Inertial Navigation and Surveying System Installation Structural Substantiation

INTRODUCTION

Static-strength substantiation of the Inertial Navigation and Surveying System installation on a DHC-8-100. The assessment follows the load path from the equipment and tray through the support angles, angle/channel beams, tee clips and aircraft attachment structure.

Classical rigid-body and beam methods are used to extract attachment demand, evaluate fastener/joint capacity, and substantiate critical structural members against applicable material allowables.

DESIGN ASSESSMENT

INSGPS Installation Components

The following component schedule identifies the load-carrying additions and the existing aircraft members used in the substantiation. Thickness and material are retained exactly from the source report.

INSGPS Installation Weight Estimation

Material Properties

Material allowables used throughout the classical checks are summarized below. Source-specific thickness ranges, longitudinal/transverse values and bearing allowables are retained.

Fasteners Allowables

Pin-Collar Fasteners

AN3 and 4 Bolts

AN3 and AN4 bolt strengths are combined with the critical-sheet bearing capacity to establish the usable joint allowables below.

LOAD CASES FORMULATION

The installation spans X370.8–X387.35. A conservative analysis station of X380.00 is used to extract the worst-case flight accelerations from the DHC-8-100 load-case data. The surrounding station values and interpolated X380.00 factors are summarized below.

The source analysis evaluates the following limit cases and incorporates the 1.5 ultimate factor in the subsequent margin-of-safety calculations. The aft case is enveloped by the forward case and is not analyzed separately.

CLASSICAL ANALYSIS

Inertial Navigation and Surveying Equipment

The bolt demand below is the maximum reaction at any one of the four equipment-mounting locations for each load case.

Governing attachment demand:

• The maximum compressive axial load is f_A,c=32.94 lb₍f₎_f (LIMIT), and it belongs to the Downward Load Case. This load will be distributed over a wider area throughout the attachment surface; therefore, it passes by inspection.

• The maximum tensile and shear loads are f_A,t=24.03 lb₍f₎_f (LIMIT), and f_s=6.21 lb₍f₎_f (LIMIT). As established in the fastener-allowables discussion above, the ultimate tensile and shear loads for the AN3-7 Bolts in this configuration are f_tu=2,210 lb₍f₎_f, and f_su=1,686 lb₍f₎_f, respectively. Therefore, it passes by observation.

The corresponding moment set is transferred primarily into the tray and is therefore carried forward into the tray member assessment below.

INSGPS Tray

Tray Attachment Reactions

Four AN4-6 bolts connect the tray to the FWD/AFT support angles. The 25.11 lb₍f₎ weighted payload is applied at (8.08, −9.12, 1.77) in and the resulting bolt reactions are resolved for every governing flight direction.

• The maximum compressive axial load is f_A,c=34.85 lb₍f₎_f (LIMIT), and it belongs to the Upward Load Case. This load will be distributed over a wider area throughout the attachment surface; therefore, it passes by inspection.

• The maximum tensile and shear loads are f_A,t=47.78 lb₍f₎_f (LIMIT), and f_s=8.06 lb₍f₎_f (LIMIT). As established in the fastener-allowables discussion above, the ultimate tensile and shear loads for the AN4-6 Bolts in this configuration are f_tu=4,080 lb₍f₎_f, and f_su=2,218 lb₍f₎_f, respectively. Therefore, it passes by observation.

Tray Member Idealization

Methodology for Simplified Tray Analysis Using Beam Assumptions

Fasteners' Reaction Forces

The maximum reaction forces of the INSGPS Equipment will act as points loads on the beams. The reaction forces are derived based on the summation of fastener forces at respective locations.

1. Beam A (L-Direction)

• The reaction force in the i-th direction at the FWD end is given by: f_i= f_i1+ f_i2

• The reaction force in the i-th direction at the AFT end is given by: f_i= f_i3+ f_i4

2. Beam B (LT-Direction)

• The reaction force in the i-th direction at the PORT end is given by: f_i= f_i1+ f_i3

• The reaction force in the i-th direction at the STARBOARD end is given by: f_i= f_i2+ f_i4

Simplification Assumptions

To simplify the analysis while preserving accuracy, the following assumptions are made:

1. Lateral Reaction Load (y-axis)
   • Reaction forces in the y-direction (f_y) are neglected in Beam A.
   • These forces are entirely accounted for in Beam B as axial loads.
2. Longitudinal Reaction Load (x-axis)
   • Reaction forces in the x-direction (f_x) are neglected in Beam B.
   • These forces are entirely accounted for in Beam A as axial loads.
3. Bending Moments around x-axis (M_x)
   • This moment acts in the YZ plane and is neglected in Beam A. The torsional effect resulting from this moment is disregarded, assuming it is absorbed by adjacent structural components.
   • It is entirely assigned to Beam B.
4. Bending Moments around y-axis (M_y)
   • This moment acts in the XZ plane and is neglected in Beam B. The torsional effect resulting from this moment is disregarded, assuming it is absorbed by adjacent structural components.
   • It is entirely assigned to Beam A.

