CE 562 Structural Design I Midterm No. 1 Closed Book Portion (25 / 100 pts)

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1 CE 56 Structural Desig I Name: Midterm No. 1 Closed Book Portio (5 / 100 pts) 1. [6 pts / 5 pts] Two differet tesio members are show below - oe is a pair of chaels coected back-to-back ad the other is a rectagular HSS sectio. If ou kow that both of the cross-sectios have the exact same et area, which of the two cross-sectios would ou expect to fail first et sectio fracture? Wh? The HSS would likel fail first, sce ol two of the sides are coectio to a supportg member. Sce two of the four sides are ot sharg the load trasfer, shear lag will reduce the amout of force that the coectio ca carr. ll elemets of the two chaels are directl attached to a supportg member, so shear lag is ot a cocer. CE 56 Sprg 009 Midterm No.1 - Closed Book Portio Page 1 of 3

2 . [3 pts / 5 pts] braced steel frame has colums with fixed bases. ll other coectios are rigid ( full-restraed or momet ) coectios. Without kowg the colum or beam sizes, what value could ou coservativel use for k the colum desigs, ad wh? k=1.0 would be a coservative choice because the frame is braced, ad that is the maximum possible value for the effective legth factor. 3. [5 pts / 5 pts] You foud that the desig of the followg colum is govered b elastic flexural bucklg. ssume it costs the same amout to either chage the colum ed coditios or to crease the ield stregth of the material b specifg a differet grade of structural steel. Which chage would be more efficiet raisg the elastic flexural bucklg capacit of the colum show? Clearl mark whether ou thk colum desig () or (B) would be the better optio, ad state wh. Elastic bucklg capacit is uaffected b material ield stregth. Therefore, chagg the ed coditios of the colum from p-p to p-fix would be more effective raisg the colum capacit. (B). 30 ft Curret Desig: F = 36 ksi P-p Proposed Desig (): F = 50 ksi P-p Proposed Desig (B): F = 36 ksi Fix-p CE 56 Sprg 009 Midterm No.1 - Closed Book Portio Page of 3

3 4. [3 pts / 5 pts] TRUE / FLSE (circle oe): extremel sleder colum that is geometricall perfect, loaded absolutel perfectl, ad has perfectl homogeous material properties is still susceptible to bucklg despite its theoretical perfectio. TRUE. 5. [3 pts / 5 pts] egeer thought he/she would make a bolted coectio betwee a tesio member ad supportg elemet extremel strog b addg lots ad lots of bolts to the coectio, thkg that the more bolts were available to take the load, the stroger the coectio would be. Was this ecessaril a good decisio? Wh or wh ot? This was ot ecessaril a good decisio. Icreasg the umber of bolts could reduce the resistg area of the steel cross-sectio, makg it more proe to failure. 6. [5 pts / 5 pts] tesio member bolted to a gusset plate is show below. There are two differet sizes of holes used the coectio. Which et sectio fracture plaes must be vestigated, without havg a further formatio? NTS EFG EFD D B E C D F G CE 56 Sprg 009 Midterm No.1 - Closed Book Portio Page 3 of 3

4 CE 56 Structural Desig I Midterm No. 1 Ope Book Portio (75 / 100 pts) Name: 1. [35 pts. / 75 pts.] Determe whether the steel tesio member show is adequate to support the followg service loads: DL = 40, LL = 40. ll bolts show are 7/8" diameter. Cosider all appropriate limit states, but ou ma eglect the stregth of the gusset plates. The tesio member is 5' log. ELEVTION VIEW OF TENSION MEMBER Gusset plates welded to supportg member face TOP VIEW OF CONNECTION ½ Gusset Plate - ½ thick Gr. 36 HSS 6 x 6 x 5/8 500 Gr. B (F=46 ksi, Fu=6 ksi ) 5" SECTION - 10" Gross Sectio Yieldg of the Tesio Member: R = F = = 538. g ksi φ R = = Net Sectio Fracture of the Tesio Member: R = F u e e = Uet U =? Case 6, Table D3.1 applies to this coectio cofiguratio. CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 1 of 10

5 x l H? Sce l=6 ad H=6, l H, therefore use U = 1 l where ( B H) B 6" x = = 4 6" + 6" = 0.75 U = 1 = " et = g holes = 11.7 holes 0.581" thick " + " + " sides = et φr = φ F = = 8.04 e u e φ R = = ksi co Block Shear Rupture of the Tesio Member: Tesio Member Block Shear Mode 1: R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t Gross area tesio, gt = 11.7 gt Net area subject to tesio, t = t gt holes t = " + " 0.581" = Gross area shear, gv = 6"(4 plaes) = 4 gv Net area subject to shear, v = v gv holes v ( )( 1 ) 8 8 = " + " 0.581" = R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t U bs = 1.0 as per Commetar to Ch. J, pg R = R = φ R = = 93.8 CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page of 10

6 Tesio Member Block Shear Mode : R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t 6 ½ Gross area tesio, gt = " 0.581" = gt Net area subject to tesio, t = t gt holes ( )( " ")( 0.581" ) t = + sides + = 8 8 Gross area shear, gv = 11"(4 plaes) = 44 gv 5" Net area subject to shear, v = v gv holes v ( 1 7 )( 1 ) 8 8 = " + " 0.581" = R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t U bs = 1.0 as per Commetar to Ch. J, pg R = R = φ R = = Tesio Member Block Shear Mode 3: R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t 6 ½ Gross area tesio, gt =.5" 0.581" =.905 gt 5" Net area subject to tesio, t = t gt holes ( )( " ")( 0.581" ) t = + sides + = 8 8 Gross area shear, gv = 11"(4 plaes) 0.581" = gv CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 3 of 10

