Joint Strength with Non-structural Shim

Non-structural Shim Joint

Joint Strength with a Non-structural Shim

Problem Statement

A non-structural shim (a filler, liner or gap shim) placed between the two sheets of a single-shear joint carries no load across the joint. It only separates the sheets. The fastener load path must now bridge the shim thickness, which increases the eccentricity between the bearing reactions in the two sheets. That eccentricity bends the fastener, and the joint strength falls below the value predicted for the same joint without the shim. This tool estimates the ultimate joint strength per fastener with the shim present.

Design guidance only. The method is a simplified strength model. Always validate a shimmed joint against test data with the shim installed, and follow company manuals and specifications that limit shim thickness.

Key Assumptions

• Single-shear joint: two metallic sheets or plates joined by one row of protruding-head fasteners.
• Non-structural shim: the shim transfers no load across the joint. It generally has fewer than three fasteners through it.
• Constant bearing distribution: the bearing stress in each sheet is uniform at the bearing allowable over the loaded length. This is why the method applies to metallic sheets only, not to composites.
• Prying balance: the full eccentric couple produced by the offset bearing reactions is reacted by bending of the fastener at its head and at its tail.
• Protruding heads: countersunk heads are excluded because the bearing distribution and head strength assumptions do not hold for them.

Geometry and Property Definitions

• t₁, t₂ = thickness of sheet 1 and sheet 2
• t_sh = shim thickness
• D = fastener shank diameter
• D_m = tail minor diameter (thread minor diameter of a bolt, or the swaged or formed tail of a pin)
• C = head-tail strength coefficient: 1.0 for a full-height head and nut; 0.84 for a shear head, a low-height head, a shear nut or a blind bolt
• F_bru1, F_bru2 = ultimate bearing strength of sheet 1 and sheet 2
• F_tu, F_ty = tensile ultimate and tensile yield strength of the fastener
• F_su = ultimate shear strength of the fastener
• P = applied load per fastener (optional, for the margin of safety)

1. Sheet Bearing Strength

With a constant bearing distribution, each sheet can carry the bearing allowable over its full thickness:

P_bru1 = F_bru1 · D · t₁

P_bru2 = F_bru2 · D · t₂

The shim does not change these values, because it carries no load.

2. Fastener Bending Modulus

The fastener bends plastically before it fails, so its bending strength is expressed with the Cozzone modulus of rupture (Bruhn, Chapter C3) rather than the elastic bending stress:

F_bu = f_m + f_o · (k − 1)

• k = 1.698 = 16 / (3π), the Cozzone shape factor of a solid circular section
• f_m = F_tu, the maximum fiber stress at ultimate
• f_o ≈ F_ty, the Cozzone intercept stress

Taking f_o equal to F_ty is a conservative approximation for most fastener materials, but it can be non-conservative for some. For an exact intercept stress from the material's stress-strain curve, use the Plastic Bending Allowable tool and compare its F_b with the F_bu shown here.

3. Moment the Fastener Can Resist

Under the prying assumption, the fastener resists the eccentric couple by bending at the head and at the tail. Each section can develop the modulus of rupture on its own section modulus, πD³/32 for a solid round section:

M_h = C · F_bu · π D³ / 32 (head side, shank diameter)

M_t = C · F_bu · π D_m³ / 32 (tail side, minor diameter)

M_tot = M_h + M_t = C · (π / 32) · F_bu · (D³ + D_m³)      (1)

The coefficient C reduces the moment a partial-strength head or tail can develop. A shear head, shear nut or blind bolt cannot react the full prying moment of a full-height head and nut.

4. Moment the Load Produces

At failure the bearing stress in each sheet equals its allowable, so the bearing reaction in sheet i occupies a length

ω_i = P_cr / (F_brui · D)

measured from the faying surface of that sheet. With a constant distribution the resultant acts at the middle of that length, so the two bearing resultants are separated by half of each loaded length plus the shim thickness. The eccentric couple is

M_tot = P_cr · (ω₁ / 2 + ω₂ / 2 + t_sh)

Substituting the loaded lengths:

M_tot = (P_cr² / (2 D)) · (1 / F_bru1 + 1 / F_bru2) + t_sh · P_cr      (2)

The shim enters only through the last term, and that term grows in direct proportion to the shim thickness. This is why shims should be kept as thin as possible.

