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155_Mechanics Homework Mechanics of Materials Solution

155_Mechanics Homework Mechanics of Materials Solution - of...

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Unformatted text preview: of point B. SOLUTION Si‘a‘i’fcs 3 Z M; “-3 O Depot-Mn i’n'ova : A IQ Epasi'm Anmp‘l sis a — 4 PM = Eli-33 : (zoowlo )(225 xlo )55 = 45* 9": 710.23Wlo'6 Q»: = 073.62%“? )(7to._2.3wo") = Since CL" [55x.|03N> fo! fink BE y 53 =- I.7r. 9 = l. 322 do" in 2.121 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 X6- mm rectangular cross section and made of a mild steel that is assumed to be elastoplastie with E = 200 GPa and as = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection 2.122 Solve Prob. 2.121, knowing that a = 1.76 m and that the magnitude of the force Q applied at B is gradually increased fi'om zero to 135 kN. 9 A = (37.5)(4) = 225' m‘ =- msxio“ m1 n q ' .6 arr skew =W5A=zc.szo°s. swam)" )(2m 9) = £7.88 xlo‘ e 6)., = PA” = slaex‘lo" 9 L33 L0 = 73.2%!0‘ 9 6‘3: "' £25 = 352 Wit)1 6 From S'i’ui‘l'cs Q. = PM“ 12:% 9m 2 P35 + L SOQWPIS = £73.8wo‘ +(ll.‘509‘)(67.88>¢io‘)]9 = l78-CZ’>(/0‘ e 9‘! av“ yieyoiina of Pink BE: 63; '-' 6”Y = 250 VIC: '-' 35‘2MIO‘i Gr 93; = A6} =(225xt0")(250xto‘) = 56.25 xro‘ N Fm... 5+4)?!“ Pm " 7‘33) (Q - PM) '1’ 52.53:“); N - P _ 52.5w? _ c = p __ SID ' f n. W - 233.3¥t0 233 M a. Fawn efluahrc amda 96s of AD 6 -‘ Letssxto‘ = 751-2?xto‘smd 1.76 (Qa- Pug-2.44 P“, = o : 2,64 9 ) $3 :7 IL 7g 6 10‘s.; = (HSxIO‘MIJc 9) me. as wo’ N ielals 68: :36} =' 250- Hpa “ 7 L322 MM " ‘7. F I - . Q t i J r‘ ' r ] __J r““' ...
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