Lecture 11 Buckling of Plates and Sections

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1 Leture Bukling of lates and Setions rolem -: A simpl-supported retangular plate is sujeted to a uniaxial ompressive load N, as shown in the sketh elow. a 6 N N a) Calulate and ompare ukling oeffiients orresponding to the four first ukling modes as a funtion of a. ) Disuss the result for the speifi value: a 6 rolem - Solution: a) Calulate and plot the ukling oeffiients orresponding to the four first ukling modes as a funtion of a a K m a m ) For a 6 K m 6 6 m m 4 K

3 rolem -: A h=0mm thik elasti flat ar stiffener is welded to a plate at the ottom. a h N N a) State the oundar onditions around all four edges. ) Calulate the total ukling load N of the stiffener, assuming that a=000mm and =00mm. (Hint: use the graphial solution for k in the notes) ) Determine the length of the ukling half-wave. d) How muh will the ukling load hange if the oundar onditions at the loaded edges are hanged from simpl supported to lamped support? rolem - Solution: a) Boundar ondition for the unloaded edges are: At the ottom, lamped oundar ondition,0 0 w x w' x, 0 0 At the top, free oundar ondition In addition, if loaded edges are: Clamped w0, wa, 0 w' 0, w'' x, 0 w''' x, 0 w'a, 0 Simpl supported

4 w 0, w a, 0 w'' 0, w'' a, 0 At the ottom, lamped oundar ondition,0 0 w x w' x, 0 0 ) Calulate the total ukling load N a 000mm 00mm Using the ukling oeffiient hart on page 4

5 We are onsidering ase D, a 5. Notie that the ases for the loaded edges eing lamped and simpl supported are almost overlapping in ukling oeffiients K. D E 00 K E 5

6 ) Determine the length of the ukling half-wave. If edges are simpl-supported: From the plot in part (), we an see that there are waves in the m-diretion. Wavelength is equal to the length of on side divided the numer of waves in that diretion a 000 mm m If edges are lamped, it is diffiult to read from the hart aout how man waves (m) exists. d) As noted in part (), the ukling oeffiient for simpl supported edges vs. lamped loaded edges is approximatel the same. Thus the effet on the ukling load is insignifiant if the oundar ondition hange from simpl supported to lamped support. 6

7 rolem -: A relativel short retangular prismati ox olumn u uh is sujeted to a uniform axial ompression. Take =40mm, =60mm, h=mm. Then eam is made of an aluminum allo with ield stress of 00Ma a) Calulate the total ukling load of the olumn. ) Consider a square olumn of the same ross-setional perimeter. Whih olumn is stronger: square or retangular? ) lot the load shorting relation for that olumn onsisting pre-ukling phase and post-ukling phase. Bonus: Consider a retangular prismati ox olumn of the same ross-setional perimeter. Whih olumn is stronger: square of retangular? \ rolem - Solution: a) Calulate the total ukling load Using the derivation in leture note page 7 In our ase, E h h K h h h h S E h 4KK Q Use the plot in leture note (see next page), when, K 4. 7 E E N 7

8 For Aluminum allo Bukling stress E 69Ga, 0. 9 E.7u u0.7 u0 8kN Q 0. V 75Ma V 00 Ma 4h This is elasti ukling. ) Crippling load of the olumn Use derivation in page of leture notes 6 where V ult.9 V 8

9 E t E V In our ase 9 t 69 u0 E 69Ga, V 00Ma, u ult.9u0.79u 6.Ma ult V ult 4h 6kN ) We know that in post-ukling phase, Young s modulus redues to half of the pre-ukling phase. E post E pre In a half-wave length, for pre-ukling phase 4 h u pre pre E pre E pre 4E pre h where E pre 69Ga, h mm for post-ukling phase 4 h u post post E post E Epre h pre 9

