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Unformatted text preview: L M 43. use that the bar in problem 43.1 is subjected to a torque
=10.000 in=lb and consists of a material that will safely
pport a maximum shear stress of 40 ksi. Based on this rion, what is the largest distance from the left end of the
at which the torque can safely be applied? a Bed Dia ram: MB
SOLUTION:
The polar moment of inertia for the bar is: J = go“ x %(0.5in)‘ = 0.098%“
.To ﬁnd the maximum allowable moment: 40900113! in2 = 0.09821}:
MM” = 7856 in—ib We see that: MA + M3 = “1000 "F113 the applied moment moves from left—toright, the moment at the righthand and increases. Knowing that
¢A '’ its: JG JG
«10,0009: — lb) — 7856i): ~lb)(14in « LR) _ (7856a  mum)
—"———70——mm _ JG
f LR = 3 in ANS LL = 11 in
E 4 ' The bar is ﬁxeduat both ends. it consists of materiai with shear
modulus G = 28 GPa and has a sotid circular cross section. : PartA is 40 mm in diameter and part B is 20 mm in diameter. H Determine the torques exerted on the bar by the waits. M“ 40mm dEU Free Bod Dia ram izooN—m 160mm 20mm die 120mm SOLUTION: y Polar moments of inertia for the two sections are:
1,, =36“ =35(o_02m)‘ :251x10’gm" J3 =£c‘ =£(0.01m)4 =15.7x10"’m“
2 2 ' 2 2
We see that: M. + M8 = 1.200 N—m [1] (0.16m)(MA) (0.12m)(MB) Since (in. = dig, we also see that: = l251x10“"m“ )5 l15.7x10‘°m‘ i5 MA = 12 MB [21 Solving equations [1] and [2] together:
V ANS ME = 110?] Nm ANS Mg = 92.3 Nm 199 Each bar is 10 in. long and has a solid circular cross section. Bar A has a diameter of 1 in. and its shear modulus is 6 x 106 Est. Bar B has a diameter of 2 in. and its shear modulus is 3.8 x 10 psi. The ends of the bars are separated by a small gap. The free end of barA is rotated 2° about the bar's axis and the bars are welded
together. What are the magnitudes of the angles of twist (in degrees)
of the two bars afterward? Fr B Dirm (“ 10in §QL§TIONz \* ,
Polar moments of inertia for the two bars are: 7;.
JA =2’5r‘ = §(o.5rn)‘ = 0.0932914 JB = gr‘ =%(1m)‘ =1.571m‘ The moment required to produce an angle of twist of two degrees in bar A is: . 4 6  2
M = ﬂ = (0.0349rad!0.0932m !6x10 lib/m i: 2056:.“ N, L lﬂin After the two bars are welded together, each of the welded ends will rotate until equilibrium is achieved. The
total of_the two deﬂection angles will be .2". Because the bars, after welding. are in contact withueach other, the
moments exerted: by each of the bars are equal. We have m = es and MA = M3, so: MA(10in) + MAIOin) =(20{3.14159rad] . :
10.0982in‘i6x10‘1b/in‘i (1.571in‘i33x1061b/m1i 180° ' MA = M3 = 1872 inlb
The angle of twist in each bar is: ANS as; =£4A_LA_= (mum—IMF”) =0.03177rad= .82°
JAGA io.09821'n‘i(6x10 [blinzi  _MBL3 __ (1872in—Ib)(10in) ANS "  4 n6  2
JBGB i1.571m R3.3x10 Ib/m i = 0.003.14rad : 0.18D ...
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This note was uploaded on 11/18/2010 for the course ENGR 213 taught by Professor Keil during the Spring '06 term at Oregon State.
 Spring '06
 Keil

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