682 CHAPTER 11 Columns. Columns with Other Support Conditions

Transcription

1 68 CHTER 11 Columns Columns with Othr Support Conditions Th problms for Sction 11.4 ar to b solvd using th assumptions of idal, slndr, prismatic, linarly lastic columns (Eulr buckling). uckling occurs in th plan of th figur unlss statd othrwis. roblm n aluminum pip column (E 10,400 ksi) with lngth 10.0 ft has insid and outsid diamtrs d in. and d 6.0 in., rspctivly (s figur). Th column is supportd only at th nds and may buckl in any dirction. Calculat th critical load cr for th following nd conditions: (1) pinnd-pinnd, () fixd-fr, (3) fixd-pinnd, and (4) fixd-fixd. d 1 d robs and Solution luminum pip column d 6.0 in. d in. E 10,400 ksi I 64 (d4 d 4 1) 3.94 in ft 10 in. (1) INNED-INNED cr (10,400 ksi)(3.94 in. 4 ) (10 in.) 35 k () FIXED-FREE (3) FIXED-INNED cr 58.7 k 4 cr k (4) FIXED-FIXED cr k roblm Solv th prcding problm for a stl pip column (E 10 Ga) with lngth 1. m, innr diamtr d 1 36 mm, and outr diamtr d 40 mm. Solution Stl pip column d 40 mm d 1 36 mm E 10 Ga I 64 (d4 d 4 1) mm 4 1. m (1) INNED-INNED cr 6. kn () FIXED-FREE (3) FIXED-INNED cr 15.6 kn 4 cr kn (4) FIXED-FIXED cr 4 49 kn

2 SECTION 11.4 Columns with Othr Support Conditions 683 roblm wid-flang stl column (E psi) of W 1 87 shap (s figur) has lngth 8 ft. It is supportd only at th nds and may buckl in any dirction. Calculat th allowabl load allow basd upon th critical load with a factor of safty n.5. Considr th following nd conditions: (1) pinnd-pinnd, () fixd-fr, (3) fixd-pinnd, and (4) fixd-fixd. 1 1 robs and Solution Wid-flang column W 1 87 E psi 8 ft 336 in. n.5 I 41 in. 4 (1) INNED-INNED allow cr n n () FIXED-FREE 53 k (3) FIXED-INNED allow k n (4) FIXED-FIXED allow k n allow 63. k 4 n roblm Solv th prcding problm for a W shap with lngth 4 ft. Solution Wid-flang column W E psi 4 ft 88 in. n.5 I 116 in. 4 (1) INNED-INNED allow cr n 166 k n () FIXED-FREE (3) FIXED-INNED allow k n (4) FIXED-FIXED allow k n allow 41.4 k 4 n

3 684 CHTER 11 Columns roblm Th uppr nd of a W 8 1 wid-flang stl column (E ksi) is supportd latrally btwn two pips (s figur). Th pips ar not attachd to th column, and friction btwn th pips and th column is unrliabl. Th bas of th column provids a fixd support, and th column is 13 ft long. Dtrmin th critical load for th column, considring Eulr buckling in th plan of th wb and also prpndicular to th plan of th wb. W 8 1 Solution Wid-flang stl column W 8 1 E ksi 13 ft 156 in. I in. 4 I 9.77 in. 4 1 XIS 1-1 (FIXED-FREE) cr k XIS - (FIXED-INNED) cr k uckling about axis 1-1 govrns. cr 9 k 1 roblm vrtical post is mbddd in a concrt foundation and hld at th top by two cabls (s figur). Th post is a hollow stl tub with modulus of lasticity 00 Ga, outr diamtr 40 mm, and thicknss 5 mm. Th cabls ar tightnd qually by turnbuckls. If a factor of safty of 3.0 against Eulr buckling in th plan of th figur is dsird, what is th maximum allowabl tnsil forc T allow in th cabls? Stl tub 40 mm Cabl Turnbuckl.1 m Solution Stl tub.0 m.0 m E 00 Ga d 40 mm d 1 30 mm.1 m n 3.0 allow cr n kn 6. kn 3.0 I 64 (d4 d 4 1) 85,903 mm 4 uckling in th plan of th figur mans fixdpinnd nd conditions. cr kn.9 m.0 m.1 m

