CE 562 Structural Design I Midterm No. 1 Closed Book Portion (30 / 100 pts)

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1 CE 56 Structural Desig I Name: Midterm No. 1 Closed Book Portio (30 / 100 pts) 1. [ pts / 30] A very sleder colum with perfectly ped eds is subjected to a ever-creasg compressive axial force. The colum has a effective slederess ratio much greater tha 4.71 E F about all axes. The colum is perfectly ideal (perfectly homogeous, perfectly isotropic, y perfectly prismatic, perfectly straight, ad loaded such that absolutely o eccetricity is geerated). Will the perfect colum buckle, or will it crush / yield? Circle oe: Buckle Yield. [ pts / 30] A colum is a oe-story ubraced frame, ad is rigidly fixed at its base. The colum is coected through a hge to a girder at its top ed. Is a value of k = 1.0 coservative whe determg the effective legth of this colum? Circle oe: Yes No 3. [ pts / 30] Some cross-sectio A is comprised oly of stiffeed elemets with idetical legths, b, ad thickesses, t. Aother cross-sectio B, has is comprised oly of ustiffeed elemets of idetical legths, b, ad thickesses, t. If cross-sectios A ad B are subjected to the same uiform compressio, which cross-sectio will be more susceptible to local bucklg? Circle oe: Cross-sectio A Cross-sectio B CE 56 Fall 007 Midterm No.1 Page 1 of 3

2 4. [4 pts / 30] What is the reliability dex, β, ad why is it importat to load ad resistace factor desig? Aswer: The reliability dex is how egeers esure structures rema safe. β>1. Specifically, β is the umber of stadard deviatios the probability of failure is away from the statistical mea. 5. [4 pts / 30] A tesio member coected to a gusset plate with six bolts (A, B, C, D, E, ad F) fails et sectio fracture o failure plae AC. What percetage of the applied force, P, was that failure plae subjected to? Aswer: 5/6 D A F E C B T 6. [5 pts / 30] What are residual stresses, ad why are they importat with relatio to colum desig? Aswer: Residual stresses are stresses that rema a member after it has bee formed to a fished product. They are the result of plastic deformatios, which ca be caused by differet sources, such as: Ueve coolg of hot-rolled shapes Cold-bedg or camberg durg fabricatio Puchg of holes or other cuttg operatios Weldg They are importat to colum desig because their presece a steel sectio meas that some of the resistace capacity has already bee used up. Therefore, uder a give loadg, a steel sectio with residual stresses may reach its capacity before expected, if residual stresses were uaccouted for. CE 56 Fall 007 Midterm No.1 Page of 3

3 7. [6 pts / 30] Two chaels are coected back-to-back, ad used as a tesio member. The tesio member is coected to a colum as show. Whe calculatg the et sectio fracture resistace of the tesio member, should a shear lag reductio factor, U, be cluded? Why or why ot? Aswer: Yes, the shear lag reductio factor should be accouted for because the web of the bedg member is ot attached to the colum. Therefore, the exchage of stress from the beam to colum will ot occur uiformly throughout the coectio. 8. [5 pts / 30] Oe of the uderlyg assumptios behd the developmet of the aligmet charts for (framed) colum effective legth factors is that behavior of the colums is purely elastic. Is this a reasoable assumptio? How does this assumptio affect the desig of real colums? Aswer: It is ot a reasoable assumptio, uless it is accouted for by usg τ A, the elastic reductio factor. Most colums fall with the elastic bucklg portio of the colum desig curve. While assumg all colums will be cotrolled by elastic bucklg is coservative, this ca result high levels of over-coservatism ad is ot geerally cosidered good practice. If a colum is desiged with a k from the aligmet charts, ad is foud to fall with the elastic bucklg portio of the colum desig curve, the he elastic reductio factor ca be used to decrease the k value to retur a more realistic (ad less overly-coservative) colum capacity. CE 56 Fall 007 Midterm No.1 Page 3 of 3

