ブックタイトル「煉瓦造建造物の保存と修復」英語版

概要

「煉瓦造建造物の保存と修復」英語版

a header bond.As seen in figures 7 to 10, it became clear that lengthadjustment in brick masonry was realized by varying thebrick dimensions.Minor discrepancies between actual measurements andestimation of planned measurements are as follows: forpilaster: A +0 to +4mm , pilaster B: +1 to +5mm , pilaster D: -3to +3mm , and pilaster E: +4 to +8mm .Pilaster C as well as those on both left and right ends arelaid following a unit based on the length of a brick stretcherand are 8 stretcher units (= 16 header units). Although fourunits of the stretcher module seems to have been plannedat 31 sun (= 939mm), the difference between the actualmeasurement and the calculated design measurement israther large, ranging from -21 to -11mm.The above descriptions can be summarized as follows:? Wall B-C, wall C-D：4 brick stretcher units is 30sun (909mm)? Wall A-B, wall D-E：4 brick stretcher units is around 31sun (939mm)? Pilaster A, pilaster B, pilaster D, pilaster E：3 brickstretcher units is 23 sun (697mm)? Pilaster C as well as those on both left and right ends：4brick stretcher units is around 31 sun (939mm)According to such calculations of the plannedmeasurement, distance between the centerlines of pilaster Aand pilaster E is 72.125 shaku (21,854mm), which does notmatch the designed measurement of 72 shaku (21,816mm).In wall A-B and wall D-E, 4 brick stretcher units is 31 sun(939mm), in other words, the same module unit is smallerthan 7.75 sun (235mm), and it can be presumed that 3brick stretcher units is 23 sun (697mm), which means that adifferent module unit of 7.67 sun (232mm) was used at thesame time. The above difference of 0.125 shaku (39mm) canbe explained if the measurement of 15 stretcher units wasfine-tuned by employing this smaller unit.That is to say,1 The length of wall A-B and wall D-E are both 25.5stretcher units and their sum is 21 units.2 If this module were uniformly 7.75 sun (235mm), thetotal length of wall A-B and wall D-E would be (51×0.775= )39.525 shaku (11,976mm).3 Of the 51 units, if 15 are of a 7.67 sun (232mm) unit and3 are of a 23 sun (697mm) unit, the total length would be(2.3×5 = ) 11.5 shaku (3,485mm).4 If the remaining 36 units are of 7.75 sun (235mm), with4 units equaling 31 sun (939mm), the sum is (3.1×9 = )27.9 shaku (8,451mm).5 The sum of 3 and 4 is (11.5 + 27.9 = ) 39.4 shaku(11,938mm).6 Adjustment as shown in 5 decreases the measurementby (39.525－39.4 = ) 0.125 shaku (39mm).Let’s assume that somewhere in wall A-B and wall D-E, 15units of the same module have been adopted. Consideringconstruction procedures, treating 3 units as a set would berational, which means that there are 5 sets and when thismeasurement is to be distributed among wall A-B and wallD-E, the following ratios are possibilities; in other words,figure 11Layout of bricks for wall A-Bfigure 12Layout of bricks for wall D-E16Chapter 2Brick Masonry Construction: Techniques and Significance