CAE303_Solution__5_F07

# CAE303_Solution__5_F07 - CAE 303 FALL 2007 STRUCTURAL...

This preview shows pages 1–4. Sign up to view the full content.

Problem 1 load P = 6 change th F t 1: As show 650 kips (80% he section wi Figure 2(a). the column b n in Figure % from dead ithin W14 gr 18 ft 22 ft A B C P x The deform buckles abou So 1, W 14 x 99 d load and 2 roup so that Fig A B C P y ation when ut x-axis. olution Set 9, F y = 36 ks 0% form live the column gure1 40 ft La Su Su sam we # 5 si steel shape e load). Che is able to ca STRUCTU ateral support at upport condition upport condition me for both stron eak axes. Figure 2(b) the column e is used for ck if the colu rry the load URAL DESIG N Instruc B) n at A. at C is the ng and A B C P y x . The deform buckles abo r column AB umn is adeq . CAE 30 N I: STEEL S tor: J. Shen, P E-mail mation when out y-axis. 1 BC. Service quate. If not, 3 – FALL 200 STRUCTURE Ph.D.,P.E., S.E l: Shen@iit.ed n 07 ES E. du

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 Solution: P r = P u = 1.2P D + 1.6P L = (1.2 x 80% +1.6 x 20%) (650) = 832 kips (LRFD) P r = P a = P D + P L = 650 kips (ASD) Column: W 14 x 99 r x =6.17 in. r y = 3.71 in. K x 1.0, K y =1.0, L y = the larger of L y1 (= 18 ft) and L y2 (= 22 ft) = 22 ft, L x = 40 ft x KL r ⎛⎞ ⎜⎟ ⎝⎠ = 1.0 40 12 77.8 6.17 ×× = (controls) y KL r = 1.0 22 12 71.2 3.71 = KL r = the larger of y KL r and x KL r = x KL r = 77.8 The column will buckle about X-Axis, as shown in Figure 2. KL 77.8 r = <4.71 29000 E 4.71 134 Fy 36 == 2 e 2 E F 47.2 (KL/r) π ksi F cr = y e F F y 0.658 F = (0.727)(36) = 26.2 ksi P n = F cr A g = (26.2) (29.1) = 761.3 kips We need to have: P c P r LRFD: P c = cn r P (0.9)(761.3) 685.2 kips < P 832 kips φ= = = ASD: P c = n r c P 761.3 455.9 kips < P 650 kips 1.67 = Ω Inadequate for both LRFD and ASD Let’s increase the weight of the W14 shape from W 14 x 99 to W 14 x 120 with F y =36 ksi Check W 14 x 120: Column: W 14 x 120 r x =6.24 in. r y = 3.74in. A g = 35.3in 2 K x = K y =1.0, L y1 = 18 ft, L y2 = 22 ft (controls) L x = 40 ft x KL r = 1.0 40 12 76.9 6.24 = (controls) y KL r = 1.0 22 12 70.6 3.74 =
3 KL r ⎛⎞ ⎜⎟ ⎝⎠ = the larger of y KL r and x KL r = x KL r = 76.9 The column will buckle about X-Axis, as shown in Figure 2(a). KL 76.9 r = <4.71 29000 E 4.71 134 Fy 36 == 2 e 2 E F 48.4 (KL/r) π ksi F cr = y e F F y 0.658 F = 26.4 ksi P n = F cr A g = (26.4) (35.3) = 932 kips We need to have P c P n LRFD: P c = cn r P (0.9)(932) 839 kips > P 832 kips ϕ= = = ASD: P c = n r c P 932 558 kips < P 650 kips 1.67 = Ω W 14 x 120 is adequate for LRFD, BUT inadequate for ASD.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This homework help was uploaded on 04/07/2008 for the course CAE 303 taught by Professor Shen during the Fall '08 term at Illinois Tech.

### Page1 / 8

CAE303_Solution__5_F07 - CAE 303 FALL 2007 STRUCTURAL...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online