Shear Centre - ecajmer.ac.in

Shear Centre

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.

Classified - Internal use

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 usedto calculate the amount of deflection.

• The flexural formula can be used to calculatethe stress in the beam.


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.


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].


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 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.



• 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= (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



• 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.


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 ∑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


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.


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= 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


2.Locating the shear centre of the cross section

shown in figure.


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


3.Determination of shear centre for a circular open

section


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


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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Shear Centre - ecajmer.ac.in