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w row!» \.,_«_5 WE gr,“ y 10.1 Knowing tha: {he torsional spring at B is Of constant K
bar AB is rigid, determine and thai the
the critical load PU. 19.4 Two rigid bars and BC are connected by a pin at C as shown.
Knowing that the torsional spring at B is of constam K, determine the critical
load PCr far the system. ‘ . £4,463
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a SAMPLE PROBLEM “10.1 An aluminum column of iength L and rectangular cross sectien has a ﬁxed end B and supports a centric load 212A. Two smooth and inunded ﬁxed plates te—
strain end A from moving in nne of the vertical planes of symmetry of the col‘
mm, but allow it to move in the other plane. (a) Determine the ratio (1/1: of
the two sides of the crass section corresponding tn the most‘efﬁcient design
against buckling. (37) Design the most efﬁcient cross section for the column,
knowing that L = 2031., E = 10.1 X 106 psi, P z: 5 kips, and that a factor ef safety of 2,5 is required. SOLUTlON h Buckling in xy Plane. Referring to Fig. 1018, we note that the effec~
live length ’ef the column with respect to buckling in this plane, is L, = 0.71.,
The radius of gyration 'rz of the cross section is obtained by writing I J‘ba3 2 .
and, since I1 "2 Ar?a r3 = = lid—b“ 2 £13 yz = a/Vﬁ The effective slendemess ratie of the column with respeet to buekling in the lane is _
W p is 0.7L _
(1} i} .7 [1/ VI 12 _ Buckling in xz Plane. The effective length of {he column with respect
to buckling in this plane is L: = 2L, and the correspnnding radius of gyration
is r, = b/V’ii, Thus, L m
“i (2). 2‘}; Wm 
n. Mast Efﬁcient Design. The most effigien: design is that for which
the cﬁtical stresses Corresponding tn the two possibie modes of buckling are
equal, Refening to Eq, (1013’), we note that this will be the ease if the two values ebtained above for the effective slendemess ratio are equal, We write 0.7L “ 2L
_ L? a/V. 12 b/x/ii
\f x a a
and, solving fur the ratio a/b, g = j)— ; é 0.35 4 I). Design for Given Dam. Since ES. = 2.5 is required,
Por = (1513,)? = (2.5)(5 Rips) = 125 kips ' Using a = 0,35b; we have A = ab = 0,3552 and 12,500 lb
035:»? Making L = 20in, in Eq. (2), we hava Lc/ry = 138.61%, Substituting for Pet
0“ = :4— = . E, LE/r, and 9'“ inte Eq, (10.13”), we write 'n'ZE _ 1250515 __ «200.1 >< 105m)
(Lg/r)2 ‘ 0.35221 (1386/11):
b = 1.6mm a = 0.355 = “0.567121: «1 l l i s' l E 10.11 A column of effective length L can b: made by gluing together identical planks in either of the arrangements shay/n. Dctcrmizxe the ratic of tha critical load using the arrangement a to the critical load using the arrange ” " mam bk 5
_ v Ii. i L. ’
‘ ' +l l4“ [1/3
i (a) Fig. 3310.11 ;
_.L_;_.c._;~JW.;.:._x.;.. .. .a.w.....vx_;_me4..._._4.;. 10,21 Caiun’m AB emits a centric load P of magnitude 15 kips. Cables
BC and 82) are tau: and pfavent motion 0f point B. in the .152: pianc, Using Buicr’s
formula and a factor of safety 0f’2.2, and neglecting the tension in the 035353,
determine the maximum allowable length L. Use E = 29 X 205 psi. ‘ g”: ” 3 L15... 11,. .3, .. L‘i’ ._r..,..___41_._ p.” y g N I t I x'
i
i
1 “MMMMMW¢Tﬂmw_ ;L+jx M— W10><22 L1 LQM l $71.;iyuviﬂjﬂ_ r..___,).w. 1...... m.. " €‘g3'r;i_§j%:i;;§;j: I I l'l ‘ w
L fBLEOx of safcsty is the same with raspect to buckling in the xz and ya planes. (b) Using
the ratio found in pan a, design tha cross section of the column so that the factor of safety will be 3.0 when P = 4.4 kN, L = 1 m, and E = ...
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 Spring '09
 MOORTHY
 Torsion, Buckling, critical load, xy plane, critical load PU

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