Air Conditioning Rack Structural Substantiation

INTRODUCTION

The substantiation demonstrates that the installed rack and its load path into the DHC-8-100 structure satisfy the applicable ultimate-strength requirements under the governing flight and emergency conditions.

DESIGN ASSESSMENT

Installation Components

The Air Conditioning Rack is composed of several structural components, as outlined in the table below:

The Supporting Structure underneath the Air Conditioning Rack is composed of several components. The newly added components that will be considered in the FE Analysis are outlined in the table below:

Installation Weight Estimation

Air Conditioning Rack

The tables below list the weights of the Modeled Structural Components, the unmodeled non-structural components, and the installed equipment along with their Center of Gravity (CoG) Locations. the figure below illustrates the payloads within the Air Conditioning Rack.

The volume of the Modeled Structural Components and the unmodeled non-structural components were generated from the 3D model. The weights of the equipment are based on their corresponding manufacturer data sheet, and their CoG locations were generated from the 3D model. It is noteworthy that an additional estimated weight (1.15 times the original weight) has been added to each equipment and component to account for the weight of the hardware, screws, nuts, and wires/cables installed.

the table below summarizes of all Nonstructural and Equipment Weights on the Air Conditioning Rack, along with the equivalent CoG location relative to the FWD inboard lower corner. It is noteworthy that the Non-Structural Components and Equipment will be omitted from the analysis. However, to account for their effect, the total weight of these components will be added to the FE model as a non-structural mass distributed over the tube structure.

The weights were estimated based on material density of 0.1 lb/in³ for AL 2024-T3 CLAD Sheet, 0.098 lb/in³ for AL 6061-T6 and T6511 Extrusion, 0.0578 lb/in³ for Silicon Rubber, 0.0379 lb/in³ for Polyurethane Rubber, 0.0596 lb/in³ for Nylon Plastic, 0.282 lb/in³ for Galvanized Steel, and 0.0452 lb/in³ for Neoprene. Moreover, the avionics box is fixed to the FWD and AFT beams through 17 ARN/4 (CR3213) at each side. It is considered as a payload on the Air Conditioning Rack FEM.

Based on the above table, the total weight of the parts to be modeled is 80.50 lb_f. However, the total weight of the model in FEM is 81.839 lb_f (1.339 lb_f more) due to the simplifications performed on the modeled parts (i.e. covering holes, removing curved edges, and other geometry simplification). Hence, the non-structural mass to be added to the model was reduced from 19.13 lb_f to 17.791 lb_f. This mass was added, as a distributed load, to the entire tube structure.

Supporting Structure

It is noteworthy that weights were estimated based on material density of 0.1 lb/in³ for AL 2024-T3 Sheet and for AL 2024-T351 Plate. Moreover, an additional estimated weight of 0.15 times the original weight (0.15x4.07=0.61 lb_f) has been applied in the FEM as non-structural mass distributed over the modeled structural elements. This additional weight is to account for the weight of the hardware, screws, and nuts installed.

Material Properties

The table below lists the mechanical properties [per MMPDS-15] of the materials used in the Air Conditioning Rack Installation and its Supporting Structure.

Fasteners Allowables

CR3212 Rivets

Rivet’s Shear Strength:

The shear strengths at the aforementioned joints depend on the CSK sheet thickness. Using [MMPDS-15-Table 8.1.3.2.2(v)], these shear strengths can be calculated as below:

• J7 [t_CSK=0.063”] : 401x(50/51)x2 = 786 lb_f. (double shear configuration)

• J8 [t_CSK=0.125”] : 594x(50/51) = 582 lb_f.

• J9 [t_CSK=0.125”] : 594x(50/51) = 582 lb_f.

• J10 [t_CSK=0.071”]: 437x(50/51) = 428 lb_f.

It is noteworthy that the values in [MMPDS-15-Table 8.1.3.2.2(v)] are for F_su=51 ksi, and the ultimate shear strength of the fastener material F_su=50 ksi [MMPDS-15-Table 8.1.1.2]. Therefore, the joint’s shear strength was scaled down by a factor of 50/51.

Joint Bearing strength:

The non-CSK sheet is thicker than the CSK sheet in this joint. Therefore, calculating the joint ultimate bearing strength is not required.

