Torsion in Structural Design. 1. Introduction

Transcription

1 Torsion in Structural Design. ntroduction.. Problems in Torsion The role o torsion in structural design is subtle, and comple. Some torsional phenomena include (a) Twist o beams under loads not passing through the shear center (b) Torsion o shats (c) Torsional buckling o columns (d) ateral torsional buckling o beams Two main tpes o situation involve consideration o torsion in design () Member's main unction is the transmission o a primar torque, or a primar torque combined with bending or aial load (Cases (a) and (b) above.) () Members in which torsion is a secondar undesirable side eect tending to cause ecessive deormation or premature ailure. (Cases (c) and (d) above.)

2 .. Development o Torsional analsis- A ew ke contributors 85 - French engineer Adhemar Jean Barre de Saint-Venant presented the classical torsion theor to the French Academ o Science A. Michell and. Prandtl presented results on leural-torsional buckling 95 - S. P. Timoshenko presented a paper on the eects o warping torsion in beams 99 - C. Bach noted the eistence o warping stresses not predicted b classical torsion theor when the shear center and centroid do not coincide H. Wagner began to develop a general theor o leural torsional buckling V. Z. Vlasov (96-958) developed the theor o general bending and twisting o thin walled beams 9 - von Karman and Christensen developed a theor or closed sections (approimate theor) 95 - Benscoter developed a more accurate theor or closed sections. umerous other contributors, these are just a ew highlights.

3 . Uniorm Torsion o Prismatic Sections Consider a prismatic shat under constant twisting moment along its length. Classical theor due to St Venant. Assume Cross-sections do not distort in plane during twisting, so ever point in the section rotates (in plane) through angle φ () about the center o twist. Out o plane warping is not constrained Out o plane warping does not var along the bar The resulting displacement ield is dφ u (, ) d dφ v φ( ) d dφ w φ( ) d

4 n plane displacements v and w are seen rom the igure below

5 Out o plane distortion (warping) o the section is assumed to var with the rate o twist d φ( ) θ constant φ( ) θ d and to be a unction o the position (,) on the crosssection onl. Several models ma be constructed Warping unction model. Conjugate Harmonic unction model St Venant's stress unction model 5

6 Warping Function Model Substituting the displ. ields into the di. eq. o equilibrium rom elasticit, we obtain with b.c. n a n a n (aplace's equation) where n is the normal direction to the boundar, and ( a, a n n ) are the components o the unit normal vector n on the boundar. t can be shown that the St Venant torsional stiness o the section is given b J A da and that the angle o twist is related to the torque b T JGφ or T dφ JG d a result that reduces to the usual polar moment o inertia when the section becomes circular, and the warping unction vanishes. Problem: The eqn. is hard to solve with the b.c given. 6

7 St Venant's Stress Function Model Assume that the non-ero stresses stress unction Ψ (, ) b τ Ψ τ Ψ τ, τ are related to a The unction Ψ (, ) automaticall satisies equilibrium. n order or the resulting displacements to be compatible (i.e. satis continuit) the d.e. Ψ Ψ dφ G Gθ d be satisied, where Gthe shear modulus. The boundar conditions or this model are Ψ(, ) constant on Γ where Γ is the boundar o the section. n man cases, it is convenient to simpl take Ψ on the boundar. Given the stress unction, it can be shown that A Ψ J (, ) dd over the section. 7

8 This d.e. is somewhat easier to solve because o the simpler b.c. We are particularl interested in rectangular sections. Consider such a section, o dimension α β, as shown. t can be shown (using a ev tpe solution) that, or this section Ψ Gφα π n,,5,... ( ) n ( n) / cosh( nπ / α) nπ cos cosh( nπβ / α) α The corresponding stresses are τ 6Gφα απ n,,5,... n ( ) ( n) / cosh( nπ / α) nπ sin cosh( nπβ / α) α τ 6Gφα απ n,,5,... n ( ) ( n) / sinh( nπ / α) nπ cos cosh( nπβ / α) α 8

