EWPAA Structural Plywood and LVL Design Manual - Engineered ...

Thermal Transmission – Worked Examplewhere:TplyT TbattpbR t = roa+ + rrf+ + r + + rk rvbrplybattpbiar oaT plyr rfT batt= outdoor surface air resistance;= thickness of plywood sheathing (m);= rigid foam resistance;= batt insulation thickness (m);r vb = vapour barrier resistance = 0;T pbr iaRRUUtt= thickness of plasterboard (m);= inside surface air resistance0.012 0.095 0.012= 0.04+ +0.5+ +0+ +0.120.13 0.05 0.22= 0.04+0.092+0.5+1.9+0+0.05+0.122 = 2.25m C/W1=R t1=2.25= 0.44W/m2 C (between studs)(15.5)If the heat flow through the studs was being considered r stud = T stud /k timber , which in thiscase would be 0.09/0.13 = 0.69, would have to be included in the calculation for R t .The Building Code of Australia (BCA) – Part J1, Building Fabric provides guidelines forminimum R – values for the various climate zones for:• various roof and ceiling types;• walls of various construction;• floors of timber and concrete construction.For a comprehensive treatment of floor insulation the reader is referred to the publication:“Insulation Solutions to enhance the Thermal resistance of SuspendedTimber Floor System in Australia”and for more details on availability see Reference 7.15.6 ConclusionThe information contained in this chapter is not meant to be exhaustive, but rather,informative. Where the designer has any doubt as to the likely outcome of the choice of theinsulation components constituting a barrier professional help should be sought. Recognitionof potential problems and implementing the correct steps towards their solution is afundamental part of the design process. Hence, the age old truism A LITTLE KNOWLEDGECAN BE DANGEROUS should forever be uppermost in ones mind.236