4 SECTION 11.4 Columns with Othr Support Conditions 685 FREE-ODY DIGRM OF JOINT T T tnsil forc in ach cabl allow comprssiv forc in tub allow T EQUIIRIUM a F vrt 0 allow T.1 m.9 m 0 OWE FORCE IN CES T allow ( allow ) 1.9 m 18.1 kn.1 m roblm Th horizontal bam C shown in th figur is supportd by columns D and CE. Th bam is prvntd from moving horizontally by th rollr support at nd, but vrtical displacmnt at nd is fr to occur. Each column is pinnd at its uppr nd to th bam, but at th lowr nds, support D is fixd and support E is pinnd. oth columns ar solid stl bars (E psi) of squar cross sction with width qual to 0.65 in. load Q acts at distanc a from column D. (a) If th distanc a 1 in., what is th critical valu Q cr of th load? (b) If th distanc a can b varid btwn 0 and 40 in., what is th maximum possibl valu of Q cr? What is th corrsponding valu of th distanc a? 35 in. D Q a C 40 in in in. E 45 in. Solution am supportd by two columns COUMN D E psi 35 in. b 0.65 in.i b in.4 1 cr lb COUMN CE E psi 45 in. b 0.65 in.i b in.4 1 CR 1859 lb (a) FIND Q cr D a IF a 1 in. Q 40 in. C CE D 8 40 Q 7 10 QQ 10 7 D CE 1 40 Q 3 10 QQ 10 3 CE If column D buckls: If column CE buckls: Q cr 600 lb (b) MXIMUM VUE OF Q CR Q 10 7 Q 10 3 oth columns buckl simultanously. D 688 lb CE 1859 lb a F vrt 0Q CR D CE 8150 lb a M 0Q CR (a) CE (40 in.) a CE (40 in.) (1859 lb) (40 in.) Q cr D CE (1859 lb) (40 in.) 9.13 in. 688 lb 1859 lb (688 lb) 8980 lb (1859 lb) 600 lb

5 686 CHTER 11 Columns roblm Th roof bams of a warhous ar supportd by pip columns (s figur on th nxt pag) having outr diamtr d 100 mm and innr diamtr d 1 90 mm. Th columns hav lngth 4.0 m, modulus E 10 Ga, and fixd supports at th bas. Calculat th critical load cr of on of th columns using th following assumptions: (1) th uppr nd is pinnd and th bam prvnts horizontal displacmnt; () th uppr nd is fixd against rotation and th bam prvnts horizontal displacmnt; (3) th uppr nd is pinnd but th bam is fr to mov horizontally; and (4) th uppr nd is fixd against rotation but th bam is fr to mov horizontally. Roof bam ip column d Solution ip column (with fixd bas) E 10 Ga 4.0 m d 100 mmi 64 (d4 d 4 1) mm 4 d 1 90 mm (3) UER END IS INNED (UT NO HORIZONT RESTRINT) (1) UER END IS INNED (WITH NO HORIZONT DISCEMENT) cr 54.7 kn 4 cr kn (4) UER END IS GUIDED (no rotation; no horizontal rstraint) () UER END IS FIXED (WITH NO HORIZONT DISCEMENT) cr kn Th lowr half of th column is in th sam condition as Cas (3) abov. cr 4() 19 kn

6 SECTION 11.4 Columns with Othr Support Conditions 687 roblm Dtrmin th critical load cr and th quation of th buckld shap for an idal column with nds fixd against rotation (s figur) by solving th diffrntial quation of th dflction curv. (S also Fig ) Solution Fixd-nd column v dflction in th y dirction DIFFERENTI EQUTION (EQ.11-3) v M M 0 vk M 0 x UCKING EQUTION.C. 3 v() 0 0 M 0 (1 cos k) cos k 1 and k v k v M 0 GENER SOUTION v C 1 sin kx C cos kx M 0 4 CRITIC OD k 4 4 cr 4 UCKED MODE SHE.C. 1 v(0) 0 C M 0 t dflction at midpoint x v C 1 k cos kx C k sin kx.c. v (0) 0 C 1 0 v M 0 (1 cos kx) y M 0 4 v M 0 1 cos k k M 0 (1 cos ) M 0 M 0 v x 1 cos

7 688 CHTER 11 Columns roblm n aluminum tub of circular cross sction is fixd at th bas and pinnd at th top to a horizontal bam supporting a load Q 00 kn (s figur). Dtrmin th rquird thicknss t of th tub if its outsid diamtr d is 100 mm and th dsird factor of safty with rspct to Eulr buckling is n 3.0. (ssum E 7 Ga.) 1.0 m 1.0 m Q 00 kn.0 m d 100 mm Solution luminum tub End conditions: Fixd-pinnd E 7 Ga.0 m n 3.0 d 100 mm t thicknss (mm) d mm t MOMENT OF INERTI (mm 4 ) I 64 (d4 d 4 1) 64 [(100)4 (100 t) 4 ] HORIZONT EM Q C (1) OWE FORCE allow cr n.046 n MOMENT OF INERTI I n allow.046 E (3.0)(.0 m) (400 kn) (.046)( )(7 Ga) () m mm 4 (3) EQUTE (1) ND (3): 64 [(100)4 (100 t) 4 ] (100 t) mm t mm t min 1. mm a a Q 00 kn comprssiv forc in tub a M c 0a Qa 0 Q Q 400 kn

