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Aviation Maintenance Technician Handbook–Airframe

FAA-H-8083-31B Version 2023

Chapter 4

Aircraft Metal Structural Repair

Certain fundamental rules must be observed if the original strength of the structure is to be maintained.

Ensure that the cross-sectional area of a splice or patch is at least equal to or greater than that of the damaged part. Avoid abrupt changes in cross-sectional area. Eliminate dangerous stress concentration by tapering splices. To reduce the possibility of cracks starting from the corners of cutouts, try to make cutouts either circular or oval in shape. Where it is necessary to use a rectangular cutout, make the radius of curvature at each corner no smaller than 1⁄2-inch. If the member is subjected to compression or bending loads, the patch should be placed on the outside of the member to obtain a higher resistance to such loads. If the patch cannot be placed there, material one gauge thicker than the original shall be used for the repair.

Replace buckled or bent members or reinforce them by attaching a splice over the affected area. A buckled part of the structure shall not be depended upon to carry its load again, no matter how well the part may be strengthened.

The material used in all replacements or reinforcements must be similar to that used in the original structure. If an alloy weaker than the original must be substituted for it, a heavier thickness must be used to give equivalent cross-sectional strength. A material that is stronger, but thinner, cannot be substituted for the original because one material can have greater tensile strength but less compressive strength than another, or vice versa. Also, the buckling and torsional strength of many sheet metal and tubular parts depends primarily on the thickness of the material rather than its allowable compressive and shear strengths. The manufacturer’s SRM often indicates what material can be used as a substitution and how much thicker the material needs to be. Figure 4-168 is an example of a substitution table found in an SRM.

Care must be taken when forming. Heat-treated and cold-worked aluminum alloys stand very little bending without cracking. On the other hand, soft alloys are easily formed, but they are not strong enough for primary structure. Strong alloys can be formed in their annealed (heated and allowed to cool slowly) condition, and heat treated before assembling to develop their strength.

The size of rivets for any repair can be determined by referring to the rivets used by the manufacturer in the next parallel rivet row inboard on the wing or forward on the fuselage. Another method of determining the size of rivets to be used is to multiply the thickness of the skin by three and use the next larger size rivet corresponding to that figure. For example, if the skin thickness is 0.040-inch, multiply 0.040-inch by 3, which equals 0.120-inch; use the next larger size rivet, 1⁄8-inch (0.125-inch). The number of rivets to be used for a repair can be found in tables in manufacturer’s SRMs or in Advisory Circular (AC) 43.13-1 (as revised), Acceptable Methods, Techniques, and Practices—Aircraft Inspection and Repair. Figure 4-169 is a table from AC 43.13-1 that is used to calculate the number of rivets required for a repair.

Extensive repairs that are made too strong can be as undesirable as repairs weaker than the original structure. All aircraft structure must flex slightly to withstand the forces imposed during takeoff, flight, and landing. If a repaired area is too strong, excessive flexing occurs at the edge of the completed repair, causing acceleration of metal fatigue.

Shear Strength & Bearing Strength

Aircraft structural joint design involves an attempt to find the optimum strength relationship between being critical in shear and critical in bearing. These are determined by the failure mode affecting the joint. The joint is critical in shear if less than the optimum number of fasteners of a given size are installed. This means that the rivets will fail, and not the sheet, if the joint fails. The joint is critical in bearing if more than the optimum number of fasteners of a given size are installed; the material may crack and tear between holes, or fastener holes may distort and stretch while the fasteners remain intact.

Maintaining Original Contour

Form all repairs in such a manner to fit the original contour perfectly. A smooth contour is especially desirable when making patches on the smooth external skin of high-speed aircraft.

Keeping Weight to a Minimum

Keep the weight of all repairs to a minimum. Make the size of the patches as small as practicable and use no more rivets than are necessary. In many cases, repairs disturb the original balance of the structure. The addition of excessive weight in each repair may unbalance the aircraft, requiring adjustment of the trim-and-balance tabs. In areas such as the spinner on the propeller, a repair requires application of balancing patches in order to maintain a perfect balance of the propeller.

