CE 562 Structural Design I Midterm No. 2 Closed Book Portion (25 / 100 pts)

Transcription

1 CE 56 Structural Design I Name: Midterm No. Closed Book Portion (5 / 100 pts) 1. ( pts) List all of the failure modes that should be checked for the followg bearg-tpe connection: P u ½ 7/8" dia bolts, tp. ½ Pu 3" 3" 3" 3" P u 6" P u 3" Bearg Bolt shear Net section fracture Block shear. ( pts) Circle one: Usg a squirter washer, a specific tpe of direct tension dicator, is one method for obtag a snug-tightened / full-tightened state for a bolted connection. 3. (3 pts) What are three parameters that affect the lateral-torsional bucklg capacit of a beam? Brace spacg Beam length Tpe of bracg Material properties Geometr of beam section Residual stresses Beam support conditions. ( pts) When might it be more appropriate to use a slip critical bolted connection as opposed to a bearg-tpe connection? One situation which slip-critical connections are advisable is when a structure is epected to undergo repeated load ccles (fatigue), as ou might fd a bridge. CE 56 Sprg 008 Midterm No. Page 1 of

2 5. (5 pts) What is the purpose of usg C b, the moment gradient coefficient, lateral-torsional bucklg? Must it alwas be used? The lateral torsional bucklg (LTB) equations AISC were determed based on a loadg producg constant moment across a beam combation with simple supports. This can be etremel conservative. Therefore, AISC allows the engeer to crease the bendg moment resistance for LTB b a factor C b, to account for different loadg and support conditions that might not have as severe of an effect. It does not, and should not, alwas be used. For eample, use of C b is prohibited for cantilevered beams and overhangs. 6. (5 pts) Do ou thk that shear strength of a W113 is more likel to be controlled b shear ieldg or shear bucklg? How about a plate girder with 16 1 flanges and a web? Epla our answer. I would epect the W113 to be more likel controlled b shear ieldg, as its web will probabl be prett stock. The plate girder will have a more slender web, and therefore more susceptible to shear bucklg. 7. ( pts) When should web ieldg and/or web cripplg be vestigated? Whenever ou have a beam loaded with large concentrated pot loads, or high forces beg transferred through the flange to the web. 8. ( pts) True or False: A nonsmmetric Z-shape is beg used as a compression member a steel framed buildg. In addition to fleural bucklg and local bucklg, torsional bucklg should be vestigated. False. CE 56 Sprg 008 Midterm No. Page of

3 CE 56 Structural Design I Midterm No. Open Book Portion (70 / 100 pts) Name: 1. (30 pts) A welded, built-up I-shape made out of A57-50 steel is beg used as a fleural member as shown. It is beg bent about its weak ais (-ais). Determe whether the beam is sufficient to support the given loads. You do not need to consider shear, web cripplg, or web ieldg for this problem. Flange PLs ea. 16" 0.75" Y X Web PL " 0.5" P u = 5 kips W u = 0.5 kips / ft 0 ft 0 ft Solution: Compute Section Properties: (7 pots) 1 1 IY = ( 0.75" )( 16" ) + ( " )( 0.50" ) = I 51.5 SY = = = c ( )( 16" ) A = 16" 0.75" + " 0.50" = 36 ( pts) c t ( 7.75" 0.75" ) + ( 5.5" 0.5" ) 7.75" 0.75".15" + 5.5" 0.5" 0.15" = = =.71" measured awa from the PNA CE 56 Sprg 008 Midterm No. Page 1 of 8

4 a = c = t =.71 () = 5. A 36 ZY = ( a) = ( 5." ) = Check Width-to-Thickness Ratios: (7 pots) Flanges (Unstiffened Elements): b b 16" f = = = = 3 t t f ( )( ") E 9,000 p = 0.38 = 0.38 = 9.15 F 50 ke c (3 pts -) rf = 0.95 FL kc = = = htw 8 Check that 0.35 kc OK rf Sce the shape is doubl smmetric, ( 0.577)( 9,000 ) ( 0.7)( 50 ) = 0.95 = 0.78 S 1.00 F 0.7F t L S = = c Sce pf < f < rf the flanges are non-compact. Web (Stiffened Element): (not entirel necessar to check if ou alread referenced Section F6) h " w = = = 8 tw 0.50" E 9,000 pw = 3.76 = 3.76 = 90.6 F 50 E 9,000 rw = 5.70 = 5.70 = F 50 Sce pw > w the web is compact. The section is non-compact. Referencg Table F1.1, Section F6 of the Specification applies. We will have to account for ieldg and compression flange local bucklg. CE 56 Sprg 008 Midterm No. Page of 8

