Masonry Anchor Embedment for a Tension (TMS 402 ASD)
The effective embedment that makes the TMS 402 masonry-breakout capacity equal a required tension.
Example
You enter
- Required tension T (lb) 5000
- Masonry strength f'm (psi) 1500
- Bolt effective tensile area Ab (in², net: 1/2 in 0.142, 3/4 in 0.334) 0.334
- Bolt yield fy (psi, A307 = 36000) 36000
You get
- Required embedment lbe 5.73366
- Steel ceiling Bas = 0.6 Ab fy 7,214 lb (adequate for T)
Checked against published worked examples from: NCMA / CMHA.
Details, formula, and sources
Scopelbe = sqrt(T / (1.25 pi sqrt(f'm))). 5,000 lb in 1,500 psi masonry needs ~5.7 in of embedment; the steel branch Bas = 0.6 Ab fy is a separate ceiling (a bolt too small yields no matter how deep). Edge distance reduces the cone. A design aid; the engineer of record's stamped design governs.
Formula or tablelbe = sqrt( T / (1.25 x pi x sqrt(f'm)) ), the masonry-breakout branch Bab = 1.25 x (pi lbe^2) x sqrt(f'm) solved for the effective embedment; steel ceiling Bas = 0.6 x Ab x fy checked separately.
Source and editionTMS 402 (Building Code Requirements for Masonry Structures) allowable-stress anchor-bolt tension provisions, as compiled in the Masonry Designers' Guide and CMHA TEK notes, by name.
Free accessCMHA TEK notes on anchor-bolt design are free public CMHA technical notes; the 1.25 x Apt x sqrt(f'm) arithmetic is public.
What governsEstimate. AHJ and licensed professional govern.
API field namesrequired_tension_lb, fm_psi, ab_in2, fy_psi, lbe_in, bas_lb
- Breakout branch inverted solves Bab = 1.25 x pi lbe^2 x sqrt(f'm) for lbe; the steel branch Bas = 0.6 x Ab x fy is a separate ceilingTMS 402 ASD
- Full-cone projected area Apt = pi x lbe^2 assumes a full breakout cone; edge distance or overlapping cones deepen the required embedmentTMS 402
- Tension only axial tension only; anchor shear (pryout) and combined shear-tension are separate checksTMS 402