4 CE 56 Structural Desig I Midterm No. 1 Ope Book Portio (70 / 100 pts) Name: C. Beett 1. [35 pts / 70 pts] Give the followg four-story steel-framed structure, determe the desig stregth of colum HI. All steel is A36. Assume that all bucklg ad bedg takes place about all sectios strog axes. All jots are rigid uless otherwise oted o the drawg. DO NOT USE THE COLUMN DESIGN TABLES AS YOUR SOLUTION. Accout for the stiffess reductio factor if appropriate (tabular value may be used). G G H I ( EI L ) = = ( 881 ) ( ) colum EI L ( EI L ) girder ( 881 ) ( ) 1 ( ) + colum = = = EI 070 L 40 girder = Usg the aligmet chart for ubraced frames, k = 1.66 CE 56 Fall 007 Midterm No.1 Page 1 of 11

5 Therefore, the effective legth for colum HI is ( kl ) = = x kl ' 33.' 33.' 1"/' 9,000 = = < 4.71 = r x 6.05" 36 therefore elastic bucklg cotrols. P = FcrAg Fy Fe cr F = F y where π E π ( 9,000) Fe = = = KL ( 65.85) r x ksi CE 56 Fall 007 Midterm No.1 Page of 11

6 ksi F cr = ( 36 ) = 8.65 P = = ksi φ P = = ksi Sce the colum was cotrolled by elastic bucklg ad the aligmet charts were used, the value for φ P is expected to be over coservative. Therefore, τ A should be applied to the effective legth of the colum to obta a more realistic colum capacity. Referecg Table 4-1: P φp u ksi Let = = = 5.79 Ag Ag 4.0 Iterpolatg, τ A = G H ( EI L ) ( )( 881 ) ( )( ) colum = = = 0.71 EI L girder G I ( EI L ) ( )( 881 ) ( )( ) 1 ( ) + colum = = = EI 070 L 40 girder CE 56 Fall 007 Midterm No.1 Page 3 of 11

7 Usg the aligmet chart for ubraced frames, k = 1.37 kl ' 1"/' 9,000 Therefore, = = < 4.71 = , ad elastic bucklg cotrols. r x 6.05" 36 P = F A cr g Fy Fe Fcr = F y where π E π ( 9,000) Fe = = = KL ( 54.35) r x ksi CE 56 Fall 007 Midterm No.1 Page 4 of 11

8 ksi F cr = ( 36 ) = 30.8 P = = ksi φ P = = ksi Coclusio: The desig stregth of colum HI for bucklg about its strog axis is Check Local Bucklg: Flages: b f λ fl = = 5.9 (1-3) tf E 9,000 λ r,fl = 0.56 = 0.56 = ( ) F 36 y Sce λ <λ fl r, local bucklg will ot occur the flages. Web: h λ web = =.4 (1-3) t w E 9,000 λ r,web = 1.49 = 1.49 = 4.9 ( ) F 36 y Sce λ <λ web r, local bucklg will ot occur the web. CE 56 Fall 007 Midterm No.1 Page 5 of 11

9 . [35 pts / 70 pts] The tesio member show below is beg used to support service dead loads of 35, service live loads of 100, ad service wd loads of +/-40. Is the member sufficiet to resist the give loads? The tesio member is 35 log. Ivestigate all pertet block shear coectio (tesio member ad gusset plate) failure modes your aalysis, additio to all other required checks. 5/8" thick gusset plate, A57 Gr. 50 Steel C1 x 5 back-to-back, A588 Gr. 50 Steel 4" " " All holes are for 1" diameter bolts. T 4" ½ 8 3" = '-0" 4 ½ Load Combatios: 1. R = γ Q = 1.4( D+ F = 1.4(35) = 49 u i i ). Ru = γ iqi = 1.( D+ F + T) + 1.6( L+ H) + 0.5( Lr or S or ) 3. Ru = γ iqi = 1.D + 1.6( Lr or S or R) + ( Lr or 0.8W ) 4. R = γ Q = 1.D+ 1.6W + L+ 0.5( L or S or ) R = 1.(35) + 1.6(100) = 0 = 1.(35) + 0.5(100) or 0.8(40) = 9 u i i r R = 5. R = γ Q = 1.D+ 1.0E+ L+ 0.S = u i i 6. R = γ Q = 0.9D+ 1.6W + 1.6H = u i i 7. R = γ Q = 0.9D+ 1.0E+ 1.6H = u i i Gross Sectio Yieldg of the Tesio Member: R = F A = = y g ksi φ = = R Net Sectio Fracture of the Tesio Member: By spectio, the rightmost le of bolts will cotrol NSF. Not oly are they subject to 100% of the force (T), there are three hole diameters without ay staggers. CE 56 Fall 007 Midterm No.1 Page 6 of 11

