Download
1 / 72
730 likes | 956 Views
Lecture 06: Tension Members. By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.edu.pk. Topics to be Addressed. Types of Steel Structures Introductory concepts Design Strength Net Area at Connection Shear Lag Phenomenon ASD and LRFD Design of Tension Members Design Examples.
E N D
Lecture 06: Tension Members By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.edu.pk
Topics to be Addressed • Types of Steel Structures • Introductory concepts • Design Strength • Net Area at Connection • Shear Lag Phenomenon • ASD and LRFD Design of Tension Members • Design Examples
Types of steel structures • The form of a tension member is governed to a large extent by • Type of structure of which it is a part • Method of joining it to connecting portions.
Introductory Concepts • Stress: The stress in an axially loaded tension member is given by Equation • The stress in a tension member is uniform throughout the cross-section except: • near the point of application of load, and • at the cross-section with holes for bolts or other discontinuities, etc.
Types of steel structures Area of bar at section a – a = 8 x ½ = 4 in2 Area of bar at section b – b = (8 – 2 x 7/8 ) x ½ = 3.12 in2 The unreduced area of the member is called its gross area = Ag The reduced area of the member is called its net area = An
Design strength • A tension member can fail by reaching one of two limit states: • Excessive deformation • Yielding at the gross area • Fracture • Fracture at the net area
Design strength Excessive deformation can occur due to the yielding of the gross section at section a-a
Design strength Fracture of the net section can occur if the stress at the net section (section b-b) reaches the ultimate stress Fu
Design strength • Yielding of the gross section will occur when the stress f reaches Fy Nominal yield strength = Pn = Ag Fy • Fracture of the net section will occur after the stress on the net section area reaches the ultimate stress Fu Nominal fracture strength = Pn = Ae Fu
Design strength • AISC/ASD Ft = 0.6 Fy on Gross Area Ft = 0.5 Fu on Effective Area • AISC/LRFD Design strength for yielding on gross area øtPn =øt Fy Ag = 0.9 Fy Ag Design strength for fracture of net section øtPn = øtFu Ae = 0.75 Fu Ae
Effective Net Area • The connection has a significant influence on the performance of a tension member. • A connection almost always weakens the member and a measure of its influence is called joint efficiency.
Effective Net Area • Joint efficiency is a function of: • (a) Material ductility • (b) Fastener spacing • (c) Stress concentration at holes • (d) Fabrication procedure • (e) Shear lag.
Effective Net Area Research indicates that shear lag can be accounted for by using a reduced or effective net area Ae CG For Bolted Connections • For bolted connection, the effective net area is Ae = U An • For welded connection, the effective net area is Ae = U Ag
Effective Net Area • For W, M, and S shapes with width-to-depth ratio of at least 2/3 and for Tee shapes cut from them, if the connection is through the flanges with at least three fasteners per line in the direction of applied load , U= 0.9 • For all other shapes with at least three fasteners per line , U= 0.85 • For all members with only two fasteners per line • U= 0.75
Net Area Example Example : A 5 x ½ bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with six 7/8 in. diameter bolts as shown in below. Assume that the effective net area Ae equals the actual net area An and compute the tensile design strength of the member.
Net Area Example Gross section area (Ag): Ag = 5 x ½ = 2.5 in2 Net section area (An): Bolt diameter = db = 7/8 in. Nominal hole diameter = dh = 7/8 + 1/16 in. = 15/16 in. Hole diameter for calculating net area = 15/16 + 1/16 in. = 1 in. Net section area = An = (5 – 2 x (1)) x ½ = 1.5 in2
Net Area Example Gross yielding design strength: ft Pn = ft Fy Ag = 0.9 x 50 ksi x 2.5 in2 = 112.5 kips Fracture design strength: ft Pn = ft Fu Ae = 0.75 x 65 ksi x 1.5 in2 = 73.125 kips Assume Ae = An (only for this problem) Therefore, design strength = 73.125 kips (net section fracture controls).
Shear Lag in Tension Members • Shear lag in tension members arises when all the elements of a cross section do not participate in the load transfer at a connection. • There are two primary phenomena that arise in these cases: • (i) Non-uniform straining of the web resulting in • biaxial stress states • (ii) Effective area reduction.
Shear Lag in Tension Members Effective area reduction
Shear Lag in Tension Members Design Bottom Line Shear lag can have a large influence on the strength of tension members , in essence reducing the effective area of the section. The amount of the reduction is related to the length of the connection and the arrangement of cross-section elements that do not participate directly in the connection load transfer.
Block Shear in Tension Members Block shear is a combined tensile/shear tearing out of an entire section of a connection.
Block Shear in Tension Members • A failure in which the member fails in tension on one section and in shear on the perpendicular section.
Block Shear in Tension Members • For such a failure to occur, there are two possible mechanisms: (1) Shear rupture + tensile yielding; and (2) Shear yielding + tensile rupturing.
Block Shear in Tension Members • AISC/ASD • Ft = 0.6 Fy on Gross Area • Ft = 0.5 Fu on Effective Area Connecting element allowable stresses where failure may be by shear Fv = 0.3 Fu Allowable block shear F = 0.3 Fu + 0.5 Fu
Block Shear in Tension Members • AISC/LRFD øtRn = 0.75(0.6 Fy Agv + Fu Ant) øtRn = 0.75(0.6 Fu Anv + Fy Agt)
Block Shear in Tension Members • Design Bottom Line As a likely limit state for connections, block shear must be considered in design. This can be accomplished by considering the strength limit states of the two failure mechanisms outlined above.