8 SECTION 11.4 Columns with Othr Support Conditions 689 roblm Th fram C consists of two mmbrs and C that ar rigidly connctd at joint, as shown in part (a) of th figur. Th fram has pin supports at and C. concntratd load acts at joint, thrby placing mmbr in dirct comprssion. To assist in dtrmining th buckling load for mmbr, w rprsnt it as a pinnd-nd column, as shown in part (b) of th figur. t th top of th column, a rotational spring of stiffnss R rprsnts th rstraining action of th horizontal bam C on th column (not that th horizontal bam provids rsistanc to rotation of joint whn th column buckls). lso, considr only bnding ffcts in th analysis (i.., disrgard th ffcts of axial dformations). (a) y solving th diffrntial quation of th dflction curv, driv th following buckling quation for this column: R (k cot k 1) k 0 in which is th lngth of th column and is its flxural rigidity. (b) For th particular cas whn mmbr C is idntical to mmbr, th rotational stiffnss R quals 3/ (s Cas 7, Tabl G-, ppndix G). For this spcial cas, dtrmin th critical load cr. (a) C y R x (b) Solution Column with lastic support at FREE-ODY DIGRM OF COUMN M H EQUIIRIUM a M 0 a M 0M H 0 H M b Ru x DIFFERENTI EQUTION (EQ. 11-3) v v M Hx vk y 0 H v k v b Ru x GENER SOUTION v dflction in th y dirction M momnt at nd angl of rotation at nd (positiv clockwis) M R H horizontal ractions at nds and v C 1 sin kx C cos kx b Ru x.c. 1 v (0) 0 C 0.C. v () 0 C 1 b Ru sin k v C 1 sin kx b Ru x v C 1 k cos kx b Ru

9 690 CHTER 11 Columns (a) UCKING EQUTION.C. 3 v () u u b Ru sin k (k cos k) b Ru Cancl and multiply by : (b) CRITIC OD FOR R 3 3(k cot k 1) (k) 0 Solv numrically for k: k cr k (k) R k cot k R Substitut k and rarrang: b R (k cot k 1) k 0 Columns with Eccntric xial oads Whn solving th problms for Sction 11.5, assum that bnding occurs in th principal plan containing th ccntric axial load. roblm n aluminum bar having a rctangular cross sction (.0 in. 1.0 in.) and lngth 30 in. is comprssd by axial loads that hav a rsultant 800 lb acting at th midpoint of th long sid of th cross sction (s figur). ssuming that th modulus of lasticity E is qual to psi and that th nds of th bar ar pinnd, calculat th maximum dflction and th maximum bnding momnt M max. 1.0 in. = 800 lb.0 in. Solution ar with rctangular cross sction b.0 in. h 1.0 in. 30 in. 800 lb 0.5 in. E psi bh3 I in.4 1 k 1.30 Eq. (11-51): Eq. (11-56): sc k in. k M max sc 1710 lb-in. roblm stl bar having a squar cross sction (50 mm 50 mm) and lngth.0 m is comprssd by axial loads that hav a rsultant 60 kn acting at th midpoint of on sid of th cross sction (s figur). ssuming that th modulus of lasticity E is qual to 10 Ga and that th nds of th bar ar pinnd, calculat th maximum dflction and th maximum bnding momnt M max. 50 mm = 60 kn 50 mm Solution ar with squar cross sction b 50 mm. m. 60 kn 5 mm E 10 Ga k I b mm 4 Eq. (11-51): sc k mm k Eq. (11-56): M max sc.03 ˇm knˇ