Rivet’s Tensile Strength:

The tensile strength value of f_tu=285 lb_f is based on 0.156” thick CSK sheet [Technical Data Sheet-Cherrymax Rivets]. Therefore, the estimated tensile strengths at the aforementioned joints are as below:

• J7 [t_CSK=0.063”] : 285x(0.063/0.156) = 115 lb_f.

• J8 [t_CSK=0.125”] : 285x(0.125/0.156) = 228 lb_f.

• J9 [t_CSK=0.125”] : 285x(0.125/0.156) = 228 lb_f.

• J10 [t_CSK=0.071”]: 285x(0.071/0.156) = 129 lb_f.

CR3213 Rivets

Rivet’s Shear Strength:

The rivets utilised in joints J3-J4 configuration are in double shear state and have D_r/t_min = 1 < 1.5, and the rivets utilised in joints J5-J6 configuration are in single shear state and have D_r/t_min = 2 < 3. Therefore, no correction factor, $\alpha$, is needed, and the ultimate shear strength remains f_su=664 lb_f.

The Static Joint Strengths at the aforementioned joints depend on the thinnest sheet thickness, and it can be calculated as below [MMPDS-15-Table 8.1.3.2.2(v), MMPDS-15-Table 8.1.3.1.2(p)]:

• J3 [t_min=0.125”] : 652x(50/51)x2 = 1,278 lb_f. (double shear configuration)

• J4 [t_min=0.125”] : 652x(50/51)x2 = 1,278 lb_f. (double shear configuration)

• J5 [t_min=0.125”] : 652x(50/51) = 639 lb_f.

• J6 [t_min=0.0625”]: 492x(50/51)  = 482 lb_f.

It is noteworthy that the values in [MMPDS-15-Table 8.1.3.1.2(p)] are for F_su=51 ksi, and the ultimate shear strength of the fastener material F_su=50 ksi [MMPDS-15-Table 8.1.1.2]. Therefore, the joint’s shear strength was scaled down by a factor of 50/51.

Joint Bearing strength:

The joint bearing strength is determined based on the bearing strength of the thinnest sheet within the joint. In cases where multiple sheets share the same minimum thickness but are composed of different materials, the calculation will use the material with the lowest bearing strength among them. Hence, the joints’ ultimate bearing strength can be calculated as below [MMPDS-15-Table 8.1.2.1(a)]:

• J3 [t_min=0.125”] : 1,606x(82/100) = 1,316 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

• J4 [t_min=0.125”] : 1,606x(82/100) = 1,316 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

• J5 [t_min=0.125”] : 1,606x(82/100) = 1,316 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

• J6 [t_min=0.0625”]: 810x(125/100)  = 1,012 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

Since the Static Joint Strengths are smaller than the Joint Bearing Strength, the former will be considered as the joint allowable.

Rivet’s Tensile Strength:

The tensile strength value of f_tu=285 lb_f is based on 0.156” thick sheet [Technical Data Sheet-Cherrymax Rivets]. Therefore, the estimated tensile strengths at the aforementioned joints are as below:

• J3 [t_min=0.125”] : 285x(0.125/0.156) = 228 lb_f.

• J4 [t_min=0.125”] : 285x(0.125/0.156) = 228 lb_f.

• J5 [t_min=0.125”] : 285x(0.125/0.156) = 228 lb_f.

• J6 [t_min=0.0625”]: 285x(0.0625/0.156) = 114 lb_f.

MS20426AD Rivets

Rivet’s Shear Strength:

The MS20426AD4 rivets have a single shear strength value of 389 lb_f [MMPDS-15-Table 8.1.2(b). MMPDS-15-Table 8.1.5(a), Analysis and Design of Flight Vehicle Structures-Bruhn-Table D1.7, Technical Data Sheet-NAS528 Fastener Codes.]. In order to take into account the reduction in rivet shear strength when it is inserted in a sheet, the Static Joint Strength is calculated, considering the CSK sheet thickness, as below:

• J25 [t_CSK=0.125”] : 532x(30/41) = 389 lb_f.

It is noteworthy that the values in [MMPDS-15-Table 8.1.2.2(r)] are for F_su=41 ksi, so the joint’s shear strength was scaled down by a factor of (30/41).