The corresponding table illustrates the total reaction forces of the grouped fasteners, for each load case, at both ends of Beams A and B.

Beam A:

Beam B:

The resulted maximum bending moments in each beam will be used to calculate the maximum bending stress that will be added to the beam’s axial stress to obtain the maximum combined stress in each beam.

Beam A:

The second moment of area about the y- and z-axes for Beam A are [Tray CrossSection Properties using FEMAP]:

• I_yy= 0.00247322 in⁴

• I_zz= 5.766871 in⁴

The maximum positive and negative bending moments are M_y =188.082 in-lb_f, M_y =-257.9 in-lb_f, M_z =3.73 in-lb_f, and M_z =-3.73 in-lb_f. The resulted maximum bending stresses, and the maximum axial stress can be calculated as below:

Conservatively, assuming that the maximum axial and bending stresses belong to the same loading case, the maximum tensile and compressive combined stresses can be calculated as below:

$$F_{comb,c} = F_{b_{z,max}} + F_{b_{y,max}} + F_{A} = 0.003 + 36.53 + 0.01014 = \boxed{36.54\ ksi}$$

$$F_{comb,t} = F_{b_{z,max}} + F_{b_{y,max}} = 0.003 + 36.53 = \boxed{36.53\ ksi}$$

Since the stress levels exceed the proportional limit (plastic range), the values resulted from the classic bending hand analysis (linear finite analysis) will be conservative and unrealistic. In this situation, plastic bending methods should be considered. In this report, Cozzone method is used. This method implies that the bending moment of the true stress distribution about the neutral axis is greater than that of a linear distribution. Hence, a trapezoidal stress distribution is used to approximate the true stress distribution. It is assumed that the outer fiber does not reach ultimate strength (maximum stress $F_{\max}$) until the material closer to the neutral axis sees an increased level of stress ($F_{o}$).

$F_{o}$ is a fictional stress which is assumed to exist at the neutral axis (at zero strain). The value of $F_{o}$ is expressed as below [Plastic Bending, Analysis & Design of Composite & Metallic Flight Vehicle Structures by Richard Abbott]:

$$\frac{F_{o}}{F_{\max}} = \frac{6}{\epsilon_{\max}^{2}}\left\lbrack \ \ \ \frac{1}{3}\left( \frac{F_{\max}}{E} \right)^{2}\ \ \ + \ \ \ \ \ \frac{n + 1}{n + 2} \times \frac{\epsilon_{p}F_{\max}^{n + 1}}{E\ F_{tu}^{n}}\ \ \ \ + \ \ \ \ \ \frac{n\ \epsilon_{p}^{2}}{2n + 1}\left( \frac{F_{\max}}{F_{tu}} \right)^{2n}\ \right\rbrack - 2$$

Where $F_{\max}$ is the maximum allowable stress, $\epsilon_{\max}\$is the maximum strain at $F_{\max}$, $\epsilon_{p}$ is the plastic strain, and $n$ is Ramberg-Osgood number. Based on the corresponding table, 0.071” thick AL 2024-T3 ALCLAD Sheet has an Ultimate Tensile Stress value of F_tu=62 ksi, Yield Tensile Stress Value of F_ty=45 ksi, Modulus of Elasticity value of E=10.5x10³ ksi, and maximum strain equals $\epsilon_{\max} = 0.15\ in/in$.

Setting $F_{\max}$ equals to the Ultimate Tensile Stress (F_tu), the plastic strain and Ramberg-Osgood number can be calculated as follows:

$$\epsilon_{p} = \epsilon_{\max} - \frac{F_{\max}}{E} = 0.15 - \frac{62}{10.5\ \times 10^{3}} = 0.1441\ in/in$$

$$n = \frac{\log\frac{\epsilon_{p}}{0.002}}{\log\frac{F_{tu}}{F_{ty}}} = \frac{\log\frac{0.1441\ }{0.002}}{\log\frac{62}{45}} = 13.347$$

Then, the value of $F_{o}$ can be computed as below:

Ultimate bending strength requires a cross-section shape factor $k$, defined as the plastic-to-elastic section-modulus ratio. The C-section geometry is evaluated in both principal directions [Page 829-830-Roark’s Formulas for Stress and Strain 9th Ed].

The resulting shape factors are:

$$k_{x} = \frac{Z_{x}}{S_{x}} = \boxed{2.00}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k_{y} = \frac{Z_{y}}{S_{y}} = \boxed{1.4163}$$

Therefore, the minimum Bending Modulus of Rupture for the tray can be calculated as below [Page C3.3-Analysis & Design of Flight Vehicle Structures by Bruhn]:

$${F_{b,y} = F_{\max} + F_{o}\left( k_{y} - 1 \right) }{= 62 + 54.79(1.4163 - 1) }{= \boxed{84.81\ ksi}}$$

The tray is fabricated from 0.071” thick AL 2024-T3 ALCLAD Sheet that has an Ultimate Tensile Stress value of F_tu=62 ksi. Therefore, the margin of safety in the tray can be computed as below:

The maximum axial and bending stress ratios can be expressed as below:

Therefore, the margin of safety can be calculated as below:

Beam B:

The second moment of area about the x- and z-axes for Beam B are:

• I_xx=(1/12)x13.42x0.071³= 0.000400264 in⁴

• I_zz=(1/12)x0.071x13.42³= 14.29995432 in⁴

The maximum positive and negative bending moments are M_x =26.52 in-lb_f, M_x =-26.52 in-lb_f, M_z =3.73 in-lb_f, and M_z =-3.73 in-lb_f. The resulted maximum bending stresses, and the maximum axial stress can be calculated as below:

The resulted maximum axial and bending stresses are very small compared to the tray’s material allowables. Therefore, it passes by observation.

FWD Support Angle

The FWD Support Angle section properties below define the centroid, extreme-fiber distances and inertias used in the axial/bending stress calculation.

FWD Support Angle section properties.

Parameter	Symbol	Value	Unit
Height	H	1.9375	in
Width	B	1.625	in
Thickness	t	0.25	in
Area	A	0.8281	in2
Centroid location along x-axis	Xc	0.4623	in
Centroid location along y-axis	Yc	0.6185	in
Extreme Fiber +y-Distance	Cy+	1.3190	in
Extreme Fiber -y-Distance	Cy-	-0.6185	in
Extreme Fiber +x-Distance	Cx+	1.1627	in
Extreme Fiber -x-Distance	Cx-	-0.4623	in
Second Moment of Inertia around x-axis	Ixx	0.296454	in4
Second Moment of Inertia around y-axis	Iyy	0.189413	in4
Product Moment of Inertia	Ixy	-0.137837	in4

the associated figure: The FWD Support Angle’s attachment points

The previous analysis of the tray has shown that the maximum reaction forces carried by one of the AN4-6 Tray Bolts are:

• Along the positive direction: f_x=3.58 lb₍f₎_f, f_y=7.93 lb₍f₎_f, and f_z=47.78 lb₍f₎_f.

• Along the negative direction: f_x=1.45 lb₍f₎_f, f_y=7.85 lb₍f₎_f, and f_z=34.85 lb₍f₎_f.

1. These loads belong to the same loading case, while these loads came from different loading cases.

2. Both bolts will carry the same maximum loads.

Due to the offset between the points of application of the applied loads and the beam’s neutral axis, bending moments will be produced as follows:

• The load f_y produces bending moments in the XY and YZ planes:

  • For the forces along the positive y-direction:

    • M_z=7.93x(1-0.4623)=4.26 lb₍f₎_f-in (C.C.W). At each bolt location.

    • M_x=7.93x0.6185=4.90 lb₍f₎_f-in(C.C.W). At each bolt location.

  • For the forces along the negative y-direction:

    • M_z=7.85x(1-0.4623)=4.22 lb₍f₎_f-in(C.W). At each bolt location.

    • M_x=7.85x0.6185=4.86 lb₍f₎_f-in(C.W). At each bolt location.

• The load f_z and f_x produce torsional moments in the XZ plane. This torsional effect is ignored, and it is assumed to be absorbed by the adjacent structural components.

the corresponding table lists the applied loads on the FWD Support Angle Beam, and the associated figure illustrates the resulted shear and bending diagrams.

Based on the associated figure, the beam’s maximum (i) reaction force, (ii) axial force, and (iii) bending moments can be summarized in the corresponding table.

Fasteners' Reaction Forces

• The total reaction forces along the x-direction:

  • Positive Direction: f_+x,total =4.29+2.87=7.16 lb₍f₎_f (LIMIT).

