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Unformatted text preview: NRE _ 3651 608
aw 8L 6011pth Problem 1 Determine the resultant internal'ioadings acting on the cross seetion
at C of the machine shaft shown in Fig. 1—5a. The shaft is supported
by bearings at A and B, which exert only vertical forces on the shaft. ; 225 N (800 N/m)(0. 150 m) = 120 N 50 mm 50 mm (b)
(a) Fig. 1—5 Soiution
We will solve this'probiern using segment AC of the shaft. Support Reactions. A free~body diagram of the entire shaft is shown
in Fig. 1—51). Since segment AC is to be considered, only the reaction
at A has to be determined. Why? 5+ 2 MB : 0; “Ay(0.400 m) + 120 N(0.125 m) m 225 N(0.100 m) : 0
A = “18.75 N y The negative sign for A), indicates that A}, acts in the opposite sense to
that shown on the freebody diagram.  FreeBody Diagram. Passing an imaginary section perpendicular to
the axis of the shaft through C yields the freebody diagram of segment AC shown in Fig. 1—5c. Equations of Equilibrium. Lszxzo; NC=O Ans.
+T2Fy=0; —18.75N—4ON—VC:0 () ° = ch—sssN Am _L+E MC = 0; MC + 40 N(0.025 in) + 18.75 N(0.2_50 m) = 0
MC : —5.69 N  in An}; Pmkﬁ em 9. The gears attached to the ﬁxed—end steel shaft are subjected to the torques
shown in Fig. 5—20a. If the shear modulus of elasticity is 80 GPa and the
shaft has a diameter of lﬁ'mm, determine the displacement of the tooth P
on gear A. The shaft toms freely within the bearing at B. Ta; = 130 Nm ﬁg: 150 Nm
150 Nm Solution Internal Torque. By inspection, the torques in segments AC, CD, and DE
are different yet constant throughout each segment. Freebody diagrams of
appropriate segments of the shaft along with the calculated internal torques t are shown in Fig. 5—2015. Using the righthand rule and the established Sign
convention that positive torque is directed away frorn the sectioned end of I' the shaft, we have _
TAG = +150Nm TCD = —130Nm TDE = ~170N'm 150N111
These results are also shown 0n the torque diagram, Fig: 540C. Angle of Twist The polar moment of inertia for the shaft is
“i _
3' =13 .003) = Harmoqm“ Applying 516 to each segment and adding the results algebraically,
we have Tm“)
__ TL i (+150 N ' m)(0.4 m)
4“ ’ 75 + (~130 N  m) (0.3 m)
5 s;;.q=t(to‘q}m4[30(109) N/m2)] + (—170N'm)(0.5 m) _r '0 “3‘ C;
 tramp—q) m4[80(109) N/m2)] ' m (C)
Since the answer is negative, by the righthand rule the thumb is directed
toward the end E of the shaft, and therefore gear A will rotate as shown in Fig. 5—20d.