Joint Bearing strength:

Since the countersunk sheet is thicker than the non-countersunk sheet, calculating the Joint Bearing strength is required, and it will be done, using the non-countersunk sheet thickness, as below [MMPDS-15-Table 8.1.2.1(a)]:

• J25 [t_{non-countersunk}=0.063”] : 810x(82/100) = 664 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

Since the Ultimate Shear Strength is smaller than the Joint Bearing Strength, the former will be considered as the joint allowable.

Rivet’s Tensile Strength:

The Ultimate Tensile Strength for 1/8” diameter AN426 (MS20426) Flush Head Rivet in 0.064” thick AL 2024 ALCLAD Sheet is 438 lb_f [Page 952-Analysis & Design of Flight Vehicle Structures-Bruhn]. Hence, the tensile strength considering the thinnest sheet in the joint can be calculated as below:

J25 [t_min=0.063”] : 438x(0.063/0.064) = 431 lb_f.

MS20470AD Rivets

The MS20470AD rivets have a single shear strength value of 389 lb_f [MMPDS-15-Table 8.1.2(b). MMPDS-15-Table 8.1.5(a), Analysis and Design of Flight Vehicle Structures-Bruhn-Table D1.7, Technical Data Sheet-NAS528 Fastener Codes].

Rivet’s Shear Strength:

The rivets in J11 and J13-J17 are in double shear state and have D_r/t_middle value as below:

The rivets in J12 and J18-J24 are in double shear state and have D_r/t_min value as below:

The D_r/t_middle in joints J13-J16 are greater than 1.5, D_r/t_min in joint J22 is greater than 3.0. Therefore, a correction factor, $\alpha$, needs to be calculated at these joints to compensate for the reduction in rivet shear strength resulting from high bearing stresses on the rivet in such cases. This factor can be calculated using the below formulas [MMPDS-15]:

• J13 [t_middle=0.05”] : α=1-0.13x(0.125/0.05 -1.5) = 0.870

• J14 [t_middle=0.025”] : α=1-0.13x(0.125/0.025 -1.5) = 0.545

• J15 [t_middle=0.05”] : α=1-0.13x(0.125/0.05 -1.5) = 0.870

• J16 [t_middle=0.025”] : α=1-0.13x(0.125/0.025 -1.5) = 0.545

• J22 [t_min=0.04”] : α=1-0.04x(0.125/0.04 -3) = 0.995

Therefore, the ultimate shear strength at joints J11-J24 are as follows:

• J11: 389x1x2 = 778 lb_f. (double shear configuration)

• J12: 389x1x1 = 389 lb_f. (single shear configuration)

• J13: 389x0.870x2 = 676 lb_f. (double shear configuration)

• J14: 389x0.545x2 = 424 lb_f. (double shear configuration)

• J15: 389x0.870x2 = 676 lb_f. (double shear configuration)

• J16: 389x0.545x2 = 424 lb_f. (double shear configuration)

• J17: 389x1x2 = 778 lb_f. (double shear configuration)

• J18: 389x1x1 = 389 lb_f. (single shear configuration)

• J19: 389x1x1 = 389 lb_f. (single shear configuration)

• J20: 389x1x1 = 389 lb_f. (single shear configuration)

• J21: 389x1x1 = 389 lb_f. (single shear configuration)

• J22: 389x1x1 = 389 lb_f. (single shear configuration)

• J23: 389x0.995x2 = 774 lb_f. (double shear configuration)

• J24: 389x1x1 = 389 lb_f. (single shear configuration)

Joint Bearing strength:

The joint bearing strength is determined based on the bearing strength of the thinnest sheet within the joint. In cases where multiple sheets share the same minimum thickness but are composed of different materials, the calculation will use the material with the lowest bearing strength among them. Hence, the joints’ ultimate bearing strength can be calculated as below [MMPDS-15-Table 8.1.2.1(a)]:

• J11 [t_min=0.063”] : 810x(82/100) = 664 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

• J12 [t_min=0.125”] : 1,606x(82/100) = 1,316 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J13 [t_min=0.05”] : 642x(82/100) = 526 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J14 [t_min=0.025”] : 321x(82/100) = 263 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J15 [t_min=0.05”] : 642x(82/100) = 526 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J16 [t_min=0.025”] : 321x(82/100) = 263 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J17 [t_min=0.125”] : 1,606x(82/100) = 1,316 lb_f. (F_bru=82 ksi for AL 6061-T6 Extrusion)