10 R = FA u e A e = UA et A = holes 0.387" thick chaels 1" + " + " = 1.07 et U = 1 x (Case from Table D3.1 applies) l 0.674" U = 1 =.97 4" A = = e R = = 81.0 ksi φ = = R Block Shear Failure of the Tesio Member: Tesio Member Block Shear Mode 1: R = 0.6FA + U FA 0.6FA + U FA u v bs u t y gv bs u t Gross area tesio, A gt A = 4" chaels 0.387" thick = gt Net area subject to tesio, A t A = A A A t gt holes t = " + " + " 0.387" = Gross area shear, A gv A = 6.5" 0.387" = 41.0 gv Net area subject to shear, A v A = A A A v gv holes v = " + " + " 0.387" = R = 0.6FA + U FA 0.6FA + U FA u v bs u t y gv bs u t CE 56 Fall 007 Midterm No.1 Page 7 of 11

11 R U bs = 1.0 as per Commetary to Ch. J, pg ( ) ( ) ( ) 0.6FA = = u v ksi U F A = = bs u t ksi 0.6FA = = y gv ksi 0.6FA + U FA = = u v bs u t 0.6FA + U FA = = y gv bs u t = R = φ = = R Block Shear Failure of the Gusset Plate: T T R = 0.6FA + U FA 0.6FA + U FA u v bs u t y gv bs u t Gross area tesio, A gt = 4" 0.65" thick =.50 Agt Net area subject to tesio, A t A = A A A t gt holes t = " + " + " 0.65" = Gross area shear, A gv A = 8.5" 0.65" = gv Net area subject to shear, A v A = A A A v gv holes v = " + " + " 0.65" = 9.97 R = 0.6FA + U FA 0.6FA + U FA u v bs u t y gv bs u t U bs = 1.0 as per Commetary to Ch. J, pg CE 56 Fall 007 Midterm No.1 Page 8 of 11

R φ R ( ) ( ) ( ) 0.6FA = 0.6 65 9.97 = 114.58 u v ksi U F A = 1.0 65 1.094 = 71.094 bs u t ksi 0.6FA = 0.6 50 35.65 = 1068.75 y gv ksi 0.6FA + U FA = 114.58 + 71.094 = 113.67 u v bs u t 0.

12 R φ R ( ) ( ) ( ) 0.6FA = = u v ksi U F A = = bs u t ksi 0.6FA = = y gv ksi 0.6FA + U FA = = u v bs u t 0.6FA + U FA = = y gv bs u t = R = = = Tesio Member Block Shear Mode : You could also have vestigated the followg mode of tesio member failure, however, you would have to accout for the area of the flage et tesio. T T Serviceability: L r m ' 1"/' r m ' ( 1"/' ) 1.6" Turg to Table 1-16 for C shapes, ote that there are o tabulated values for a separatio of 5/8. Learly terpolatg betwee the values for 3/8 ad 3/4 of separatio, r m, 3/8 = 1.16 r m, 3/4 = 1.31 Therefore, r m, 5/8 1.6 = CE 56 Fall 007 Midterm No.1 Page 9 of 11

13 Coclusio: Net sectio fracture of the tesio member cotrols, adφ R 616 =. However, the tesio member does ot satisfy serviceability, ad should be resized L such that 300. r m CE 56 Fall 007 Midterm No.1 Page 10 of 11

14 [3 pts. Extra credit] Almost evitably, structors fail to ask you a exam all those thgs for which you so carefully prepared. As it happes, geeratg good questios is almost as difficult as geeratg adequate aswers. Thus, to give you your tur o the pitcher's moud, if you have the time ad the clatio, write a origal exam questio pertet to material that was fair-game to appear o this exam. You will receive betwee 0 ad 3 pots depedg upo the quality of your questio. JUST THE QUESTION PLEASE - DON'T SUPPLY THE ANSWER. CE 56 Fall 007 Midterm No.1 Page 11 of 11

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

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