UNIVERSITY OF AKRON Department of Civil Engineering 4300:401-301 July 9, 2013 Steel Design Sample Quiz 2 1. The W10 x 54 column shown has both ends pinned and consists of A992 steel (F y = 50 ksi, F u = 65 ksi). a) Check for slender elements. b) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using Table 4-22 of c) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using Table 4-1 of d) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using the column formulas from Specification Chapter E. Use the recommended design K value for the support conditions shown. W10 x 54 (A = 15.8 in 2, b f /2t f = 8.15, h/t w = 21.2, r x = 4.37, r y = 2.56 ) K = 1.00 (AISC Table C-A-7.1, p. 16.1-511) a) Check for slender elements (AISC Table B4.1a). Flanges (Case 1): λ r = 0.56 (E/F y ) 1/2 = 0.56 (29,000/50) 1/2 = 13.49 > b f /2t f = 8.15 Web (Case 5): λ r = 1.49 (E/F y ) 1/2 = 1.49 (29,000/50) 1/2 = 35.88 > h/t w = 21.2 The section has no slender elements. b) Using the critical stress values in Table 4-22 of the AISC Manual: (KL/r) max = (KL/r) y = 1.00 (16) (12 / )/2.56 = 75.0 By interpolating from Table 4-22 of the AISC Manual, LRFD: φ c F cr = 29.8 ksi ASD: F cr /Ω c = 19.8 ksi φ c P n = (φ c F cr ) A g = 29.8 (15.8) = 470.8 kips P n /Ω c = (F cr /Ω c ) A g = 19.8 (15.8) = 312.8 kips
c) Using Table 4-1 of the AISC Manual: KL = 1.00 (16) = 16.0 LRFD design compression strength: φ c P n = 471 kips ASD allowable compression strength: P n /Ω c = 314 kips d) Using Specification Chapter E: Since there are no slender elements, use Specification Section E3. (KL/r) max = (KL/r) y = 1.00 (16) (12 / )/2.56 = 75.0 (same as before) Determine the flexural buckling stress F cr. F e = π 2 E/(KL/r) 2 = π 2 (29,000)/(75.0) 2 = 50.88 ksi KL/r y = 75.0 < 4.71 (E/F y ) 1/2 = 4.71 (29,000/50) 1/2 = 113.43 Use AISC Eq. E3-2 F cr = [0.658 Fy/Fe ] F y = [0.658 50/50.88 ] 50 = 33.14 ksi φ c P n = φ c (F cr A g ) = 0.90(33.14)(15.8) = 471.3 kips P n /Ω c = (F cr A g )/Ω c = 33.14 (15.8)/1.67 = 313.5 kips 0:08
2. The HSS 10 x 10 x 3/8 column shown has both ends pinned and consists of A500 Grade B steel (F y = 46 ksi, F u = 58 ksi). a) Check for slender elements. b) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using Table 4-22 of c) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using Table 4-4 of d) Determine the LRFD design strength φ c P n and the ASD allowable strength P n /Ω c using the column formulas from Specification Chapter E. Use the recommended design K value for the support conditions shown. HSS 10 x 10 x 3/8 (A = 13.2 in 2, b/t = 25.7, r x = r y = 3.92 ) K = 1.00 (AISC Table C-A-7.1, p. 16.1-511) a) Check for slender elements (AISC Table B4.1a). Walls (Case 6); λ r = 1.40 (E/F y ) 1/2 = 1.40 (29,000/46) 1/2 = 35.15 > b/t = 25.7 The section has no slender elements. b) Using the critical stress values in Table 4-22 of the AISC Manual: (KL/r) max = (KL/r) y = 1.00 (18) (12 / )/3.92 = 55.10 By interpolating from Table 4-22 of the AISC Manual, LRFD: φ c F cr = 33.77 ksi ASD: F cr /Ω c = 22.48 ksi φ c P n = (φ c F cr ) A g = 33.77 (13.2) = 445.8 kips P n /Ω c = (F cr /Ω c ) A g = 22.48 (13.2) = 296.7 kips c) Using Table 4-4 of the AISC Manual: KL = 1.00 (18) = 18.0 LRFD design compression strength: φ c P n = 446 kips ASD allowable compression strength: P n /Ω c = 296 kips d) Using Specification Chapter E: Since there are no slender elements, use Specification Section E3. (KL/r) max = (KL/r) y = 1.00 (18) (12 / )/3.92 = 55.10 (same as before)
Determine the flexural buckling stress F cr. F e = π 2 E/(KL/r) 2 = π 2 (29,000)/(55.10) 2 = 94.27 ksi KL/r y = 55.10 < 4.71 (E/F y ) 1/2 = 4.71 (29,000/46) 1/2 = 118.26 Use AISC Eq. E3-2 F cr = [0.658 Fy/Fe ] F y = [0.658 46/94.27 ] 46 = 37.50 ksi φ c P n = φ c (F cr A g ) = 0.90(37.50)(13.2) = 445.5 kips P n /Ω c = (F cr A g )/Ω c = 37.50 (13.2)/1.67 = 296.4 kips 0:08
