32l8 Thursday 515 pm A tension member is to carry 15 k dead

32l8 Thursday 5:15 pm A tension member is to carry 15 k dead load plus 60 k live load. It is proposed to use an L5x single angle tension member of A992 steel to be connected to a -inch A992 gusset plate use bolted connection with a single gage line of 34-inch diameter A325-X (threads excluded from shear plane) high strength bolts in standard size holes. Using LRFD (a) Compute the factored tension load T, and verify that the gross section yielding strength T, is adequate according to AISC Specs. D2(a). (b) Determine the number of bolts required for a bearing type connection according to AISC J3.6. (See AISC Table J3-2 for Fa, thear strengths) (c) Consider minimum preferred bolt spacing according to AISC Specs, J3.3 and use the minimum preferred spacing of 3d. Check the net section fracture strength OT, to determine if it is adequate to carry T, according to AISC Specs. D2(b). (d) Assuming deformation of the bolt hole is not tolerated at service load levels check bearing strength and using 1.5 inch end spacing check gth at last bolt hole according to J3.10 and if necessary increase the end spacing. (e) Check block shear rupture of L5x3x3/8 according to J4.3 and if required increase the bolt sp (f) Sketch and detail your connection with dimensions. (g) Determine the number of bolts required if the connection is slip-critical at the required strength level and with class B surfaces (AISC 3.8)

Solution

*Gross area of angle = Ag = 2.86 in² (from table of AISC)

*Net sectional area = A_n

-bolt diameter = 3/4 in

- nominal hole diameter = 3/4 + 1/16 = 13/16 in.

- Hole diameter for calculating net area = 13/16 + 1/16 = 7/8 in.

- net sectional area = Ag - (7/8 * 3/8) = 2.86 - (7/8 * 3/8) = 2.53 in²

* effective net area = A_e = 0.85 * 2.53

= 2.15 in²

Gross yielding design strength = phi * Ag * Fy

   phi = 0.9( resistance factor for yielding )

   for A992 steel, Fy = 50 ksi and Fu = 65 ksi

= 0.9 * 2.86* 50 = 128.7 kips

* Net section fracture = 0.75 Ae.Fu

(Resistance factor for fracture is 0.75)

= 0.75 * 2.15*65

= 104.81 kips

Design strength = 104.81 kips ( smaller of 128.7 and 104.81)

* ultimate load acting on the tension member, Pu

Dead load D = 15 kips

Live load L = 60 kips

The applicable load combinations are :

1.4 D = 1.4 × 15 = 22 kips

1.2 D + 1.6 L = (1.2 *15)+(1.6 *60) = 114 kips

The design of tension member is based on the larger of the two, ie, 114 kips

* compare the design strength with the ultimate load

- the design strength ( 104.81 kips) is less than the ultimate load (114 kips)

Hence the angle section made of A992 steel is not adequate for carrying the factored loads.