By American Institute of Steel Construction
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Additional resources for Design Guide 1: Base Plate and Anchor Rod Design (Second Edition)
0 kips/in. 8 kips/in. 02 in. 46 in. 4 in. Therefore, e < ecrit, and the design meets the criteria for the case of a base plate with small moment. where e is the effective steel reinforcement lap required to develop the load in the reinforcing steel. 2 in. 5. Select 25-in. embedment for anchors. 4. Determine bearing length, Y. 50) = 14 in. , respectively. 7 in. 2 in. The ratio of the concrete to base plate area is unity; Fy of the base plate is 36 ksi and f ′c of the concrete is 4 ksi. Pu = 376 kips/14 in.
Therefore, as noted above, q ≤ qmax. 4, it follows that fp ≤ fp(max). 8 is For moment equilibrium, the line of action of the applied load, Pu, and that of the bearing force, qY must coincide; that is, e = ε. 6) exceeds the maximum value that ε can attain, the applied loads cannot be resisted by bearing alone and anchor rods will be in tension. In summary, for values of e less than εmax, Y is greater than Ymin and q is less than qmax, and obviously, fp is less than fp(max). For values of e greater than εmax, q = qmax.
1. Compute the required strength. 6(1,500) = 3, 600 kip-in. ASD Pa = 100 + 160 = 260 kips M a = 1, 000 + 1,500 = 2 ,500 kip-in. 8 = 355 Since 315 > 306, the inequality Since 315 > 306, the inequality is not satisﬁed. is not satisﬁed. Hence, a larger plate dimension Hence, a larger plate dimension is required. is required. As the second iteration, try a 20 × 20 plate. The increased dimensions cause a modiﬁcation in the maximum bearing pressure, qmax, f, and ecrit. 2 kips/in. 5 in. 5 in. 75 in. 22 in.