Review Module Steel Design TENSION MEMBERS PDF

Preview

Summary

on the gross section.```CONCENTRICALLY LOADEDCONNECTIONSA. BOLTED AND RIVETED CONNECTION:Determine the permissible tensile load “P” under the following conditions using NSCP 2001: 1 on the shear capacity of bolts in kN. 2 on the bearing capacity of the plates in kN. 3 on the block shear strength in ...

Description

```

CONCENTRICALLY LOADED CONNECTIONS A. BOLTED AND RIVETED CONNECTION: NSCP 2001/2010/2015 TENSILE STRENGTH 1.For tensile yielding in the gross section: R n = FyA gt ϕ = 0.90 (LRFD)

Ω = 1.67 (ASD)

2.For tensile rupture in the net section: Rn = FuAe ϕ = 0.75 (LRFD)

Ω = 2.00 (ASD)

Determine the permissible tensile load “P” under the following conditions usin NSCP 2001: 1.Based on the shear capacity of bolts in kN. 2.Based on the bearing capacity of the plates in kN. 3.Based on the block shear strength in kN. 4. Determine the joint efficiency in percentage.

EFFECTIVE NET AREA The effective area of tension members shall be determined as follows: Ae = AntU NOTE: Section 510.4.1.(2) limits An to a maximum of 0.85Ag connection design for splice plates with holes.

As shown in the bolted lap splice connection above, 5.Determine the design strength based on gross area yielding, in kN. 6.Determine the design strength based on net area rupture, in kN. 7.Determine the design strength based on block shear, in kN.

The diameter of standard sized holes is taken as 3.2 mm (2001) 4 mm (2010/2015) Larger than the nominal diameter of the bolt.

Effect of Staggered Hole / Staggered Connections Critical Net Area: An = bnet * thickness s = [ bgross - ∑ diameter of holes + ∑ 2 ] * thickness

If the nominal diameter of the hole is given, we add 1.6 mm (2001) 2 mm (2010/2015) To get the effective/standard hole diameter. SHEAR STRENGTH 1. For shear yielding of the element: Rn = 0.6FyAgv ϕ = 1.00 (LRFD)

Ω = 1.50 (ASD)

2. For shear rupture of the element: Rn = 0.6FuAnv ϕ = 0.75 (LRFD)

Ω = 2.00 (ASD)

4g

Where: s 2 = gage space 4g

s = longitudinal center to center spacing (pitch) of any consecutive holes in mm. g = transverse center to center spacing (gage) between fastener gage lines in mm. Note: Choose smallest An for design and safety purposes.

BEARING STRENGTH For bearing strength: Rn = 2.4dbtFu ϕ = 0.75 (LRFD)

Ω = 2.00 (ASD)

BLOCK SHEAR STRENGTH Rn = 0.6 FuAnv + UbsFuAnt ≤ 0.6FyAgv + UbsFuAnt ϕ = 0.75 (LRFD) Ω = 2.00 (ASD) NOTE: Where tensile stress is uniform, Ubs = 1 Where tensile stress is non-uniform, Ubs = 0.5 Where: db = Bolt diameter, mm t = Thickness of the connected material, mm Agt = Gross area subject to tension, mm2 Agv = Gross area subject to shear, mm2 Ant = Net area subject to tension, mm2 PROBLEM NO. 1: subject to shear, mm2 Anv = Net area Two A36 steel plates each with thickness t = 20 mm are bolted together U = Shear lag factor with 9 – 20 mm diameter bolts forming a lap connection. Bolts spacing are as follows: a = 75 mm, b = 400 mm, c = 125 mm, d = 40mm. Allowable Stress: Tensile stress on gross area of the plate = 0.6FY Tensile stress on net area of the plate

. . . PROBLEM NO. 2: A single-angle tension member has two gage lines in its long leg and one in the short leg for 20 mm bolts arranged shown. Area of the angular section is 5100 mm2. Use A36 steel FY = 248 MPa and FU = 400 MPa. Standard nominal hole diameter for 20 mm bolt = 23 mm. 1. Determine the net area (mm2) of the angular section. 2. Determine the allowable tension strength (kN) of the angular sectionon based the gross section. 3.Determine the allowable tensile strength (kN) of the angular section based on the effective net area. U = 0.75. 4. Determine the

= 0.5FU Shear stress of the bolt: FV = 120 MPa Bearing stress of the plate: FP = 1.2FU

B. WELDED CONNECTION:

ultimate tensile strength (kN) of the angular section. U = 0.75.

```

NSCP 2001/2010/2015 5.Determine the shear lag factor, U = 1 − x0 ⁄ℓ . 6.Determine the ultimate load based on net area of the angle. 7.Determine the ultimate load based on the strength of the fillet welds. 8.Determine the ultimate load based on the block shear strength of the gusset plate. NOTE:. For fillet weld groups concentrically loaded and consisting of elements that are oriented both longitudinally and transversely to the direction of applied load Rn = Rwl + Rwt ECCENTRICALLY LOADED Using NSCP 2010/2015,

STRENGTH OF FILLET WELDS Rn = 0.60 FEXX(0.707wL) ϕ = 0.75 (LRFD)Ω = 2.00 (ASD)

CONNECTIONS or

.

(510.2-9 Rn = 0.85Rwl + 1.5Rwt(51

PROBLEM NO. 4: The figure shows an eccentrically Where: bracket. P=15 shear, kN. mm2 t = Thro Ae = Effective area ofloaded fillet weld resisting Boltweld, diameter = =28Leg mm. L = Total length of the mm w of weld/ Size of we 1.If θ =metal, 0, determine the= shearing FEXX = Strength of the weld MPa EXX Electrode class stress acting on the mostloaded criticalfillet w Rwl = Total nominal strength of longitudinally bolt.

Examples of commonly used electrode: E60xx : FE60XX = 415 MPa E70xx : FE70XX = 485 MPa E80xx : FE80XX = 550 MPa

2. If θ = 0 and the shearing stress on each bolt is limited to 50 MPa, determine the maximum force P that can be applied. 3.If � = 20° and total ultimate load 20 kN, determine the shearing stress acting on the most critical bolt.

PROBLEM NO. 5: The figure shows a bracket that is welded to a plate and carries a load P. 1. Determine the polar moment of inertia (106 mm4/mm) of the welds. 2.If the load P is 150 KN, determine the maximum force carried by the welds per mm.

PROBLEM NO. 3: Two L100x100x10 are welded to a gusset plate with thickness of 16mm as shown and loaded concentrically. Use E70xx fillet weld, A36 steel for the gusset plate and the angular section. Given: Angular Section: Gross area of a single angle bar, Ag = 1380 mm². Distance from the outermost part of the angle leg to the centroid of the angle bar, x0 = 28 mm Weld: c = 100 mm e = 20 mm w = 8 mm 1.Determine the allowable tensile force “P” based on yielding of the gross area. 2.Determine the total length “L” in mm. 3.Determine the length of weld A in mm.

3.If FV = 150 MPa, determine P (KN) that the welds can carry if thickness of welds is 9.5 mm.

PR Re

4. Determine the length of weld B in mm.

Given: P = 360 kN a = 0.20 m b = 0.50 m Allowable stress, Fvw = 93 MPa 1.Calculate the average vertical force per unit length of weld (N/mm) assuming that distance a is zero. 2.Calculate the resultant force per unit length of weld (N/mm) due to the eccentric load P. 3.If the resultant force per unit length is 750 N, find the required weld thickness (mm).

m

h of t...