Statics and Mechanics of materials- Chapter 10
January 21, 2017 | Author: Matthew Scott | Category: N/A
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F
-.f=Ee*
(1o*loo) = GqtN,nn-{ $t*t**= *(o,atnf Problam10,2
-= 97.3xros Pe 97.? MPa
fRO?[IfT.tnY IilTIRIAL. O ztll ThcMcCtt-lH Ccpir* lF. Afl dtb rxrd. No Ft dtb Menrnlmry bc diilph!64 nnoo*m, r fu|r:l b q Lr q V :l ru, *h p;* lb Fh rft of b l*5rr, r rrd bqila b H ffiitn b Fhors ard €fucr0or!pr*a It Mcf.fl*'-Hill for tr€fuiadividrdcou,tnpryantion. $tudarils rnrngthir melrul$'crurrl it witM pcrmirlbn.
probrem 10s
;iffi;; lfl*J"di$*ljl1JilffiiTfl::;}XffiHffiffi|i;ilH'#1il hown is the same as in part a.
=(*Yf'6)
= o'8in'
t = (t[a't) = i.e in, tJ-c,t)= E(-2, - o-8")= 2.c,sgi,,r oo lL'ft -€ zt 6 o o !1. ;n
(o) T^*=ts" W
= qetlr=; = q.1? ksr'
L*
(b)
Tc
/=
J
c3 = 3I ffi
PfOblem
.\rr = I2'* .+ "
= =
e,oo\ = -(R\(zr Tr (gg t? )
ffl#*%ffi"H.tottd
J
to)frfl
es"/,Jsl',a{l}
S = I.t T*,*=
F
zT IT3E:
1.39666 i,,,3
10.a (a) Determinethe torquethat can be appliedto a solid shaftof 3.6 in. outer diameterwithout exceedingan allowableshearingstressof l0 ksi. (D) Solve pafi a,
10.4
: l
.:.
-L d ==
c=+A=
shaftisreplaced byahollowshaftofthesam"*assanaorr.O- :
({X3.6)=
t.g in.
= { (r-e)t= 9-reoq ,.ng (fO)(t.tGoq)= gl.Go?*ip.i", or :f = H= |
( b)
Il"//"w s['of*: fi'r
cr = *d, = (tX 3-e) =
equo,l mcgses *4.
J : T ( a r t - c , u )=
W=
or
= 7.63 Hf .ft
t-B in-
6pqss sec,*r',,no.! a.re.Ls -u"l
A = Trez = Tr(c..- c,. ) ez=ffi=2..sq$ti.,
T=T'
{
b. .|uol.
cr = {c,rc
+q.+6q in+
t?'i.33kip.i,' T : t6.17k,F.ff-
PROPRIETAR'YMATERIAL' o 201I The.McGraw-Hill Companics,Irrc.All rightsreserved. No part of this Manualmay bc displayed,reproduced, or distributedin any.fo11 o1by any means,withou! tlrc prior p.trission of the publisher,or usedbeyondth€ limited distribution to teachprsand educators permittedby McGraw-Hillror theirindividualcou^"",ritlrn pi"p"lttn. studentsusingthis manualareusingit witlnut permission.
PfOblgm
ri
I0.5 (e) For thc 3-in.-diametersolid cylinder rnd loding *rown, detenninethc maximumshoring stress. (D) Determincthc inner dimreter of the hollow rylinder, of 4-in. outerdiurutor, for which the manimum$tr€ssis the sarne8s in part a.
10.5 T,
sh4+ti
T
(b) Hollow sLr*t;
,
c=*Jt= |(ao1 = l,5in.
= -4, = (L)(ryI -- 7,5+sksi (f.s)t Tcg (b)
!,
I
cz=+d = {(to) = z.o i,,.
{ce = $(c.{l- c,t) = T c" t** (2)(tlox3.o') q.2s + inl = = 1 = ?,,01 )or .- JIS. cr = LTrT^*>
ffi
Ct = l- 7+ 315 ir, PfOblem
i
4,= 2C, ?
g , 1 ? i n-
10.6 (a) Determinethe torquethat canbe appliodto a solid shaftof 0.75-in.diameter
10.6
trJiHffff tf;ffitH?$:"*T:i:ffi-"il:1H.:?*:'H*il#,ffiH and with an inncr diameter equal to half of its owll outer diamden.
SOLUTION
(d Sr.0;J sho{t: s = +"1= (*Xo.zs) -- o.3?.fin. |.*,.,= lo !ts,' J'= * c* i { (a-e7s)" = O-ogloig ir,r
T= +
=(o€?l?FX,*-. = o.sask,.p.i,ror sitElL.in <
(b) llo0!o, sh$tt E.
tlne sa-n-e. ur
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