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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—wt 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(>mmf {‘3} MW 1 PROBLEM 21 .19 'Fhe force F exerted an the bar is 20% 283 .333: {3%)}. "£339
piane P is para] 23 to the yz 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 Gaza9034
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 xdirectio ZF= M20533 gamma): 29233 633930 555?: 3‘5}
ANS gm, = 39 a in CT)
X i”? Us ...
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 Spring '06
 Keil

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