Audio and Power Equipment Racks Installations Structural Substantiation

INTRODUCTION

The substantiation establishes the rack load path from installed equipment through the rack structure and attachment studs into the aircraft, then checks the critical members and joints against applicable ultimate-strength criteria.

DESIGN ASSESSMENT

General Layout

Audio Equipment Rack Components

The rack combines AL 6061-T6 tube/extrusion load-carrying members with AL 2024-T3 sheet components. The following tables retain the source geometry, thicknesses and material allowables used in the substantiation.

Fasteners Allowables

In rivets, the standard methods of stress analyses of riveted joints consider two primary types of failure, namely, the (i) shear of the rivet’s shank, and the (ii) bearing or compressive failure of the metal at the point where the rivet bears against the connecting sheet or plate.

The tension on rivets shall be restricted to conditions in which tension load is minor compared with the shear load, which is the main purpose of the rivet. If this is not the case, the rivet tension allowable should be determined.

MS20426AD4 rivet

In the Audio Equipment Rack assembly, the tensile load carried by the MS20426AD4 rivets within all joints are minor compared with the shear load. Therefore, the tension failure is not critical for the MS20426AD4 rivets and will not be checked. The shear and bearing allowables for the MS20426AD4 rivets will be calculated, and the lowest among them will be chosen as the joint allowable for the MS20426AD4 rivet-sheet configuration.

The MS20426AD4 rivet is utilised at various locations within the Audio Equipment Rack assembly with different configurations. As illustrated in the corresponding table, the rivet was used in a single shear state at joints J₃-J₉ and in a double shear state at joints J₁ and J₂.

(i) Shear of the Rivet’s Shank:

As shown in the corresponding table, the MS20426AD4 is made from AL 2117-T3 alloy that has an ultimate shearing strength F_su = 30 ksi, and rivet single shear strength value of f_su=389 lb_f (for 1/8 rivet size). As illustrated in the corresponding table, the rivets in the joints J₁ – J₂, and J₃ - J₈ are in double and single shear state, respectively, and the CSK sheet in these joints is made from AL 2024-T3 ALCLAD sheet. Therefore, the joints’ shear strength is determined using [MMPDS-15- Table 8.1.2.2(o)] (considering the CSK sheet thickness) and tabulated in the corresponding table.

In joint J₉, the CSK sheet is made from 0.125” thick AL 6061-T6 Extrusion. There is no table for the Static Joint Strength for this material type. Although the CSK sheet is thick enough to consider the ultimate single shear strength value as the joint allowable, the ultimate single shear strength value is scaled down, conservatively, by a factor of 1.5. As a result, the estimated ultimate shear strength of that joint is 389/1.5=259 lb_f.

(ii) Joint Bearing strength:

When considering a countersunk rivet joining two sheets of different thicknesses, especially when the countersunk portion does not engage with the thinner sheet(s), the calculation of ultimate joint stress with a focus on bearing stress becomes particularly important since the mechanical interlock provided by the rivet is primarily with the thicker sheet. Moreover, the load transfer mechanism relies heavily on the bearing stress between the rivet shank and the hole in the thinner sheet. The ultimate bearing stress calculation became critical to ensure that the material of the thinner sheet around the rivet hole can withstand the compressive load exerted by the rivet without yielding or failing.

The non-CSK sheet is thinner than the CSK sheet in joints J₃ and J₅. Also, the joints J₇, J₈, and J₉ involve equipment flange that were conservatively omitted. For these joints the Joint Bearing Strength must be calculated using the non-CSK sheet thickness and compared with the Static Joint Strength calculated previously and the smaller value will be considered as the joint allowable. The non-CSK sheets in joints J₃, J₅, J₇, and J₈, are AL 2024-T3 ALCLAD, and it has an ultimate bearing strength of 121 ksi (refer to the corresponding table). The non-CSK sheet in the joint J₉ is AL 6061-T6 Extrusion, and it has an ultimate bearing strength of 82 ksi (refer to the corresponding table). Hence, the ultimate bearing strength for these joints can be calculated using [MMPDS-15- Table 8.1.2.1(a)] by multiplying the Joint Bearing strength for F_br=100 ksi by the ratio of actual bearing strength to 100 ksi. The results are listed in the corresponding table. It is noteworthy that the rest of the joints have a non-CSK sheet thickness ≥ the CSK sheet thickness. Hence, calculating the joint ultimate bearing strength is not required for these joints.