9 Ψ is shown below or two sections Assume overall dimensions bα, tβ. Assume that b>t. n general, it is useul to write τ ma J d ktg φ d k t Then, since τ ma b T k t τ ma b T, we ma write 9

10 Values o the coeicients are tabulated below or dierent aspect ratios Table Coeicients or Torsion o a Rectangular Section b/t k k k t is seen that or the thin section, the response awa rom the ends is almost independent o. Hence, a simpliied model takes the orm d Ψ d Gθ subject to Ψ( ±β)

11 k k k * This leads to the simple solution ( ) Gθβ with resulting shear stress β Ψ ( β) τ Gθβ / Hence, at t/, τ ma dφ Gtθ Gt d ntegrating this approimate Ψ unction over the area, J bt

12 consistent with the limit o the -D solution as the width to thickness ratio approaches. The shear stresses in a narrow rectangular beam are as shown below Thin-walled open section beams Some thin walled open sections: angles, channels, W and S sections, T sections, etc. Ke result. The section torsional stiness J is approimatel equal to the sum o the J's or the constituent thin walled plates. Thus, J n i b i t i Ke result. The maimum shear stress is estimated as τ ma Tt ma J i ecept that larger stresses ma occur at the corners. Ke result. The shear low around the section, caused b St Venant torsion, is as shown below.

13 That is, the stresses var linearl through the thickness on an o the constituent plates, achieving a maimum on each edge.

14 . Shear Stresses due to bending - thin walled open sections Provides important background needed or analsis o torsion Three items are needed shear stresses and shear low concept shear center sectorial moments These items can be obtained rom a generaliation o the analsis studied in undergraduate Mechanics o Materials. We summarie the results o the analsis below. ormal Stresses and Resultants From generalied analsis o beams, σ M M M M where

15 da A A A da product o da inertia the moments o inertia. From equilibrium o the slice o the section cut above, Q ( s) Q ( s) ( s) q τt V Q Q ( s) V where 5

16 Q da tds, Q da A* ( s) s A*( s) s tds q is the shear low on the section. the shear resultants act through the shear center, there is no twist. Summing moments about the centroid, it can be shown that the coordinates o the shear center are e e n the above, S S Q Q ( s) rds ( s) rds S S Q ( s) d( s) Q ( s) d( s) are the sectorial products o inertia o the section, and r( s) ds d( s) deine the sectorial area (the area shaded in the igure above.) r(s) is the perpendicular distance rom the origin to the orce q ds needed in calculating moments. 6

17 . Bending and torsion o open sections For combined bending and torsion, a number o additional sectorial area properties are needed or the section. η d v ds d w ds rφ et point A undergo displacements v, w, and rotation φ. A dierent point s, located at coordinates (,) displaces b v S w s v ( a w ( a ) φ ) φ This can be resolved into a normal, and a tangential displacement component. The tangential component o displacement is The shear strain in the tangential plane is given b u η γ s (Wagner's assumption) s From which the out o plane displacement is approimatel dv dw dφ u ( s) ( s) ( s) u ( ) d d d n the above eqn. 7

18 ( s) s rds s d( s) is the sectorial area at point s. The resulting normal stress is then σ u d v d w d φ du ( ) d d d d ntegrating the normal stress over the area leads to the aial and bending resultants. These ma be written as du d φ A S d d d v d w M d d d v d w M d d n the above, the irst sectorial moment S ( s) da A d φ d d φ d has been used. The sectorial products o inertia were deined in connection with the shear center., Solving or the displacements, 8

19 du S d φ d A A d d v M M d ( d w d M ( M ) ) d φ d d φ d The coeicients multipling φ" are the distances rom A to the shear center, so i point A is taken at the shear center, these terms disappear, and the equations or v and w reduce to d v d d w d M ( M ( M M ) ) The equation or u " can be written in alternative orm as du d φ d A d where the sectorial area between the outside edge o the section and the sectorial centroid, a point or which S vanishes. the s origin were taken at a sectorial centroid, instead o at an outside edge, then this term would disappear. 9