10 SECTION 11.5 Columns with Eccntric xial oads 691 roblm Dtrmin th bnding momnt M in th pinnd-nd column with ccntric axial loads shown in th figur. Thn plot th bnding-momnt diagram for an axial load 0.3 cr. Not: Exprss th momnt as a function of th distanc x from th nd of th column, and plot th diagram in nondimnsional form with M/ as ordinat and x/ as abscissa. x M 0 v robs , , and y M 0 Solution Column with ccntric loads Column has pinnd nds. Us EQ. (11-49): v tan k sin kx cos kx 1 From Eq. (11-45): M v M or tan sin x cos 1.707x M 1.16 sin 1.71x cos 1.71x (Not: k and kx ar in radians) M tan k sin kx cos kx ENDING-MOMENT DIGRM FOR 0.3 cr FOR 0.3 cr : From Eq. (11-5): k cr 0.3 M x roblm lot th load-dflction diagram for a pinnd-nd column with ccntric axial loads (s figur) if th ccntricity of th load is 5 mm and th column has lngth 3.6 m, momnt of inrtia I mm 4, and modulus of lasticity E 10 Ga. Not: lot th axial load as ordinat and th dflction at th midpoint as abscissa. Solution Column with ccntric loads Column has pinnd nds. Us Eq. (11-54) for th dflction at th midpoint (maximum dflction): sc 1 R cr (1) DT 5.0 mm 3.6 m E 10 Ga I mm 4 CRITIC OD cr kn

11 69 CHTER 11 Columns MXIMUM DEFECTION (FROM EQ. 1) (5.0) [sc ( ) 1] Units: kn mm angls ar in radians. SOVE EQ. () FOR : arccos 5.0 R () OD-DEFECTION DIGRM (kn) cr (mm) roblm Solv th prcding problm for a column with 0.0 in., 1 ft, I 1.7 in. 4, and E psi. Solution Column with ccntric loads Column has pinnd nds Us Eq. (11-54) for th dflction at th midpoint (maximum dflction): sc 1 R cr (1) SOVE EQ. () FOR : arccos 0. R OD-DEFECTION DIGRM DT 0.0 in. 1 ft 144 in. E psi I 1.7 in. 4 CRITIC OD (kips) cr cr k MXIMUM DEFECTION (FROM EQ. 1) (in.) (0.0) [ˇsc (0.0894) 1ˇ] Units: kips inchs ngls ar in radians. () roblm wid-flang mmbr (W 8 15) is comprssd by axial loads that hav a rsultant acting at th point shown in th figur. Th mmbr has modulus of lasticity E 9,000 ksi and pinnd conditions at th nds. atral supports prvnt any bnding about th wak axis of th cross sction. If th lngth of th mmbr is 0 ft and th dflction is limitd to 1/4 inch, what is th maximum allowabl load allow? W 8 15

12 SECTION 11.5 Columns with Eccntric xial oads 693 Solution Column with ccntric axial load Wid-flang mmbr: W 8 15 E 9,000 psi 0 ft 40 in. Maximum allowabl dflction 0.5 in. ( ) innd-nd conditions nding occurs about th strong axis (axis 1-1) From Tabl E-1: I 48.0 in. 4 CRITIC OD 8.11 in. cr 38,500 lb in. MXIMUM DEFECTION (EQ ) max sc 1 R cr 0.5 in. (4.055 in.) [sc( ) 1] Rarrang trms and simplify: cos( ) arccos (Not: ngls ar in radians) Solv for : 11,300 lb OWE OD allow 11,300 lb roblm wid-flang mmbr (W 10 30) is comprssd by axial loads that hav a rsultant 0 k acting at th point shown in th figur. Th matrial is stl with modulus of lasticity E 9,000 ksi. ssuming pinnd-nd conditions, dtrmin th maximum prmissibl lngth max if th dflction is not to xcd 1/400th of th lngth. = 0 k W Solution k in.1 Column with ccntric axial load Wid-flang mmbr: W innd-nd conditions. nding occurs about th wak axis (axis -). 0 k E 9,000 ksi lngth (inchs) Maximum allowabl dflction ( ) 400 From Tabl E-1: I 16.7 in in..905 in. DEFECTION T MIDOINT (EQ ) sc k 1 (.905 in.) [sc ( ) 1] 400 Rarrang trms and simplify: sc( ) in. 0 (Not: angls ar in radians) Solv th quation numrically for th lngth : in. MXIMUM OWE ENGTH max in. 1.5 ft

