HW35 - 10.1 Knowing that the torsional spring at B is of...

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Unformatted text preview: 10.1 Knowing that the torsional spring at B is of constant K and that the bar AB is PROBLEM 10-1 rigid, determine the critical load Pu. W SOLUTION 1.3+ 9 be 'He angle Jamie of bar AB. M=Ke, ersrnefi’ul—e gap/18:0 M-Px=O Ke—PLe=o (K-PL)9 =0 E,= K/L 4 E i 10.2 KnowingthatthcspringatAisofconstantkandthaxthebarABis rigid, PROBLEM m'z determine the critical load Pa. SOLUTION Le‘l’ 9 Le +110 await cLamae a? bar AB. F 7 kX = kl. 5-319 92mm FLme — Px = o P kL‘smecose - PLst = 0 05.343- sxnexe Wt 5059 xi kL‘e —PL9 =0 (RB—P09 = o p“: kL .. 10.11 Determine (a) the critical load for the brass strut, (b) the dimension d for which the aluminum strut will have the same critical load, (c) the weight of the aluminum strut as a percent of the weight of the bus: strut. PROBLEM 10. 11 SOLUTION (a) Brass slur} I = 7’;(2°i(2033 = 13.’.3.~33nt.'«33 Immy : |3, 333 Into" My P _ 1715.15 _ Tr‘gmoxlofibsjssno‘“) cr ‘ L1- ‘ (‘_‘)1 r [3.06 x/o’ N = l3.06 kN .- (5) Aluminum s‘l'ru‘l _ 11251, = TI‘EJdl/Il) Brass Aluminum Pcr- ‘ L1 L2. E = 120 CPa E = 70 CPa p = 5740 kg/m3 p = 2710 kg/m3 ., 123,1} _ (\2\(I3.oen!o‘)(LliL — _ -1 *4 cl - 172E“- - 171(k) x10.) — 27%?»‘(0 m d = 22.7 )‘IO-zm = 22-9mm -— m Y‘ Lal" Y‘ a 1 271 ‘ 227 a C —E— T -—“——— : A :I o _.._.‘_— t '1' ‘ ( 3 mb Tu. Lair,1 (HM-7;) (87qo){2° 10.28 Column AB carries a centric load P of magnitude 72 W. Cables BC and BD pROBLEM 10,23 are taut and prevent motion of point B in the xz plane. Using Euler's formula and a factor of safety of 2.3, and neglecting the tension in the cables, determine the maximum allowable length L. Use E = 200 GPa. SOLUTION W 250): 32.7 L: us.avlo‘w..*= Id.— 4.73wlo‘mM“= P: 72 no" N Ram-"=(ES.)(P)=]6$'.3>~IJ N Bookfiing in XZ—fvpamef L; = ‘O.7L P - we. a - (05W? __ L = J. Ely = 1W'73*’°41 0-7 a, 0.7 Issuer—103 ‘: 10.74 M TIZEI’I Lc 1' 2—!— Pct"= W _ 1 QoouolXfiLfiiiLQli). : I2 08 m "' ‘ 2 155.3XL03 ' 10.29 An axial load P is applied to the l.25-in.-square aluminum bar ABC as shown. When P = 3.8 kips, the horizontal deflection at end C is 0.16 in. Using E = 10.1 x 10" psi, determine (a) the eccentricity e of the load, (1:) the maximum stress in the rod. SOLUTION I = Hi 0-25)“ -‘ 0.203q5in" A :.n-zs‘= 1.5425 in‘ Lg = z in L: = 2L = $0 in, - ELI. _ Tr‘(Ikoto‘)(o.2os~+s) _ p s Pw‘ Let - (30 )1 ‘ 3.1121)!» 11:. ? _ 3:3»10’ _ P“. 3mm“), - 0.4e2u2 (on y...“ = el- 342%)- I] = e[5eo(-'}Jo.*+s‘ng l-1 = e[$ec(l.o75oa)- I] = {.1023 e - _ 0-16 _ e ‘ maze. ' 1.1023 ' O'HS' "‘- 4 (1,) MW; P(e+ y...) = (3.8xlo‘)(o.H51 + 0.14) = 1.15957 Ib-m _ Mc 3.8103 (l.!5‘r57)lo.62$)_ 3 .g \ GM" 311* ‘1— r TTErTZ—s—‘fl‘L 0.203%“ ‘ S'qq’lo 1"" 5'” k" "A ...
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HW35 - 10.1 Knowing that the torsional spring at B is of...

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