This load will be assumed to be carried by the 14xHL18PB/HL70 (YA5) Pin-Collar Fasteners. Based on the fastener-allowables discussion above, these fasteners at this joint have an ultimate tensile load value of f_tu=1,400 lb₍f₎_f. Therefore, it passes by observation.

• Negative Direction: f_-x,total =2.15+0.75=2.9 lb₍f₎_f (LIMIT).

This load will be distributed over a wider area throughout the attachment surface; therefore, it passes by inspection.

• The total reaction forces along the y- and z-directions

  • Positive Directions: f_+y,total=2x7.85=15.70 lb₍f₎_f (LIMIT) and f_+z,total=34.04+35.66=69.7 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.7)\ }^{2} + (69.7)^{2}} = 71.45\ {lb}_{f}\ \ \ (LIMIT)$.

• Negative Directions: f_-y,total=2x7.93=15.86 lb₍f₎_f (LIMIT) and f_-z,total=46.96+48.6=95.56 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.86)\ }^{2} + (95.56)^{2}} = 71.48\ {lb}_{f}\ \ \ (LIMIT)$.

This load will be assumed to be carried by the 14xHL18PB/HL70 (YA5) Pin-Collar Fasteners as a shear load. Based on the fastener-allowables discussion above, these fasteners at this joint have an ultimate shear load value of f_su=2,005 lb₍f₎_f. Therefore, it passes by observation.

• Fastener Tension in the FWD Support Angle Beam as a Tension Clip:

The fastener-line moment is represented by a couple between the fastener and a triangular bearing reaction under the outstanding flange. Rotational fixity is conservatively taken at the fastener line for this local tension-clip idealization.

Equivalent moment at the fastener line:

$$M = \frac{f \times e}{2}$$

Additional tensile force created by the couple reaction:

$$f_{add} = \frac{M}{\frac{2}{3}B} = \frac{3}{4}\frac{f \times e}{B}$$

Therefore, the total maximum tensile forces on the AN4‑6 Bolts and on the 18PB/HL70 (YA5) Pin-Collar Fasteners can be calculated as below:

The $f_{t,\ AN46}$ load will be carried by the two AN4-6 Bolts, and the $f_{t,\ YA5}$ load will be carried by 14xHL18PB/HL70 (YA5) Pin-Collar Fasteners. As shown in the relevant discussion, the ultimate tensile load for the AN4-6 Bolts, and the 18PB/HL70 (YA5) Pin-Collar Fasteners in this configuration are 4,080 lb₍f₎_f and 1,400 lb₍f₎_f, respectively. Therefore, it passes by observation.

The FWD Support Angle Beam

Based on the corresponding table,:

• The maximum bending moments in the:

  • YZ plane are M_x= 81.45 in-lb_f and -110.57 in-lb_f.

  • XY plane are M_z= 10.72 in-lb_f and -5.9 in-lb_f.

• The maximum axial loads are f_a= 15.86 lb₍f₎_f and -15.70 lb₍f₎_f.

These bending moments and axial load result in maximum combined tensile and compressive stress values of 0.94 ksi (LIMIT) and -0.73 ksi (LIMIT), respectively. The FWD Support Angle Beam is made from 0.25” thick AL 6061-T6511 Extrusion which has an Ultimate Tensile Strength and Yield Compressive Strength value of F_tu=38 ksi and F_cy=34 ksi, respectively (referencing the corresponding table). Therefore, the beam pass by observation.

• Beam Bending for the FWD Support Angle Beam as a Tension Clip:

The allowable applied load to yield the angle in bending per one inch of angle can be expressed as below:

$$f_{b,o} = \alpha\ S_{F}\frac{2M_{o}}{e} = \frac{2\alpha\ S_{F}F_{ty}I}{ey}$$

Where

• $M_{o}:$ is the allowable moment to yield

• $\alpha:$ is the minimum factor from yield allowable to ultimate allowable. For extruded aluminium, $\alpha = 1.167$.

• $S_{F}$: is the shape factor for a rectangular section, namely $S_{F} = 1.5$.

• $e:$ is the eccentricity of the clip, i.e. the distance between the bolt’s axis and the flange’s outer surface.

• $F_{ty}:$ is the yield tensile strength of the material

• $I:$ is the second moment of area for a unit length of angle flange; namely, t³/12

• $y:$ is the distance from the flange cross section neutral axis to the outer surface; namely, t/2

Therefore, the ultimate allowable applied load (on the YZ and XY flanges) per one inch of angle can be expressed as below:

As shown previously, the maximum load on the YZ and XY flanges are:

• f_YZ= f_+z,total=34.04+35.66=69.7 lb₍f₎_f (LIMIT)

• f_XY= f_+x,total =4.29+2.87=7.16 lb₍f₎_f (LIMIT).