CR3212 rivet

In the Audio Equipment Rack assembly, the tensile load carried by the CR3212 rivets within all joints are minor compared with the shear load. Therefore, the tension failure is not critical for the CR3212 rivets and will not be checked. The shear and bearing allowables for the CR3212 rivets will be calculated, and the lowest among them will be chosen as the joint allowable for the CR3212 rivet-sheet configuration.

As illustrated in the corresponding table, the CR3212 rivet is utilised at two locations within the Audio Equipment Rack assembly with different configurations, but in both configurations the rives are used in a single shear state.

(i) Shear of the Rivet’s Shank:

Based on the corresponding table, the CR3212 is made from AL 5056 alloy that has an ultimate shearing strength F_su = 50 ksi, and rivet single shear strength value of f_su=664 lb_f (for 1/8 rivet size). As illustrated in the corresponding table, the rivets in the joints J₁ – J₂ are in single shear state, and the CSK sheet in these joints is made from AL 2024-T3 ALCLAD sheet. Therefore, the joint’s shear strength using the CSK sheet thickness is determined using [MMPDS-15-Table 8.1.3.2.2(v)] and tabulated in the corresponding table. It is noteworthy that the values in [Table 8.1.3.2.2(v)] are for F_su=51 ksi, so the joint’s shear strength was scaled down by a factor of 50/51.

(ii) Joint Bearing strength:

The non-CSK sheet is thicker than the CSK sheet in both joints. Therefore, calculating the joint ultimate bearing strength is not required [MMPDS-15].

MS20470AD4 rivet

In the Audio Equipment Rack assembly, the tensile load carried by the MS20470AD4 rivets within all joints are minor compared with the shear load. Therefore, the tension failure is not critical for the MS20470AD4 rivets and will not be checked. The shear and bearing allowables for the MS20470AD4 rivets will be calculated, and the lowest among them will be chosen as the joint allowable for the MS20470AD4 rivet-sheet configuration.

The MS20470AD4 rivet is utilised at various locations within the Audio Equipment Rack assembly with different configurations. As illustrated in the corresponding table, the rivet was used in a single shear state at all joints, namely, J₁-J₆.

(i) Shear of the Rivet’s Shank:

As illustrated in the corresponding table, the rivets at all joints are in single shear state and have D_r/t_min ratios < 3, except for J₅ where D_r/t_min ratio is 3.91. Therefore, a correction factor, $\alpha$, must be calculated for J₅ to compensate for the reduction in the rivet shear strength resulting from hight bearing stresses on the rivet in this case [MMPDS-15]. The correction factor in a single shear joint can be expressed as below:

$$\boxed{\alpha_{ss} = 1 - 0.04\left( \frac{D_{r}}{t_{\min}} - 3 \right)}$$

Therefore, the correction factors for the joints J₅ can be calculated as below:

Based on [MMPDS-15-Table 8.1.2(b)], the Ultimate Single Shear Strength, f_su, for the MS20470AD4 rivet can be tabulated as in the corresponding table.

(ii) Joint Bearing strength:

For these joints shown in the corresponding table, the Joint Bearing Strength will be calculated using the thinnest sheet material. Then it will be compared with the Ultimate Single Shear Strength calculated previously and the smaller value will be considered as the joint allowable.

The thinnest sheet in the joints J₁, J₂, J₅, and J₆ is AL 2024-T3 ALCLAD that has an ultimate bearing strength of 121 ksi (refer to the corresponding table). The thinnest sheet in the joints J₃ and J₄ is AL 6061-T6 Extrusion that has an ultimate bearing strength of 82 ksi (refer to the corresponding table). Hence, using [MMPDS-15-Table 8.1.2.1(a)], the ultimate bearing strength for these joints can be calculated by multiplying the values from that table by the ratio of actual bearing strength to 100 ksi. The results are listed in the corresponding table.

MS24693-S273 Screw

The MS24693-S273 Screw is utilised at various locations within the Audio Equipment Rack assembly with different configurations. As illustrated in the corresponding table, the screw was used in single shear states at joints J₁ and J₃, and in double shear state in joint J₂. Note that in joints J₂ and J₃, the screw passes through the equipment flanges EF1 and EF2, conservatively, these flanges will be omitted while determining the screw allowable. Therefore, the joint J₂ will be considered as single shear joint.

Based on the corresponding table, the #10-32 MS24693-S273 screw has the below specifications:

• Material: Cadmium Plated Carbon Steel

• Minimum Tensile Strength F_tu=60 ksi.

