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Masonry Tie Bar Length Calculator

Total length, count and weight, with seismic embedment rules.

Quick answer

With the default inputs: Wall thickness 240 · Seismic intensity 7 · Wall segment length 4 etc.Total length per tie bar 1238 mm

Edit the inputs below to recalculate.
Spacing sRun L
Total = embedment into the wall + anchorage into the column (+ hook). In seismic design the wall leg is at least wall length/5 and no less than 700 mm.

Inputs

pcs

Result

Total length per tie bar
1238mm
Embedment into wall1000mm
Anchorage into column200mm
End hook (6.25d)37.5mm
Rows along height7rows
Bars per wall14bars
Length per wall17.3m
Weight per wall3.8kg

Per GB 50011, infill walls need ties of 2Φ6 at about 500 mm vertical spacing (3Φ6 where the wall exceeds 240 mm). At intensity 6-7 the embedment into the wall is at least 1000 mm; at 8-9 it is at least one fifth of the wall length and not less than 700 mm. Anchorage into the column must satisfy anchorage rules, or use post-installed bars with pull-out testing. Design drawings govern.

Formula

Intensity 6-7: embed ≥ max(1000 mm, wall/5)
Intensity 8-9: embed ≥ max(700 mm, wall/5)
Total = embed + anchorage + hookhook 6.25d
Count = (height ÷ spacing + 1) × bars per row
240 mm wall, intensity 7, 4 m long, 3 m high, 200 mm anchorage, Φ6 at 500 mm with 2 bars per row: embedment max(1000, 800) = 1000 mm plus 200 anchorage and a 37.5 mm hook gives 1237.5 mm per bar; 7 rows means 14 bars, 17.3 m and 3.8 kg.

How to work it out

  1. Set the masonry embedment L1 by seismic intensity: at least 700 mm and a fifth of the wall segment at 6 and 7; at least 1,000 mm and a fifth at 8 and 9.
  2. Embedment into the column or wall L2 follows the design, typically 15d-20d or 200 mm, or the resin anchor depth.
  3. Total bar length = L1 + L2 + hook, normally 6.25d.
  4. Courses = ⌊wall height ÷ vertical spacing⌋, with a minimum of two.
  5. Total weight = bars × length × 0.00617 × d².

Where it goes wrong

Material and cost

Φ6 ties run 0.9 per metre; resin anchoring 3-5 per bar. On 1,000 m² of wall at 1 m lengths that is about 2,000 bars — 2,000 in steel but 8,000 in resin, since the anchoring usually costs more than the bar.

On site

Basis

FAQ

Why must the tie extend a metre into the wall?
The tie anchors the infill to the frame so it cannot fall outward in an earthquake. One metre provides enough bond with the masonry to hold the panel; shorter ties pull out once the wall cracks and the panel still collapses. This detailing was learned repeatedly from the Tangshan and Wenchuan earthquakes.
Post-installed bars or cast-in starter bars?
Cast-in starters are more reliable (formed in one pour, no adhesive durability question) but awkward on site and easily bent during formwork. Post-installed bars are convenient and accurately positioned, but depend heavily on hole depth, cleaning and adhesive quality, so pull-out testing is mandatory - typically 1% of each batch, minimum three bars.
Why three bars for 240 mm walls?
To spread the anchor points across the thickness. With only two bars near the middle, the masonry faces are poorly restrained and can split through the thickness in an earthquake. Three bars - one central and one near each face - confine the full depth. At 120 mm, two suffice.

Cite this page

Use this format in reports, method statements or coursework. The link is permanent.

Building Calculator. Masonry Tie Bar Length Calculator. 2026-10-25. https://jianzhujisuan.com/en/tie-bar-length

Results are estimating aids only and are not a design basis. Construction and final accounts must follow the drawings, the applicable standards and an approved method statement. Anything affecting structural safety, shoring or falsework must be checked by a suitably qualified engineer.