13 694 CHTER 11 Columns roblm Solv th prcding problm (W 10 30) if th rsultant forc quals 5 k. Solution Column with ccntric axial load Wid-flang mmbr: W innd-nd conditions nding occurs about th wak axis (axis -) 5 k E 9,000 ksi lngth (inchs) Maximum allowabl dflction ( ) 400 From Tabl E-1: I 16.7 in in..905 in. k in.1 DEFECTION T MIDOINT (EQ ) sc k 1 (.905 in.) [sc( ) 1] 400 Rarrang trms and simplify: sc( ) in. 0 (Not: angls ar in radians) Solv th quation numrically for th lngth : 1.6 in. MXIMUM OWE ENGTH max 1.6 in. 10. ft roblm Th column shown in th figur is fixd at th bas and fr at th uppr nd. comprssiv load acts at th top of th column with an ccntricity from th axis of th column. ginning with th diffrntial quation of th dflction curv, driv formulas for th maximum dflction of th column and th maximum bnding momnt M max in th column. x y (a) (b) Solution Fixd-fr column ccntricity of load dflction at th nd of th column v dflction of th column at distanc x from th bas DIFFERENTI EQUTION (EQ. 11.3) v M ( v)k v k ( v) v k v k () GENER SOUTION v C 1 sin kx C cos kx v C 1 k cos kx C k sin kx.c. 1 v(0) 0 C.C. v (0) 0 C 1 0 v ()(1 cos kx).c. 3 v() ()(1 cos k) or (sc k 1) MXIMUM DEFECTION (sc k 1) MXIMUM ENDING MOMENT (T SE OF COUMN) M max () sc k NOTE: v ()(1 cos kx) (sc k) (1 cos kx)

14 SECTION 11.5 Columns with Eccntric xial oads 695 roblm n aluminum box column of squar cross sction is fixd at th bas and fr at th top (s figur). Th outsid dimnsion b of ach sid is 100 mm and th thicknss t of th wall is 8 mm. Th rsultant of th comprssiv loads acting on th top of th column is a forc 50 kn acting at th outr dg of th column at th midpoint of on sid. What is th longst prmissibl lngth max of th column if th dflction at th top is not to xcd 30 mm? (ssum E 73 Ga.) t b Sction - robs and Solution Fixd-fr column dflction at th top Us Eq. (11-51) with rplacd by : (sc k 1) (1) (This sam quation is obtaind in rob ) SOVE FOR FROM EQ. (1) sc k 1 cos k k arccos 1 k arccos arccos k () NUMERIC DT E 73 Ga b 100 mm t 8 mm 50 kn 30 mm b 50 mm I 1 1 [b4 (b t) 4 ] mm 4 MXIMUM OWE ENGTH Substitut numrical data into Eq. () m 0.65 arccos radians max (.4717 m)( ).1 m roblm Solv th prcding problm for an aluminum column with b 6.0 in., t 0.5 in., 30 k, and E ksi. Th dflction at th top is limitd to.0 in. Solution Fixd-fr column dflction at th top Us Eq. (11-51) with rplacd by : (sc k 1) (1) (This sam quation is obtaind in rob ) SOVE FOR FROM EQ. (1) sc k 1 cos k k arccos 1 k arccos arccos k () NUMERIC DT E ksi b 6.0 in. t 0.5 in. 30 k.0 in. b 3.0 in. I 1 1 [b4 (b t) 4 ] in. 4

15 696 CHTER 11 Columns MXIMUM OWE ENGTH Substitut numrical data into Eq. () in arccos radians max ( in.)(0.9730) in ft roblm stl post of hollow circular cross sction is fixd at th bas and fr at th top (s figur). Th innr and outr diamtrs ar d 1 96 mm and d 110 mm, rspctivly, and th lngth 4.0 m. cabl CD passs through a fitting that is wldd to th sid of th post. Th distanc btwn th plan of th cabl (plan CD) and th axis of th post is 100 mm, and th angls btwn th cabl and th ground ar Th cabl is prtnsiond by tightning th turnbuckls. If th dflction at th top of th post is limitd to 0 mm, what is th maximum allowabl tnsil forc T in th cabl? (ssum E 05 Ga.) C = 4.0 m Cabl = 100 mm d = = D d 1 d Solution Fixd-fr column dflction at th top comprssiv forc in post k Us Eq. (11-51) with rplacd by : (sc k 1) (1) (This sam quation in obtaind in rob ) SOVE FOR FROM EQ.(1) sc k 1 cos k k arccos k Squar both sids and solv for : arccos arccos () MXIMUM OWE COMRESSIVE FORCE Substitut numrical data into Eq. (). allow 13,63 N 13,63 kn MXIMUM OWE TENSIE FORCE T IN THE CE T T Fr-body diagram of joint : a F vrt 0 T sin 0 T sin N 8 NUMERIC DT E 05 Ga 4.0 m 100 mm 0 mm d 110 mm d 1 96 mm I 64 (d4 d 4 1) mm 4