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Concrete In Australia : June 2013
Concrete in Australia Vol 39 No 2 35 Figure A1. Analysing inter plank shear. Figure A2. A typical example of a finite width solution. Figure A3. A loaded plank at the edge. erefore, if we know v1 at either side of the loaded plank and r, we know all the inter plank shears. e finite width solution may then be found by applying a shear correction to make v edge = 0, and reflecting this from edge to edge until no further correction is required. Figure A2 shows a typical example of this process taking v1 = 100, and r = 0.5. Figure A3 shows another example with the same arbitrary values for v1 and r, but with the loaded plank at the edge. To fi nd r, consider the following typical compatibility requirement: [(v n -v n+1)(L/π)4]/EI + [(v n+ v n+1)B2(L/π)2]/GJ =[(v n+1-vn+2)(L/π)4]/EI - [v n+1 + v n+2)B2(L/π)2]/GJ erefore: [(vn --2vn+1+vn+2)(L/π)2]/EI=[(-vn -2vn+1-vn+2)(B2)]/GJ -vn-2vnr--vnr2=(L/π)2GJ=Q vn-2vnr+vnr2B2EI -(r+1)2=Q (r-1)2 r=√Q-1 (1) √Q+1 Equation 1 permits r to be determined where Q = (L/π)2 GJ B2 EI And G = Shear modulus of plank concrete (= E/2(1+v) & v = Poisons ratio = 0.15, say) J = Torsional inertia of plank section E = Modulus of elasticity of plank concrete I = Moment of inertia of plank concrete L=Span B = Half the plank width To fi nd v1 at locations a and b, as shown in Figure A4, with l1 applied at any eccentricity within the plank, consider the symmetrical and anti-symmetrical cases shown in Figure A4. en the general case may be found by combining these symmetrical and anti-symmetrical cases such that: P=l1 , andM=Pe =l1EB(whereE=e/B=0for no eccentricity, & 1 for maximum e =B) en for the symmetrical case, v = va = vb : Forcom patibility (P-2v)(L/π)4 /EI = (v-rv)(L/π)4 /EI + (v+rv) B2 (L/π)2 /GJ Which simplifies to v = QP/(1+r +Q[3-r]) Or v = Q l1 /(1+r+Q[3-r]) for P=l1 (2) And for the anti-symmetrical case, v = vb = -va Forcom patability (M-2vB)B(L/π)2 /GJ = (v-rv)(L/π)4 /EI + (v+rv) B2 (L/π)2 /GJ Which simplifies to v = M / (Q[1-r] +3+r) Or v = EBl1 / (Q[1-r]+3+r) for M=EBl1 (3) en v1 values may be determined from the previously determined values of Q and r. A more realistic example is considered by analysing the 16 m span superstructure with 10 hollow plank units connected via elastomeric shear keys, as shown in Figure A5. manual computations. erefore, the feasibility of manual computations is maintained by adopting a different approach in this Appendix. Consider one line of wheels on one plank (the solution for multiple lines of wheels being available by a simultaneous superimposition process which is demonstrated in this Appendix) within an infinitely wide deck. en by inspection of Figure A1, each successive inter plank shear is a constant proportion r of the previous one. atisvn+1=rvn vn+2=r2vnetc.