Post on 23-Jan-2022
1
3. S
truc
ture
s
2142
111
Stat
ics,
201
1/2
©D
epar
tmen
tofM
echa
nica
l©
Dep
artm
ent o
f Mec
hani
cal
Engi
neer
ing,
Chu
lalo
ngko
rn U
nive
rsity
Obj
ectiv
es S
tude
nts
mus
t be
able
to #
1
Cou
rse
Obj
ectiv
e
Ana
lyze
stru
ctur
es (t
russ
es a
nd fr
ames
/mac
hine
s) in
equ
ilibr
ium
C
ht
Obj
tiC
hapt
er O
bjec
tives
D
iffer
entia
te c
onne
ctin
g bo
dies
in e
quili
briu
m in
to
fram
es/m
achi
nes
and
truss
es
For 2
D/3
D fr
ames
/mac
hine
s
Sta
te a
ppro
pria
te a
ctio
n/re
actio
ns b
etw
een
join
ts/m
embe
rs a
nd
disa
ssem
ble
the
stru
ctur
e by
dra
win
g FB
Ds
of m
embe
rs a
nd
impo
rtant
join
ts
2
impo
rtant
join
ts
Iden
tify
stru
ctur
es w
ith s
ymm
etry
con
ditio
n, e
xpla
in th
e ef
fect
s of
sy
mm
etry
and
iden
tify
the
equa
l-val
ues
and
zero
-com
pone
nts
due
to s
ymm
etry
A
naly
ze s
truct
ures
for u
nkno
wn
load
s/re
actio
ns b
y ap
prop
riate
FB
Ds
Obj
ectiv
es S
tude
nts
mus
t be
able
to #
2
Cha
pter
Obj
ectiv
esFo
r 2D
/3D
trus
ses
D
escr
ibe
char
acte
ristic
s an
d ap
prox
imat
ion
just
ifica
tion
for
mem
bers
as
2-fo
rce
mem
bers
A
naly
ze fo
r for
ces
in m
embe
rs b
y th
e m
etho
d of
join
ts a
nd
met
hod
of s
ectio
ns w
ith a
ppro
pria
te F
BD
s
Exp
lain
the
2 ca
uses
of z
ero-
forc
e m
embe
rs a
nd d
eter
min
e ze
ro-fo
rce
mem
bers
in tr
usse
sId
tift
ithth
tdi
tid
li
th
3
Id
entif
y tru
sses
with
the
sym
met
ry c
ondi
tion
and
expl
ain
the
effe
cts
of s
ymm
etry
on
the
prob
lem
Con
tent
s
Ana
lyse
s of
Fra
mes
and
Mac
hine
s
A
naly
ses
of T
russ
es
Trus
s M
etho
d of
Joi
nts
M
etho
d of
Sec
tions
4
2
Equi
libriu
m O
bjec
ts
S
o fa
r, yo
u an
alyz
e si
mpl
e rig
id b
odie
s
Dra
w F
BD
A
pply
equ
atio
ns o
f equ
ilibr
ium
W
hat’s
you
are
goi
ng to
do
with
com
plex
sys
tem
s
Dra
w F
BD
of e
ach
mem
bers
Is
olat
ed m
embe
r
Iden
tify
two
and
thre
e-fo
rce
mem
bers
5
A
dd e
xter
nal a
nd s
uppo
rt lo
ads
A
pply
equ
atio
ns o
f equ
ilibr
ium
Fo
r tw
o fo
rce
mem
bers
, als
o de
term
ine
the
inte
rnal
fo
rces
.
Equi
libriu
m S
truc
ture
s
M
embe
rs a
re in
terc
onne
cted
par
ts in
stru
ctur
es.
S
truct
ures
are
any
con
nect
ing
syst
em o
f mem
bers
that
is
built
to s
uppo
rt or
tran
sfer
forc
es a
nd s
afel
y w
ithst
and
the
appl
ied
load
s.
W
e w
ill s
tudy
sta
tical
ly d
eter
min
ate
Fr
ames
& M
achi
nes
T
6
Tr
usse
s
Fram
es a
nd M
achi
nes
C
onsi
der s
truct
ures
of
inte
rcon
nect
ed m
embe
rs
that
do
not s
atis
fy th
e
F&M
tat
doot
sats
yt
ede
finiti
on o
f a tr
uss.
Fr
ames
rem
ain
stat
iona
ry a
nd s
uppo
rt lo
ads.
7
M
achi
nes
are
desi
gned
to
mov
e an
d ap
ply
load
s.
Fram
es &
Mac
hine
s Pr
oced
ures
A
naly
ze th
e en
tire
stru
ctur
e an
d fin
d as
man
y re
actio
n fo
rces
at t
he s
uppo
rts a
s yo
u ca
n.
