University:
Michigan State University
Course:
ERG 100 | Introduction to Engineering Design
Academic year:

2021
Views:

314

Pages:

1
Author:

folclord7m

PROBLEM 1.50 Determine the factor of safety for the cable anchor in Prob. 1.49 when P = 3 kips, knowing that a = 2 in. and b = 7.5 in. PROBLEM 1.49 A steel plate 165 in, thick is embedded in a horizontal concrete slab and is used to anchor a high-strength vertical cable as shown. The diameter of the hole in the plate is 43 in., the ultimate strength of the steel used is 36 ksi, and the ultimate bonding stress between plate and concrete is 300 psi. Knowing that a factor of safety of 3.60 is desired when P = 2.5 kips, determine (a) the required width a of the plate, (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the lower end of the plate.) SOLUTION Based on tension in plate: A = (a − d )t 3 ·§ 5 · § = ¨2 − ¸¨ ¸ = 0.3906 in 2 4 ¹© 16 ¹ © PU = σ U A = (36)(0.3906) = 14.06 kips F. S. = PU 14.06 = = 4.69 3 P Based on shear between plate and concrete slab: 5 · § (7.5) A = perimeter × depth = 2( a + t) b = 2 ¨ 2 + 16 ¸¹ © A = 34.69 in 2 τ U = 0.300 ksi PU = τU A = (0.300)(34.69) = 10.41 kips F. S. = 10.41 PU = = 3.47 3 P Actual factor of safety is the smaller value. F .S . = 3.47 W