shear centre

A seminar on,
“Shear Centre”
Prepared by,
M.prasannakumar
Venkatesha.A
(1RV13CSE06)
Under the guidance of,
M.V.Renuka devi
Professor
Dept. of Civil Engineering, RVCE, Bangalore
Classified - Internal use

Introduction
• Most examples of beam bending involve
beams with the symmetric cross sections.
• However, there are an ever increasing number
of cases where the cross section of a beam is
not symmetric about any axis.
• If the cross section of the beam does not have a
plane of symmetry, the displacements of the
beam get increasingly complicated.

Fig.1 Effect of loading at shear center

Effect of loading at shear center
• Case1: The displacement consists of both
translation down and anticlockwise twist.
• Case2: The displacement consists of both
translation down and clockwise twist.
• Somewhere in-between these two extremes we
would expect a point that we could apply the
load and produce only a twisting .This point is
called “shear center” .

Fig. 2. Effects of loads on unsymmetrical section.

Effect of loading at shear center
• The flexural formula σ=My/I is valid only if the
transverse loads which give rise to bending act in
a plane of symmetry of beam cross section.
• In this type of loading there is obviously no
torsion of the beam.
• In more general cases the beam cross section will
have no axis of symmetry and the problem of
where to apply the load so that the action is
entirely bending with no torsion arises .

Advantages of loading the beam at
the shear center
• The path of any deflection is more obvious.
• The beam translates only straight downward.
The standard deflection formulas can be used
to calculate the amount of deflection.
• The flexural formula can be used to calculate
the stress in the beam.

Shear Center for Axis Symmetry
• Every elastic beam cross-section has a point
through which transverse forces may be
applied so as to produce bending only, with no
torsion of the beam. The point is called the
“shear center” of the beam.
• The shear center for any transverse section of
the beam is the point of intersection of the
bending axis and the lane of the transverse
section. Shear center is also called center of
twist.

Flexure axis or bending axis
• Flexural axis of a beam is the longitudinal axis
through which the transverse bending loads must
pass in order that the bending of the beam shall
not be accompanied by twisting of the beam.
• In Fig.3 ABCD is a plane containing the principal
centroidal axis of inertia and plane AB’C’D is the
plane containing the loads. These loads will cause
unsymmetrical bending. In Fig.3 AD is the flexural
axis [2].

Fig.3 Flexural axis or bending axis

Classification on the basis of symmetry
• Double symmetrical section
• Single symmetrical section
• Unsymmetrical section

Fig.4 Two axis symmetry[7]

Fig.5 One axis symmetry & unsymmetrical section[7]

Shear center or Center of flexure
• Beam carries loads which are transverse to the
axis of the beam and which cause not only
normal stresses due to flexure but also
transverse shear stresses in any section.
• Consider the cantilever beam shown in Fig.7
carrying a load at the free end. In general, this
will cause both bending and twisting.

Fig. 6 Cantilever beam loaded with force P

• Let Ox be centroidal axis .The load, in general will,
at any section, cause:
1. Normal stress 휎푥 due to flexure;
2. Shear stresses 휏푥푦 and 휏푥푧 due to the
transverse nature of the loading; and
3. Shear stresses 휏푥푦 and 휏푥푧 due to torsion.
4. To arrive at the solution, we assume that
휎푥=-
−푃 퐿−푥 푦
퐼푧
,휎푦=휎푧=휏푦푧=0
This is known as St.Venant’s assumption.

• A position can be established for which no
rotation occurs .
• If a transverse force is applied at this point, we
can resolve it into two components parallel to the
y and z axis and note from the above discussion
that these components do not produce the rotation
of centroidal elements of the cross sections of the
beam. This point is called the shear center of
flexure or flexural centre.

Fig. 7 Load P passing through shear centre

Shear center for a Channel section
Fig. 8 Beam of Chanel cross section.

• F1= (휏a/2) bt , and sum of vertical shear
stresses over area of web is,
ℎ/2
F3= −ℎ/2
휏 푡 푑푦
• F1h=Fe and F=F3
• e =
퐹1ℎ
퐹
=
1
2
휏푎푏1푡ℎ
퐹
=
푏1푡ℎ
2퐹
퐹3푄
퐼푡
=
푏1 푡ℎ 퐹3 푏1푡(
ℎ
2
)
2퐹 .퐼.푡
e=
2
4퐼
푏1
ℎ2 푡

• I= Iweb + 2 Iflange =
1
12
tℎ3 + 2[ (1/12) b1 푡3 + b1 t (h/2)^2
• =(1/12) t ℎ2 (6b + h)
• So finally, e=
3
6푏푡ℎ
2 =
푏1
푏1
2+
ℎ
3푏푡
• Here ‘e’ is independent of the magnitude of applied force F as well
as of its location along the beam.
• The shear center for any cross section lies on a longitudinal line
parallel to the axis of the beam.
• The procedure of locating the shear center consists of determining
the shear forces , as F1 and F3 at a section and then finding the
location of the external forces as F1 and F3,at a section and then
finding the location of the external force necessary to keep these
forces in equilibrium.

