1. Tension Members

Tension members 3.2-1. The tension member shown in Figure P3.2-1 is a PL 1 2  8 of A36 steel. The member 1 is connected to a gusset plate with 1 8 -inch-diameter bolts. It is subjected to the dead and live loads shown. Does this member have enough strength? Assume that Ae  An .

PL

1

2

8

D = 95k L = 9k

Solution Load combination: 1. Pu  1, 2 D  1, 6 L  (1, 2  95)  (1, 6  9)  128, 4 kips 2. Pu  1, 4 D  1, 4  95  133 kips So Pu  133 kips

1 Ag   8  4 in.2 2 Required gross area: 1  1  9 in.  in. in.  3,375 in. 8  2  8 Required net area: Pu  133 kips > t Fy Ag  0,9  36  4  129, 6 kips  Not ok An  Ae  Ag  Aholes  4 in.2  

Pu  133 kips > t Fu Ae  0, 75  58 ksi  3, 375 in.2  146,812 kips

Both conditions appear to be unsatisfied with the factored load So, the member does not have enough strength

 Not ok

7 3.2-2. A C8 11,5 is connected to a gusset plate with 8 -inch-diameter bolts as shown in Figure P3.2-2. The steel is A572 Grade 50. If the member is subjected to dead load and live load only, what is the total service load capacity if the live-to-dead load ratio is 3? Assume that Ae  0,85 An . C 8  11,5

Solution Fy  50 ksi Fu  65 ksi Gross area: Net area:

Ag  3,38 in.2

An  Ag  Aholes

(From the manual) 7 1  3,38  2(  )0, 22  2,94 in.2 8 8

2 Effective area: Ae  0,85 An  0,85  2,94  2, 45 in. P  min  (t Pn ) yield ; (t Pn ) fracture   min  t Fy Ag ; t Fu Ae  The factored load: u Pu  min  0,9  50  3,38 ; 0, 75  65  2, 45  min  152,1 ; 119, 437

Pu  119, 437 kips Ultimate load: Pu  1, 2 D  1, 6 L  6 D P 119, 437  D u   19,906 kips 6 6 Service load: Ps  D  L L  3  L  3D But: D  Ps  4 D  4 19,906  Ps  79, 624 kips

3.2-2. A C12  30 is connected with a line of 1-inch-diameter bolts in each flange as shown is F  50 ksi Figure P3.2-3. If y and Fu  65 ksi , what is the tensile design strength? Assume that Ae  0,9 An .

Solution