Question

Select double angle tension member to resist a factored load of 100 kips. Member is connected with 2 lines of 3⁄4-inch diameter bolts placed at usual gage distances (see AISC manual specification J3.3). There will be more than 2 bolts in each line. Member is 15’-0” long. Hint: Try an angle with 6 inch legs back to back and an outstanding length of 3 1⁄2 -inches. Choose required thickness. Assume U=0.85 for design.

Answer 1

Ans) Given,

Factored load = 100kips

Diameter of bolts = 3/4"

Yield strength of steel(f_y) = 50 ksi

Ultimate strength of steel(f_u) = 65 ksi

Since, it is given that number of bolts in a line should be more then 2 so assuming 3 bolt in a line.

Using angle of 6" x 6" x 5/16" having properties as follows:

A = 3.65 in² , weight = 14.9 lb/ft

I =13 in⁴ , = 1.616 in

Required tensile strength = 100kips

Strength for 2 angles (double angle) = 2 f_y A

= 2 x 50 x 3.65

= 365 kips > 100kips

Here ,assuming stress reduction factor = 0.90

Hence stength = 0.9 x 365

= 328.5 kips > 100kips

Available tensile rupture strength(Pn) ,

Pn = F_u x A_e

A_e = U x An

An = A - 2(D_b + 1/18" ) t

Putting values,

An = (2x3.65) - 2(3/4 + 1/8)5/16

= 7.3 - 0.546

= 6.753 in²

Therefore, Ae = 0.85 x 6.753

= 5.74 in²

Hence, Pn = 65 x 5.74

= 373.1 kips

Strength reduction factor() = 0.75,

Hence, Pn =  0.75 x 373.1

= 279.825 kips > 100 kips

Also , we know,

U = 1 - / (n x s) , where n is number of bolts and s is spacing

0.85 = 1 - 1.616/(3 x s)

Therefore, s = 3.6 in

where, n = number of bolts

s = spacing of bolts

Hence, use two 6" x 6" x 5/16" angle section and 3/4" diameter of bolts @ 3.6"