Design of anchors according to ACI 349.2R97 appendix B Appendix B of ACI 349 was developed to better define the design req...
Check Anchor head bearing Area of the anchor head (Ah ) (including the area of the tensile stress component) is at least 2...
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Mathcad anchors (2)

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Anchors deisgn to
ACI 349.2R97 appendix B

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Mathcad anchors (2)

  1. 1. Design of anchors according to ACI 349.2R97 appendix B Appendix B of ACI 349 was developed to better define the design requirements for steel embedmnts revisions are periodically made to the code as a result of on-going research and testing. As with other concretebuilding codes, the design of embedments attempts to assure a ductile failure mode sothat the reinforcement yields before the concrete fails. In embedments designed for direct loading, the concrete pullout strength must be greater than the tensile strength of the steel. Vu  6kip fy  50000psi fut  60000psi fc  4000psi Must be in psi Anchor diameter d s  0.5in Anchor head diameter d h  1in Anchor head thickness Th  0.312in Anchor length Ld  10 d s  5  in ϕ  0.85 μ  0.9 Vu Vn   7.059  kip ϕDetermine the required area of steel of the anchorUse the shear friction provision of Section 11.7 with = 0.85, = 0.9.Equate the external (required strength) and internal (design strength)forces and solve for the required steel area for the stud. Vu 2 Avf   0.157  in ϕ μ fy 2 π d s 2 As   0.196  in 4   check1  if As  Avf "ok" "no good"  "ok"
  2. 2. Check Anchor head bearing Area of the anchor head (Ah ) (including the area of the tensile stress component) is at least 2.5 times the area of the tensile stress component. 2 π dh 4 2 Ah   5.067  10 m 4 Ah 4 Must be more than 2.5 As dh  ds c1   0.25 in 2  check2  if Th  c1 "ok" "no good"  "ok" Determine the embedment of the stud The design pullout strength of the concrete, Pd , must exceed the minimum specified tensile strength (Asfut ) of the tensile stress component. As fut  11.781 kip 0.65 4 psi fc  164.438  psi Acp = [(Ld + dh /2)^2– (dh /2)^2]   dh  2 2   dh  Pc  π  Ld        0.65 4  psi fc  15.498 kip  2  2   check3  if Pc  As fut "ok" "no good"  "ok"  By Khaled Eid, PE email khaled_eid@yahoo.com

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