• J18 [t_min=0.0625”]: 810x(82/100) = 664 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J19 [t_min=0.0625”]: 810x(82/100) = 664 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J20 [t_min=0.05”] : 642x(82/100) = 526 lb_f. (F_bru=142 ksi for AL 7075-T62 Sheet)

• J21 [t_min=0.0625”]: 810x(82/100) = 664 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J22 [t_min=0.04”] : 514x(82/100) = 421 lb_f. (F_bru=142 ksi for AL 7075-T62 Sheet)

• J23 [t_min=0.0625”]: 810x(82/100) = 664 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

• J24 [t_min=0.0625”]: 810x(82/100) = 664 lb_f. (F_bru=125 ksi for AL 2024-T3 Sheet)

For joints J12, J17 through J22, and J24, the Static Joint Strengths are lower than the Joint Bearing Strengths; therefore, the Static Joint Strengths will govern and be used as the allowable values. Conversely, for joints J11, J13 through J16, and J23, the Joint Bearing Strengths are lower than the Static Joint Strengths, and thus will be used as the allowable values for those joints.

Rivet’s Tensile Strength:

The tensile strength considering the thinnest sheet in the joints can be calculated as below [Page 952-Analysis & Design of Flight Vehicle Structures-Bruhn]:

• J11 [t_min=0.063”] : 495x(0.063/0.064) = 487 lb_f.

• J12 [t_min=0.125”] : 495 lb_f. (For 0.072” thick sheet due to lack of data)

• J13 [t_min=0.05”] : 471x(0.05/0.051) = 461 lb_f.

• J14 [t_min=0.025”] : 197 lb_f.

• J15 [t_min=0.05”] : 471x(0.05/0.051) = 461 lb_f.

• J16 [t_min=0.025”] : 197 lb_f.

• J17 [t_min=0.125”] : 495 lb_f. (For 0.072” thick sheet due to lack of data)

• J18 [t_min=0.0625”]: 495x(0.0625/0.064) = 483 lb_f.

• J19 [t_min=0.0625”]: 495x(0.0625/0.064) = 483 lb_f.

• J20 [t_min=0.05”] : 471x(0.05/0.051) = 461 lb_f.

• J21 [t_min=0.0625”]: 495x(0.0625/0.064) = 483 lb_f.

• J22 [t_min=0.04”] : 353 lb_f.

• J23 [t_min=0.0625”]: 495x(0.0625/0.064) = 483 lb_f.

• J24 [t_min=0.0625”]: 495x(0.0625/0.064) = 483 lb_f.

MS24694 Screws

Screw’s Shank Shear Strength:

The Ultimate Single Shear Strength of the screw is f_su=2,125 lb_f [MMPDS-15-Table 8.1.5(a)].

Joint Bearing strength:

The thinnest sheet in both joints is AL 2024-T351 Plate that has an ultimate bearing strength of 119 ksi (refer to the table below). Hence, the joints’ ultimate bearing strength is 1,197x(119/100)= 1,424 lb_f [MMPDS-15-Table 8.1.5.1].

Since the Ultimate Shear Strength is greater than the Joint Bearing Strength, the later will be considered as the joint allowable.

Screw’s Tensile Strength:

Ultimate Tensile Strength of the screw is f_tu=2,500 lb_f [Technical Data Sheet-MS24694].

AN3 and AN4 Bolts

Screw’s Shank Shear Strength

The Ultimate Shear Strength for the bolt is 3,680 lb_f [Technical Data Sheet-NASM3-20].

Joint Bearing strength

The thinnest sheet in joint J30 is AL 2024-T3 Sheet that has an ultimate bearing strength of 125 ksi (refer to the table below). Hence, the joints’ ultimate bearing strength is 1,575x(125/100)= 1,968 lb_f [MMPDS-15-Table 8.1.5.1]. Since the Ultimate Shear Strength is bigger than the Joint Bearing Strength, the later will be considered as the joint allowable.

Screw’s Tensile Strength:

The Ultimate Tensile Strength for the bolt is 4,080 lb_f [Technical Data Sheet-NASM3-20].

FE200744 Stud and Lock Fitting

Stud’s Shear Strength:

The FE200744 fitting has an Ultimate Shear Strength of 4,200 lb_f when it is utilized in the aforementioned track [Technical Data Sheet-FE200744].

