How to calculate wind load on security camera poles

Calculation method of wind load on security camera poles

 

1. Design Parameters

• Pole dimensions:

  • Height: 10 m

  • Base diameter: 300 mm (0.3 m), Top diameter: 180 mm (0.18 m)

  • Thickness: 5 mm (0.005 m), Material: Q235 steel (σ_yield = 235 MPa)

• Solar panel: 3 m² at 5 m height.

• Location: Beijing (50-year return wind pressure: w0=0.45 kN/m2w0​=0.45kN/m2).

• Standards: GB 50009 (Wind loads), GB 50011 (Seismic).

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2. Wind Load Calculation

 

2.1 Basic Wind Pressure Adjustment

• Height coefficient (μzμz​): For 10m height (Terrain Category B, Beijing):

  μz=1.0μz​=1.0

• Gust factor (βgβg​): βg=1.7βg​=1.7.

• Shape coefficient (μsμs​):

  • Pole (cylindrical): μs=0.8μs​=0.8.

  • Solar panel (flat plate): μs=1.3μs​=1.3.

 

2.2 Wind Load on Pole

The pole is tapered; divide it into 2 segments for simplified analysis:

• Segment 1 (0–5m): Avg. diameter D1=0.3+(0.3−0.012×5)2=0.27 mD1​=20.3+(0.3−0.012×5)​=0.27m.

• Segment 2 (5–10m): Avg. diameter D2=0.27+0.182=0.225 mD2​=20.27+0.18​=0.225m.

Wind force per segment:

Fpole=βg⋅μz⋅μs⋅w0⋅Davg⋅LsegmentFpole​=βg​⋅μz​⋅μs​⋅w0​⋅Davg​⋅Lsegment​

• Segment 1:

  F1=1.7×1.0×0.8×0.45×0.27×5=0.826 kNF1​=1.7×1.0×0.8×0.45×0.27×5=0.826kN

• Segment 2:

  F2=1.7×1.0×0.8×0.45×0.225×5=0.688 kNF2​=1.7×1.0×0.8×0.45×0.225×5=0.688kN

• Total pole wind force:

  Fpole,total=0.826+0.688=1.514 kNFpole,total​=0.826+0.688=1.514kN

 

2.3 Wind Load on Solar Panel

Fsolar=βg⋅μz⋅μs⋅w0⋅A=1.7×1.0×1.3×0.45×3=3.02 kNFsolar​=βg​⋅μz​⋅μs​⋅w0​⋅A=1.7×1.0×1.3×0.45×3=3.02kN

 

2.4 Total Bending Moment at Base

• Pole wind forces act at mid-segment heights:

  Mpole=F1×2.5+F2×7.5=0.826×2.5+0.688×7.5=7.36 Mpole​=F1​×2.5+F2​×7.5=0.826×2.5+0.688×7.5=7.36

• Solar panel moment:

  Msolar=Fsolar×5=3.02×5=15.1 Msolar​=Fsolar​×5=3.02×5=15.1

• Total moment:

  Mtotal=7.36+15.1=22.46 Mtotal​=7.36+15.1=22.46

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3. Strength Verification

3.1 Section Modulus at Base

• Outer diameter D=0.3 mD=0.3m, Inner diameter d=0.29 md=0.29m.

• Moment of inertia (II):

  I=π64(D4−d4)=π64(0.34−0.294)=6.74×10−5 m4I=64π​(D4−d4)=64π​(0.34−0.294)=6.74×10−5m4

• Section modulus (WW):

  W=ID/2=6.74×10−50.15=4.49×10−4 m3W=D/2I​=0.156.74×10−5​=4.49×10−4m3

 

3.2 Bending Stress

σ=MtotalW=22.46×1034.49×10−4=50.0 MPaσ=WMtotal​​=4.49×10−422.46×103​=50.0MPa

• Allowable stress:

  [σ]=σyield1.5=2351.5=156.7 MPa[σ]=1.5σyield​​=1.5235​=156.7MPa

• Verification: 50.0 MPa<156.7 MPa50.0MPa<156.7MPa ✔️

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4. Seismic Verification (Simplified)

• Seismic acceleration (Beijing, 8-degree zone): a=0.2ga=0.2g.

• Horizontal seismic force:

  Feq=0.2×(Wpole+Wsolar)Feq​=0.2×(Wpole​+Wsolar​)

  Assume pole weight = 500 kg, solar panel = 100 kg:

  Feq=0.2×(500+100)×9.81=1.18 kNFeq​=0.2×(500+100)×9.81=1.18kN

• Moment from seismic force:

  Meq=Feq×10=11.8 Meq​=Feq​×10=11.8

• Combined stress (wind + seismic):

  σtotal=22.46+11.84.49×10−4×10−3=76.4 MPa<156.7 MPa✔®σtotal​=4.49×10−422.46+11.8​×10−3=76.4MPa<156.7MPa✔R◯

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5. Conclusion

• Wind load: 22.46 kN·m (governs).

• Safety factor: 3.13 (against yield).

• Compliance: Meets GB 50009 and GB 50011 requirements.

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Note: Detailed dynamic analysis (e.g., natural frequency, vortex shedding) is recommended for critical applications.