• Minimum Shear Strength F_su=36 ksi.

• Length: 0.625”

• Shank Diameter: 0.19”

• Thread: UNF-2A

By referring to [MMPDS-15-Table 8.1.5(a) and MS24693-Specs], the Ultimate Single Shear Strength and the Ultimate Tensile Strength of the screw are f_su=992x36/35=1,020 lb_f and f_tu=1,200 lb_f, respectively.

To determine the Joint Bearing Strength, the thinnest sheet in the joints must be used. Note that the thinnest sheet in:

• J₁ and J₃: is made from AL 2024-T3 ALCLAD. Its ultimate bearing strength F_bru=121 ksi (refer to the corresponding table).

• J₂: Both sheets are made from different materials that have the same thicknesses. The sheet with the lowest F_bru value will be used, namely, AL 6061-T6 Extrusion that has an ultimate bearing strength F_bru=82 ksi (refer to the corresponding table).

Hence, using [MMPDS-15-Table 8.1.5.1], the ultimate bearing strength for these joints can be calculated by multiplying the values from that table by the ratio of actual bearing strength to 100 ksi. The results are listed in the corresponding table.

MS24693-S6 Screw

The MS24693-S6 Screw is utilised at the USB Port. It passes through the Port flange and 0.156” thick AL 6061‑T6 Extrusion. Conservatively, only the AL 6061‑T6 Extrusion will be considered in calculating the joint allowable.

Based on the corresponding table, the #4-40 MS24693-S6 CSK screw has the below specifications:

• Material: Cadmium Plated Carbon Steel

• Minimum Tensile Strength F_tu=60 ksi.

• Minimum Shear Strength F_su=36 ksi.

• Length: 0.5”

• Shank Diameter: 0.112”

• Thread: UNC-2A

By referring to [MMPDS-15-Table 8.1.5(a) and MS24693-Specs], the Ultimate Single Shear Strength and the Ultimate Tensile Strength of the screw are f_su=345x36/35=354 lb_f and f_tu=360 lb_f, respectively.

The screw passes through 0.156” thick AL 6061-T6 Extrusion that has an ultimate bearing strength F_bru=82 ksi (refer to the corresponding table). Hence, the ultimate bearing strength for the joints can be calculated as below:

$$f_{br} = F_{br} \times A_{br}$$

Where: $F_{br}$ is the bearing strength of the sheet material [ksi]. $A_{br}$ is the bearing area, which is the sheet thickness times the bolt shank diameter (${t \times D}_{s}$).

$$f_{br} = 82,000 \times 0.156 \times 0.112 = 1432\ {lb}_{f}$$

Since the Ultimate Shear Strength is less than the ultimate bearing strength, the former will be the joint allowable.

FE200744 STUD

Based on the corresponding table, the FE200744 stud is 3/8-24 UNRF screw that is made from Zinc Plated Carbon steel. It has a vertical (tension) ultimate load of 5,000 lb_f, and a horizontal (shear) ultimate load of 2,000 lb_f.

The FE200744 stud is utilized in a Seat Track Stud that is made from 0.125” thick AL 2024-T351 Plate. Conservatively, the Bearing Strength of 0.25” thick AL 2024-T351 Plate will be considered. Based on MMPDS-15-the corresponding table.2.4.0(b₂)], the bearing strength is F_bru=119 ksi. Therefore, based on [MMPDS-15-Table 8.1.5.1], the Unit Bearing Strength of the joint is f_bru=4688*119/100 = 5,578.

Since the Ultimate Shear Strength is less than the ultimate bearing strength, the former will be the joint allowable.

Track tooth allowable:

Based on [Ancra Clear Medium-Duty Aircraft Track-Anodized-40456-10-144-Specs], the seat track’s vertical load allowable is 4500 lb_f. Hence, the vertical load allowable per tooth is 4500/2 = 2,250 lb_f (this is the allowable reaction force exerted on the rail lip by the FE200744 lock head)

UPPER ATTACHMENT STUD

The customized Upper Attachment Stud is made from the ANCRA Threaded Stud. However, its load capacity is not available. Hence, ANCRA Threaded Stud (P/N 40351-10) [40351 and 40352 Threaded Stud-Specs] will be used to determine the Upper Attachment Stud allowable.

