Estimating wind loads on banners

Share This Article

  • Del.icio.us
  • Digg
  • Facebook
  • Google Bookmark
  • StumbleUpon
  • Technorati
  • Twitter

Related Articles

Tags

Wind loads on banners can be estimated and a simple example is shown below. A 90 mph, (40m/s) three second gust has a basic wind pressure of 15 psf (720 pascal or 0.72kN/m2). Codes adjust this with a coefficient of pressure and for a square fabric sign is a minimum of Cp = 1.45. For a banner 10 ft. by 10 ft. (3.05m by 3.05m) the total force on the banner is 10 x 10 x 15 x 1.45 = 2,175 lbs. (9.7kN)

With a one way span top to bottom, half of the load to the top and half to the bottom equates to 1,100 lbs. (4.9kN) horizontal. The fabric will deflect as shown and tension internal to the fabric banner adds an additional vertical force of approximately 300 plf. (4.4kN/m) or 3,000 lbs, (13.4kN) top and bottom. The total resultant load of 3,200 lbs. (14.3kN) is dominated by the tension or catenary forces in the banner.

This estimated calculation should be used for preliminary purposes only and final design should be verified by a licensed professional engineer.

The International Building Code 2006 is used as a model code in most states and wind loads and pressure coefficients are calculated using the ASCE 7-05 Standard.

Comments

There are not yet any comments.
You can submit a comment using the form below.


Submit a Comment

Required. Will appear next to your comment.
Required. Will not be displayed on site or used to send unsolicited messages.
If applicable. A link to your site will appear with your comment.
Optional. Will appear in bold type above your comment.