Appendix L: BOLT LOAD AND MOTOR TORQUE CALCULATIONS

CASE A – 3 × M5 BOLTS IN A VERTICAL LINE

Description

This case considers a 15 kg load acting downward at an eccentricity of 86 mm from a bolt group made up of 3 × M5 bolts arranged vertically. The bolts are spaced 15 mm apart, with 13 mm thread engagement.

Given

Load mass
m = 15 kg

Gravity
g = 9.81 m/s²

Weight force
W = m × g
W = 15 × 9.81
W = 147.15 N

Eccentricity
e = 86 mm
e = 0.086 m

Bolt arrangement
3 × M5 bolts in a vertical line

Bolt spacing
15 mm between adjacent bolts

Bolt positions relative to centroid
Top bolt: y = +15 mm = +0.015 m
Middle bolt: y = 0
Bottom bolt: y = -15 mm = -0.015 m

Thread engagement
13 mm

1.1 Moment on Bolt Group

The load acts at an offset from the bolt group, creating a moment:

M = W × e

M = 147.15 × 0.086

M = 12.65 N·m

Moment on bolt group
M ≈ 12.65 N·m

1.2 Direct Shear Force

Assuming the direct vertical load is shared equally by all 3 bolts:

V_total = W = 147.15 N

V_per_bolt = 147.15 / 3

V_per_bolt = 49.05 N

With safety factor 1.5:

V_design = 49.05 × 1.5

V_design = 73.58 N

Direct shear per bolt
49.05 N

Direct shear per bolt with SF 1.5
73.58 N

1.3 Bolt Tension Due to Moment

Using the elastic bolt-group / second moment method:

F = (M × y) / I

where

I = Σ(y²)

I = 0.015² + 0² + 0.015²
I = 0.000225 + 0 + 0.000225
I = 0.00045 m²

For the top bolt:

F_top = (12.65 × 0.015) / 0.00045

F_top = 421.9 N

For the middle bolt:

F_middle = 0 N

For the bottom bolt:

F_bottom = -421.9 N

Negative sign indicates compression.

Maximum bolt tension
421.9 N

Maximum bolt tension with SF 1.5
421.9 × 1.5 = 632.9 N

1.4 Summary – Case A

Weight force
147.15 N

Eccentricity
0.086 m

Moment on bolt group
12.65 N·m

Direct shear per bolt
49.05 N

Direct shear per bolt with SF 1.5
73.58 N

Maximum bolt tension
421.9 N

Maximum bolt tension with SF 1.5
632.9 N

1.5 Safety Assessment – Case A

For the bolts themselves, this case is safe. The direct shear is very small, and the maximum design tension of approximately 0.633 kN is well below the typical tensile capacity of an M5 steel bolt.

Typical M5 steel bolt tensile capacity
approximately 6 to 9 kN

Conclusion
Safe for bolt strength, assuming proper wall material and thread quality.

2. CASE B – 4 × M5 BOLTS ON A CIRCULAR PATTERN

Description

This case considers the same 15 kg load, but with a much larger eccentricity of 488.393 mm. The load is supported by 4 × M5 bolts arranged on a circular pattern with a bolt circle diameter of 44.1 mm. Thread engagement is 20 mm.

Given

Load mass
m = 15 kg

Gravity
g = 9.81 m/s²

Weight force
W = 15 × 9.81
W = 147.15 N

Eccentricity
e = 488.393 mm
e = 0.488393 m

Bolt arrangement
4 × M5 bolts on a circular pattern

Bolt circle diameter
D = 44.1 mm
D = 0.0441 m

Bolt circle radius
r = D / 2
r = 22.05 mm
r = 0.02205 m

Thread engagement
20 mm

2.1 Moment on Bolt Group

M = W × e

M = 147.15 × 0.488393

M = 71.87 N·m

Moment on bolt group
M ≈ 71.87 N·m

2.2 Direct Shear Force

Assuming equal sharing by 4 bolts:

V_total = 147.15 N

V_per_bolt = 147.15 / 4

V_per_bolt = 36.79 N

With safety factor 1.5:

