Shear Lag Phenomenon eg Angle with connection to one leg Member has elements

Shear lag phenomenon eg angle with connection to one

This preview shows page 7 - 13 out of 15 pages.

Shear Lag Phenomenon (e.g. Angle with connection to one leg) Member has elements not in common plane and connectors are not provided in all elements Number of fasteners per gage line may not be sufficient to develop full section strength Tensile force is not uniformly distributed over net area o Non-Uniform Tensile Stress concentration of shear stress → shear distortion → shear lag in vicinity of connection AISC D3 (pg. 16.1-27): Effective net area A e = UA n where U depends on member shape, and number of fasteners per gage line. An empirical measure for the shear lag reduction coefficient 𝑈 = 1 − 𝑐?? 𝐿 𝑐?? ??? : distance measured from the plane of connection to the centroid of cross sectional area (tabulated in the section tables -AISC Part I) L con : length of connection Note: For Splice plates, A e = A n ≤ 0.85A g (AISC J4.1). Members such as single angles, double angles and WT sections shall have connections proportioned such that U ≥ 0.6.
Image of page 7
35
Image of page 8
36 Examples
Image of page 9
37 Example : Determine the strength of the tension member in a truss Member: W8×24 and Grade 50 steel Two lines of ¾-in diameter bolts Three bolts per line Solution : Properties of W8×24: A g = 7.08 in; d =7.93 in; b f =6.5 in; t f =0.40 in Based on gross area: P u = φ t F y A g = 0.9(50)(7.08) = 319 kips Based on effective net area: P u = φ t F u A e where A e = UA n 𝐴 ? = 𝐴 ? − 4 ( 7 8 ) 0.40 = 7.08 − 1.40 = 5.68 in 2 Since 𝑏 ? = 6.5 in > 2 3 𝑑 (= 0.67 × 7.93 = 5.31 in), 𝑈 = 0.90 1 A e = UA n = 0.9(5.68) = 5.11 in 2 φ t F u A e =0.75(65)(5.11) = 249 kips The design strength of this member is 249 kips as controlled by rupture failure based on least net area. 1 Suppose we assume bolt-to-bolt separations are 3 in, then 𝑈 = 1 − ? ̅ 𝐿 = 1 − 0.695 6 = 0.88. (see WT4×12 page 1-66 for ? ̅ ).
Image of page 10
38 Example : Design a 9 feet single angle subject to 30 kips dead load and 40 kips live load. Member: A36 steel 7/8-in diameter bolts At least 4 bolts per line Solution : Load combinations: P u =1.4(30) = 42 kips P u =1.2(30) + 1.6(40) = 100 kips Based on gross area: min 𝐴 ? = 𝑃 ? 𝜑 ? 𝐹 ? = 100 0.9(36) = 3.09 in 2 Based on effective net area 2 : U =0.80 since o bolts connected to one leg only o at least 4 bolts per gage line o only one bolt at any one cross section min 𝐴 ? = 𝑃 ? 𝜑 ? 𝐹 ? 𝑈 = 100 0.75(58)(0.80) = 2.87 in 2 min A g = min A n A due to one bolt where Δ A =( d bolt + 1/8) t = t . Slenderness ratio: min 𝑟 = 𝐿 300 = 12(9) 300 = 0.36 in Section selection: Angle t (in) Hole area ΔA Gross Area Required max ( 𝑃 ? 𝜑 ? 𝐹 ? ; 𝑃 ? 𝜑 ? 𝐹 ? 𝑈 + ∆𝐴 ) Lightest Angles ( A (in 2 ), r (in)) 5/16 0.312 max(3.09, 2.87+0.312) = 3.18 6x6x 5 / 16 ( A =3.67, r =1.19) 3/8 0.375 max(3.09, 2.87+0.375) = 3.25 6x3 1 / 2 x 3 / 8 ( A =3.44, r =0.762) 7/16 0.438 max(3.09, 2.87+0.438) = 3.30 5x3x 7 / 16 ( A =3.31, r =0.644) 4x4x 7 / 16 ( A =3.30, r =0.777) 1/2 0.500 max(3.09, 2.87+0.500) = 3.37 4x3x 1 / 2 ( A =3.25, r =0.633) NG 3 1 / 2 x3 1 / 2 x 1 / 2 ( A =3.25, r =0.679) NG 5/8 0.625 max(3.09, 2.87+0.625) = 3.50 4x3x 5 / 8 ( A =3.99, r =0.631) Use L5x3x 7 / 16 or L4x4x 7 / 16 (with minimum area (weight)) 2 Assume L6 x 6 x 5 / 16 (see page 1-42) and 3 in. bolt-to-bolt separations, 𝑈 = 1 − 𝐿 = 1 − 1.6 9 = 0.82.
Image of page 11
39 Block Shear : Tearing out (fracture) failure (AISC J4.3)
Image of page 12
Image of page 13

You've reached the end of your free preview.

Want to read all 15 pages?

  • Spring '14
  • GregoryG.Deierlein
  • Strength of materials, Tensile strength, in2, net area

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes