(a) Riveted patch seams are calculated as follows:

1. (1) Determine the length, L, of the patch from the area of the defect to be repaired (Figures 1, 2, 3, and 4). The normal efficiency, e, of the single-riveted seam that is to be used in the patch should then be determined from Table 1. This is governed by the thickness of plate and diameter of rivet holes.

   Table 1

   EFFICIENCIES OF SINGLE-RIVETED SEAMS

   Plate Thickness, t

   Rivet Hole Diameter, d

   Pitch of Rivet, p

   Efficiency of Seam, e

   ¼

   ¹¹/₁₆

   1 ⅞

   63.6

   ⁹/₃₂

   ¾

   1 ⅞

   60.3

   ⁵/₁₆

   ¾

   1 ⅞

   60.0

   ¹¹/₃₂

   ¹³/₁₆

   1 ¹⁵/₁₆

   58.0

   ⅜

   ¹³/₁₆

   1 ¹⁵/₁₆

   57.0

   ¹³/₃₂

   ⅞

   2 ¹/₁₆

   57.5

   ⁷/₁₆

   ¹⁵/₁₆

   2 ¼

   56.0

   ¹⁵/₃₂

   ¹⁵/₁₆

   2 ⅛

   55.5

   ½

   1

   2 ¼

   55.7

   ⁹/₁₆

   1 ¹/₁₆

   2 ⅜

   53.0

   ¹⁹/₃₂

   1 ¹/₁₆

   2 ¼

   52.8

   ⅝

   1 ¹/₁₆

   2 ¼

   50.5

   ²¹/₃₂

   1 ⅛

   2 ⁵/₁₆

   51.4

   ¹¹/₁₆

   1 ⅛

   2 ⁵/₁₆

   51.4

   Tensile strength assumed at 55,000 psi and shearing strength at 44,000 psi.

   After determining the length, L, of a patch, determine its width girthwise, W. This is found by multiplying the length by the constant, C, as shown in Table 2 or 3, depending upon the type of boiler to be repaired. These tables give a constant, C, for a given efficiency, e, of patch and efficiency, E, of the longitudinal seam.

2. (2) To determine the longitudinal efficiency of an existing patch, the length, L, and width, W, the pitch, P, and the diameter, d, of rivet, should be measured. “W” divided by “L” will give the constant, C. Table 1 will give “e.” Then under “e” in Table 2 or 3, depending upon the type of boiler to be repaired, find the constant, C. Whatever “E” is in the first column is the longitudinal or allowed efficiency of the patch seam.