20 Shear Stresses and Resultants For a section under bending and twisting, the total shear stress at an point is where τ ( ) s τs SV q t ( τ St Venant torsion stresses q t s ) SV shear stresses caused b shear low on the section. ntegrating the twisting moments caused b the shearing stresses, T dφ GJ T d where T is the torque caused b constrained warping induced shear low. t is sometimes convenient to epress this in terms o applied torque per unit length t (). Here, t( ) dt d d φ GJ d dt d the ree edge is taken as the origin o the s coordinate, the shear low is given b s σ q( s) da d

21 Hence, q( s) Q Q V Q Q where W σ ( ) da the bimoment V V C A d d V W σ( ) da d d Q C s A ( ) da the irst sectorial moment A ( ) da the warping constant ntegrating q over the section, and recogniing that the bending related shear low vanishes i the load passes through the shear center (which we ma alwas impose) we obtain, ater some manipulation T C w d φ d rom which the equation or twist o a thin-walled section d φ d φ C w GJ t( ) d d is obtained. ample Consider an beam, simpl supported at each end, ree to warp at the ends, loaded b a twisting moment in the middle. Q

22 The conditions at the let and right end are similar to those at the end o a simpl supported beam in bending. Hence φ() φ"() φ( ) φ"( ) Rather than use the ourth order dierential equation which has discontinuous load at midspan, use the third order dierential equations C C φ''' GJφ ' T / φ''' GJφ ' T / / / < l The homogeneous eqn. C has the solution φ' '' GJφ'

23 φ h( ) C C sinh( λ) C cosh( λ) GJ where λ. Since the r.h.s. o the nonhomogeneous equations are constant, and the lowest C order derivative o φ in the o.d.e. is one, assume a particular solution φ p α Substituting into the o.d.e.'s, GJα T / α T / GJ GJα T / α T / GJ / < Hence, the general solution is given b φ C φ C C C sinh( λ) C sinh( λ) C / cosh( λ) T / GJ cosh( λ) T / GJ / < / The two solutions must be identical at midspan, and b smmetr, φ' ( / ). Consider the let hal o the beam irst. For this portion o the beam, the b.c. at the origin are φ() C φ"() λ C C

24 Hence, C C, and the solution reduces to φ C sinh( λ) T / GJ Then, the smmetr condition at midspan ields φ '( / ) λc cosh( λ / ) T / GJ C T. GJλ cosh( λ/ ) Hence, or the let hal o the beam, T sinh( λ) φ λ JGλ cosh( λ/ ) For the right hand side o the beam, we ma place the origin at the right end, running to the let. Then, it can be shown that the resulting solution (ater converting to the original coordinate sstem) is φ T sinh( λ( )) λ( ) JGλ cosh( λ/ ) et, consider the relative magnitudes o the two components o torsion over the span. We have, or the let hal o the span, T SV T GJφ ' cosh( λ) cosh( λ / )

25 T W C φ''' T cosh( λ) cosh( λ / ) For a tpical W section, the λ values range rom roughl. or a heav column section to less than. or some ver slender beams. Contributions o the warping torsion are shown below or a ' long beam. When combined bending and twist o a beam occurs, it is necessar to include all three components o shear low Shear Flow due to bending St Venant Torsion Warping Torsion to evaluate the total shear stresses. 5

26 The shear low rom bending is calculated in the usual wa, as are the St Venant torsion shear stresses. The shear low rom torsion ma be calculated as q Q V C However, it is more direct to calculate the shear low rom the act that W d φ d C V C d φ d d φ So, q Q provides the easiest calculation, given φ. d How do we calculate Q? B deinition, Q s ( ) da 6

