shear centre - ecajmer.ac.in
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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.
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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” .
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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 .
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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 usedto calculate the amount of deflection.
• The flexural formula can be used to calculatethe stress in the beam.
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Shear Center for Axis Symmetry
• Every elastic beam cross-section has a pointthrough which transverse forces may beapplied so as to produce bending only, with notorsion of the beam. The point is called the“shear center” of the beam.
• The shear center for any transverse section ofthe beam is the point of intersection of thebending axis and the lane of the transversesection. Shear center is also called center oftwist.
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Flexure axis or bending axis
• Flexural axis of a beam is the longitudinal axisthrough which the transverse bending loads mustpass in order that the bending of the beam shallnot be accompanied by twisting of the beam.
• In Fig.3 ABCD is a plane containing the principalcentroidal axis of inertia and plane AB’C’D is theplane containing the loads. These loads will causeunsymmetrical bending. In Fig.3 AD is the flexuralaxis [2].
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Classification on the basis of symmetry
• Double symmetrical section
• Single symmetrical section
• Unsymmetrical section
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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.
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Shear center or Center of flexure
• Let Ox be centroidal axis .The load, in general will,at any section, cause:
1. Normal stress 𝜎𝑥 due to flexure;
2. Shear stresses 𝜏𝑥𝑦 and 𝜏𝑥𝑧 due to thetransverse nature of the loading; and
3. Shear stresses 𝜏𝑥𝑦 and 𝜏𝑥𝑧 due to torsion.4. To arrive at the solution, we assume that
𝜎𝑥=-−𝑃 𝐿−𝑥 𝑦
𝐼𝑧,𝜎𝑦=𝜎𝑧=𝜏𝑦𝑧=0
This is known as St.Venant’s assumption.
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Shear center or Center of flexure
• 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.
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Shear center for a Channel section
• F1= (𝜏a/2) bt , and sum of vertical shear
stresses over area of web is,
F3= −ℎ/2ℎ/2
𝜏 𝑡 𝑑𝑦
• F1h=Fe and F=F3
• e =𝐹1ℎ
𝐹=
1
2𝜏𝑎𝑏1𝑡ℎ
𝐹=𝑏1𝑡ℎ
2𝐹
𝐹3𝑄
𝐼𝑡=𝑏1 𝑡ℎ 𝐹3 𝑏1𝑡(
ℎ
2)
2𝐹 .𝐼.𝑡
e=𝑏12
4𝐼ℎ2 𝑡
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Shear center for a Channel section
• I= Iweb + 2 Iflange =1
12tℎ3 + 2[ (1/12) b1 𝑡3 + b1 t (h/2)^2
• =(1/12) t ℎ2 (6b + h)
• So finally, e=3
6𝑏𝑡ℎ𝑏12 =
𝑏1
2+ℎ
3𝑏𝑡
• Here ‘e’ is independent of the magnitude of applied force F as wellas of its location along the beam.
• The shear center for any cross section lies on a longitudinal lineparallel to the axis of the beam.
• The procedure of locating the shear center consists of determiningthe shear forces , as F1 and F3 at a section and then finding thelocation of the external forces as F1 and F3,at a section and thenfinding the location of the external force necessary to keep theseforces in equilibrium.
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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 ∑MA=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
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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 b2 𝑏2
3)
• 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.
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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.
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Fig .10. Location of shear centre for inverted T-
section and angle section.
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Fig.11 Twisting effect on some cross-sections if load is
not applied through shear center.
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1. Determination of the shear centre for the
channel section shown in figure below.
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Solution
• e= 𝑏1
2+𝑤ℎ
3𝑏𝑡
• Here b1 =10-1=9cm
• h =15-1=14cm
• w =1cm
• t =1cm
• e = 1
2.9
1+
12.9
16.(1 𝑋14)(9𝑋 1)
=3.57 cm
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Solution
• H2 = 𝜏 𝑑𝐴
= 𝐹𝐴𝑦
𝐼 𝑡dA =
𝐹
𝐼 𝑡 𝐴 𝑦 dA
=𝐹
𝐼 𝑡 0
32 3 − 𝑥 6.5 2 𝑑𝑥 = 58.5(F/I)
• H1 =𝐹
2𝐼 0
42 4 − 𝑥 𝑋 6.5 𝑋 2𝑑𝑥 =104(F/I)
• 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
• e =2𝑋 45.5 𝑋 6.5
𝐼=591.5/I
• I = 2[7𝑋23
12+ 14 X 6.52 ]+
1𝑋113
12• =1303.251 𝑐𝑚4
• e = 591.5 / 1303.251 = 0.454 cm
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Solution
• The static moment of the crossed section is,
Qz = 0
𝜃𝑅 𝑑𝜑 𝑡 𝑅 𝑠𝑖𝑛𝜑
• =R^2 . t (1-cos𝜃)
• Iyz=0,
• 𝜏𝑥𝑧 = 𝐹𝑄𝑧
𝑡𝐼𝑧= ( F/t Iz ) R^2 t (1-cos𝜃 )
• But Iz = 𝜋. R^3. T
• Hence𝜏𝑥𝑧 =𝐹
𝜋𝑅𝑡(1 − 𝑐𝑜𝑠𝜃)
• When 𝜃 = 180°,• 𝜏𝑥𝑧 = 2F / (𝜋𝑅𝑡)
• M= 02𝜋𝜏𝑥𝑧 𝑅𝑑𝜃 𝑡 𝑅
• = 𝐹
𝜋𝑅𝑡 02𝜋𝑅2 𝑡 (1 − cos 𝜃 ) 𝑑𝜃
• = 2 FR
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Conclusion
• The shear center is having practical significancein the study of behavior of beams with sectioncomprising of thin parts, such as channels, angles,I-sections, which are having less resistance totorsion but high resistance to flexure.
• To prevent twisting of any beam cross-section, theload must be applied through the shear centre.
• It is not necessary, in general, for the shear centreto lie on the principal axis, and it may be locatedoutside the cross section of the beam.
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