As shown in [40351 and 40352 Threaded Stud-Specs], the load capacity for ANCRA Threaded Stud (P/N 40351-10) is 4000 lb_f in any direction. Moreover, both ANCRA Threaded Studs (P/N 40352-10 & 40351-10) are made from the same material. However, only P/N 40352-10 is heat treated to 180-200 ksi per AMS-H-6875 and is made per MIL-S-6049 that has ultimate tensile strength of 125ksi.

Stud Bending:

At the head of the fastener there is region that increases in diameter, and its length is approximately 0.18”. Moreover, there is a curved portion at the tip of the stud that can be discounted since the worst (and still conservative) case would be a concentrated load at the beginning of the full diameter. Consequently, the bending arm can be calculated as follows:

$$L_{bending - arm} = 1.77 - 0.19 - 0.18 = 1.4\ in$$

Hence, the maximum allowable shear load before bending failure occurs, can be calculated as follows:
$$f_{s,\ max} = \frac{125 \times 10^{3} \times \pi \times {0.38}^{4}}{64 \times 1.4 \times 0.38/2} = 481\ {lb}_{f}$$

Track tooth allowable:

Based on [Ancra Clear Medium-Duty Aircraft Track-Anodized - 40456-10-144-Specs], the seat track’s vertical load allowable is 4500 lb_f. Hence, the vertical load allowable per tooth is 4500/2 = 2,250 lb_f (this is the allowable reaction force exerted on the rail lip by the stud fastener head)

SUMMARY OF ALLOWABLES

As we mentioned previously, the joint allowable is established by identifying the lowest value between f_su and f_bru. Based on the previous calculations, we can set the below Joint Allowables as a benchmark in our static stress analysis.

WEIGHTS

The Non-Structural Exterior Removable Panels, the Non-Structural Assembly Parts, and Tiny Equipment Units/Parts (shown in the corresponding table) will be added to the FE model as a non-structural mass distributed over the tube structure. Its value will be W_NSM=17.918 + 22.913+ 0.426 = 41.257 lb_f

Tube Structure (TOP) Equipment Units (BOTTOM) in the Audio Equipment Rack Assembly.

the corresponding tables show the weights of the Audio Equipment Rack’s installed equipment unit, based on their specification sheets. It is noteworthy an additional estimated weight (1.5 times the component weight) has been added to each equipment to account for the weight of the hardware and wires/cables installed.

STATIC STRESS ANALYSIS

LOAD CASES FORMULATION

These are the limit loads values. However, the ultimate load values must be used for the static stress analysis purposes. The ultimate load values ($N_{z,u}$) can be obtained by multiplying the limit load values by 1.5. the corresponding table lists these values.

Hence, the load cases the Audio Equipment Rack will be checked against are listed in the corresponding table.

FINITE ELEMENT ANALYSIS (FEA)

MODEL

LOADS AND CONSTRAINTS

ANALYSIS

FEA RESULTS

Tube Structure

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 corresponding figure illustrates the distribution of the Axial Force and Bending moment, within the tube structure. This figure highlights that the highest Tensile and Compression Axial Forces and Bending moment are $f_{A,t} = 715\ {lb}_{f}$, $f_{A,c} = 622.5\ {lb}_{f}$, and $M = 1573\ in - {lb}_{f}$, respectively. It is noteworthy that these maximum loads occur due to the load from the 9g FWD case. This scenario represents a critical load condition that must be thoroughly evaluated to ensure the structural resilience and safety of the structure under extreme operational loads. 

Based on the maximum Axial Force and Bending Moment values on the tube structure, the resulted axial and bending stresses at the tubes can be calculated as below:

$$F_{A,\ t}^{tube} = \frac{f_{A,t}}{A_{tube}} = \ \frac{715.7}{0.44}\ \rightarrow \boxed{f_{A,\ t}^{tube} = 1.627\ ksi}$$

$$F_{A,\ c}^{tube} = \frac{f_{A,c}}{A_{tube}} = \ \frac{622.5}{0.44}\ \rightarrow \boxed{f_{A}^{tube} = 1.415\ ksi}$$

$$F_{b}^{tube} = \frac{M\ c_{tube}}{I_{tube}} = \ \frac{1573\ \times \ \frac{1}{2}}{0.056966}\ \rightarrow \boxed{f_{b}^{tube} = 13.807\ ksi}$$

