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“Shear Centre”

A seminar on,

Prepared by, M.prasannakumar

Venkatesha.A(1RV13CSE06)

Under the guidance of,M.V.Renuka devi

Professor

Dept. of Civil Engineering, RVCE, Bangalore


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 due to flexure;2. Shear stresses and due to the transverse

nature of the loading; and3. Shear stresses and due to torsion.4. To arrive at the solution, we assume that - ,===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= (a/2) bt , and sum of vertical shear stresses over area of web is,

F3=• F1h=Fe and F=F3

• e = = = = e=



• I= Iweb + 2 Iflange = t + 2[ (1/12) b1 + b1 t (h/2)^2 • =(1/12) t (6b + h) • So finally, e= = • 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 ∑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)• 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= • Here b1 =10-1=9cm• h =15-1=14cm• w =1cm• t =1cm• e = =3.57 cm


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


Solution

• H2 = = dA = dA = = 58.5(F/I)• H1 = =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 = =591.5/I• I = 2[+ 14 X ]+ • =1303.251 • 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 =• =R^2 . t (1-cos)• Iyz=0,• = ( F/t Iz ) R^2 t (1-cos )• But Iz = . R^3. T• Hence) • When • = 2F / (• M= • = • = 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.


References1. 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.

http://gaia.ecs.csus.edu/-ce113/steel%20-shear.pdf

http://www.me.mtu.edu/-mavable/Spring03/chap6.pdf

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