11.4 Worked Examples · ECSS-E-HB-32-23A

11.4.1 Fatigue of a Threaded Fastener Example

11.4.1.1 Overview

A critical threaded fastener joint design uses an M8 diameter Titanium fastener to the following loading spectrum during each mission:

• 50 cycles at a stress amplitude of 160 MPa
• 4000 cycles at a stress amplitude of 50 MPa
• 880 cycles at a stress amplitude of 60 MPa
• 300 cycles at a stress amplitude of 130 MPa
• 100 cycles at a stress amplitude of 90 MPa

What is the life of the fastener in numbers of missions?

11.4.1.2 Analysis

From the S-N curve in Figure 11-12 the cycles to failure for each spectrum block are as follows:

Table 11-3Analysis of cycles to failure

Alternating Stress, σ_a (MPa)	Cycles per Mission, n	Cycles to Failure, N
160	50	10³
50	4000	below endurance limit
60	880	5 x 10⁵
130	300	7 x 10³
90	100	8 x 10⁴

From Section 11.2.2, using the Palmgren - Miner Rule, the fatigue damage per mission is given by,

f = 4\Omega

\begin{aligned} &= 4 \sum_{i=1}^{m} \frac{n_{i}}{N_{f,i}} \\ &= 4 \times \left[ \frac{50}{1e^{3}} + \frac{880}{5e^{5}} + \frac{300}{7e^{3}} + \frac{100}{8e^{4}} \right] \\ &= 0.37 \end{aligned}

The fatigue life of the fastener is therefore 1 / 0,37 = 2,7 missions.

11.4.2 Threaded Fastener Fracture Mechanics Example

This example is based on an older version of the NASGRO software (Ref. [11.3] or [11.4]) as well as other literature and will not be fully up to date with the latest NASGRO version (Ref. [11.9]). Cracked fastener models are still being further developed.

11.4.2.1 Problem Description

A M8 Titanium fastener (TA28 (Ti6Al4V)) is used to clamp together two 10mm thick flanges made from Aluminium. The external axial load applied to the joint is 18 kN and the fastener has a 0,5mm deep thumbnail crack at the first thread into the nut.

Determine the static strength of the fastener.

The following assumptions and properties apply:

• The yield stress for a TA 28 fastener is 1100 N/mm²
• The stress area of an M8 fastener = 32,8 mm²
• The fastener is preloaded to between 60 and 65% of yield
• The fastener sees no bending loads
• The flanges are sufficiently wide to allow fully developed compression cones
• No thermal effects apply.

11.4.2.2 Solution

The yield load of the fastener is,

F_{b,y} = A_{S} \times \sigma_{y} = 1100 \times 32.8 = 36.1 \text{ kN}

Thus, the maximum fastener preload is

Now the force ratio is calculated. Since $L_{j} > 2d$ ,

D_{\text{lim}} \approx 3d_{uh,brg}

It is assumed for the sake of this example that the force ratio is,

$\Phi \approx 0.47$

The loading plane is assumed to be at the mid-point of the joint members, i.e. n = 0,5.

Thus,

Therefore, since $F_{A} = 18\text{kN}$ , the load increment in the fastener due to the external load is given by,

F_{b,A} = 18\text{ kN} \times 0,235 = 4,2\text{ kN}

The maximum fastener tensile load is,

The gross axial stress is,

\sigma_{0} = \frac{F_{b,\max}}{A_{s}} = \frac{27.7}{32.8} \times 1000 = 8

For the thumbnail crack with an initial assumed length a = 0,5mm and c = 0,5mm the stress intensity correction factor for tensile stress, F0 , can be read from Table 11-1 to be 0,84.

Thus, the stress intensity factor can be calculated with Equation,

K_{1} = \sqrt{\pi a} \sigma_{0} F_{0} = \sqrt{\pi \times 0.5} \times 844 \times 0.84 = 889 \text{ MPa}\sqrt{\text{mm}}

K_{1} = \sqrt{\pi a} \sigma_{0} F_{0} = \sqrt{\pi \times 0,5} \times 844 \times 0,84 = 889 \text{ MPa}\sqrt{\text{mm}}

Note that assuming such small crack size (0,5 mm) would require special NDT procedure, acc. Reference [11.14].

For TA28 fasteners a typical value for the critical stress intensity factor, K1C , is 2780 MPa√mm (see Reference 11.3), therefore the Margin of Safety on fastener fracture is,

Note: Using KIC (instead of K1e )will generally be conservative. The value of 2780 MPa√mm is one of the highest in the NASGRO database for Ti6Al4V, so it needs to be confirmed for the actual fastener material grade used. Depending on the nature of the loading (sustained or varying) it can be appropriate to use a reduced toughness value to cover sustained load cracking phenomena (see section 11.3.1).

Hence, given the assumptions in this calculation, it can be concluded that a thumbnail crack of 0,5 mm length (or depth) does not represent a critical defect and can be tolerated.

To find the critical defect size for this joint it is necessary to repeat this calculation for a number of crack lengths until KIC is exceeded and interpolate for the critical length.