Joint Bearing strength:

The studs of the FE200744 fittings are utilized in joints J1 and J2, and it bears with 0.25” thick AL 6061-T6511 Extrusion that has an ultimate bearing strength of 69 ksi (refer to the table below). Hence, the joints’ ultimate bearing strength can be calculated as f_bru=F_bru x A_br = 69,000 x 3/8” x 0.25” = 6,468 lb_f.

Since the Ultimate Shear Strength is less than the Joint Bearing Strength, the former will be considered as the joints’ shear allowable.

Screw’s Tensile Strength:

The FE200744 fitting has an Ultimate Tensile Strength of 5,700 lb_f when it is utilized in the aforementioned track [Technical Data Sheet-FE200744].

The Heavy Duty Anodized Aircraft Track has a vertical load allowable of 6,000 lb_f [Technical Data Sheet-ANCRA Aircraft Track]. This represents the allowable reaction force exerted on the rail lip by the FE200744 stud head.

Since the Ultimate Tensile Strength of the FE200744 fitting is less than the Ultimate Tensile Strength of the rail, the former will be considered as the joints’ tensile allowable.

Summary of the Fasteners Allowables

LOAD CASES FORMULATION

The rack center is taken at X563.58. Directional flight accelerations are obtained from the DHC-8-100 load-case charts and interpolated at the rack station before comparison with the emergency-landing envelope.

It is noteworthy that the acceleration values at X563.58 were estimated using linear interpolation.

These are the limit loads values. However, the ultimate load values must be used for the static stress analysis purposes. The ultimate load values ($N_{u}$) can be obtained by multiplying the limit load values by 1.5. Alternatively, a factor of 1.5 will be utilized in calculating the safety margins later in this report.

The limit load cases that the Air Conditioning Rack Installation will be checked against are listed in the table below:

FINITE ELEMENT MODEL (FEM)

Air Conditioning Rack Model

Loads and Constraints

As illustrated previously, five cases are to be examined. These cases were examined through FEMAP by utilizing body limit loads in the direction of each case.

As shown in the above figure, the boundary conditions for the static stress analysis were applied to the following nodes:

• Node ①: AFT Inboard node.

• Node ②: FWD Inboard node.

• Node ③: AFT Outboard corner node.

• Node ④: FWD Outboard corner node.

• Node ⑥: FWD node at the attachment angle.

• Node ⑦: Center node at the attachment angle.

• Node ⑧: AFT node at the attachment angle.

The restrained degrees of freedom (DoFs) at each of the previous nodes varies per the load case. This ensures that the model accurately simulates the real-world constraints experienced by the Air Conditioning Rack. The table below lists the DoFs restrained at each node for each load case.

For the “FWD” loading case, point ③ has no upward restraint capability

For the “Inboard” loading case, points ③ and ④ have no upward restraint capability

For the “Upward” load case, points ③ and ④ have no upward restraint capability.

Air Conditioning Rack’s Supporting Structure Model

Loads and Constraints

Five load cases were analyzed on the Main Air Conditioning Rack model using FEMAP, with body limit loads applied in the direction corresponding to each case. The resulting reaction forces at the attachment points of the air conditioning rack, obtained from each load case, were then extracted and used as applied loads at the corresponding locations in the model of the rack’s supporting structure. This approach ensures load path consistency between the primary rack and its supporting structure.

The translational DoFs (T_xT_yT_z) were restrained at the following locations:

• All Nodes at the Floor Edges.

• All Nodes at the Doubler Edges.

• Three Nodes at the AFT Side of the AFT Clip.

• Fifteen Nodes at the Bottom Side of the Intercostal Brace.

Analysis

FEA RESULTS

Beams

Tube Structure

the figure below provides a cross-sectional view of a tube weld, where the weld extends 1/16” outward from the tube's surface and penetrates inward by 1/32”. The estimation of stress within the welds assumes a conservative methodology, accounting for stress redistribution across the weld section. This approach ensures that the evaluation of the weld's structural integrity remains on the safe side by considering potential stress concentrations and the redistribution that occurs in the surrounding material. 