Shear center for I-section:
Fig. 9 Beam of Chanel cross section.

Shear center for I-section:
• Assuming an I section of cross-sections mentioned in
figure,
• For equilibrium,F1 + F2 =F
• Likewise to have no twist of the section,From ΣMA=0,
Fe1 =Fe2h & Fe2= F1h,
• Since the area of a parabola is (2/3) of the base times
the maximum altitude.
• F2= (2/3) b2 (q2) max
• Since V=F
• (q2) max= VQ/I =FQ/I

Shear center for I-section
• Where Q is the statical moment of the upper half of
the right hand flange and I is the moment of inertia of
the whole section. Hence
• Fe=F2h= (2/3) b2 (q2) max
• e1= (2hb2Q)/(3I)= (2h b2/3I) (b2 t2 /2 ) (b2/4) = h/I (t2
3)
b2 푏2
• Where I2 is the moment of inertia of the right hand
flange around the central axis.
• e2=h ( I1/ I )
• Where I1 is the moment of inertia of left hand flange.

Shear Centers for a few other sections
• Thin walled inverted T-section, the distribution
of shear stress due to transverse shear will be
as shown in Fig.11 .
• The moment of this distributed stress about C
is obviously zero. Hence, the shear center for
this section is C.

Fig .10. Location of shear centre for inverted T-section
and angle section.

Fig.11 Twisting effect on some cross-sections if load is
not applied through shear center.

1. Determination of the shear centre for the
channel section shown in figure below.

Solution
• e=
푏1
2+
푤ℎ
3푏푡
• Here b1 =10-1=9cm
• h =15-1=14cm
• w =1cm
• t =1cm
• e =
1
2
.9
1+
1
2
.9
1
6
.
(1 푋14)
(9푋 1)
=3.57 cm

2.Locating the shear centre of the cross section
shown in figure.

Solution
• H2 = 휏 푑퐴
=
퐹퐴푦
퐼 푡
퐹
퐼 푡
dA =
퐴 푦 dA
퐹
퐼 푡
=
3
2 3 − 푥 6.5 2 푑푥 = 58.5(F/I)
0
퐹
2퐼
• H1 =
4
2 4 − 푥 푋 6.5 푋 2푑푥 =104(F/I)
0
• Taking moments about point D, we get
• FR e = 2 (H1 –H2) 6.5
• =2(104-58.5)6.5 X (F/I)
• Now FR= F
2푋 45.5 푋 6.5
• e =
퐼
=591.5/I
• I = 2[
7푋23
12
+ 14 X 6.52 ]+
1푋113
12
• =1303.251 푐푚4
• e = 591.5 / 1303.251 = 0.454 cm

3.Determination of shear centre for a circular open
section

Solution
• The static moment of the crossed section is,
휃
Qz = 0
푅 푑휑 푡 푅 푠푖푛휑
• =R^2 . t (1-cos휃)
• Iyz=0,
• 휏푥푧 = 퐹
푄푧
푡퐼푧
= ( F/t Iz ) R^2 t (1-cos휃 )
• But Iz = 휋. R^3. T
• Hence휏푥푧 =
퐹
휋푅푡
(1 − 푐표푠휃)
• When 휃 = 180°,
• 휏푥푧 = 2F / (휋푅푡)
2휋
• M= 0
휏푥푧 푅푑휃 푡 푅
• =
퐹
휋푅푡
2휋
0
푅2 푡 (1 − cos 휃 ) 푑휃
• = 2 FR

Conclusion
• The shear center is having practical significance
in the study of behavior of beams with section
comprising of thin parts, such as channels, angles,
I-sections, which are having less resistance to
torsion but high resistance to flexure.
• To prevent twisting of any beam cross-section, the
load must be applied through the shear centre.
• It is not necessary, in general, for the shear centre
to lie on the principal axis, and it may be located
outside the cross section of the beam.

References
1. Alok Gupta (2004) , “Advanced strength of materials’’, Umesh
Publications, First Edition.
2. LS Srinath, “Advanced Mechanics of solids”, 15th edition, Tata McGraw
Hill.
3. Timoshenko & JN Goodier (1997), “Mechanics of solids”, Tata McGraw
Hill.
4. S.S.Bhavikatti,“Structural Analysis” ,Vol.2.
5. Vazrani and Ratwani“ Analysis of structures”,Vol 2.
6. B.C.Punmia & A.K.Jain . “Strength of materials and Theory of
structures”, Vol.2Laxmi Publications (P) Ltd.
7. James Gere & Barry Goodno, “ Mechanics of materials”, Google Books.
8. A.C. Ugral ,“Advanced Mechanics of Materials and Applied
Elasticity”,Fifth Edition”, Safari Books Online.
9. http://gaia.ecs.csus.edu/-ce113/steel -shear.pdf
10. http://www.me.mtu.edu/-mavable/Spring03/chap6.pdf
11. Jaehong Lee, “Centre of gravity and shear centre”,
www.elsevier.com/locate/compstruct.

shear centre

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shear centre