10 Bonus: For a retangular prismati ox, assume 48mm, =mm, so that 0.67, using the plot K 5 E h h K h E h h h E h K.90 7 E Compare with square prismati ox olumn.70 7 E Square prismati ox olumn is stronger 0

11 rolem -4: Bukling of a ox setion A relativel slender, thin-walled square ox olumn is sujeted to aial ompression. The olumn is simpl supported at its ends. What is the omination of the geometrial parameters (length L, wall thikness h and width of the fledge ) so that the ritial Euler ukling load will e equal to the ritial loal plate ukling load. Explain all the assumptions that ou made in our derivation rolem -4 Solution: The ritial Euler ukling load is S EI Euler l where I h For a slender olumn, a!!, so k 4, ritial plate ukling load is 4k plate D Eh where D Equating two ritial loads plate Euler, this gives us Rearrange the aove equation, we have Finall S E h 4k Eh l Q. Q hl 0..5 hl.5hl

12 rolem -5: Thin walled prismati ox Consider a thin-walled prismati ox struture of length a =, where is the width of the ross-setion. The ox is put in the universal testing mahine and is sujeted to a ompressive load. a) Calulate the ultimate ompressive load of the ox struture. ) Would the ultimate load hange if the memer were twie as long? ) What will the most weight effiient wa to inrease the ultimate load a fator of Would our solution depend on the /t ratio or on the magnitude of the effetive width? rolem -5 Solution: a) Von Karman theor of effetive width eff.9 E t If eff, we have eff eff eff eff.9 E t The ultimate ompressive load is 4t 7.6t E ult ult or eff 4t 4t ult ult ult eff 4t If eff, we have ult 4 t To sum up ult eff 4 t, 4 t, eff eff

13 Note: when, the expression for eomes eff eff ult eff 4t V eff eff 4V t So is ontinuous at ult eff ) The ultimate load is independent of length a. So the ultimate load ult won t hange if memer is twie as long. ) If eff, the most effiient wa is to inrease length until reahes eff If eff, the most effiient wa is to inrease thikness t Reall ult 7.6t EV, inrease t to t will inrease ult a fator of

14 rolem -6: Bukling of a hannel Consider a plain hannel (as supposed to lipped hannel) with sides of equal width ( ). Both loaded edges are simpl supported. The length of the olumn (a) is equal to 5. Consider the three ases shown in the figure elow, all setions are thin-walled. a) Calulate the total ukling load of open setion; figure (). ) Calulate the ukling load if two lips are added figure (). ) Finall, alulate the ukling load of the setion where two lips are welded, meaning that it eomes a square prismati setion; figure (). d) Disuss the results found in (a), () and (). rolem -6 Solution: We will use the aove figure in the following dedutions 4

15 Assumption Upon uniform ompression, the stresses on the adjaent plates (flanges) are the same V =V a) For a plain hannel, use ase Ĺ in the figure, we also know in this ase 4.8H k E o E o where E o 5H, H in our ase Critial stresses on the adjaent plates H, E o 4 k 0.95 D k h k D h Total ukling load is the sum of ritial loads on all plates h V V h We know,, h h h V h h h D k Eh 0.95 Eh 0. 5

16 ) For a lipped hannel, use ase ķ in the figure, we also know in this ase H k.8 7 H H Similar to part a), total ukling load is the sum of ritial loads on all plates D k Eh 4 Eh ) For ox setion k 4k 4 D Eh. d) Let s alulate load-earing per unit volume in a unit perimeter length for eah ase Eh 0. Case a): h h Case ): Case ): Eh 4h 4 4h Eh. 4h 4 4h Eh The ox setion is the stronger while the plain hannel is the weakest. Eh 0.5 Eh 0. 6

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Compression Members Local Buckling and Section Classification

Compression Members Local Buckling and Section Classification Compression Memers Loal Bukling and Setion Classifiation Summary: Strutural setions may e onsidered as an assemly of individual plate elements. Plate elements may e internal (e.g. the wes of open eams