27 S n the integral,, where S A da was deined A earlier. n these integrals, is a continuous unction o position on the section, starting rom at the sectorial origin. n general, calculating these integrals is mess, but we ma oten use a simpler approach. 7

28 ample: Suppose the above problem involves a W section with properties shown. Since the beam is doubl smmetrical, the shear center coincides with the centroid. Taking this point as the origin o sectorial areas, and as the sectorial center, the sectorial areas (taken positive when rotated counterclockwise) are as shown below. The negative and positive contributions cancel, so S. ntegrating the sectorial areas over the langes, (rom the outside edges), the Q unctions are as shown. Within the web, the and - areas rom the langes cancel, so the warping torsion induced shear low vanishes. 8

29 Then, q Q d φ d TQ C cosh( λ) cosh( λ/ ) n the above, C h b t q ma T hb ( ) cosh( λ) cosh( λ / ) The same result can be achieved in a much more intuitive manner. The sketch below illustrates the procedure The lateral translation v introduced into the lange b the rotation φ causes a bending moment unless φ' is constant. 9

30 M v'' V v' '' h φ' '' b t h φ''' This shear produces a torque () h b t h T V φ''' A similar torque develops on the bottom lange, so T T () T () b t h φ''' But the warping torsion is given as T C φ''' C b t h in agreement with the previous result. Within the top lange, bending introduces a shear low

31 VQ q From the above sketch, t b b t Q So ' ' ' ' ' ' φ φ b t h t b b t h t b q

32 For the beam analed previousl, / ) cosh( ) cosh( ''' C T λ λ φ so / ) cosh( ) cosh( ) / cosh( ) cosh( ) / cosh( ) cosh( h b T b T h t b b t h C T b t h q λ λ λ λ λ λ The peak value o warping shear low occurs at the middle o the lange () at the midspan o the beam (/), where h b T q The third part o the sketch on p. TO. re-establishes compatibilit b rotating the langes, and the web about an angle φ about their respective centroids. This introduces the St Venant torsion, as well as a secondar warping torsion caused b the plate bending o the individual lange and web plates. We usuall ignore the secondar warping ecept or certain torsionall weak sections.

33 Secondar Warping: For certain sections, the onl resistance to warping is the secondar warping resistance. Several such sections are shown below The secondar warping torsional resistance can oten be computed or such sections using an analog to the approach used or the lange above. Consider a plate, rotated about its base through an angle φ which varies with. At, v '''( ) φ''' dv s ( ) s ( ) φ''' t d φ''' dt s ( ) dv s ( ) t d φ''' ntegrating over the height o the plate,

34 T s C h s p t t h 6 p d φ''' p t h 6 φ''' All o the cases shown above can be constructed b variations o the above. For eample, a plate o height h rotated about its centroid is equivalent to two plates o h/ rotated about their respective bases. Hence, t ( h / ) C s 6 t h ikewise, the T section has warping stiness h t b C s t 6 and the cruciorm shown has warping stiness t h C s t h 7 (You should calculate the corresponding warping stiness or an angle with unequal legs.) the plates are thin, the warping stiness ma be relativel small. This can have an important eect, as we'll see.

35 Design Approimations or Torsion Analsis using the dierential equation or torsion is complicated, even or simple loads, so various approimations have been introduced. dea: n warping torsion, the langes act like beams bending laterall. A (conservative) approimate analsis might be. convert the torque into couples acting on the langes b dividing b the beam height.. Appl the "equivalent" lateral loads as orces on the langes.. Anale the langes as beams under these loads. 5

36 ample: For the beam analed beore, the torque converts into two lateral loads P T / h applied at the top and bottom langes. The resulting shear is hal o this, on each end V T / h The shear low is thus q VQ / b hb T 6