Therefore, the beam pass by observation against bending.

AFT Support Angle

The AFT Support Angle section properties below define the centroid, extreme-fiber distances and inertias used in the axial/bending stress calculation.

AFT Support Angle section properties.

Parameter	Symbol	Value	Unit
Height	H	1.5000	in
Width	B	2.4400	in
Thickness	t	0.25	in
Area	A	0.9225	in2
Centroid location along x-axis	Xc	0.3791	in
Centroid location along y-axis	Yc	0.8491	in
Extreme Fiber +y-Distance	Cy+	1.5909	in
Extreme Fiber -y-Distance	Cy-	-0.8491	in
Extreme Fiber +x-Distance	Cx+	1.1209	in
Extreme Fiber -x-Distance	Cx-	-0.3791	in
Second Moment of Inertia around x-axis	Ixx	0.552035	in4
Second Moment of Inertia around y-axis	Iyy	0.160102	in4
Product Moment of Inertia	Ixy	-0.169703	in4

the associated figure: The AFT Support Angle’s attachment points

The previous analysis of the tray has shown that the maximum reaction forces carried by one of the AN4-6 Tray Bolts are:

• Along the positive direction: f_x=3.58 lb₍f₎_f, f_y=7.93 lb₍f₎_f, and f_z=47.78 lb₍f₎_f.

• Along the negative direction: f_x=1.45 lb₍f₎_f, f_y=7.85 lb₍f₎_f, and f_z=34.85 lb₍f₎_f.

1. These loads belong to the same loading case, while these loads came from different loading cases.

2. Both bolts will carry the same maximum loads.

Due to the offset between the points of application of the applied loads and the beam’s neutral axis, bending moments will be produced as follows:

• The load f_y produces bending moments in the XY and YZ planes:

  • For the forces along the positive y-direction:

    • M_z=7.93x(0.5-0.3791)=0.96 lb₍f₎_f-in (C.C.W). At each bolt location.

    • M_x=7.93x0.8491=6.73 lb₍f₎_f-in(C.W). At each bolt location.

  • For the forces along the negative y-direction:

    • M_z=7.85x(0.5-0.3791)=0.95 lb₍f₎_f-in (C.W). At each bolt location.

    • M_x=7.85x0.8491=6.67 lb₍f₎_f-in(C.C.W). At each bolt location.

• The load f_z and f_x produce torsional moments in the XZ plane. This torsional effect is ignored, and it is assumed to be absorbed by the adjacent structural components.

the corresponding table lists the applied loads on the AFT Support Angle Beam, and the associated figure illustrates the resulted shear and bending diagrams.

Based on the associated figure, the beam’s maximum (i) reaction force, (ii) axial force, and (iii) bending moments can be summarized in the corresponding table.

Fasteners' Reaction Forces

• The total reaction forces along the x-direction:

  • Positive Direction: f_+x,total =3.77+3.39=7.16 lb₍f₎_f (LIMIT).

This load will be assumed to be carried by the 14xHL18PB/HL70 (YA5) Pin-Collar Fasteners. Based on the fastener-allowables discussion above, these fasteners at this joint have an ultimate tensile load value of f_tu=1,400 lb₍f₎_f. Therefore, it passes by observation.

• Negative Direction: f_-x,total =1.64+1.26=2.9 lb₍f₎_f (LIMIT).

This load will be distributed over a wider area throughout the attachment surface; therefore, it passes by inspection.

• The total reaction forces along the y- and z-directions

  • Positive Directions: f_+y,total=2x7.85=15.7 lb₍f₎_f (LIMIT) and f_+z,total=36.18+33.52=69.7 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.7)\ }^{2} + (69.7)^{2}} = 71.45\ {lb}_{f}\ \ \ (LIMIT)$.

• Negative Directions: f_-y,total=2x7.93=15.86 lb₍f₎_f (LIMIT) and f_-z,total=49.13+46.43=95.56 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.86)\ }^{2} + (95.56)^{2}} = 96.87\ {lb}_{f}\ \ \ (LIMIT)$.

This load will be assumed to be carried by the 14xHL18PB/HL70 (YA5) Pin-Collar Fasteners as a shear load. Based on the fastener-allowables discussion above, these fasteners at this joint have an ultimate shear load value of f_su=2,005 lb₍f₎_f. Therefore, it passes by observation.

