1.051 Structural Engineering Design Prof O. Buyukozturk Fall 2003 1.051 Structural Engineering Design Problem Set 5 Solutions TENSION MEMBERS 1. Effective Net Area Effective net area of the MC12x31 shown in Fig. 1. The holes are for 1-in bolts. A B C D E Figure 1 For MC12x31, Area, A = 9.12 in2, web thickness, tw = 0.37 in flange thickness, tf = 0.7 in Two possible failure paths: ABDE & ABCDE For ABDE: æ An = Ag - 2detw = 9.12 - (2) ççç1 + è For ABCDE An = Ag - 3detw + S = 8.08 in2 1 ö÷ 1 2 + ÷(0.37) = 8.29 in ø 16 16 ÷ s2 22 tw = 9.12 - (3) (1.125)(0.37) + (2) (0.37) 4g (4)(3.5) ABDE controls Effective area: Ae = UAn where U=0.85 (M shape with bf > (2/3)d) Ae = 0.85(8.08) = 6.87 in2 1.051 Structural Engineering Design Prof O. Buyukozturk Fall 2003 2. Tensile and Block Shear Strength of Bolted Members 7x4x3/8 angle, three 1-in bolts, A36 steel, determine block shear strength and compare with the tensile design strength. Figure 2 For 7x4x3/8 angle, Ag = (7 + 4 - 3/ 8)(3/ 8) = 3.98 in2 3 8 3 Agt = (3)( ) = 1.125 in2 8 Agv = (10)( ) = 3.75 in2 3 8 Anv = 3.75 - 2.5(1.125)( ) = 2.7 in2 Ant = 1.125 - 1.125 3 = 0.91 in2 2 8 Fu Ant = (58)(0.91) = 52.8 kips 0.6Fu Anv = (0.6)(58)(2.7) = 94 kips Fu Ant < 0.6Fu Anv Use LRFD Equation J4-3b f Rn = f éêë0.6Fu Anv + Fy Agt ùûú£ f [Fu Anv + Fu Ant ] = (0.75) [(0.6)(58)(2.7) + (36)(1.125) ]£ (0.75) [(58)(2.7) + (58)(0.91) ] = 100.8 £ 157.0 = 100.8 kips Tensile strength Yielding of gross section 1.051 Structural Engineering Design Prof O. Buyukozturk Fall 2003 f t Pn = f t Fy Ag = (0.9)(36)(3.98) = 129 kips Fracture of the net section ö 1 öæ 1 2 ÷ çç3 ÷ + ÷ ÷= 3.6 in øç8 ÷ ø 16 16 ÷ x 0.87 U = 1= 1= 0.89 Note that U can be taken L 8 æ An = 3.98 - (1) ççç1 + è conservatively as 0.85 as allowed by LRFD. Ae = UAn = (0.89)(3.6) = 3.2 in2 f t Pn = f t Fu Ae = (0.75)(58)(3.2) = 139.2 kips Block shear controls failure: f Rn = 100.8 kips 3. Tensile and Block Shear Strength of Welded Members Tensile design strength of the 6x6x1/2 angle, Fy=50 ksi, Fu=65 ksi. Consider block shear as well as the tensile strength. Figure 3 For the angle: Ag = 5.75 in2 , x = 1.68 in 1.051 Structural Engineering Design Prof O. Buyukozturk Fall 2003 Gross section yield: f t Pn = f t Fy Ag = (0.9)(50)(5.75) = 258.8 kips Fracture of the net section: An = Ag = 5.75 in2 Use average value of l: l = 3+ 9 = 6 in 2 w = l Þ U = 0.75 Ae = UAn = (0.75)(5.75) = 4.3 in2 f t Pn = f t Fu Ae = (0.75)(65)(4.3) = 209.6 kips (Controls) Block shear strength Fu Ant = (65)(5.75) = 373.8 kips 0.6Fu Anv = (0.6)(65)(9 + 3)(0.5) = 234 kips Fu Ant > 0.6Fu Anv Þ Use LRFD Eq. J4-3a f Rn = f éêë0.6Fy Agv + Fu Ant ùûú£ f [0.6Fu Anv + Fu Ant ] = (0.75) [(0.6)(50)(6) + (65)(3) ]£ (0.75) [(0.6)(65)(6) + (65)(3) ] = 281.2 £ 321.8 = 281.2 kips Design strength: f t Pn = 209.6 kips

1.051 Structural Engineering Design, Problem Set 5 Solutions

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