HW 10

HW 10 - —-—- _-_. -._ ' ' '\- ——I'I'I 7.11—1 ....

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Unformatted text preview: —-—- _-_. -._ ' ' '\- ——I'I'I 7.11—1 . (a) Tension member gress section strength P” = FyAg == 50(2.50) = 125.0 kips- 95:13., = 0.90025) = 112.5 kips Net seetien strength: U: 1 — ~35 = 1 — = 0.9671 A. = AgU= 2.50(0.9671) = 2. 418 in.2 P... = FHAE = 65(2.418) = 157. 2 kips 6:13,. = 0.75(157.2) = 117. 9 kips Weld strength is Rn = 0.707w(0.6Fs.sr) = 0.707(3/16)(0.-6 x 70) = 5. 568 Rips/in. 95R” = 0.75(5.568) = 4. 176 kips/in. Base metal shear strength of the plate: Yielding: 6R” 2 ¢(O.6Fyr) 2 1.00(0.6)(36)(3/8) = 8. 1 kips/in. :> 4. 176 kips/in. Rupture: 6R” = ¢(0.6Fut) = 0.75(0.6)(58)(3/8) = 9. 788 kips/in. )- 4. 176 kips/in. Base metal shear strength of the tension member: Yielding: 95]?" = ¢(O.6Fyr) = 1.00(0.6)(50)(O. 179) = 5. 37 kipsfiin. 22> 4. 176 kips/in. Rupture: 6R... :- ¢(O.6F..r) == 0.75(0.6)(65)(0. 179) = 5. 236 kips/in. > 4. 176 hips/in. [7-41] The weld strength is smaller than the base metal strength. Total weld strength = 4.176(13 + 13) = 108. 6 kips Weld strength controls overall: PMI = 108. 6 kips. Let 1.2D +1.6(2.5D) = 108.6, Solution is: {D = 20. 88} P = D+L = 20.88 +2.5(20.88) = 73.1kips P = 73.1kips 7.11-5 From AISC Table J2.4. the minimum weld size is 3X16 inoh (based on the angle thickness). Maximum size = 5/16 — 1/16 = 1/4 in. (a) LRFD solution: P... = 1.2D+ 1.6L = 1.2(30) + 1.6(75) = 156.0 kips Try w m ‘A in... 113R” = 1.392 x 4 sisteenths = 5.563 kips/in. The base metal shear yield strength (gusset plate controls) is 0.651.: = 0.6(36)(%) = 3. 1 kips/in. Shear rupture strength is 0.45F..t = 0. 45(53)(%) = 9. 733 kips/in. Base metal shear yield strength for the angles is _ _5__ = . . 0.617.: _ 0.6(50)( 16 x 2) 13. 7s laps/m. and the shear rupture strength is 0.4517“: = 0.45(55)(% x 2) = 21. 94 kips/in. The weld strength of 5.563 hips! in. governs. Both longitudinal and transverse welds will be used. To determine the required length of the longitudinal welds. investigate the two options specified in AISC 12.4(0). First. assuming the same strength for both the longitudinal and transverse welds. t _ 156 = t total requlred length of weld — 5. 563 23. 02 111. ———23-022‘5 == 11. 51 in. For the second option. the strength of the longitudinal welds is O.35(5.563) = 4. 733 kips/in. length of longitudinal welds = and the strength of the transverse weld is 1.5(5.563) = 3. 352 kips/in. The load to be carried by the longitudinal welds is 156 -— 5(3.352) = 114. 2 kips so the required length of the longitudinal welds is 114.2 = - $204.73” 12. 06 1n. The first Option requires shorter longitudinal welds. Try a 5—inch transverse weld and twol2-inoh longitudinal welds. Check the block shear strength of the gusset plate. 21.. = A... = 2 x %(12) = 9.0 int 21...: %(5) = 1. 3751s.2 R... = name... + UbsFuAm = 0.6(53)(9.0)+1.0(53)(1.375) = 422. 0 kips [7-47] with an upper limit of 0.6FyAg1, + UbgFuAm = 0.6(36)(9. 0) + 1.0(58)(1. 875) = 303. 2 kips (eentrels) The design strength is $12” = 0.75(303.2) = 227 kips 2:» 156 kips (OK) 12H—.| i._______.._..._._ l I Use IAt-ifl. fillet welds as shown. sags I-r-‘j—"fl---- 7.11-9 (a) LRFD Solution P...I = 1.2(12)+1.6(36) = 72.0 kips or 72/2 = 36 kipsiangle a __ PH : = ' 2 Req dAgm 0‘90Fy . 03086) 1.11m. :- _ PH = = ' 2 RequE‘ 0.75F. 0.75(53) “231” . __ L _ 16112} _ . 1x«’i[1n.ru—-——-~300 — 300 —O.641n. Try 212%: x 21/2 >< V4, Ag = 1.19in.2 (for one angle) :1 1.11 in.2 (OK) From the properties table for the double-angle section, run-n = 0.764 in. Net section: Assume U m 0. 80 : A. = Ago = 1.19(0.30) = 0.952 in.2 e 0.3231111? (OK) Weld size: min. w = 1/8 in. and max w :— -1~ — —l—— = ~35 in. 4 16 Try w = 118 in., of?” = 1.392 x 2 sixteenths = 2.734 kips/in. The base metal shear yield strength (angle controls) is 0.653.: = owed-Elf) = s. 4 kips/in. Shear rupture strength is 0.4517”! = O.45(58)(—3T) = 6. 525 kips/in. The weld strength of 2.734 kips/in. governs. Required length = fin = 12. 93 in... try two 61/2—in. longitudinal welds. [7—55] Check assumed value of U : U = 1 — = 1 — m = 0. 391 2.- 213311de value Muse (OK) Use ELEV: x 21/2 x 1%, welded with i-in. E70 fillet welds as shown. 61/2" "nun-'fll "nun-m ...
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This note was uploaded on 04/18/2009 for the course CE 456 taught by Professor Kargahi during the Fall '07 term at USC.

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HW 10 - —-—- _-_. -._ ' ' '\- ——I'I'I 7.11—1 ....

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