For a given bending moment ($M$), the ratio of weld bending stress ($F_{b}^{weld}$) to the tube bending stress ($F_{b}^{tube}$) can be calculated as follows:

$$\frac{F_{b}^{weld}}{F_{b}^{tube}} = \frac{M\ c_{weld}}{I_{weld}} \times \frac{I_{tube}}{M\ c_{tube}}\ = \frac{c_{weld}I_{tube}}{I_{weld}c_{tube}}$$

Hence, the weld bending stress ($F_{b}^{weld}$) can be expressed as follows:

$${F_{b}^{weld} = \frac{c_{weld}I_{tube}}{I_{weld}c_{tube}}f_{b}^{tube}\ = \ \frac{\frac{9}{16}\ (0.056966)}{\ \frac{1}{2}(0.069111)}f_{b}^{tube} }\boxed{F_{b}^{weld} = 0.927\ F_{b}^{tube}}$$

For a given axial load ($f_{A}$), the ratio of weld axial stress ($F_{A}^{weld}$) to the tube axial stress ($F_{A}^{tube}$) can be calculated as follows:

$$\frac{F_{A}^{weld}}{F_{A}^{tube}} = \frac{f_{A}}{A_{weld}} \times \frac{A_{tube}}{f_{A}}\ = \frac{A_{tube}}{A_{weld}}$$

Hence, the weld axial stress ($F_{A}^{weld}$) can be expressed as follows:

$$F_{A}^{weld} = \frac{A_{tube}}{A_{weld}}F_{A}^{tube}\ = \frac{1^{2} - \left( \frac{3}{4} \right)^{2}}{\left( \frac{9}{8} \right)^{2} - \left( 1 - \frac{2}{32} \right)^{2}}F_{A}^{tube}$$

$$\boxed{F_{A}^{weld} = 1.131\ F_{A}^{tube}}$$

The table below lists the maximum Axial Force and Bending Moments values for each load case within the tube structure. As shown in the figure below, the global maximum values occur at different locations. Therefore, the tensile and compressive combined stresses (due to axial and bending moments loads) were calculated for each element within the tube structure using excel sheet.

As shown in the above figure, the maximum tensile and compressive combined stresses occur at element No. 9049 and they belong to the forward load case. The stress calculations at Element No. 9049 are as follows:

• The axial and bending stresses at the tubes:

• Therefore, the global maximum axial and bending stresses at the welds can be estimated as below:

• Hence, the maximum tensile and compression combined stresses on the tubes can be calculated as below (the bold values are the global maximums):

• Hence, the maximum tensile and compression combined stresses on the welds can be calculated as below (the bold values are the global maximums):

The tube structure is fabricated from 1”x1”x0.125” AL 6061-T6 extrusion. This material selection is noted for its high Ultimate Tensile Stress value of $F_{tu}^{tube} = 38\ ksi$ (refer to the table below), and its Compression Yield Stress value is $F_{cy}^{tube} = 34\ ksi$. Adjacent to the weld areas, Ultimate Tensile Stress and the Compression Yield Stress values are $F_{tu}^{weld} = 24\ ksi$ and $F_{cy}^{weld} = 15\ ksi$, respectively [Table 2-19W, Aluminum Design Manual 2010]. This reduction in strength accounts for the weakening effects of welding, which includes alterations in microstructure and potential introduction of stress concentrators. Therefore, the margin of safety in the Tube Structure can be computed as below:

Column-Buckling allowable:

The column buckling allowable can be expressed as below:

$$P_{cr} = \frac{n^{2}K\pi^{2}E_{c}I}{L^{2}}$$

Where, K is the buckling coefficient. E_c is the compressive modulus of elasticity of the material. I is the minimum moment of inertia of the column. L is the total length of column.

Since the tubes are welded and we are considering bending at the welds they are closer to fixed supports. A reasonable approximation would be to calculate the column allowable for the fixed-fixed case (K=4) as well as the pinned-pinned (K=1) case, then the average of the resulted critical values will be our benchmark.

