Possible remedies include using a larger diameter bolt, higher grade of bolt or more bolts.Ģ) Edge Tear-Out: This occurs when the bolt is located too close to the edge of the plate in the direction of load. The design shear strength is dictated in AISC Table J3.2 p. It occurs when the applied load exceeds the shear capacity through the bolt. Possible Bolted Shear Failure Mechanisms: There are 4 basic types of failure mechanisms for bolted connections under shear: 1) Bolt Shear: This is probably the most obvious failure mode. 16.1-102) AISC Part 7 AISC Part 9 AISC Part 10
The LRFD references the design of bolted connections in the following: ĪISC Spec. ASTM A325 and A490 bolts are typically used. In order to achieve the friction capacity, these bolts are tensioned to 70% of the ultimate tensile strength of the material according to the table below.
#AISC 14TH EDITION TABLE J3.4 PLUS#
2) High-strength bolts – These bolts are used for high-load connections and obtain their total strength from the shear strength across the diameter of the bolt PLUS the friction developed between the nut and joined steel surfaces. The typical carbon steel bolt used in structural steel buildings is ASTM A307 and F1554 for use in anchor rods. They are relatively low-strength and are used primarily for low-load applications such as for anchor rods. 2-41: 1) Carbon steel bolts – These bolts achieve their total strength from shear (or tension) strength across the diameter of the bolt. Lecture 12 – Bolted Connections Below is a typical bolt and the terms given to the parts of a bolt:īolts used in structural steel fasteners fall within 2 categories – see AISC Table 2-5 p.
= 5.96” Step 4 – Determine minimum base plate thickness, tmin: t min = L ΦcPp = Design bearing strength of concrete = 0.6Pp = 0.6(0.85f’cA1) Re-arranging to solve for A1: A1 =Ī1 = 457.5 in2 Step 2 – Determine “Optimized” base plate dimensions: Step 1 – Determine required base plate area, A1 to avoid conc. It bears on a steel base plate using A36 steel. 14-6: t min = LĮxample (LRFD) GIVEN: A W14x82 A992 column has a factored axial load Pu = 700 KIPS. Tmin = minimum base plate thickness per AISC p. Where Ωc = 2.50 ASD Ωc where: f’c = specified compressive strength of concrete, KSI A1 = area of steel base plate concentrically loaded on conc, in2 = BN (where B and N use whole inches if possible) B bf The design of a base plate involves the following steps: Pp = Nominal bearing strength of concrete = 0.85f’cA1 Design Bearing strength of concrete: φcPp where φc = 0.60 LRFD Pp The design of steel base plates is based on the following: ĪISC Spec. 14-9 Steel base plate Base Plate thickness An intermediary steel base plate is used to distribute this column load without crushing the concrete. Lecture 10 – Column Base Plates Columns must transmit vertical loads to the concrete footing.