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

$$f_{A,t}^{weld} = 1.131\ \times 1.627\ \rightarrow \boxed{f_{A}^{weld} = 1.840\ ksi}$$

$$f_{A,c}^{weld} = 1.131\ \times 1.415 \rightarrow \boxed{f_{A}^{weld} = 1.6\ ksi}$$

$$f_{b}^{weld} = 0.927\ \times 13.807 \rightarrow \boxed{f_{b}^{weld} = 12.799\ ksi}$$

Hence, the maximum tensile and compression stresses on the tubes and welds are as below:

$$f_{t}^{tube} = f_{b}^{tube} + f_{A,\ t}^{tube}\ \rightarrow \boxed{f_{t}^{tube} = 15.434\ ksi}$$

$$f_{c}^{tube} = f_{b}^{tube} + f_{A,\ c}^{tube}\ \rightarrow \boxed{f_{c}^{tube} = 15.222\ ksi}$$

$$f_{t}^{weld} = f_{b}^{weld} + f_{A,\ t}^{weld}\ \rightarrow \boxed{f_{t}^{weld} = 14.639\ ksi}$$

$$f_{c}^{weld} = f_{b}^{weld} + f_{A,\ c}^{weld}\ \rightarrow \boxed{f_{c}^{weld} = 14.399\ ksi}$$

The tube structure is fabricated from 1”x1”x0.125” aluminum extrusion, specifically using 6061-T6 alloy as per AMS-WW-T-700/6 TYPE II specifications. This material selection is noted for its high Ultimate Tensile Stress value of $F_{tu}^{tube} = 38\ ksi$ as documented in [MMPDS-15-Table 3.6.2.0(g)], 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, [Aluminum Design Manual 2010-Table 2-19W]. Since 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 Welded Tube Structure can be computed as below:

Column-Buckling allowable:

The column buckling allowable can be expressed as below:

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

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 longest beam in the tube structure has a length of 35.7 in. Therefore, the critical buckling stresses can be calculated as below:

$$F_{cr}^{pin - pin} = \frac{1 \times \pi^{2}(10.1 \times 10^{6}) \times 0.056966}{0.4375 \times {35.7}^{2}}\ \rightarrow F_{cr}^{pin - pin} = 10.18\ ksi$$

$$F_{cr}^{fix - fix} = \frac{4 \times \pi^{2}(10.1 \times 10^{6}) \times 0.056966}{0.4375 \times {35.7}^{2}} \rightarrow F_{cr}^{fix - fix} = 40.72\ ksi$$

$$\boxed{F_{cr} = \frac{10.18 + 40.72}{2} = 25.45\ ksi}$$

As previously shown, the maximum compression stress is $f_{c}^{tube} = 15.222\ ksi$ and it occurs at the tubes. Therefore, the margin of safety in the Welded Tube Structure can be computed as below:

Structural Sheet Metals and their Attachments

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 corresponding table.

As shown in the corresponding table and the corresponding figure, the Maximum Tensile Principal Stress for the Structural Sheet Metals reaches 6.02 ksi under the downward loading case at the bottom side of the plate. In addition, the Maximum Compressive Principal Stress reaches 6.03 ksi under the same loading case at the top side of the plates.

Based on the corresponding table, the structural sheet metals are made from 0.071” thick Aluminum 2024-T3 ALCLAD sheet. Based on the corresponding table, The allowable ultimate tensile strength (f_tu) and allowable yield compressive strength (f_cy) are 62 ksi and 37 ksi, respectively. The minimum 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:

Loads on Rivets:

As shown in the corresponding figure. the Maximum Plate Membrane Forces per unit length in the structural sheet metals are as follows:

1. For the top panel: $N_{xy} = 20.17\ lb/in$, $N_{x} = 14.6\ lb/in$, and $N_{y} = 52.61\ lb/in$.

2. For the inboard step panel: $N_{xy} = 40.92\ lb/in$, $N_{x} = 72.7\ lb/in$, and $N_{y} = 45.76\ lb/in$

All these values belong to the forward load case. Conservatively, assume that these forces are equal along all the structural sheet metals edges, the maximum membrane forces at the longitudinal and lateral structural sheet metals edges, $P^{Panel,\ \ L}\$and $P^{Panel,\ \ LT}$ for both the top (D: 20.8” x 23”) and inboard step (D: 20.8” x 4.8”) panels, can be calculated as below:

The top panel is fixed to tube structure using:

1. To the Tube Structure through 24 x CR3212 rivets along the lateral direction at the fwd and aft sides (12 on each side).

