Question on "Economical Design of Shelf Angles" Paper 2

[starcasm]

Hello,

I am developing a spreadsheet that will design shelf angles. Everywhere I looked pointed me to the same resource below:

"Economical Design of Shelf Angles," by R. H. R. Tide and N. V. Krogstad (1993, Proceedings of the Symposium on Masonry: Design and Construction, Problems and Repair, STP 1180, American Society for Testing and Materials (ASTM), Philadelphia, PA.

So I purchased a copy. I am having trouble understanding the explanation of the line of reaction. I’ve quoted the two sections I am referring to below:

From Page 65 (Under Structural Requirements):

“1. The section of shelf angle between bolts is assumed to be loaded by a uniformly distributed load, the height of which is equal to one-half the bolt spacing. The remainder of the masonry wall load, is distributed to the shelf angle at each bolt location over a distance calculated by extending a 45 degree angle from the top corner of the shims
to the assumed reaction point of the supported masonry. The derivation of this distance will be presented in the finite element model (FEM) section. The line of reaction is assumed to be located 1/2 in. (13mm) in front of the back face of the masonry.”


From Page 70 (The Case Study Design Example):

“Step 2 - Assume 3 in. finger shims behind anchors. Masonry located near edge of horizontal leg resulting in 3 in. eccentricity (the masonry reaction is assumed to act 1/2 in. in front of back face of masonry.) Effective tributary length of outstanding leg of shelf angle is one-half of the shim width plus height of vertical leg and eccentricity (45 degree angle from edge of finger shim.) Z = 1/2 (3) + 6 + 3 = 10.5 in. (Use 10 in.).”


I am focusing on the text put in bold. If the paper is talking about the same thing I am thinking about, I always thought the reaction line of the masonry was located at the center of its width. The paper is stating something different. I am reading that as the reaction force is 1/2" in front of the back face of the masonry width. If you consider the 3” eccentricity from the example, wouldn’t the overhang be very large on say a 4” horizontal leg? This doesn’t match the paper’s suggestion that the masonry have 2/3’s of its width in bearing.

I have a feeling that the reaction of the masonry load on the angle is different than the line of reaction the paper is talking about. Could someone point me in the right direction?

Thank you,
Daniel…

 

[JAE]

Daniel,

Perhaps it is this: when the masonry bears on a horizontal leg of an angle, there is a natural rotation of the angle under load such that the application of force drifts inward, toward the root of the angle let (towards the back of the brick).

Since the brick is rigid, and the angle leg flexible, the resultant of the load won't be at the mid-width of the brick, but rather back towards the angle where the angle deflection is smaller.

 

[DaveAtkins]

JAE is correct. I have a copy of a publication that recommends assuming the load application is not at the middle of the veneer, but at the third point, closest to the heel of the shelf angle.

This is the first time I have heard of the 1/2" assumption, but it also seems reasonable.

DaveAtkins

 

[starcasm]

Thank you JAE and DaveAtkins for the comments. It does make more sense now with your explanations.

DaveAtkins, what is the name of the publication that uses the third point? I'd like to read that one too.

Thank you again,
Daniel...