7 Net area subject to shear, v = v gv holes v ( 1 7 )( 1 ) 8 8 = " + " 0.581" = R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t U bs = 1.0 as per Commetar to Ch. J, pg R = φ R = = 610. Tesio Member Block Shear Mode 4: R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t 6 ½ Gross area tesio, gt = 3.669" sides 0.581" +... gt ( thick )( sides) " = Net area subject to tesio, t = t gt holes ( 1 7 )( " ")( 0.581" ) 6.01 t = + sides + = 8 8 Gross area shear, gv = 11"( plaes) 0.581" = 1.78 gv 5" Net area subject to shear, v = v gv holes v ( 1 7 )( 1 ) 8 8 = " + " 0.581" = R = 0.6F + U F 0.6F + U F u v bs u t gv bs u t U bs = 1.0 as per Commetar to Ch. J, pg R = φ R = = 610. Factored Loads: 1. Ru γ iqi ( D) R u = = = 1.4 = = 56 8 CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 4 of 10

8 . Ru γ iqi ( D) ( L) φ R R = u = = = = = R = 11 u, therefore the member is satisfactor for stregth. Serviceabilit: L r m 300 ( 5' )( 1 ) '' '.17" = , therefore the tesio member is OK for serviceabilit. Coclusio: NSF of the tesio member cotrols the desig, with φ R = CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 5 of 10

9 . [0 pts. / 75 pts.] Cosider a sgle stor, sgle ba structure braced as show the figure below. Each of the four colums is a W1x7 made of 57 Gr. 4 steel, orieted as show the figure. Determe the service dead load ad service live load the colums ca each support if the total service load is 5% dead load ad 75% live load. Note: The colums do ot have F = 50 ksi, therefore the colum desig charts are ot applicable. 10 ft 10 ft Cotrollg Slederess: ( 0' " )( 1 ' ) KL = = 45.0 Bucklg about the x-axis cotrols r x 5.31" " KL ( 10' )( 1 ' ) = = r 3.04" Colum Desig Stregth: Compare maximum slederess ratio agast E 9, = 4.71 = 13.8 F 4 E 4.71 F to determe how the colum will buckle. F KL E Fe Sce = Fcr F r = = F Ielastic bucklg govers. CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 6 of 10

10 P = F cr g F Fe Fcr = F where π E π ( 9,000) Fe = = = KL ( 45.0) r ksi F cr = ( 4 ) = P = = ksi φ P = = Service Load Capacit: Ru = γ iqi = 1. D L Let φ P = 1.( D) + 1.6( L) ( P ) ( P ) ( P ) = =.8 P 51.6 service = P = = 6.8 service DL ksi service service service P = = service LL Local Bucklg Check: W1x7 Flages: b f λ fl = = 8.99 (1-3) tf E 9,000 λ r,fl = 0.56 = 0.56 = 14.7 ( ) F 4 Sce λ fl <λ r, local bucklg will ot occur the flages. ksi Web: h λ web = =.6 (1-3) t w E 9,000 λ r,web = 1.49 = 1.49 = ( ) F 4 Sce λ web <λ r, local bucklg will ot occur the web. CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 7 of 10

11 3. [0 pts. / 75 pts.] thick steel plate made of 36 steel is used as a colum a structure. The plate has cross-sectioal dimesios of x 10. The base of the colum is perfectl fixed. The top of the colum is supported agast lateral traslatio b frictioless kife edge supports oe directio, as show. The kife edges do permit rotatio. Lateral traslatio ad rotatio are both allowed the other plae. Determe the desig P () capacit of the colum. Check whether local bucklg ma be a issue, ad if it is, still determe the capacit of the colum disregardg local bucklg effects. Cross-Sectioal Properties: rea: = 10" " = 0 0 ft Bedg Momets of Iertia: x 1 3 Ix = " 10" + ad = I = 10" " + ad = Radii of Gratio: 4 I x rx = = =.89 4 I 6.67 r = = = Cotrollg Slederess: (.0)( 0' " )( 1 ' ) KL = = r.89" x ( 1.0)( 0' " )( 1 ' ) KL = = Bucklg about the -axis cotrols r 0.577" CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 8 of 10

12 Colum Desig Stregth: Compare maximum slederess ratio agast E 9, = 4.71 = F 36 KL E > 4.71, therefore, elastic bucklg govers, ad Fcr = 0.877Fe r F E 4.71 F to determe how the colum will buckle. P = F cr g Fcr = 0.877Fe where π E π ( 9,000) Fe = = = 1.65 KL ( ) r ksi ksi F = = 1.45 cr P = = 9.0 ksi φ P = = 6.11 ksi Local Bucklg: b 10" λ fl = = = 0 t " λ E 9,000 r = 0.45 = 0.45 = 1.77 F 36 The stregth of this colum should be reduced further to accout for local bucklg. CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio Page 9 of 10

13 BONUS 1 (1 pt.): How tall is Cole ldrich? (Must be with +/- 1. of correct aswer) BONUS (1 pt.): What is the state bird of Kasas? CE 56 Sprg 009 Midterm No. 1 - Ope Book Portio 10 Page 10 of