• Fastener Tension in the AFT Support Angle Beam as a Tension Clip:

The fastener-line moment is represented by a couple between the fastener and a triangular bearing reaction under the outstanding flange. Rotational fixity is conservatively taken at the fastener line for this local tension-clip idealization.

Equivalent moment at the fastener line:

$$M = \frac{f \times e}{2}$$

Additional tensile force created by the couple reaction:

$$f_{add} = \frac{M}{\frac{2}{3}B} = \frac{3}{4}\frac{f \times e}{B}$$

Therefore, the total maximum tensile forces on the AN4‑6 Bolts and on the HL18PB/HL70 (YA5) Pin-Collar Fasteners can be calculated as below:

The $f_{t,\ AN46}$ load will be carried by the two AN4-6 Bolts, and the $f_{t,\ YA5}$ load will be carried by 19xHL18PB/HL70 (YA5) Pin-Collar Fasteners. As shown in the relevant discussion, the ultimate tensile load for the AN4-6 Bolts, and the 18PB/HL70 (YA5) Pin-Collar Fasteners in this configuration are 4,080 lb₍f₎_f and 1,400 lb₍f₎_f, respectively. Therefore, it passes by observation.

Angle Beam

The Angle Beam section properties below define the centroid, extreme-fiber distances and inertias used in the axial/bending stress calculation.

Angle Beam section properties.

Parameter	Symbol	Value	Unit
Height	H	3.00	in
Width	B	1.00	in
Thickness	t	0.1875	in
Area	A	0.7148	in2
Centroid location along x-axis	Xc	0.2003	in
Centroid location along y-axis	Yc	1.2003	in
Extreme Fiber +y-Distance	Cy+	1.7997	in
Extreme Fiber -y-Distance	Cy-	-1.2003	in
Extreme Fiber +x-Distance	Cx+	0.7997	in
Extreme Fiber -x-Distance	Cx-	-0.2003	in
Second Moment of Inertia around x-axis	Ixx	0.659383	in4
Second Moment of Inertia around y-axis	Iyy	0.039998	in4
Product Moment of Inertia	Ixy	-0.084289	in4

the associated figure: The Angle Beam’s attachment points

The previous analysis of the FWD Support Angle Beam has shown that the maximum reaction forces carried by the all 14x18PB/HL70 (YA5) Pin-Collar Fasteners are:

• Along the positive direction: f_x=32.62 lb₍f₎_f (Taking into consideration the tension clip bending), f_y=15.7 lb₍f₎_f, and f_z=69.7 lb₍f₎_f.

• Along the negative direction: f_x=2.9 lb₍f₎_f, f_y=15.86 lb₍f₎_f, and f_z=95.56 lb₍f₎_f.

Due to the offset between the points of application of the applied loads and the beam’s neutral axis, bending moments will be produced as follows:

• The load f_y produces bending moments in the XY and YZ planes:

  • For the forces along the positive y-direction:

    • M_z=15.7x(0.2)=3.14 lb₍f₎_f-in (C.C.W). At the load distribution midpoint (13.86-4.4)/2 + 4.4 =9.13 in.

    • M_x=15.7x(((2.39-1.6)/2 + 1.6)-1.2)=12.48 lb₍f₎_f-in(C.C.W). At the load distribution midpoint (13.86-4.4)/2 + 4.4 =9.13 in.

  • For the forces along the negative y-direction:

    • M_z=15.86x(0.2)=3.17 lb₍f₎_f-in (C.W). At the load distribution midpoint (13.86-4.4)/2 + 4.4 =9.13 in.

    • M_x=15.86x(((2.39-1.6)/2 + 1.6)-1.2)=12.61 lb₍f₎_f-in(C.W). At the load distribution midpoint (13.86-4.4)/2 + 4.4 =9.13 in.

• The load f_z and f_x produce torsional moments in the XZ plane. This torsional effect is ignored, and it is assumed to be absorbed by the adjacent structural components.

the corresponding table lists the applied loads on the Angle Beam, and the associated figure illustrates the resulted shear and bending diagrams.

Based on the associated figure, the beam’s maximum (i) reaction force, (ii) axial force, and (iii) bending moments can be summarized in the corresponding table.

Fasteners' Reaction Forces

• The total reaction forces along the x-direction:

  • Positive Direction: f_+x,total =1.30+1.64=2.94 lb₍f₎_f (LIMIT).

This load will be assumed to be carried by the 14xHL40/HL70 (ARV5) Pin-Collar Fasteners. Based on the fastener-allowables discussion above, these fasteners at this joint have a minimum ultimate tensile load value of f_tu=1,350 lb₍f₎_f. Therefore, it passes by observation.