37 The ma. shear low at is q ma T hb which clearl is an upper bound to the actual shear low. This approach is oten overl conservative, because The normal stresses caused b the bimoment are most important quantitativel, and ma be as large as the bending stresses or a beam in combined bending and torsion The lange shear low predicted b the bending analog will over-estimate the actual warping torsion shear low over much o the beam. Since, in the same wa that V dm d M ( ) M () V ( ξ) dξ We also have V dw W( ) W() V( ξ) dξ d 7

38 For eample, b the beam analog, The beam analog ields T M rom which h σ T b / h However, the actual shear is V T cosh( λ) h cosh( λ / ) Hence, the integrated lateral lange moment is M / T h cosh( λ) d cosh( λ / ) T β h < T h where β is a reduction actor, dependent upon λ The end conditions The speciics o the applied torque S & J give values o β or some common load cases. 8

39 Are the stresses caused b torsion large enough to be a problem? et's see. ample: A W87 beam spanning t. is loaded with a concentrated load o kips at midspan. The load acts '' awa rom the Z ais. The beam is ied at both ends. (Variation on S&J eample.) For the beam, the properties are b t h C (7.65") (.8")(8.7".8") 6,685 in J bt (7.65")(.8") (8.7.6)(.95").9 in [ ] / G ( ν) (.).6 9

40 λ JG C.9 (,685) in. For the ied-ied case, we can show that T φ JGλ [ λ sinh( λ) ( cosh( λ / ) ) ( cosh( λ)) sinh( λ / ) so T ( cosh( λ / ) ) φ ' cosh( λ) sinh( λ) JG sinh( λ / ) Tλ ( cosh( λ/ ) ) φ" sinh( λ) cosh( λ) JG sinh( λ / ) (a) Saint-Venant Torsion τ SV Gt φ' Tt [ cosh( λ) J ( cosh( λ / ))sinh( λ) /sinh( λ/ )] Here, sinh( λ / ) 5. 77, cosh( λ / ). 567, so

41 τ (" k)(.8") [ cosh(.668).88sinh(.668)] (.9 in ) SV For the ied-ied case, the Saint-Venant torsion is ero at both support and centerline, and is maimum at /, where τ SV. ksi (b) Warping Torsion Shear: For the W section, ollowing the above reasoning, τ ma b h 6 φ''' From the solution, ( cosh( λ / ) ) Tλ φ''' cosh( λ) sinh( λ) JG sinh( λ / ) so b ht cosh( λ / ) τ cosh( ) sinh( ) ma C λ λ sinh( / ) λ.556 cosh(.668).88sinh(.668) [ ] τ ma reaches its ma. amplitude at the end, and at the centerline, where the warping restraint is greatest. There τ ( ) τ ( / ).556

42 t is smallest at the quarter points where warping restraint is least. (c) ateral bending stresses introduced b warping torsion The bending stress caused b warping is given b σ d φ d From the above analsis, the largest magnitude stress occurs at the lange tips, where b h /. Thus, b h σ φ" b h λt [ sinh( λ) cosh( λ / ) cosh( λ) /sinh( λ/ G J at the lange tips ( ) ] Substituting the numerical values leads to

43 σ (7.65")(7.66") (.668/ in.)( "k) (.6) (.9 in ) [ sinh(.668).88cosh(.668) ] 8.65[ sinh(.668).88cosh(.668)] The maimum values o σ occur at the ends, and at midspan, where σ 7.9 ksi (compressive on one side o the lange, tensile on the other. The maimum normal stress on one edge o the lange is shown over a hal beam length. Maimum warping restraint eists at the support, and at midspan. (d) Bending analog For this case

44 h 7.66", T "k P kips For a ied-ied beam, ma moments are M P 8 (.65 kips)(88") 8 8.5"k Then, i S l t b 6 (.8")(7.65") in σ equiv 8.5"k in..6 ksi As noted above, this is overl conservative. From table 8.6., the interpolated β value is β (.8)(.68) (.)(.76).696 Then, σ.696(.6 ksi) 7. ksi which is prett close to the maimum value obtained rom the closed orm solution. Observations:

45 The Saint-Venant torsion shear stresses are somewhat larger than the warping torsion shear stresses, but both are relativel small. The warping normal stresses ma be signiicant. (e) To see how signiicant, let's compute the bending stresses M P 8 ( kips)(88") 8 7"k S 7 in σ bending 7"k /7 in 5.67 ksi n this case, the warping normal stresses are larger than the bending stresses! 5

46 6 5. Other Approaches: Calculating φ ma be simpliied b tabulated solutions or certain idealied boundar conditions, or a matri stiness equation φ φ φ φ ' ' a a a a a a a a a a a a a a a a JG W T W T where λ λ λ λ λ ) cosh( ) sinh( a a a a λ λ λ λ λ λ λ λ λ ) sinh( ) sinh( ) cosh( ] ) [cosh( ) sinh( ma be useul. This stiness matri ma be used to replace the torsional stiness submatri φ φ JG T T commonl used in the beam (or rame) equations, and incorporates the end conditions commonl encountered.

47 This approach has the advantage that it can produce reasonable torsional stinesses or applications involving torsional moments caused b eccentricall connected beams (See S&J, section 8.7) The associated solution φ( ) ( ) φ ( ) φ' ( ) φ ( ) φ' can then be used to obtain all derivatives o φ needed to calculate the shear low, where [ a( ) a( cosh( λ)) a sinh( λ)/ λ] / [ a a( cosh( λ)) ( a a)sinh( λ)/ λ] [ a( sinh( λ)/ λ) a( cosh( λ)) ]/ [ a ( cosh( λ) a ( sinh( λ)/ λ] / are the "shape unctions" or the problem. Distributed torques can also be accomodated. / 7

48 S&J suggest that the best approach or determining the applied torque is one that establishes compatibilit between the twisting rotation φ o the beam supporting a torque producing member, and the end bending rotation o the attached element. The note that Appling the shear at the ace o the web underestimates the torsion Appling the shear at the centerline o the connection overestimates the torsion Appling the shear so as to achieve rotational continuit with the end o the beam raming in tends to be about right. 8

49 6. RFD or Torsional Moments Assumption: Torsion stresses add to bending stresses, but lateral torsional instabilit is prevented. i.e. lateral restraint eists, but laterall restraining members ma induce sel-limiting torsion beore restraint becomes eective. Other problems eist as well, or which torsional rotation ma not be sel limited. ASC Philosoph: (consistent with ASD) Assume the limit state occurs when ielding is reached. lastic analsis is applicable. Compute actored stresses using w D, w, etc, e.g. wu.wd. 6w, Once the actored design moment and actored design lateral lange moments are known, the stresses are superpositioned with the limit state. Formall, M b S M S φ b F or, in terms o the equivalent lateral lange moment M b S M S φ b F n the above, φ b. 9resistance actor. 9

50 Since λ is a section propert that inluences the M term, through the β actor, or through the d.e. solution, iterative solution is necessar. Guess a value o λ in the right approimate range. Use the corresponding value o λ, to determine the value o M to be used with the design bending moment. Use the bounds to estimate a section sie. Use the section sie to update the estimated value o λ., etc. ample: An A6 beam with torsionall simpl supported ends must carr two kip loads acting at an eccentricit o 6" (5 Kips D, 5 Kips ) (Variation on S&J eample) P u.(5 kips).6(5 kips) kips Guess w beam 5 pl wu.(5) 8 pl Bending moments: 5

51 Torsion: T u ( 6")( kips) 8"k. Using beam analog P Tu / h 8" k / h M P / (8"k)(")/ h 5,9 in k / h The β reduction actor depends upon λ. Guess λ. 5, (or a majorit o W sections,. < λ <. ) Thus, λ. 5(") 6.8 β. 6 5

