Strength of Materials

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1 Strngth of Matrials Sssion Column 08 ctur not : ramudiyanto, M.Eng.

2 Strngth of Matrials STBIITY OF STRUCTURE

spcifications E spc ftr ths dsign calculations, may discovr that

3 In th dsign of columns, oss-sctional ara is slctd such that - allowabl strss is not xcdd all - dformation falls within spcifications E spc ftr ths dsign calculations, may discovr that th column is unstabl undr loading and that it suddnly bcoms sharply curvd or buckls.

4 Considr modl with two rods and torsional spring. ftr a small prturbation, K rstoring momnt sin dstabilizing momnt Column is stabl (tnds to rturn to alignd orintation) if K 4K

5 ssum that a load is applid. ftr a prturbation, th systm sttls to a nw quilibrium configuration at a finit dflction angl. sin K 4K sin Noting that sin <, th assumd configuration is only possibl if >.

6 Considr an axially loadd bam. ftr a small prturbation, th systm rachs an quilibrium configuration such that d y dx d y dx M EI EI EI y 0 Solution with assumd configuration can only b obtaind if EI y E r E r

7 Th valu of strss corrsponding to th itical load, rcding analysis is limitd to cntric loadings. r E r E r EI slndrnss ratio itical strss

8 column with on fixd and on fr nd, will bhav as th uppr-half of a pin-connctd column. Th itical loading is calculatd from Eulr s formula, EI E r quivalnt lngth

10 Sampl roblm 1 n aluminum column of lngth and rctangular oss-sction has a fixd nd at B and supports a cntric load at. Two smooth and roundd fixd plats rstrain nd from moving in on of th vrtical plans of symmtry but allow it to mov in th othr plan. a) Dtrmin th ratio a/b of th two sids of th oss-sction corrsponding to th most fficint dsign against buckling. = 0 in. E = 10.1 x 10 6 psi = 5 kips FS =.5 b) Dsign th most fficint oss-sction for th column.

11 SOUTION: Th most fficint dsign occurs whn th rsistanc to buckling is qual in both plans of symmtry. This occurs whn th slndrnss ratios ar qual. Buckling in xy lan: z 1 1 I ba r z z ab, z 0.7 r a 1 3 a 1, y ry b / 1 a 1 Buckling in xz lan: r y I y 1 1 ab ab 3 b 1 r r z y b 1 Most fficint dsign:, z, y r r z 0.7 a 1 a b b / 1 y 0.7 a b 0.35

12 = 0 in. E = 10.1 x 10 6 psi = 5 kips FS =.5 a/b = 0.35 Dsign: r y b 1 FS.55 kips 1500 lbs 0.35b b E r 1500 lbs 0.35b b 0 in b b b 1.60 in. a 0.35b in b psi 1.5 kips b 6 psi

13 Strngth of Matrials ECCENTRIC ODING: THE SECNT FORMU

14 Eccntric loading is quivalnt to a cntric load and a coupl. Bnding occurs for any nonzro ccntricity. Qustion of buckling bcoms whthr th rsulting dflction is xcssiv. Th dflction bcom infinit whn = d y y dx EI y max sc Maximum strss max 1 1 y max r c sc r 1 1 c E r EI

15 max 1 c sc r Y 1 E r

Sampl roblm E 910 6 psi. Th uniform column consists of an 8-ft sction of structural tubing having th oss-sction shown.

16 Sampl roblm E psi. Th uniform column consists of an 8-ft sction of structural tubing having th oss-sction shown. a) Using Eulr s formula and a factor of safty of two, dtrmin th allowabl cntric load for th column and th corrsponding normal strss. b) ssuming that th allowabl load, found in part a, is applid at a point 0.75 in. from th gomtric axis of th column, dtrmin th horizontal dflction of th top of th column and th maximum normal strss in th column.

17 SOUTION: Maximum allowabl cntric load: - Effctiv lngth, 8 ft 16 ft 19 in. - Critical load, EI 6.1 kips psi 8.0 in 19 in - llowabl load, all FS all 6.1 kips 31.1 kips 3.54 in all 31.1kips 8.79 ksi

18 Eccntric load: - End dflction, y m y m sc in sc in. 1 - Maximum normal strss, c m 1 sc r 31.1 kips in 0.75 in in 1.50 in sc m.0 ksi

19 Strngth of Matrials DESIGN OF COUMNS UNDER CENTRIC OD

20 rvious analyss assumd strsss blow th proportional limit and initially straight, homognous columns Exprimntal data dmonstrat - for larg /r, follows Eulr s formula and dpnds upon E but not Y. - for small /r, s is dtrmind by th yild strngth s Y and not E. - for intrmdiat /r, s dpnds on both s Y and E.

21 Structural Stl mrican Inst. of Stl Construction For /r > C c E / r all FS FS 1.9 For /r > C c / 1 r Y Cc FS / r C c 1 / r 8 Cc 3 all FS t /r = C c 1 Y C c E Y

22 luminum luminum ssociation, Inc. lloy 6061-T6 /r < 66: all /r > 66: all ksi / / r r ksi Ma / r / r Ma lloy 014-T6 /r < 55: all /r > 66: all ksi / r / r ksi Ma / r / r Ma

23 Sampl roblm Evaluat slndrnss ratio and vrify initial assumption. Rpat if ncssary. Using th aluminum alloy014-t6, dtrmin th smallst diamtr rod which can b usd to support th cntric load = 60 kn if a) = 750 mm, b) = 300 mm SOUTION: With th diamtr unknown, th slndrnss ration can not b valuatd. Must mak an assumption on which slndrnss ratio rgim to utiliz. Calculat rquird diamtr for assumd slndrnss ratio rgim.

24 For = 750 mm, assum /r > 55 Dtrmin cylindr radius: all 6010 c r N Ma Ma m c/ c mm c cylindr radius r radius of gyration I 4 c c 4 c Chck slndrnss ratio assumption: r c / 750mm mm assumption was corrct d c 36.9 mm

For = 300 mm, assum /r < 55 Dtrmin cylindr radius: 6010 all c 3 c 1.00 mm 1 1.585 N r Ma 0.3 m 1 1.

25 For = 300 mm, assum /r < 55 Dtrmin cylindr radius: 6010 all c 3 c 1.00 mm N r Ma 0.3 m / c Chck slndrnss ratio assumption: r c / 300 mm 1.00 mm assumption was corrct d c 4.0 mm a

26 Strngth of Matrials DESIGN OF COUMNS UNDER N ECCENTRIC OD

cntric load and coupl, max cntric Mc I bnding llowabl strss

27 n ccntric load can b rplacd by a cntric load and a coupl M =. Normal strsss can b found from suprposing th strsss du to th cntric load and coupl, max cntric Mc I bnding llowabl strss mthod: Mc all I Intraction mthod: Mc all cntric all I bnding 1

28 That s for now THNK YOU

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS 00 Th McGraw-Hill Companis, Inc. ll rights rsrvd. T Edition CHTER MECHNICS OF MTERIS Frdinand. Br E. Russll Johnston, Jr. John T. DWolf Columns ctur Nots: J. Walt Olr Txas Tch Univrsit 00 Th McGraw-Hill