• Negative Direction: f_-x,total =16.16+16.48=32.64 lb₍f₎_f (LIMIT).

This load will be distributed over a wider area throughout the attachment surface; therefore, it passes by inspection.

• The total reaction forces along the y- and z-directions

  • Positive Directions: f_+y,total=2x7.93=15.86 lb₍f₎_f (LIMIT) and f_+z,total=47.06+48.50=95.56 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.86\ )\ }^{2} + (95.56)^{2}} = 96.87\ {lb}_{f}\ \ \ (LIMIT)$.

• Negative Directions: f_-y,total=2x7.85=15.7 lb₍f₎_f (LIMIT) and f_-z,total=34.16+35.56=69.73 lb₍f₎_f (LIMIT).

The total resultant shear load is $\sqrt{{(15.7)\ }^{2} + (69.72)^{2}} = 71.47\ {lb}_{f}\ \ \ (LIMIT)$.

This load will be assumed to be carried by the 14xHL40/HL70 (ARV5) Pin-Collar Fasteners as a shear load. Based on the fastener-allowables discussion above, these fasteners at this joint have a minimum ultimate shear load value of f_su=968 lb₍f₎_f. Therefore, it passes by observation.

Moreover, all these loads will be transmitted to the Keel Beam FWD Angle and the Strut-Floor Beam existing structures. Since these loads are insignificant, these existing structures pass by observation.

• Fastener Tension in the Angle Beam as a Tension Clip:

The fastener-line moment is represented by a couple between the fastener and a triangular bearing reaction under the outstanding flange. Rotational fixity is conservatively taken at the fastener line for this local tension-clip idealization.

Equivalent moment at the fastener line:

$$M = \frac{f \times e}{2}$$

Additional tensile force created by the couple reaction:

$$f_{add} = \frac{M}{\frac{2}{3}B} = \frac{3}{4}\frac{f \times e}{B}$$

Therefore, the total maximum tensile forces on the 14xHL40/HL70 (ARV5) Pin-Collar Fasteners can be calculated as below:

The $f_{t,\ ARV5}$ load will be carried by 14xHL40/HL70 (ARV5) Pin-Collar Fasteners. As shown in the relevant discussion, the ultimate tensile load for the HL40/HL70 (ARV5) Pin-Collar Fasteners in this configuration is 1,350 lb₍f₎_f. Therefore, it passes by observation.

The Angle Beam

Based on the corresponding table:

• The maximum bending moments in the:

  • YZ plane are M_x= 329.25 in-lb_f and -241.90 in-lb_f.

  • XY plane are M_z= 11.48 in-lb_f and -111.90 in-lb_f.

• The maximum axial loads are f_a= 15.7 lb₍f₎_f and -15.86 lb₍f₎_f.

These bending moments and axial load result in maximum combined tensile and compressive stress values of 19.92 ksi (LIMIT) and -13.87 ksi (LIMIT), respectively. The Angle Beam is made from 0.1875” thick AL 6061-T6511 Extrusion which has an Ultimate Tensile Strength and Yield Compressive Strength value of F_tu=38 ksi and F_cy=34 ksi, respectively (referencing the corresponding table). Therefore, the beam’s Margin of Safety can be calculated as below:

• Beam Bending for the Angle Beam as a Tension Clip:

The allowable applied load to yield the angle in bending per one inch of angle can be expressed as below:

$$f_{b,o} = \alpha\ S_{F}\frac{2M_{o}}{e} = \frac{2\alpha\ S_{F}F_{ty}I}{ey}$$

Where

• $M_{o}:$ is the allowable moment to yield

• $\alpha:$ is the minimum factor from yield allowable to ultimate allowable. For extruded aluminium, $\alpha = 1.167$.

• $S_{F}$: is the shape factor for a rectangular section, namely $S_{F} = 1.5$.

• $e:$ is the eccentricity of the clip, i.e. the distance between the bolt’s axis and the flange’s outer surface.

• $F_{ty}:$ is the yield tensile strength of the material

• $I:$ is the second moment of area for a unit length of angle flange; namely, t³/12

• $y:$ is the distance from the flange cross section neutral axis to the outer surface; namely, t/2

Therefore, the ultimate allowable applied load (on the XY flange) per one inch of angle can be expressed as below:

As shown previously, the maximum load on the XY flange is f_XY= f_-x,total =16.16+16.48=32.64 lb₍f₎_f (LIMIT). Therefore, the beam pass by observation against bending.

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.