2. To the Tube Structure through 9 x CR3212 rivets along the longitudinal direction at the inboard side.

3. To the Top Support Angle through 8 x CR3212 rivets along the longitudinal direction at the outboard side.

Top Panel attachment points

Conservatively, it is assumed that the load is carried through 8 rivets at each edge. Assuming that the load calculated previously will be evenly distributed between the 8 rivets at each edge, the load carried by each rivet will be as follows:

The rivet that is in the corner will carry the highest shear load because it carries loads due to N_x, N_y, N_xy, and N_yx. The total shear load carried by the corner rivet can be calculated as follows:

$$P_{s,corner - rivet\ }^{TopPanel\ } = \sqrt{{(52.44 + 151.25)}^{2} + {(37.96 + 57.99)}^{2}} = 225.16\ {lb}_{f}$$

Based on the corresponding table, the CR3212 rivet has an ultimate shear load of 401 lb_f. Therefore, the minimum margin of safety can be expressed as below:

the corresponding figure illustrates the previous steps followed to calculate the minimum margin of safety of the Top Panel’s rivets.

Top Support Angle and Upper Support Clips

As illustrated in the corresponding figure, the Upper Support Clip is made from 0.05” thick AL 2024-T3 ALCLAD Sheet. It is attached to the tube structure through two CR3213 rivets, and to the Top Support Angle though another two CR3213 rivets.

Based on the maximum Axial Force and Bending Moment values on the Top Support Angle (shown in the corresponding figure), the resulted axial and bending stresses can be calculated as below:

$${F_{A,\ t}^{Supp.Angle\ } = \frac{f_{A,t}}{A_{Supp.Angle}} = \ \frac{203.4}{0.44}\ \rightarrow \boxed{F_{A,\ t}^{Supp.Angle\ } = 1.67\ ksi}$$$$}{F_{A,\ c}^{Supp.Angle\ } = \frac{f_{A,c}}{A_{Supp.Angle}} = \ \frac{209.7}{0.44}\ \rightarrow \boxed{F_{A,\ c}^{Supp.Angle\ } = 1.72\ ksi}$$$$}{F_{Bending,\ t}^{Supp.Angle\ } = \frac{M\ c_{t}}{I_{Supp.Angle}} = \ \frac{172.6\ \times \ 0.0326943}{0.0119587}\ \rightarrow \boxed{F_{Bending,\ t}^{Supp.Angle\ } = 0.47\ ksi}$$$$}{F_{Bending,\ c}^{Supp.Angle\ } = \frac{M\ c_{c}}{I_{Supp.Angle}} = \ \frac{172.6\ \times \ 0.967306}{0.0119587}\ \rightarrow \boxed{F_{Bending,\ c}^{Supp.Angle\ } = 13.96\ ksi}}$$

Therefore, the maximum tensile and compressive stresses on the Top Support Angle are as below:

$$F_{\ t}^{Supp.Angle\ } = F_{Bending,\ t}^{Supp.Angle\ } + F_{A,\ t}^{Supp.Angle\ }\ \rightarrow \boxed{F_{\ t}^{Supp.Angle\ } = 2.14\ ksi}$$

$$F_{c}^{Supp.Angle\ } = F_{Bending,\ c}^{Supp.Angle\ } + F_{A,\ c}^{Supp.Angle\ }\ \rightarrow \boxed{F_{c}^{Supp.Angle\ } = 15.68\ ksi}$$

The Top Support Angle is made from 0.063" AL 6061-T6 Extrusion. Based on the corresponding table, it has an Ultimate Tensile Strength value of $F_{tu}^{Supp.Angle} = 38\ ksi$, and a Compressive Yield Strength value is $F_{cy}^{Supp.Angle} = 34\ ksi$. Therefore, the margin of safety can be computed as below:

Column-Buckling allowable:

The column buckling allowable can be expressed as below:

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

Where, K is the buckling coefficient. Ec 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 Top Support Angle is attached to the Upper Support Clips though 4 rivets (2 at each side), this angle can be considered as pin-pin supported (K=1).

