It is seen that the stress T xy varies along the depth of the beam following a parabolic lawThe maximum and minimum values of t x, in the web of the beam are obtained by putting y,=0 andbh22Vhybilz88bh?bh,2Vbilz88maxancwhenb1issmallincomparisonwithb,thereisnogreatdifferencebetween(andthedistributionoftheshearingstressesoverthecrosssectionofthewebispracticallyuniform.Agood approximationfor(txy)max isobtainedbydividing thecompleteshearingforceVbythecross-sectionalareaofthewebalone.Thisfollowsfromthefactthattheshearingstressesdistributedoverthecrosssectionof theweb yieldaforcewhichisnearlyequal toV,whichmeansthattheweb takes nearly all the shearing force and theflanges have only a secondarypart in itstransmission.11
xy in the web of the beam are obtained by putting y1=0 and xy It is seen that the stress varies along the depth of the beam following a parabolic law. The maximum and minimum values of 1 1 1 y = 2 h 8 8 min 8 8 max 2 1 2 1 1 2 1 2 1 bh bh b I V b b bh h b I V Z x y Z x y xy and the distribution of the shearing stresses over the cross section of the web is practically uniform. when b1 is small in comparison with b, there is no great difference between xy max and xy min A good approximation for ( ) max is obtained by dividing the complete shearing force V by the cross-sectional area of the web alone. This follows from the fact that the shearing stresses distributed over the cross section of the web yield a force which is nearly equal to V, which means that the web takes nearly all the shearing force and the flanges have only a secondary part in its transmission. 11
ShearFlowTxy=Tvxtlztlzm is the first moment about the neutral axis of the cumulative section area starting from the“open"endofthesection.q=tt-shearflowandhasspecialsignificance inthetorsionofthinwalledsectionsandhassomeanalogiestotheflowofanidealfluidwithinaclosedpipe.Thesumoftheshearflowsenteringandleavingthejunctionmustbezeroq=ttVmV&Iareconstantfortheentiresectionandheregcm itsvaluedoes not changewithlocalthicknessas doest.12
Shear Flow m tI V Ay tI V ydA tI V Z Z Z xy yx . m is the first moment about the neutral axis of the cumulative section area starting from the “open” end of the section. q= t – shear flow and has special significance in the torsion of thin walled sections and has some analogies to the flow of an ideal fluid within a closed pipe. The sum of the shear flows entering and leaving the junction must be zero q m q t I V V & I are constant for the entire section and here q m its value does not change with local thickness as does t. 12
ShearFlowinChannel&I-beamTheconceptof shearflowisgivenbyg=ttis veryuseful sinceit maybe shownthat shearflowsbehave likeafluid inthesensethatthe sumofall theshearflowsentering andleavinga junctionmustbezero,ieZNqi=0i=1The above equation may be proved by multiplying the shear stresses at a junction by thecorresponding wall thicknesses to givethe longitudinal shearforces ontheedges of members whichmeetatajunction.Longitudinaleguilibriumdemandsthatthesumofalltheseshearforcesequalzerointheabsenceofconcentratedlongitudinal loadsata junction.13
Shear Flow in Channel & I-beam The concept of shear flow is given by q = tx is very useful since it may be shown that shear flows behave like a fluid in the sense that the sum of all the shear flows entering and leaving a junction must be zero, ie 0 1 i N q i The above equation may be proved by multiplying the shear stresses at a junction by the corresponding wall thicknesses to give the longitudinal shear forces on the edges of members which meet at a junction. Longitudinal equilibrium demands that the sum of all these shear forces equal zero in the absence of concentrated longitudinal loads at a junction. 13
Shear CentreConsidertheaboveopensection.Wecan seethatthe shearflow ineachelementgivesrisetoaresultantforceactinginthedirectionof theshearflow.Theresultantofalltheseforcestendstotwistthecross-section(inthiscaseanti-clockwise)andisseentobeactingthroughapointPontheneutralaxis(NA).ThistendencytotwistcanbecounteractedbysomeexternalforcesystemwhichtwiststhesectioninanoppositesenseorbyapplyingtheverticalshearloadatthepointP.ThepointPisknownastheshear centre.Vi.s.h/Theshearflowg(forceperunitlength)istheflangeatpointsfromedge/21Theforces inthetwo flanges will be in oppositedirections providing a couple aboutthe centre C.Ifthereisto benotwist insection,then the force causing shearing must be applied off centre from thewebbytheamounte,ieVhtb?V.eh41h2b214114
Shear Centre t s h b P V e Consider the above open section. We can see that the shear flow in each element gives rise to a resultant force acting in the direction of the shear flow. The resultant of all these forces tends to twist the cross-section (in this case anti-clockwise) and is seen to be acting through a point P on the neutral axis (NA). This tendency to twist can be counteracted by some external force system which twists the section in an opposite sense or by applying the vertical shear load at the point P. The point P is known as the shear centre. The shear flow q (force per unit length) is the flange at point s from edge 2 . . h I V t s The forces in the two flanges will be in opposite directions providing a couple about the centre C. If there is to be no twist in section, then the force causing shearing must be applied off centre from the web by the amount e, ie I h b t e h I Vhtb V e 4 4 . 2 2 2 14
Forthechannelsection62th3+2xhxt1242th2h+b)62h2b21242th2(%+b)b262hh/+2b3bbh3bThus'eis independentoftheforcesandis onlyafunctionofsectional propertiesForsectionswhichhavenoplanesofsymmetry,theshearflowsmustbecalculatedforplanesparalleltotheprincipalplanesofthesection,thecorrespondingforcesdeducedandthemomentoftheseforcesaboutsomepoletobetakeninordertodeterminetheshearcentre15
For the channel section ) 6 ( 2 4 2 12 2 2 3 2 b h th b b t th I Thus ‘e’ is independent of the forces and is only a function of sectional properties. For sections which have no planes of symmetry, the shear flows must be calculated for planes parallel to the principal planes of the section, the corresponding forces deduced and the moment of these forces about some pole to be taken in order to determine the shear centre. 2 3 2 3 2 3 6 2 4 2 2 2 2 2 2 b h b b h b b b h b b h th h b t e 15