The basic crippling stress equation is given by:

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

Where

• t: Angle thickness, in

• b: Minimum leg length, in

• h: Maximum leg length, in

• F_cy: Material compressive yield Strength, psi

• E: Material young’s modulus, psi

• $C_{e}$: End Fixity Coefficient, $C_{e} = \left\{ \begin{aligned} 0.316,\ \ & two\ edges\ free \\ 0.342,\ \ & one\ edge\ free \\ 0.366,\ \ & no\ edge\ free \end{aligned} \right.\$

Therefore, the crippling stress for the Angle Beam’s cross-section can be calculated as below:

$$F_{Crippling} = \frac{0.316\sqrt{34 \times 10^{3} \times 9.9 \times 10^{6}}}{\left( \frac{\frac{3 + 1}{2}}{0.1875} \right)^{0.75}} = \boxed{31.06\ ksi}$$

Therefore, the margin of safety for the Angle Beam against crippling can be computed as below:

Tee Clip

The previous analysis of the Channel Beam has shown that the maximum reaction forces carried by the all 5x18PB/HL70 (YA5) Pin-Collar Fasteners are:

• Along the positive direction: f_x=32.61 lb₍f₎_f, f_y=15.86 lb₍f₎_f, and f_z=95.58 lb₍f₎_f.

• Along the negative direction: f_x=2.91 lb₍f₎_f, f_y=15.70 lb₍f₎_f, and f_z=69.71 lb₍f₎_f.

For conservatism, it is assumed that:

1. The loads belong to the same loading case, while these loads came from different loading cases.

2. Fasteners No. 3, 4, and 5 at the top of the tee clip are ignored.

3. Fasteners No. 1 and 3 at the leg of the tee clip are ignored.

Fasteners' Reaction Forces

• The shear force along the z-axis (f_z) carried by each of the fasteners:

  • No. 1 and 2 at the top of the tee clip is 95.58/2=47.79 lb₍f₎_f.

  • No. 2 and 4 at the leg of the tee clip is 95.58/4=23.90 lb₍f₎_f (at the FWD and AFT sides).

• The moment produced by the shear force (f_z) around x-axis is M_x= 95.58x1=95.58 lb₍f₎_f-in. Therefore, the shear force carried by each of the fasteners:

  • No. 1 and 2 at the top of the tee clip is 95.58/0.836=114.33 lb₍f₎_f (Along the y-axis).

  • No. 2 and 4 at the leg of the tee clip is 95.58/2.73=35.01 lb₍f₎_f (Along the x-axis).

• The moment produced by the tensile force (f_x) around z-axis is M_z= 32.61x1=32.61 lb₍f₎_f-in. Therefore, the shear force carried by each of the fasteners:

  • No. 1 and 2 at the top of the tee clip is 32.61/0.836=39.0 lb₍f₎_f (Along the y-axis).

  • No. 2 and 4 at the leg of the tee clip is 32.61/2.73=11.95 lb₍f₎_f (Along the x-axis).

Hence, the total reaction forces at each of the fasteners:

• No. 1 and 2 at the top of the tee clip are:

  • R_x=32.61/2=16.31 lb₍f₎_f.

  • R_y=114.33+39.0+15.86/2=161.26 lb₍f₎_f.

  • R_z=47.79 lb₍f₎_f.

• No. 2 and 4 at the leg of the tee clip are:

  • R_x=35.01+11.95+32.61/4=55.11 lb₍f₎_f.

  • R_y= 15.86/4=3.97 lb₍f₎_f.

  • R_z=23.90 lb₍f₎_f

Therefore, the resultant shear force carried by each of the fasteners:

• No. 1 and 2 at the top of the tee clip is $\sqrt{{47.79}^{2} + {161.26}^{2}}$ = 168.19 lb₍f₎_f.

• No. 2 and 4 at the leg of the tee clip is $\sqrt{{23.9}^{2} + {55.11}^{2}}$ = 60.07 lb₍f₎_f.

Based on the fastener-allowables discussion above, the fasteners the top and leg of the tee clip have a minimum ultimate tensile and shear load value of f_tu=1,400 lb₍f₎_f and f_su=1,601 lb₍f₎_f, respectively. Therefore, it passes by observation.

Moreover, all these loads will be transmitted to the existing stringers. Since these loads are insignificant, the stringers pass by observation.

The Tee Clip

The tee clip is made from 0.125” thick AL 6061-T6511 Extrusion which has an Ultimate Shear Strength value of F_su=26 ksi, and a Yield Compressive Strength value of F_cy=34 ksi (referencing the corresponding table). Therefore, the tee clip pass by observation.

REFERENCES