Figure 4-168. Material substitution.
Figure 4-168. Material substitution.
Formula: Figure 4-168 formula reference: Aircraft-sheet material-substitution lookup: replacement gauge = original gauge × material replacement factor. Example: replacing 0.040-inch 7075-T6 with clad 7075-T6 uses 0.040 × 1.10 = 0.045 inch. Source text: Shape Initial Material Re M pl a a t c e e r m ial e nt ; Clad 2024–T42 0F Clad 2024–T3 ; 2024–T3 ; Clad 7075–T6 0A ; 7075–T6 0A ; Sheet ; 0.016 to 0.125 Clad 2024–T3 2024–T3 ; Clad 7075–T6 0A ; 7075–T6 0A ; Clad 7075–T6 7075–T6 ; Formed or 2024–T42 0F 7075–T6 0A ; Extruded Section 0B ; Sheet Material Replacement Factor ; M to a b te e r ial 7075–T6 C 70 la 7 d 5 –T6 2024–T3 C 20 la 2 d 4 –T3 F 2 2 0 0 2 2 4 4 – – T T 4 4 2 F C C l l a a d d 2 2 0 0 2 2 4 4 – – T T 4 4 2 ; Replaced ; C C H D E D E D E D E ; 7075–T6 1.00 1.10 1.20 1.78 1.30 1.83 1.20 1.78 1.24 1.84 ; Clad 7075–T6 1.00 1.00 1.13 1.70 1.22 1.76 1.13 1.71 1.16 1.76 ; 2024–T3 1.00 0A 1.00 0A 1.00 1.00 1.09 1.10 1.00 1.10 1.03 1.14 ; Clad 2024–T3 1.00 0A 1.00 0A 1.00 1.00 1.00 1.00 1.00 1.00 1.03 1.00 ; 2024–T42 1.00 0A 1.00 0A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.14 ; Clad 2024–T42 1.00 0A 1.00 0A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ; 7178–T6 1.28 1.28 1.50 1.90 1.63 2.00 1.86 1.90 1.96 1.98 ; Clad 7178–T6 1.08 1.18 1.41 1.75 1.52 1.83 1.75 1.75 1.81 1.81 ; 5052–H34 0G0 H 1.00 0A 1.00 0A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 ; Notes ; • All dimensions are in inches, unless given differently. 0A These materials cannot be used as replacements for the ; initial material in areas that are pressured. ; • It is possible that more protection from corrosion ; will be necessary when bare mineral is used to 0B They also cannot be used in the wing interspar structure at ; replace Clad material. Refer to 51-10-2. the wing center section structure. ; • It is possible for the material replacement factor 0C Use the next thicker standard gage when you use a formed ; to be a lower value for a specific location on the section as a replacement for an extrusion. ; airplane. To get that value, contact Boeing for a ; case by case analysis. 0D For all gages of flat sheet and formed sections. ; • Refer to Figure 3 for minimun bend radii. 0E For flat sheet less than 0.071 thick. ; • Example: 0F For flat sheet 0.071 thick and thicker, and for formed sections. ; To refer 0.040 thick 7075–T6 with Clad 7075–T6, ; multiply the gage by the material replacement factor 0G 2024–T4 and 2024–T42 are equivalent. ; to get the replacement gage 0. 040 x 1.10 = 0.045. ; 0H A compound to give protection from corrosion must be ; applied to bare material that is used to replace 5052–H34.

When flight controls are repaired and weight is added, it is very important to perform a balancing check to determine if the flight control is still within its balance limitations. Failure to do so could result in flight control flutter.

Flutter & Vibration Precautions

To prevent severe vibration or flutter of flight control surfaces during flight, precautions must be taken to stay within the design balance limitations when performing maintenance or repair. The importance of retaining the proper balance and rigidity of aircraft control surfaces cannot be overemphasized.

Figure 4-169. Rivet calculation table.
Figure 4-169. Rivet calculation table.
Formula: Figure 4-169 formula reference: Rivet calculation table: use sheet thickness and rivet diameter to find the number of 2117-T4 (AD) protruding-head rivets required per inch of joint width, or the corresponding number of AN-3 bolts. Source text: No. of 2117–T4 (AD) protruding head rivets required ; Thickness per inch of width “W” No. of ; “t” in Bolts ; inches Rivet size ; 3/32 I 1/8 I 5/32 I 3/16 I 1/4 AN–3 ; .016 6.5 I 4.9 I - - I - - I - - - - ; .020 6.5 4.9 3.9 - - - - - - ; .025 I 6.9 I 4.9 I 3.9 I - - I - - I - - ; .032 8.9 4.9 3.9 3.3 - - - - ; .036 I 10.0 I 5.6 I 3.9 I 3.3 I 2.4 I - - ; .040 11.1 6.2 4.0 3.3 2.4 - - ; .051 I - - I 7.9 I 5.1 I 3.6 I 2.4 I 3.3 ; .064 - - 9.9 6.5 4.5 2.5 3.3 ; .081 I - - I 12.5 I 8.1 I 5.7 I 3.1 I 3.3 ; .091 - - - - 9.1 6.3 3.5 3.3 ; .102 I - - I - - I 10.3 I 7.1 I 3.9 I 3.3 ; .128 - - - - 12.9 8.9 4.9 3.3 ; Notes ; a. For stringer in the upper surface of a wing, or in a fuselage, 80 percent of the number of rivets shown in the table ; may be used. ; b. For intermediate frames, 60 percent of the number shown may be used. ; c. For single lap sheet joints, 75 percent of the number shown may be used. ; Engineering Notes ; a. The load per inch of width of material was calculated by assuming a strip 1 inch wide in tension. ; b. Number of rivets required was calculated for 2117–T4 (AD) rivets, based on a rivet allowable shear stress equal to ; percent of the sheet allowable tensile stress, and a sheet allowable bearing stress equal to 160 percent of the sheet ; allowable tensile stress, using nominal hole diameters for rivets. ; c. Combinations of shoot thickness and rivet size above the underlined numbers are critical in (i.e., will fail by) bearing ; on the sheet; those below are critical in shearing of the rivets. ; d. The number of AN–3 bolts required below the underlined number was calculated based on a sheet allowable tensile ; stress of 55.000 psi and a bolt allowable single shear load of 2.126 pounds.