ISMnfor
,Structural and Stress n n n
Analysis
Second Edition (2005)
n
by
Dr. T.H.G. Megson
n n
,Solutions Manual n
S o l u t i o n s n to n C h a p t e r n 2n P r o b l e m s n
S.2.1
(a) Vectorsnrepresentingnthen10nandn15nkNnforcesnarendrawnntonansuitablenscalenasnshow
nninnFig.nS.2.1.nParallelnvectorsnACnandnBCnarenthenndrawnntonintersectnatnC.nT
henresultantnisnthenvectornOCnwhichnisn21.8nkNnatnannanglenofn23.4◦n tonthen15nkNnfo
rce.
B C
10nkN
R
60
u
O 15nkN A FIGUREnS.2.1
(b) Fromn Eq.n (2.1)n andn Fig.n S.2.1
R2n =n152n+n102n+n2n×n15n×n10ncosn60◦
whichngives
Rn =n 21.8nkN
Also,nfromnEq.n(2.2)
tannθ
= 10nsinn 60◦n1
5n+n10ncosn60◦
sonthat
θn =n23.4◦.
3
, 4 • SolutionsnManual
S.2.2
(a) Then vectorsn don notn haven ton ben drawnninn anyn particularn order.n Fig.n S.2.2n showsn th
envectorndiagramnwithnthenvectornrepresentingnthen10nkNnforcendrawnnfirst.
12nkN
8nkN
u 10nkN
20nkN R
FIGUREnS.2.2
ThenresultantnRnisnthennequalnton8.6nkNnandnmakesnannanglenofn23.9◦n tonthennegativ
endirectionnofnthen10nkNnforce.
(b) Resolvingn forcesn inn then positiven xn direction
Fxn=n10n+n8ncosn60◦n−n12ncosn30◦n−n20ncosn55◦n=n−7.9nkNn
Then,nresolvingnforcesninnthenpositivenyndirection
Fyn=n8ncosn30◦n+n12ncosn60◦n−n20ncosn35◦n=n−3.5nkNnTh
enresultantnRnisngivennby
R2n =n (−7.9)2n +n(−3.5)2
sonthat
Rn =n 8.6nkN
Also
3.5
tannθ
=
7.9
whichngives
θn =n 23.9◦.
,Structural and Stress n n n
Analysis
Second Edition (2005)
n
by
Dr. T.H.G. Megson
n n
,Solutions Manual n
S o l u t i o n s n to n C h a p t e r n 2n P r o b l e m s n
S.2.1
(a) Vectorsnrepresentingnthen10nandn15nkNnforcesnarendrawnntonansuitablenscalenasnshow
nninnFig.nS.2.1.nParallelnvectorsnACnandnBCnarenthenndrawnntonintersectnatnC.nT
henresultantnisnthenvectornOCnwhichnisn21.8nkNnatnannanglenofn23.4◦n tonthen15nkNnfo
rce.
B C
10nkN
R
60
u
O 15nkN A FIGUREnS.2.1
(b) Fromn Eq.n (2.1)n andn Fig.n S.2.1
R2n =n152n+n102n+n2n×n15n×n10ncosn60◦
whichngives
Rn =n 21.8nkN
Also,nfromnEq.n(2.2)
tannθ
= 10nsinn 60◦n1
5n+n10ncosn60◦
sonthat
θn =n23.4◦.
3
, 4 • SolutionsnManual
S.2.2
(a) Then vectorsn don notn haven ton ben drawnninn anyn particularn order.n Fig.n S.2.2n showsn th
envectorndiagramnwithnthenvectornrepresentingnthen10nkNnforcendrawnnfirst.
12nkN
8nkN
u 10nkN
20nkN R
FIGUREnS.2.2
ThenresultantnRnisnthennequalnton8.6nkNnandnmakesnannanglenofn23.9◦n tonthennegativ
endirectionnofnthen10nkNnforce.
(b) Resolvingn forcesn inn then positiven xn direction
Fxn=n10n+n8ncosn60◦n−n12ncosn30◦n−n20ncosn55◦n=n−7.9nkNn
Then,nresolvingnforcesninnthenpositivenyndirection
Fyn=n8ncosn30◦n+n12ncosn60◦n−n20ncosn35◦n=n−3.5nkNnTh
enresultantnRnisngivennby
R2n =n (−7.9)2n +n(−3.5)2
sonthat
Rn =n 8.6nkN
Also
3.5
tannθ
=
7.9
whichngives
θn =n 23.9◦.