3. Select the lightest available W10 section to support the axial compression loads P D = 120 kips and P L = 250 kips if KL = 16 feet and A992 Grade 50 steel is used. Use the trial and error procedure in which the stress φ c F cr is determined from AISC Table 4-22, required areas calculated, and trial sections selected and revised as necessary until the lightest section is determined. Assume an initial value for KL/r of 50. This problem is to be solved using the LRFD procedure only. P u = 1.2 D + 1.6 L = 1.2 (120) + 1.6 (250) = 544.0 kips Assume KL/r = 50. From AISC Table 4-22: φ c F cr = 37.5 ksi A g = P u /φ c F cr = 544.0/37.5 = 14.51 in 2 Try W10 x 54 (A = 15.8 in 2, r x = 4.37, r y = 2.56 ) KL/r y = 16(12)/2.56 = 75.00 From AISC Table 4-22: φ c F cr = 29.80 ksi φ c P n = φ c F cr A g = 29.80 (15.8) = 470.8 kips < P u = 544.0 kips Try W10 x 60 (A = 17.7 in 2, r x = 4.39, r y = 2.57 ) KL/r y = 16(12)/2.57 = 74.71 From AISC Table 4-22: φ c F cr = 29.92 ksi φ c P n = φ c F cr A g = 29.92 (17.7) = 529.6 kips < P u = 544.0 kips Try W10 x 68 (A = 19.9 in 2, r x = 4.44, r y = 2.59 ) KL/r y = 16(12)/2.59 = 74.13 From AISC Table 4-22: φ c F cr = 30.15 ksi φ c P n = φ c F cr A g = 30.15 (19.9) = 600.0 kips > P u = 544.0 kips NG NG OK Select W10 x 68 0:09
4. Select the lightest available wide flange section (W8, W10, W12 or W14) to support the axial compression loads P D = 220 kips and P L = 350 kips if KL = 18 feet and F y = 50 ksi. Compare the available strength in axial compression for the selected member with the required strength for the column. Use the column tables in Part 4 of This problem is to be solved by both the LRFD and the ASD procedures. LRFD P u = 1.2 D + 1.6 L = 1.2 (220) + 1.6 (350) = 824.0 kips Possible selections from AISC Table 4-1: W8 None W10 x 112 φ c P n = 921 kips > P u = 824.0 kips OK W12 x 96 φ c P n = 888 kips > P u = 824.0 kips OK W14 x 90 φ c P n = 929 kips > P u = 824.0 kips OK Select W14 x 90 ASD P a = D + L = 220 + 350 = 570.0 kips Possible selections from AISC Table 4-1: W8 None W10 x 112 P n /Ω c = 613 kips > P a = 570.0 kips OK W12 x 96 P n /Ω c = 591 kips > P a = 570.0 kips OK W14 x 90 P n /Ω c = 618 kips > P a = 570.0 kips OK Select W14 x 90 0:05
5. Design a square column base plate for a W12 x 35 column with P D = 180 kips and P L = 400 kips using A36 steel for the plate and f c = 4.0 ksi for the concrete in a 6 x 6 footing. Specify the plate length and width to the nearest inch; specify the plate thickness to the nearest 1/4 inch. This problem is to be solved by the LRFD procedure only. W12 x 35 (d = 12.5, b f = 6.56 ) P u = 1.2 D + 1.6 L = 1.2 (180) + 1.6 (400) = 856.0 kips Determine the base plate area (φ c = 0.65). The area of the supporting concrete is much greater than the base plate area, such that (A 2 /A 1 ) 1/2 = 2.0. A 1 = P u /[φ c (0.85 f c )(A 2 /A 1 ) 1/2 ] = 856.0/[0.65(0.85)(4.0)(2.0)] = 193.7 in 2 The base plate must be at least as large as the column. b f d = 6.56 (12.5) = 82.0 in 2 < 193.7 in 2 OK Determine the base plate dimensions. Simplify the base plate dimensions by making it square. B = N = (193.7) 1/2 = 13.92 (Say 14 x 14 ) Check the bearing strength of the concrete (φ c = 0.65). φ c P p = φ c 0.85 f c A 1 (A 2 /A 1 ) 1/2 = 0.65(0.85)(4.0)(14 x 14)(2.0) = 866.3 kips > P u = 856.0 kips OK Compute the required base plate thickness. m = (N 0.95d)/2 = [14 0.95(12.5)]/2 = 1.063 n = (B - 0.80b f )/2 = [14 0.80(6.56)]/2 = 4.376 n = ¼ (d b f ) 1/2 = ¼ [12.5 (6.56)] 1/2 = 2.264 l = max (m, n, or λn ) = 4.376 t min = l (2 P u /0.9 F y B N) 1/2 = 4.376 [2(856.0)/0.90(36.0)(14)(14)] 1/2 = 2.272 (Say 2.50 ) Use PL 2½ x 14 x 14 (A36 steel) 0:07