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Bar AB .in Problem 32.14 is made of a material that will safely
support a normal stress (in tension or ompression) of 5 GPa.
Based on this criterion. through what a gle in degrees can bar
CD safely be rotated relative to the po ition shown. i
Fre 0d i r m: C, )———40th—1 §QLQIIONz
0' 5x1 091w»:2 Maximum allowable strain in the material is: a = — = ————9——7 = 0.049
l E 102x10 N / 771'
Maximum allowable change in length for H is: AH = 1.5 = (0.4mX049): 0.0196»: In the diagram, maximum allowable distance d is: d = (AH) (cos 30°) = (0.0196m) (cos 30°)
0 = 0.01697 m
The angle through which bar AB may rotate is: :9 = L—d— = M = :0.049rad
AB : 0.3m
‘\ /sin 60°)
ANS 9:22.53:  ghg"; ——t If an upward force is applied at H that causes bar GH to rotate 41
0.02 degrees in the counterclockwise direction, what are the ~
axial strains in bars AB, DC, and EF'? (You can neglect the
defamation of bar GH.) W:
The vertical displacement at point H is:
0 = (0.02% 80°) (3.14159 radians) = 349 x 1045 radians The vertical displacement of points 8, D and F are: VB = (349 x 10‘E recess} 1:400 mm) = 0.14 mm
vD = (349 x 10’: radians; {800 mm) = 0.28 mm vF = (349 x 10‘5 radians) moo mm) = 0.42 mm
The strains in each of the vertical bars is: ANS a =3 3” = 000035 = i‘jgm’" = 0.00070 = ”"3””! = 0.00105
—. mm _' mm PROBLEM 3 .33 Two aluminum bars (EAL : 10.0 x 105 psi) are attached to a rtgid support at the tett and a crossbar at the right. An iron
bar {Egg = 28.5 x 105 psi} is attached to the rigid support at
the ieft and there ts a gap b between the right end of the iron
bar and the crossbar. The crosssecttonal area of each bar
and L = 19 in. The tron bar is stretched until it
contacts the crossbar and welded to it. Afterward, the axial
strain of the iron bar is measured and determined to ‘be is A = 8.5 in2 «53:5 = 6002. What was the size of the gap 13? Free 80:33 tiliagram: SOLUTION: Fm ”‘4”;
Fre< I
E“ ' The toresttottening of the two aluminum bars pins the lengthening of the steel bar must equal the gap, 13. The tangthening of the steel bar is (approximately): Calcutating the force in the steet bar: This same force of: 5,: ~ A b: AL The total engined gap is: ANS b = 9.0485 in WM @3119 in Probtem 33.9, the iron will safely support a tensile stress of 100 ksi and the aluminum wiél safely support a compressive
stress of 40 ksé. What is the largest safe value of the gap b? Free Body Diagram: SOLQTiON: We see from 32 (28.5uuib}(10in))
AE zinsmz it ext 06151;»? 5 + 5;:g = 0.0285 in + 0.02 in ;
ht rpeﬁm—s—j PFE = EEA = (0.002) (28.5 x tos lbllnz} (0.5 inz) = 23,500 lb 5“; = (0.002) (10 in) = 0.02 in is compressing the W0 aluminum bars. The aluminum bars are shortened by an amount = 0.02851): i
Li the F80 that Fpe 3': 2 FM. Maximum allowable toad in the aluminum bars is: (Jim gm. 240,0061b/z'222 = F31
0.5022 Maximum altewabte load in the iron bar ts: {Guam = 100,00010 z‘ w = The maximum atlowaole stresses show that the allowable load in the aluminum is the controlling criterion. 0.51712 " (FF9)MALY ~_) _> (FN )MAX : 20,000 :b (FEM/ax = 50,000 lb Using a toad of 29,980 to tn the aluminum and 40,000 in the iron:
{20,0901bx10in) 5:; ANS bet) a”; + FRI, .088 in Mm A134, .45“, ‘ {0.5in2)(10x306lb/in2) (40,0001b)(10in)
(0.523»?)(22;.5x10611;/m2 ) 112 ”‘2‘ rn {in “‘0‘: 3‘11 3L At S m x = 0 to x = 10 in., the crosssecti‘ nai area of the bar y
‘3'! Problem 34.5 is A =1 inz. From x = 10 in. to x = 20 in., l
A = (0.1 x) inz. The modulus of elasici of the material is
=12 x106 psi. There is a gap b = 0. 2 in. between the
tight end of the bar and the rigid wall. it a 40—kip axial force ard the right is applied to the bar at it = 10 in., what is the »
multing normdl stress l the left half of the bar? :1
20in. b , l ' f——10in———[
0‘3 m P<—i:::l —> P
l A= i in2 /
‘ r——10in—f
P<— >P
'ON' A=(O.1x)in’ / rst determine whether the 40,000lb load is sufﬁcient to close the 0.02—in gap. (40,0001b)(10in) W = 0,033m, which is larger than the 0.02in gap. so there WILL be reaction at
n x m 5 =
the righthand end of the bar. tion of the left'hand portion of the bar is: at = W = 0.03331): — 8.333x10‘7RR
(11;: )(leIO [M m )
, tion of the righthand portion oli the bar is:  RR
2: 12x10’ [ln(20) — mam] = —5.776x10'7 RRm 2" — dex  RR 2&9:
0 6 = “—1—:
" ,3[(0.1xin2)(12x1061b/in) 12;:10‘l expression for total deformation of the bar is: ta + 5x = 0.02m
(0.0333in — 8.333x10‘7RR)— 5.776x10’7 RR = 0.02;: RR = 9,430 lb
RL = 3o,570 lb 1‘ stress in the lefthand portion of the'bar is: a' = _R__42_ ___ 30,57201b lin lin 0' = 30 7o lb/inz 131 ...
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 Spring '06
 Keil

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