The most critical beam in the tube structure has a length of 40 in. Which is the longest beam in the tube structure that is only fixed at its ends. Therefore, the critical buckling stresses can be calculated as below:

$$F_{cr}^{pin - pin} = \frac{1^{2} \times 1 \times \pi^{2}(10.1 \times 10^{6}) \times 0.056966}{0.4375 \times 40^{2}}\ \rightarrow F_{cr}^{pin - pin} = 8.11\ ksi$$

$$F_{cr}^{fix - fix} = \frac{1^{2} \times 4 \times \pi^{2}(10.1 \times 10^{6}) \times 0.056966}{0.4375 \times 40^{2}} \rightarrow F_{cr}^{fix - fix} = 32.45\ ksi$$

$$\boxed{F_{cr} = \frac{8.11 + 32.45}{2} = 20.28\ ksi}$$

As previously shown, the maximum compressive combined stress is ${F_{Comb(EndB)}^{tube}}_{MAX.COMP} = 5.96\ ksi$ and it occurs at the tubes. Therefore, the margin of safety in the Tube Structure can be computed as below:

Flange Crippling

Crippling is a mode of failure that occurs due to compression effects. Typically, this is a check that is applied to thin-walled columns where the local stability of the cross section may not allow the column to achieve its full column strength.

To calculate the Crippling Allowable of the Tube Structure, Needham and Gerard Method is used. Needham's method divides the structural section into angle elements whose individual crippling strengths are determined experimentally and summed to obtain the total section strength. Instead of determining crippling strengths experimentally, semi-empirical equations were used as proposed in the relevant discussion in [Stress Analysis Manual-Air Force Flight Dynamics Laboratory-Wright-Patterson]. Moreover, the tube structure cross-section is modified, as shown below to simplify the analysis.

Angle #	h	th	b	tb	An	Ce	Fccn	AnFccn
1	0.500	0.125	0.500	0.125	0.125	0.366	75.075	9.384
2	0.500	0.125	0.500	0.125	0.125	0.366	75.075	9.384
3	0.500	0.125	0.500	0.125	0.125	0.366	75.075	9.384
4	0.500	0.125	0.500	0.125	0.125	0.366	75.075	9.384
				Σ	0.500			37.537
					FCrippling [ksi]	75.07

C_e=0.316 for 2-edges free, 0.342 for 1-edge free, and 0.366 for no edge free. F_ccn is expressed below:

$$F_{ccn} = \frac{C_{e}\sqrt{F_{cy}E}}{\left( \frac{h + b}{2t_{\min}} \right)^{0.75}}$$

Where: h and b are the angle’s height and width, respectively. t_min is the minimum thickness in the angle. F_cy is Material compressive yield Strength. E is Material young’s modulus

As previously shown, the maximum compressive combined stress is ${F_{Comb(EndB)}^{tube}}_{MAX.COMP} = 5.96\ ksi$ and it occurs at the tubes. Therefore, the Tube Structure passes by observation against crippling.

Structural Sheet Metals in the Air Conditioning Rack Assembly

In this section, the structural sheet metals will be assessed based on the maximum tensile and compressive principal stresses. The maximum tensile principal stress will be checked against the allowable ultimate tensile strength (F_tu) of the material. On the other hand, the maximum compressive principal stress will be checked against the allowable yield compressive strength (F_cy) of the material. The maximum principal stress (F₁) and the minimum principal stress (F₂) on both sides of the plates were reviewed for all cases, and it is summarized in the below table

AL 2024-T3 CLAD Sheet (0.063” – 0.128” thick)

In the AL 2024-T3 CLAD Sheets, a tensile and compressive threshold values are set to 35 ksi and -22 ksi, respectively. The table below lists some of the flagged elements that have principal stress values falling out of the set thresholds.

Excluding these elements from the failure assessment, a new maximum and minimum principal stresses for the structural sheet metal are determined and summarized in the below table. The maximum tensile and compressive principal stresses are 28.08 ksi and 21.81 ksi, respectively, and they belong to the forward load case.

Based on the table below, the allowable ultimate tensile strength (F_tu) and yield compressive strength (F_cy) for AL 2024-T3 CLAD Sheets are 62 ksi and 37 ksi, respectively. The margin of safety will be the lowest value between the tensile load MS_T and the compressive load MS_C. These margins of safety can be expressed as below:

AL 6061-T6 and T6511 Extrusion (Thickness ≤1”)

In the AL 6061-T6 and T6511 Extrusions that have a thickness value less than or equal 1 inch, a tensile and compressive threshold values are set to 23 ksi and -21 ksi, respectively. The figure below shows the flagged elements that have principal stress values falling out of the set thresholds.

Excluding these elements from the failure assessment, a new maximum and minimum principal stresses for the structural sheet metal are determined and summarized in the below table. The maximum tensile and compressive principal stresses are 23.00 ksi and 20.95, respectively, and they belong to the forward load case.