder shows the cross-section of a steel beam fabricated by welding a 180
mm wide by 12 mm thick plate to the top flange of a doubly symmetric 200 mm * 100 mm
I-section. The cross-sectional area of the I-beam is 4160 mm2 and its second moment of area
IXX is 25.78 * 106 mm4. It is subjected to a shearing force of 50 kN.
(a) Determine and sketch the distribution of vertical shear stress over the depth of the
section, showing all principal values.
(b) Compare the maximum shear stress in the web from the analysis in (a) with the average
shear stress based on the assumption that the total shear load is carried uniformly by the
web alone.
(c) Determine the load per unit length of the beam that must be resisted by the welded
connection

S6
Apply the shear stress equation is repeated where there is a change in geometry and at the neutral
axis, as marked

Part Area (A) y Ay ICC h Ah2

(mm2) (mm) (mm3*103) (mm4*106) (mm) (mm4*106)
Plate 2160 206 444.96 0.03 69.77 10.51
I-beam 4160 100 416.0 25.78 36.23 5.46
6320 860.96 25.81 15.97

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The shear stresses are computed thus:

At level b Aỳ τ
(mm) (*103 mm3) (N/mm2)
1 100 156.28 1.88
2 10 156.28 18.75
3 10 233.44 28.01
4 10 220.03 26.40
5 100 220.03 2.64
6 100 150.70 1.81
7 180 150.70 1.00

The shear stress distribution is as shown hereunder. Take note of the abrupt changes of stress at
the intersection of the web and flanges and where the top flange and flange plate meet

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