Question

Please answer the following Steel Structure Design question. You may need to referance a steel design manual.

A3 sbelts are used in a Cder yield, rupture CU Lo in is e and block skar ropture Determin the design tension by LRFD. (25%) Block Shepr AI し5"

Answer 1

Solution :

LRFD Method :The design philosophy of Load and Resistance Factor Design (LRFD) is primarily based on a consideration of failure conditions rather than working load conditions.

The equation format for the LRFD method is stated as: ΣγiQi = φ Rn

Where: Qi = a load (force or moment)

γi = a load factor (LRFD Specification)

Rn = the nominal resistance, or strength, of the component under consideration

φ = resistance factor (for bolts given )

Design Tension Strength

φRn = φ Ab F n

Where: φ = Resistance Factor = 0.75

Ab = Nominal unthreaded body area of bolt or threaded part, in2

Fn = Nominal tensile strength, Ft = 0.75 Fu

Fu = minimum tensile strength of bolt material,

The ¾-in bolts require a minimum edge distance of 1 ¼-in per LRFD Table(at a sheared edge),

the recommended spacing is taken as 3 x bolt diameter = 2.25 in., let’s use 3”

The minimum width of the tension members can be found as:

W = 2(1.25) + 2.25 = 4.75 in Considering no additional constraints,

let’s try a width = 5 in. for these members

Design strength of connecting elements in tension (LRFD)

Design the tension members for yielding in the gross section,

the design tension strength in yielding is φRn with φ = 0.90

Equating φRn to the applied load,

where Rn = FyAg 0.90FyAg = 0.9(36)(5t) = 121.6 kips

Solving for the thickness required, t = 121.6 / (2 x 0.90 x 180) = 0.375 in

Therefore, the thickness required based on yielding of the gross section is 3/8 in.

Check the plates for the limit-state of tension fracture in the net section: φ = 0.75

φRn = φFu An = 0.75(58)1.22 x 2 = 106.1 kips < 121.6 kips NG

Where, An = 5(0.375) – 2(3/4 + 1/8) 0.375 = 1.22 in2

Note: LRFD Chapter B, section B2, requires that in computing the net area for tension and shear, the width of the bolt hole be taken as 1/16-in greater than the nominal bolt hole. Thus increase the plate thickness or the width of the narrow plate, Let’s increase the width to 6”

The revised, An = 6(0.375) – 2(3/4 + 1/8) 0.375 = 1.59 in2 < 0.85Ag

φRn = φFu An = 0.75(58)1.59 x 2 = 138.3 kips > 121.6 kips

∴OK Use 3/8” x 6” Plates for the Tension Members