5. Fastener Bending Strength with the Shim

Equating (1) and (2) gives a quadratic in the failure load:

a · P_cr² + b · P_cr + c = 0

• a = (1 / F_bru1 + 1 / F_bru2) / (2 D)
• b = t_sh
• c = − C · (π / 32) · F_bu · (D³ + D_m³)

Because a is positive and c is negative, the two roots have opposite signs. The positive root is the fastener bending strength with the shim, P_bu. Without a shim (b = 0) the root reduces to P_bu = √(−c / a).

If the loaded length ω_i computed at P_bu exceeds the sheet thickness t_i, the sheet cannot develop that bearing load and bearing governs. That condition is the same as P_brui < P_bu, and the tool reports it.

6. Fastener Shear Strength

The fastener shank also has to carry the load in single shear:

P_su = F_su · π D² / 4

7. Critical Strength and Margin of Safety

The joint strength per fastener is the lowest of the four failure loads:

P_cr = min(P_bru1, P_bru2, P_bu, P_su)

When an applied load is entered, the margin of safety is reported against that critical mode:

• Ultimate load: M.S. = P_cr / P − 1
• Limit load with a factor of safety FS: M.S. = P_cr / (FS · P) − 1

8. Worked Example (the tool's default inputs)

Two 2024-T3 clad sheets, 0.090 in and 0.160 in thick, are joined by a 3/16 in protruding-head low-alloy steel bolt with a full-height head and nut, through a 0.060 in non-structural shim. The applied ultimate load is 1,600 lb per fastener.

• Sheet 1: t₁ = 0.090 in, F_bru1 = 125 ksi (A basis, L, e/D = 2.0, 0.063–0.128 in band, MIL-HDBK-5J Table 3.2.3.0(e))
• Sheet 2: t₂ = 0.160 in, F_bru2 = 127 ksi (same table, 0.129–0.249 in band)
• Shim: t_sh = 0.060 in
• Fastener: D = 0.190 in, D_m = 0.151 in (10-32 thread minor diameter), C = 1.0; F_tu = 160 ksi, F_ty = 142 ksi, F_su = 96 ksi (160 ksi strength level, Table 2.3.1.0(g))

Sheet bearing:

P_bru1 = 125,000 · 0.190 · 0.090 = 2,138 lb

P_bru2 = 127,000 · 0.190 · 0.160 = 3,861 lb

Fastener bending modulus:

F_bu = 160 + 142 · (1.698 − 1) = 259.1 ksi

Quadratic coefficients:

a = (1/125,000 + 1/127,000) / (2 · 0.190) = 4.177 × 10^-5 in/lb

b = 0.060 in

c = −1.0 · (π/32) · 259,100 · (0.190³ + 0.151³) = −262.1 lb·in

Fastener bending strength with the shim:

P_bu = [−0.060 + √(0.060² + 4 · 4.177 × 10^-5 · 262.1)] / (2 · 4.177 × 10^-5) = 1,887 lb

Check of the eccentricity: ω₁ = 1,887 / (125,000 · 0.190) = 0.079 in and ω₂ = 1,887 / (127,000 · 0.190) = 0.078 in, both shorter than their sheet thickness, so the bearing lengths fit. The couple arm is 0.079/2 + 0.078/2 + 0.060 = 0.139 in and 1,887 · 0.139 = 262 lb·in, which equals M_tot.

Fastener shear:

P_su = 96,000 · π · 0.190² / 4 = 2,722 lb

Critical strength and margin:

P_cr = min(2,138, 3,861, 1,887, 2,722) = 1,887 lb, fastener bending with the shim

M.S. = 1,887 / 1,600 − 1 = +0.18 (ultimate)

What the shim did: without the shim (b = 0) the bending strength would be √(262.1 / 4.177 × 10^-5) = 2,505 lb, above the sheet 1 bearing strength of 2,138 lb, so bearing would govern. The 0.060 in shim lowers the bending strength to 1,887 lb and makes fastener bending the critical mode, a 12 percent loss of joint strength from a shim only 0.060 in thick.

Limitations

• Metallic sheets and plates only. The constant bearing distribution is not valid for composite laminates.
• Single-shear joints with protruding-head fasteners only. Countersunk heads are not covered.
• Fastener bending uses the approximate intercept stress f_o = F_ty unless a better value is supplied through the Plastic Bending Allowable tool.
• The model gives no credit to clamp-up or to friction, and it assumes the shim carries nothing.
• Test data with the shim installed takes precedence over this prediction.