5 Yieldg: (5 pots) Mn = Mp = FZ 1.6FS AISC (F6-1) 3 3 M = M = =, F S = = 5,1 k k n p k k ft Mn Mp,875 = = = 06.3 Compression Flange Local Bucklg: (7 pots) Compression flange local bucklg for sections with noncompact flanges is given b Section.3 the Specification. M = M M 0.7FS n p p rf pf pf ( ) 3 M,875, k =,530.8 Mn k k n = AISC (F6-) n k k k ft φ M = 0.90,531 =,078 = 30 Factored Moment Demand, M u : (7 pots) wl u PL u Mu = + 8 k/ft k ( 0.5 )( 0' ) ( 5 )( 0' ) Mu = + = k ft Conclusion: The compression flange local bucklg strength governs, and the beam has a design strength of 30 k-ft, which is greater than the factored moment demand of 100 k-ft. The beam section is adequate for the given loads. Recognizg that LTB is not a valid failure mode pots CE 56 Sprg 008 Midterm No. Page 3 of 8

6 . (35 pts) Two C130 channels are used as a built-up column as shown. Determe the capacit of the built-up column. The steel is A57-50, and the effective length is 17-ft with respect to all aes. Sce the lacg spacg dimension, a, has not et determed, please determe an appropriate value for this dimension. The lacg is connected to the channels with snug-tightened bolts X Solution: Check Local Bucklg: (5 pts) Y Flanges: Web: b 3.17" f = = = tf 0.501" < r,fl 6.33 E 9,000 r,fl = 0.56 = 0.56 = 13.9 F 50 Sce, local bucklg will not occur. d t 1" 0.501" f web = = = tw 0.510" 1.56 E 9,000 r,web = 1.9 = 1.9 = F 50 Sce web < r, local bucklg will not occur the web. a 10. Compute Section Properties: (5 pts) g A = 8.81 = 17.6 I = I = 16 = 3 I = " 3.17" " = 10.7 I 31 A 17.6 r = = =.7" I 10.7 r = = =.6" A 17.6 / KL 17' 1" ' = = 7.78 r.7" / KL 17' ( 1" ') = = r.6" CE 56 Sprg 008 Midterm No. Page of 8

7 Connector Spacg: (10pts) The maimum connector spacg is given b: a 3 KL ma = ri r ma where: / KL 17' ( 1" ') = = r ma.6" 3 a ma = ( 0.76" ) ( 77.86) =.50" The angle between the lacg and the ais of the member is limited to 60 degrees b the Specification, Section E. tan 30 a = d a = ( 8" ) tan( 30) = 9." Use a = 9 Modified Slenderness Ratio: (6 pts) Sce bucklg will be controlled b the -ais, the connectors ma be placed shear. Therefore, a modified slenderness ratio should be used to account for the additional demand. KL KL a = + r r r m 0 i (AISC Equation E6-1) KL 10" = ( 77.86) + = r 0.76" m Column Design Capacit: ( pts) F KL E Fe Sce = = Fcr = F r F Inelastic bucklg governs. F Fe Fcr = F where CE 56 Sprg 008 Midterm No. Page 5 of 8

8 m ( 9,000) ( 78.96) π E π Fe = = = 5.91 KL r F cr = ( 50 ) = φ P = = 50.6 n kips Conclusion: The design capacit of the column is 503 kips. CE 56 Sprg 008 Midterm No. Page 6 of 8

9 3. Determe the design fleural-torsional capacit of a C130 made of A36 steel loaded uniform compression the light steel frame shown below. You ma assume that all connections are pned, cludg the column base. Please onl solve for the fleural-torsional bucklg capacit. 15' 30 15' KL for the ais of smmetr = (KL) = (KL) = 30 = 360 Fe F ez FeFez H + Fe = 1 1 H ( Fe + Fez ) where: π ( 9,000) ()( 1 360" ) π E Fe = = = 0.65 KL r.9" π w F GJ 11, ( KL z ) Ag + ( r0 ) ( 9,000)( 151 ) ( ) π EC 1 1 ez = + = + F = ez ( ) CE 56 Sprg 008 Midterm No. Page 7 of 8

10 ( + ) Fe = Fe = F = 0.1 e Sce n F 0.F F = F F Fe e cr F cr = ( 36 ) =.75 φ P = = 196. kips CE 56 Sprg 008 Midterm No. Page 8 of 8