52 The β value or the torques in the middle o the beam aren't eactl known, but the estimates tend to be somewhat conservative. Guessing a " beam, the reduced lange moment is M.6(5,9 in - k)/" 667"k For a midsied W, S / S.6, taking both langes into account, so 67"k 667"k S S /.6 S 5in.9(6 ksi) Tr W59. For this section, S /. 6, so the guess S is o.k. or now. Also, or this section λ C GJ 68. (p. -8 o ASC-RFD) so λ "/ This implies a revised β. 7 ow, 5

53 M So.7(5,9) /.79" 695"k S,67 "k. ksi (695"k)()./.6 57 in The W59 is not quite big enough, since S 5 in. (actuall, it's probabl o.k., in view o the approimations in calculating β, but without a more precise calculation, we can't be sure.) Tr a W76 Revise the moment upward to,7"k to account or the etra dead load. From the tables, λ /6.9 λ / S S / S.6 Revised β. 5 h.9" M. 5(5,9)/.9 65"k ksi <. ksi 8 8/.6 o.k. A W76 is more than adequate here 5

54 7. Presence o Aial oads (Torsional nstabilit) n presence o aial loads, or under certain lateral load conditions, instabilit can occur. n this case, ormulation in the displaced state is necessar. The relevant displacements are v, w and φ. So, the irst equation (or u ) is not changed. Formulation is complicated, (as is the result) so, we'll just present the resulting equations or v, w and φ. The complete equations are nonlinear. A linear version which omits the non-linear terms is presented here. ( [ ( V M l V d v ) d M l dφ ) d q ( e d v d q e d w d d φ )] d 5

55 ( ( [ C M l d w ) d M l dφ V V ) q d d φ GJ d A ( e d w d q e d v d d φ )] d d v d w ( M l e ) ( M e ) V l d d CV C V H C V C M l C dφ t( ) d M l H C dv d W V d φ d dw d n these equations, a number o additional quantities are to be deined. M, M l l Moments caused b transverse loads onl ) ( e e A polar moment o inertia about shear center Three new geometrical quantities are 55

56 H H H A A A ( ) ( ( da ) da )( ) da From these, C C H H H H e e The equations are three simultaneous equations in three unknowns. Although the geometrical properties are constant, the load coeicients M, M, V, V, W V all var with. l l, The d.e. coeicients are variable. Oten, the d.e.s can be simpliied, b taking advantage o special properties o dierent classes o problems. We will want to do this in practice, since the original set o equations is unwield. 56

57 First, assume that the - coordinates are principal coordinate aes. Then, and the eqns reduce to d v d v d φ dφ ( M l e ) V d d d d q d w d w d φ dφ ( M l e ) V d d d d q C d φ GJ d A C M C d v d w ( M l e ) ( M e ) l d d l V V dw d H dφ CV CV V t( ) C d, in addition, the transverse loads q, q are ero, and the transverse moments are constant, then the shears V, V are also ero, and the additional simpliication d v d v d φ ( M e ) l d d d M l H C W dv d d φ d d w d d w d d φ ( M l e ) d 57

58 58 ) ( ) ( ) ( t d w d e M d v d e M d d W d d C H d d M C M C A GJ d d C l l l l φ φ φ is possible. ow, assuming that the bar is subjected onl to an aial compressive load through the centroid, the moments due to transverse loads are ero, the applied torque vanishes, and i the loading is applied so that the bimoment is ero, the equations become φ d d e d v d d v d φ d d e d w d d w d φ φ d w d e d v d e d d A GJ d d C These equations represent a generaliation o the elastic column buckling problem. (ote sign change in because o compression.) Assume a pinned end column or which the appropriate boundar conditions are " " " φ φ w v w v

59 59 Then a solution o the orm ) / sin( ) / sin( ) / sin( w w v v π φ φ π π where,, φ w v are the values o φ,w, v at midspan, will satis the resulting equations provided ) ( φ π e v ) ( φ π e w φ π A GJ C w e v e or, in matri orm φ π π π w v A GJ C e e e e on-trivial solutions eist i the determinant o coeicients is ero. Rather than solve or the general case, let's consider two special cases.