The Top Support Angle has a length of 18.9 in. Therefore, the critical buckling stresses can be calculated as below:

$$F_{cr}^{pin - pin} = \frac{1 \times \pi^{2}(10.1 \times 10^{6}) \times 0.0119587}{0.122031 \times {18.9}^{2}}\ \rightarrow F_{cr}^{pin - pin} = 27.35\ ksi$$

Therefore, the margin of safety in against buckling can be computed as below:

Mid Shelf Horizontal Angles

The two Mid Shelf Horizontal Angles are made from 0.125” thick AL 6061-T6 Extrusion. Each angle is attached to the tube structure through four CR3213 rivets (two at each side). Moreover, each angle is attached to two Vertical Support Angles that is made from the same material but 0.25” thick. This has been done through eight MS20470AD4 rivets (four for each Vertical Support Angle).

Based on the corresponding figures:

1. The maximum Tensile and Compressive Axial Forces in the Mid Shelf Horizontal Angle are 36.57 lb_f and 39.89 lb_f, respectively.

2. The maximum positive and negative bending moment values in plane 1 are 192.4 in- lb_f and 209.8 in- lb_f, respectively.

3. The maximum positive and negative bending moment values in plane 2 are 206 in- lb_f and 214 in- lb_f, respectively.

These maximum values belong to different loading cases scenarios and occur at different locations in the beams. Hence, by considering these values in calculating the maximum tensile and compressive combined stresses, the resulted value (calculated below) will be conservative.

• The tensile and compressive bending stresses due to the maximum moments in plane 1 are:

$${F_{1}^{+}}_{Bending,\ t}^{Mid.Hor.Angle\ } = \frac{M_{1}^{+}\ c_{t}}{I_{zz}} = \ \frac{192.4\ \times \ 1.285326}{0.150349}\ \rightarrow \boxed{{F_{1}^{+}}_{Bending,\ t}^{Mid.Hor.Angle\ } = 1.64\ ksi}$$

$${F_{1}^{+}}_{Bending,\ c}^{Mid.Hor.Angle\ } = \frac{M_{1}^{+}\ c_{c}}{I_{zz}} = \ \frac{192.4\ \times \ 0.214674}{0.150349}\ \rightarrow \boxed{{F_{1}^{+}}_{Bending,\ c}^{Mid.Hor.Angle\ } = 0.27\ ksi}$$

$${F_{1}^{-}}_{Bending,\ t}^{Mid.Hor.Angle\ } = \frac{M_{1}^{-}\ c_{t}}{I_{zz}} = \ \frac{209.8\ \times \ \ 0.214674}{0.150349}\ \rightarrow \boxed{{F_{1}^{-}}_{Bending,\ t}^{Mid.Hor.Angle\ } = 0.30\ ksi}$$

$${F_{1}^{-}}_{Bending,\ c}^{Mid.Hor.Angle\ } = \frac{M_{1}^{-}\ c_{c}}{I_{zz}} = \ \frac{209.8\ \times 1.285326}{0.150349}\ \rightarrow \boxed{{F_{1}^{-}}_{Bending,\ c}^{Mid.Hor.Angle\ } = 1.79\ ksi}$$

Hence, the maximum tensile and compressive bending stresses due to the moment in plane 1 are:

$\boxed{{F_{1}}_{Bending,\ t}^{Mid.Hor.Angle\ } = 1.64\ ksi}$ AND $\boxed{{F_{1}}_{Bending,\ c}^{Mid.Hor.Angle\ } = 1.79\ ksi}$

• The tensile and compressive bending stresses due to the maximum moments in plane 2 are:

$${F_{2}^{+}}_{Bending,\ t}^{Mid.Hor.Angle\ } = \frac{M_{2}^{+}\ c_{t}}{I_{yy}} = \ \frac{206\ \times \ 0.785326}{0.0263256}\ \rightarrow \boxed{{F_{2}^{+}}_{Bending,\ t}^{Mid.Hor.Angle\ } = 6.15\ ksi}$$

$${F_{2}^{+}}_{Bending,\ c}^{Mid.Hor.Angle\ } = \frac{M_{2}^{+}\ c_{c}}{I_{yy}} = \ \frac{206\ \times \ 0.214674}{0.0263256}\ \rightarrow \boxed{{F_{2}^{+}}_{Bending,\ c}^{Mid.Hor.Angle\ } = 1.68\ ksi}$$

$${F_{2}^{-}}_{Bending,\ t}^{Mid.Hor.Angle\ } = \frac{M_{2}^{-}\ c_{t}}{I_{yy}} = \ \frac{214\ \times \ \ 0.214674}{0.0263256}\ \rightarrow \boxed{{F_{2}^{-}}_{Bending,\ t}^{Mid.Hor.Angle\ } = 1.75\ ksi}$$