×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Masonry Anchorage

Masonry Anchorage

Masonry Anchorage

(OP)
Regarding ASD and Strength Design for Masonry (ACI 530-02):

For ASD the allowable tension (when masonry controls) is Ba = 0.5 * Ap * f’m^0.5

For Strength Design the nominal axial tensile strength (when masonry controls) is ? Ban = 4 * 0.5 * Ap * f’m^0.5

The difference is a factor of 4.

With trying to compare apples to apples (I realize ASD to LRFD is not apples to apples) a load combination of 0.6DL+WL for ASD and 0.9DL+1.6WL for Strength Design say WL = 1 and DL = 1.

For ASD this results 1.6
For Strength Design this results in 2.5

Now the question/concern why is the Strength Design factor of difference of 4 when computing tensile strength (when masonry controls) allowing so more capacity particularly when the load combination for Strength Design is only ~1.5 times as much as ASD?

RE: Masonry Anchorage

I think this is a case where the "old" ASD formulae have not been looked at or improved in many years and the LRFD has been fined tuned and tightened up to be less conservative and more beneficial to the masonry industry.

Also - there is a tendency to push engineers toward LRFD.

RE: Masonry Anchorage

The ASD provisions in the recent codes have puzzled me:

The 2002 code (2.1.3.3.2) says the “Nominal Strength” = 2.5 x allowable stress value and that the “Design Strength” = Nominal Strength x “phi” (2.1.3.3.3).  For axial load, “phi” = 0.8.

So, for axial loads, the allowable design stress = 2.5 x 0.8 x allowable stress = 2.0 x fa

The 1995 code says allowable stress = fa.

The load combinations (per the masonry code) didn’t change, and the axial compression equations didn’t change.

Am I missing something here?

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources