Chap. 3-Pbs

Problem 3.2-2 3.2-2  A PL 12 ×200 tension member is connected with 6 M24-mm-diameter  bolts, as shown in the figure. figure. The steel ha hass F  y = 248 MPa, and F u = 400 MPa. Assume that Ae = An and compute the following. Take Take the hole diameter; h = 24

+3.2 = 27.2 mm. a. The Th e des desig ign n sstr tren engt gth h ffo or LRF LRFD. D.  b. The allowable strength for ASD.

1

Problem 3.2-2 (Solution)  Ag  = 200(12) = 2400 mm mm 2  An = Ag − 2ht = 2400 − 2(27.2)(12) = 1747.2 mm 2 = Ae

a. LRFD:  φ t P n = min {0.9 Fy Ag ; 0.75Fu Ae } = m   in {0.9(248)(2400); 0.75(400)(1747.2)}

kN φ t P n = min {535680; 524160} = 524.16 kN

b.

ASD:

Pn / Ω = min {0.6 Fy Ag ; 0.5Fu Ae } = m   in {0.6(248)(2400); 0.5(400)(1747.2)} Pn  / Ω = min {357120; 349440} = 349.44 kN 2

Problem 3.2-3 3.2-3  A UPN300 is connected with M24 diameter bolt in each flange, as shown in the figure. The steel has F  y = 248 MPa, and F u = 400 MPa. Assume that Ae = 0.90 An and compute the following.

a. The Th des desiign str streng en gth for for for LRF LRASD. FD.  b. Thee allowable strength Take the hole diameter; h = 24 +3.2 = 27.2 mm.

3

Problem 3.2-3 (Solution)  Ag = 58.8 cm 2 , t f  = 16 mm  An = Ag − 2ht f  = 58.8 − 2( 2.72)(1.6) = 50.096 cm 2  Ae = 0.90 An = 0.90(50.096) = 45.0864 cm cm 2

a. LRFD:  φ t P n = min {0.9 Fy Ag ; 0.75Fu Ae } = m   in {0.9(248)(5880); 0.75(400)(4508.64)}

φ t P n = min {1312416; 1352592} = 1312.416 kN

b.

ASD:

Pn / Ω = min {0.6 Fy Ag ; 0.5Fu Ae } = m   in {0.6(248)(5880); 0.5(400)(4508.64)} Pn  / Ω = min {874944; 901728} = 874.944 k kN N 4

Problem 3.2-5 3.2-5  The tension member shown is a PL 12×200 of a steel having F  y  = 248 MPa, and F u  = 400 MPa. The member is connected to a gusset plate with M30  bolts. It is subjected to the dead and live loads shown. Does the member have n e enough a. Usestrength. LRFD. Assume that A  = A .  b. Use ASD. Take the hole diameter; h = 30 +3.2 = 33.2 mm.

5