V_design = 36.79 × 1.5

V_design = 55.18 N

Direct shear per bolt
36.79 N

Direct shear per bolt with SF 1.5
55.18 N

2.3 Bolt Tension Due to Moment

For the circular bolt pattern, using the conservative top-bottom bolt case:

F = (M × y) / I

with

y = r = 0.02205 m

I = Σ(y²)

For the top and bottom bolts only:

I = r² + r²
I = 2r²

I = 2 × (0.02205²)

I = 2 × 0.0004862025

I = 0.000972405 m²

Now calculate top bolt tension:

F_top = (71.87 × 0.02205) / 0.000972405

F_top = 1629.6 N

Bottom bolt carries the same magnitude in compression:

F_bottom = -1629.6 N

The two side bolts are approximately 0 N from this bending axis.

Maximum bolt tension
1629.6 N

Maximum bolt tension with SF 1.5
1629.6 × 1.5 = 2444.4 N

2.4 Summary – Case B

Weight force
147.15 N

Eccentricity
0.488393 m

Moment on bolt group
71.87 N·m

Direct shear per bolt
36.79 N

Direct shear per bolt with SF 1.5
55.18 N

Maximum bolt tension
1629.6 N

Maximum bolt tension with SF 1.5
2444.4 N

2.5 Safety Assessment – Case B

The direct shear per bolt is very small. The governing load is the tension created by the large eccentricity. Even with a 1.5 safety factor, the design bolt tension is about 2.44 kN.

Typical M5 steel bolt tensile capacity
approximately 6 to 9 kN

Conclusion
Safe for bolt strength, provided the threads and mounting material are adequate. The bolt itself is not the limiting factor.

3. CASE C – MOTOR TORQUE CALCULATION

Given

Dish mass
m = 15 kg

Gravity
g = 9.81 m/s²

Weight force
W = m × g
W = 15 × 9.81
W = 147.15 N

Offset distance from axis
e = 488.393 mm
e = 0.488393 m

Motor rated torque
T_motor = 10.2 N·m

Gear ratio
G = 72 : 1

3.1 Required Torque to Hold the Dish

The required static holding torque is:

T_required = W × e

T_required = 147.15 × 0.488393
T_required = 71.87 N·m

Required torque
= 71.87 N·m

3.2 Available Motor Torque

The motor is rated at a maximum torque of:

T_available = 10.2 N·m

This value is taken as the effective output torque available at the gimbal axis.

3.3 Torque Comparison

To evaluate whether the motor is sufficient, the available torque is compared against the required torque:

Safety factor:

SF = T_available / T_required

SF = 10.2 / 71.87
SF = 0.142

This indicates that the motor can only provide approximately 14.2% of the required torque.

Torque deficit:

T_deficit = T_required − T_available

T_deficit = 71.87 − 10.2
T_deficit = 61.67 N·m

3.4 Summary – Case C

Weight force
147.15 N

Offset distance
0.488393 m

Required torque
71.87 N·m

Available motor torque
10.2 N·m

Safety factor
0.142

Torque deficit
61.67 N·m

3.5 Safety Assessment – Case C

The required torque to hold the dish is significantly higher than the available motor torque. The motor is therefore not capable of sustaining the load under static conditions, and will experience difficulty in maintaining position or achieving smooth motion.

Conclusion
The motor torque is insufficient to support the dish assembly at the given offset. A counterweight or equivalent balancing mechanism is required to reduce the effective torque acting on the system.

4. COMPARISON TABLE

Case A – 3 bolts vertical
Moment = 12.65 N·m
Shear per bolt = 49.05 N
Max bolt tension = 421.9 N
Max bolt tension with SF 1.5 = 632.9 N

Case B – 4 bolts circular
Moment = 71.87 N·m
Shear per bolt = 36.79 N
Max bolt tension = 1629.6 N
Max bolt tension with SF 1.5 = 2444.4 N

Case C – Motor torque
Required torque = 71.87 N·m
Ideal output torque = 734.4 N·m
Realistic output torque = 587.52 N·m
Safety factor = 8.17 to 10.22