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# HW 2 - The cross~sectionai area of the bar is A z m2 =...

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Unformatted text preview: The cross~sectionai area of the bar is: A z m2 = 7: {6.65 mf «2 9.807854 13'? The average normai stress is {he 308d distribu? : {4009 M/{aeoraaa m2} exam 2. piane F? We see from ihe FBD forces: 193081.591! 243 The prismatic bar has ' SOLUTION: A: Piane Pg, the average name} Ans At Hana P2, the average norma! hemogeneou v materiat is 2800 kgima. Determine the average normai sttess at the ptane P, where x ts the distance from the battens of the bar in meters. STATEGY - Straw a freabody diagram of the part of the bar baiowwthe piane P. Frge bgdy diagram (”93> / A T “It "t it” W The was ght of the cy tinder for any distance x from Ets bottom :5: w: p g 7: t2 x= (288i) kglm ){9 81 raises”) 2: to. 03 m) x w = 7155 x N The stress produced in supporting this weight is: a :— WIA = (177.66 x N)! 7: (0.03 an? ANS <3 = 22,467 x Pa 3 27.5 x kPa PROBLEM 2 1.9 The bar is m de of materiat 14a thick. its width varies iineariy from 2 tn. at its ieft and to 4 in‘ at its right end. if the axial toad P = 200 it}, what is the average normai stress (a) at plane P7; 8)) at plane Pg. Free body diagram 69 \$2 1 r n ; s Mew/F9 808 {to 5:3 / aid!) to r’” 1—; CE 900 b” 911 <——t gees mma I N“ i—o—w-t aft—““1 s [w———-———8 W SOLUTiOﬁ: The width of the material at P3 and P; is: W1 = 2 in + (VS) {2 in) = 246627511 W2 = 2 in + (2/3) {2 in) = 3.333 in The cross—seationa! areas at P1 and P2 are: A = {2.667 in) ( t in) = 2.567 in; A2 = (3.333 mu 1 in) = 3.333 m2 The average normat stress at P1 and P2 are: m = {200 Sb) / (2.66? inz; ANS g1 : 75 983 52=(2Gt>tb)i(3.333in2) ”‘3, TE 'v , ANS gg=BGQsi “Mg... W Figure (2) as a diagram cf the bones and hi seeps muscie of a person ’5 arm supporting a mass ﬁgure {23) is a bi 0273963338! model cf the arjm in which {he biceps mussie AB as represented by a bar with pin supports The suspended mass is am: 2 kg and ﬁne we§ght of the forearm 28 9 N.1f%he cross— Meet Hana area of the ten or: cannecting the biceps \$0 the forearm at A is 28 mm2 wh Us the average nomai siress m the tandem Free Body Diag‘ ram 9238 3 132 I; O 2‘ f 8032 “.71.“ \$ng 2330 mm——-— = «.3: 3:150 ; ‘3" 150 mm~—<—§ mm ‘3 3 {EN ’ 9.82N ‘ SOLUTION: Summing moménis about 330m C: ZMﬁUm-(WQMZNXOBSm 3—{9NX‘375mE- mmizaxze 33333995333) ! ANS gjvgmss“: MPa zoomm«-~§-—35(>mm-f {‘3} MW 1 PROBLEM 2-1 .19 'Fhe force F exerted an the bar is 20% 283 .333: {3%)}. "£339 piane P is para] 23 to the y-z piane andi 33 5 an {mm 22332 ? origin 0. The hafs cross secti onai arse a: P :5 Q 853:: " ' What as the average normal stress m ﬁne bar a: PCP. Free aody Diagram //\\A\ x {73 ‘1 if Gaza-9034 5 \ SOLUUON: The average n crime! stress a: point P .s exerted an! y by the x romponen: a? the am? Eed farce Summing forces in the x-directio ZF= M20533 gamma): 29233 633930 555?: 3‘5} ANS gm, = 39 a in CT) X i”? Us ...
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