SOLUTIONS FOR PROBLEMS IN SYSTEM DYNAMICS, 4TH EDITION – COMPLETE ANSWER KEY FOR ALL B-PROBLEMS | OGATA | STEP-BY-STEP SOLUTIONS

All 20 Chapters Covered
m m m




SOLUTION MANUAL
m

, www.konkur.in




Chapterm1

Problemsm1-1mthroughm1-6maremformstudentmresearch.mNomstandardmsolutionsmaremprovided.

1-7 FrommFig.m1-
2,mcostmofmgrindingmtomm0.0005minmism270%.mCostmofmturningmtomm0.003minmism60
%.
Relativemcostmofmgrindingmvs.mturningm =m270/60m=m 4.5mtimes Ans.


1-8 CAm=m CB,

10m+m0.8mPm=m60m+m0.8mPmm0.005mPm2

Pm2m=m50/0.005  Pm=m100mpartsm Ans.


1-9 Max.mloadm=m1.10mPm
Min.maream=m(0.95)2A
mMin.mstrengthm= m0.8
5mS
Tomoffsetmthemabsolutemuncertainties,mthemdesignmfactor,mfrommEq.m(1-1)mshouldmbe

1.10
nm m Ans.
m
m1.43
0.850.95
2
d




1-10 (a)m X1m+mX2:
x1m mx2m m X1m me1m mXm2m me2
error memmmx1mmx2mmmmX1mmXm2m
 me1mme2 Ans.
(b) X1mmX
2:
x1mmx2mmX1mme1mmmXm2mme2

m Ans.
(c) X1mX2: emmmx1mmx2mmmmX1mmXm2mm
me1mme2

x1x2mmmX1mme1mmXm2mme2m
emmx1x2m mX1mXm2m m X1e2m mXm2e1m me1e2
m e em m 

Shigley’sm MED,m10thmedition
Chapterm1mSolutions,mPagem1/
12

, www.konkur.in



m Xm e mXm m Xm m Ans.

m m 1m m

mmm 2 m m
e X
1m 2 2m 1 1m m 2
 X1 Xm2m 




Shigley’sm MED,m10thmedition
Chapterm1mSolutions,mPagem1/
12

, www.konkur.in







(d) X1/X2:
x Xm me Xm m1mem m Xm m 
m 1mm
mm m 1 1mm
mm m 1mmm 1 1m m

x2 Xm2m me2 Xm2m m1 Xm2m 
me2
1
 em m  e m1 mem m Xm  em  em m  e e

1m  m12 mm then m 1 1m mmm1mm m 1m m 1 mm m 2m mmm1mm m 1m m mm m 2m
m m 2m m
m m

 Xm2m  Xm2 m1me2 Xm2m   X1 m  Xm2 X1 Xm2
x X X m e e   m 
Thus, emmmmm 1m m mm m 2m m m
m 1mm m m 1mm m m 1m
Ans.
m m 
x2 Xm2 Xm2m m Xm2m 
X1
 





1-11 (a) x1m= 7 =m2.645m751m311m1
X1m=m2.64 (3mcorrectmdigits)
x2m= 8 =m2.828m427m124m7
X2m=m2.82 (3mcorrectmdigits)
x1m+mx2m=m5.474m178m435m8
e1m=mx1mm X1m=m0.005m751m311m1
e2m=mx2mm X2m=m0.008m427m124m7
em=me1m+me2m=m0.014m178m435
m8 mSumm=mx1m+ mx2m=mX1m+ mX

2m+me
=m2.64m+m2.82m+m0.014m178m435m8m=m5.474m178m435m8 Checks
(b) X1m=m2.65,m X2m=m2.83m m (3mdigitmsignificantmnumbers)
e1m=mx1mm X1m=mm0.004m248m688m9
e2m=mx2mm X2m=mm0.001m572m875m3
em=me1m+me2m=mm0.005m821m5
64m2mSumm=mx1m+mx2m=mX1m+
mX2m+me

=m2.65m+2.83mm0.001m572m875m3m=m5.474m178m435m8 Checks


S 
25 103m 
32m1000

1-12 m     dm m1.006m Ans.
in
nd 2.5
md
3
m

TablemA-17: dm=m 14m1m in Ans.

Factor mofmsafet
S
nm  

25 103m  Ans.
y: m4.79

Shigley’sm MED,m10thmedition
Chapterm1mSolutions,mPagem1/
12