60 (a) The section is doubl smmetric. Then e e, and the equations reduce to π π C π GJ A v w φ The corresponding determinantal equation π π π C GJ A is alread actored, and has the three solutions π P (buckling about ais) π P (buckling about ais) A π C GJ P T (Torsional, or twist buckling) The rd solution corresponds to a new mode o buckling, shown below, or a cruciorm section. 6

61 Dividing b A and noting that p or a doubl smmetrical section ields the critical stress π σ cr C GJ p this critical stress is less than the minor ais uler stress, then torsional buckling will preceed uler buckling o the section, and must be accounted or in design. 6

62 ample: Consider an equal leg cruciorm section with h6",t/". Section properties: th( h t ) (6")(.5")[(6") (.5") ].58 in p 9.6 in A th t (6")(.5") (.5").98 in r r.9" C t h 7 (.5") (6") in 6 J ( h / ) t ht (6")(.5").65 in For uler buckling, both aes have the same, so ( uler) π π (,) 5, 5 σ cr (.9") ( / r) For torsional buckling, 6

63 σ ( Torsion) cr p ( C π GJ ) π [(,)(.688) 9.6,5 8 (,5)(.65)] (The GJ term dominates the torsional buckling equation unless the column is ver short.) ( Torsion) ( uler) Torsion controls i σ < σ, so cr cr 5 5,5 8 < < 75." n a smmetrical section, torsional buckling tends to control onl or short columns. The thinner the walls o the column, relative to the plate widths (Thin plates) the more likel it is that torsional buckling will control. (ncreasing the wall thickness to /8" in the above calculations decreases the length column or which torsion controls to 5".) t ma be easible to ignore the warping torsion term in the above in the practical range o interest. For the torsion buckling, it is appropriate to replace with /( ν ) consistent with dominant plate action. (b) The section is singl smmetrical. the ais is an ais o smmetr, then e, and 6

64 6 φ π π π w v A GJ C e e From which, the determinantal equation is π π π e A GJ C Beore solving this problem, it is useul to substitute the relations P π P π GJ C P A T π Then, the determinantal eqn. takes the orm ( ) ( )( ) T e P P A P This is an important case, as T beams, one section commonl used or compression elements (e.g. truss members) is singl smmetrical. The leading actor is just uler column buckling in the plane o smmetr. The other actor leads to a coupled leural-torsional buckling mode. The ormal solution is

65 65 T T T Ae P P Ae P P P P ) ( ) ( Tpicall, this critical load will be smaller than either the twist buckling load, or the leural buckling mode.

66 Dividing the equation b the area permits the relationship to be written in terms o the critical stresses or the individual buckling modes. σ ( FT ) σ ( ( ) cr σ ( T ) cr ) ( σ ( ) cr σ ( T ) cr Ae cr where ) Ae σ π ( T ) π J σ σ cr ( / r ) cr C ( ν) ( ) cr σ ( T ) cr One can use the latter equation to deine an equivalent radius o gration or torsional buckling. quating π ( / r ) C π J ( ν) ields the resulting equivalent r r C J π ( ν) 66

67 ample: Consider a WT7.5: A 8. in 7 in.5 in r.8" r." e.8" J σ σ.588 in C 6.76 in 7 in.5 in (8.in )(.8") cr, (.") / r ) ( T ) cr 7 in π π 5,7 ( π J C ( ν),77.6, π.76 7 Substituting into the critical stress eqn ields σ ( FT ) cr 55,5 55,5,89,,77 (.) (.).. This result is shown below in two plots. First plot shows the interaction o the critical stresses over a large theoretical range Second plot shows the interaction o the critical stresses over a practical range 67

68 Fleural-Torsional interaction signiicantl reduces the buckling load or this column, etending well into the elastic range! 68

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