The displacement of tooth P on gear A is
SP 2 qur = {.1161 rad)(100 mm) : lblmm Ans. Remember that this analysis is valid only if the shear stress does not
exceed the proportional limit of the material. Probkm 3 I'Ihe member shown in Fig. 8—5a has a rectangular cross sectiOn. Determine
5 the state of stress that the loading produces at point C. [6.45 kN ' I Solution Intemal Loadings. The support reactions on the member have been
: determined and are shown in Fig. 8—515. If the left segment AC of the mem—
‘ her is considered, Fig. 8—5c, the resultant internal loadings at the section consist of a normal force, a shear force, and a bending moment. Solving, 7‘ N = 16.45 kN V = 21.93 kN M = 32.89 kNm To SJQQOFK’ feaddoﬂﬁ m Cb) Congreef 3:} EEC = 0:. Ax ~ (35‘X125‘ \ch «x 0)
+72 if :0: Ag(thzswﬁ + (gm C13 +5 Engo = 075 margin ’— moss») — Aécsm C3) Normal Force 64.5 MPa CFC: 1.32 MPa 66:63.15 MPa Shear Force Bending Moment (e) ' (0 ﬁg. 8~5 (cont) Stress Components. Normal force. The uniform normalstress distribution acting over the
cross section is produced by the normal force, Fig. 8—5d. At point C, P 16.45 kN
“C i Z i (0.050 m)(0.250 m) _ 1'32 MP3 Shear force. Here the area A’ = 0, since point C is located at the top
of the member. Thus Q 2 FA' 2 0 and for C, Fig. 8—52, the shear stress TC=0 Bending moment. Point C is located at y = C = 125 mm from the
neutral axis, so the normal stress at C, Fig. 8—5f, is Me _ (32.89 kN‘ m)(0.125 m) 0' : — = 63.15 MPa
C [50.050 m)(0.250)3] Superposition} The shear stress is zero. Adding the normal stresses
determined above gives a compressive stress at C having a value of crc : 1.32 MPa + 63.15 MI’a = 64.5 MPa Ans. ' This result, acting on an element at C; is shown in Fig. $55K "55km '
We COL Shown RACKS cu éiamfcr oi \5m (on wkod arc«M: magﬁvbée and orien+c~¥ion crew ‘MO'XXMW nowm
_ glass ark \ooiNr A? 303 WWGr ‘1‘: 4w: MaﬁMgm shear Shes: Mm? 8001b ([4 in.) = 11200 lbin. 3001b 500 [b
)6 800 1b (10 in.) :3000 lbin. “r 500 lb (141m): 7000'lbin
800 lb 500 lb
(13) elation . Fig 36 Mama! Loading; The rod is sectioned through point A. Using the free—
pdy diagram of segment AB, Fig. 8—6b, the resultant internal loadings can ‘der the equal but opposite resultant? acting on AC, Fig. 8260 800 lb 112001bin. _‘ 21.13 ksi
Normal force Shear force Bending moment Bending moment Torsional moment
Combined loading (500 1b) (800 lb) (8000 lbin.) (7000 ibin.) (11200 lbin.)
' (c) (a) (e) m (g) (h) S tress Components. Normal force. The normal—stress distribution is shown in Fig. 86d. For point A, we have
0.604 ksi + 16.90 ksi P 5001b  _ .
0.283 ksi+2Ll3 ksi “A : X : 7“0‘75 my a 283 981 _ 0283 km
Shearforce. The shearstress distribution is shown in Fig. 8456. For point
or 17.5 ksi A, Q is determined from the shaded semicircular area. Using the table
$2M ksi on the inside front cover, we have
_ 4(0.75 in.) 1 , 2 _ 3
Q = y'A' = % —7r(0.75 1n.) 2 0.28131n
so that
Fig 34‘ “on” _ VQ " 8001b(0.2813 in3) T ’ M = 604 si = 0,604 ksi
A It [$740.75 in.)4]2(0.75 in.) p .Bending moments. For the 8000”:  in. component, point A lies on the
neutral axis, Fig. 8~6f, so the normal stress is 014:0 For the 700071b  in. moment, c = 0.75 in., so the normal stress at point
A, Fig. 8—6g, is MC 7000 lb ~ in.(0.75 in.) _ .
2 w = ——1——~—‘—4—— = 21126p51: 21.13 ks:
I [31r(0.751n.) ] Torsional Moment. At point A, p A = c = 0.75 in, Fig. 8—6h. Thus the
shear stress is Te 11 200 lb  in.(0.75 in.) . _
TA = 7 : ———1—w—IT— : 16901 p31 = 16.901991
. [517(035 1n.) ] ,
Superposition. When the above results are superimposed, it is seen that an element of material at A is subjected to both normal and shear stress
components, Fig. 8—611 “A [_._.._.
2 L \/___> mukbi
s1 ,11 :: 3L2 GMax 3 3L1 ‘23}; .6“ =— 993" A
“(1%: a : (it) + (as) \m = 2L0 L ...
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This note was uploaded on 04/11/2008 for the course AAE 352 taught by Professor Chen during the Spring '08 term at Purdue.
 Spring '08
 Chen

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