NASGRO can be used more conveniently. In version 9.2 the result for a 0,5 mm crack in an M5 bolt at 844 MPa is: 785 MPa√mm.

11.4.1 Fatigue of a Threaded Fastener Example

11.4.1.1 Overview

A critical threaded fastener joint design uses an M8 diameter Titanium fastener to the following loading spectrum during each mission:

• 50 cycles at a stress amplitude of 160 MPa
• 4000 cycles at a stress amplitude of 50 MPa
• 880 cycles at a stress amplitude of 60 MPa
• 300 cycles at a stress amplitude of 130 MPa
• 100 cycles at a stress amplitude of 90 MPa

What is the life of the fastener in numbers of missions?

11.4.1.2 Analysis

From the S-N curve in Figure 11-12 the cycles to failure for each spectrum block are as follows:

Table 11-3Analysis of cycles to failure

Alternating Stress, σ_a (MPa)	Cycles per Mission, n	Cycles to Failure, N
160	50	10³
50	4000	below endurance limit
60	880	5 x 10⁵
130	300	7 x 10³
90	100	8 x 10⁴

From Section 11.2.2, using the Palmgren - Miner Rule, the fatigue damage per mission is given by,

f = 4\Omega

\begin{aligned} &= 4 \sum_{i=1}^{m} \frac{n_{i}}{N_{f,i}} \\ &= 4 \times \left[ \frac{50}{1e^{3}} + \frac{880}{5e^{5}} + \frac{300}{7e^{3}} + \frac{100}{8e^{4}} \right] \\ &= 0.37 \end{aligned}

The fatigue life of the fastener is therefore 1 / 0,37 = 2,7 missions.

11.4.2 Threaded Fastener Fracture Mechanics Example

11.4.2.1 Problem Description

Determine the static strength of the fastener.

The following assumptions and properties apply:

• The yield stress for a TA 28 fastener is 1100 N/mm²
• The stress area of an M8 fastener = 32,8 mm²
• The fastener is preloaded to between 60 and 65% of yield
• The fastener sees no bending loads
• The flanges are sufficiently wide to allow fully developed compression cones
• No thermal effects apply.

11.4.2.2 Solution

The yield load of the fastener is,

F_{b,y} = A_{S} \times \sigma_{y} = 1100 \times 32.8 = 36.1 \text{ kN}

Thus, the maximum fastener preload is

Now the force ratio is calculated. Since $L_{j} > 2d$ ,

D_{\text{lim}} \approx 3d_{uh,brg}

It is assumed for the sake of this example that the force ratio is,

$\Phi \approx 0.47$

The loading plane is assumed to be at the mid-point of the joint members, i.e. n = 0,5.

Thus,

Therefore, since $F_{A} = 18\text{kN}$ , the load increment in the fastener due to the external load is given by,

F_{b,A} = 18\text{ kN} \times 0,235 = 4,2\text{ kN}

The maximum fastener tensile load is,

The gross axial stress is,

\sigma_{0} = \frac{F_{b,\max}}{A_{s}} = \frac{27.7}{32.8} \times 1000 = 8

For the thumbnail crack with an initial assumed length a = 0,5mm and c = 0,5mm the stress intensity correction factor for tensile stress, F0 , can be read from Table 11-1 to be 0,84.

Thus, the stress intensity factor can be calculated with Equation,

K_{1} = \sqrt{\pi a} \sigma_{0} F_{0} = \sqrt{\pi \times 0.5} \times 844 \times 0.84 = 889 \text{ MPa}\sqrt{\text{mm}}

K_{1} = \sqrt{\pi a} \sigma_{0} F_{0} = \sqrt{\pi \times 0,5} \times 844 \times 0,84 = 889 \text{ MPa}\sqrt{\text{mm}}

Note that assuming such small crack size (0,5 mm) would require special NDT procedure, acc. Reference [11.14].

For TA28 fasteners a typical value for the critical stress intensity factor, K1C , is 2780 MPa√mm (see Reference 11.3), therefore the Margin of Safety on fastener fracture is,

Hence, given the assumptions in this calculation, it can be concluded that a thumbnail crack of 0,5 mm length (or depth) does not represent a critical defect and can be tolerated.

To find the critical defect size for this joint it is necessary to repeat this calculation for a number of crack lengths until KIC is exceeded and interpolate for the critical length.

NASGRO can be used more conveniently. In version 9.2 the result for a 0,5 mm crack in an M5 bolt at 844 MPa is: 785 MPa√mm.

11.4Worked Examples

11.4.1 Fatigue of a Threaded Fastener Example

11.4.1.1 Overview

11.4.1.2 Analysis

11.4.2 Threaded Fastener Fracture Mechanics Example

11.4.2.1 Problem Description

11.4.2.2 Solution

11.4Worked Examples

11.4.1 Fatigue of a Threaded Fastener Example

11.4.1.1 Overview

11.4.1.2 Analysis

11.4.2 Threaded Fastener Fracture Mechanics Example

11.4.2.1 Problem Description

11.4.2.2 Solution