F&M
A
naly
ze in
divi
dual
mem
bers
D
isas
sem
ble
stru
ctur
e an
d is
olat
e m
embe
rs
Iden
tify
two-
forc
e m
embe
rs
Iden
tify
impo
rtan
t joi
nts
whi
ch c
onne
ct th
ree
or m
ore
mem
bers
/sup
ports
Al
i/
if
i
8
A
pply
act
ion/
reac
tion
forc
es o
n an
y tw
o co
ntac
ting
mem
bers
R
eass
embl
e m
embe
rs to
che
ck fo
r err
ors.
All
inte
rnal
an
d ac
tion/
reac
tion
forc
es m
ust c
ance
l out
.
3
Exam
ple
F &
M1
#1
Det
erm
ine
the
mag
nitu
des
of
horiz
onta
l and
ver
tical
com
pone
nts
of fo
rce
whi
ch th
e pi
n at
Bex
erts
on
F&M
oo
cec
te
pat
ee
tso
mem
ber CB
.
9
Exam
ple
F &
M1
#2 3 eq
uatio
ns, 4
unk
now
nsD
o yo
u re
ally
una
ble
to fi
nd re
actio
ns?
F&M
yy
10
=−
=
=
+−
=
=
+°
−+
°=
0
0
020
00 N
0
0(2
m)(2
000
N)
(3si
n60
m)
(4 m
3co
s60
m)
0
xx
x
yy
y
Cx
y
FA
C
FA
C
MA
A
Exam
ple
F &
M1
#3
Con
side
r FB
D o
f BC
F&M
App
roac
h 1
=
−
=
=
=
+
=
0
(2 m
)(200
0 N
)4
0
1000
NA
ns
0 (200
0N
)0
C
y
y yM
B
B F BC
11
−+
=
=
=
−
=
(200
0 N
)0
1000
N
0
0(1
)
yy
y x xx
BC
C F BC
Exam
ple
F &
M1
#4
Con
side
r FB
D o
f
00
AB
FA
B
F&M
App
roac
h 1
=−
=
=
=°
−°
=
= =
00
1000
N
0(3
sin
60 m
)(3
sin
60 m
)0
1000
/3
N57
7 N
Ans
yy
y
y
Ax
y
x x
FA
B
A
MB
B
B B
12
=−
=
=
=
0
0
1000
/3
N
From
(1),
1000
/3
N
xx
x
x
x
FA
B
A
C
4
Exam
ple
F &
M1
#5
is a
two-
forc
e m
embe
r.AB
F&M
App
roac
h 2
=−
°=
Con
side
r FB
D o
f
0(2
m)(
2000
N)
(4m
)si
n60
0
BC
MAB
13
==
=
=°=
=°=
0
(2 m
)(20
00 N
)(4
m)
sin
600
2000
/3
N is
the
forc
e th
at p
in
exe
rts o
n ,
Hor
izon
tal c
ompo
nent
cos6
057
7 N
Ver
tical
com
pone
ntsi
n60
1000
NA
ns
CM
AB
AB
AB
BBC
AB
AB
Exam
ple
F &
M2
#1
Dis
asse
mbl
ing
the
stru
ctur
e
F&M
14
Exam
ple
F &
M2
#2
Dis
asse
mbl
ing
the
stru
ctur
e
F&M
15
Exam
ple
F &
M2
#3
Isol
ate
the
the
mem
ber ACED
F&M
16
5
Exam
ple
F &
M2
#4
Dis
asse
mbl
ing
the
stru
ctur
e
F&M
17
Exam
ple
F &
M2
#5
Dis
asse
mbl
ing
the
stru
ctur
e
F&M
18
Exam
ple
Hib
bele
r Ex
6-10
#1
F&M
A co
nsta
nt te
nsio
n in
the
conv
eyor
bel
t is
mai
ntai
ned
by u
sing
the
devi
ce s
how
n.D
raw
the
free-
body
di
agra
ms
of th
e fra
me
and
the
cylin
derw
hich
supp
orts
19
the
cylin
der w
hich
sup
ports
th
e be
lt. T
he s
uspe
nded
bl
ock
has
a w
eigh
t of W
.
Exam
ple
Hib
bele
r Ex
6-10
#2
F&M
20
6
Exam
ple
F &
M4
#1
The
mec
hani
sm is
use
d to
wei
gh
mai
l. A
pac
kage
pla
ced
at A
caus
es
the
wei
ghte
dpo
inte
rto
rota
teth
roug
h
F&M
the
wei
ghte
d po
inte
r to
rota
te th
roug
h an
ang
le a
. Neg
lect
the
wei
ghts
of t
he
mem
bers
exc
ept f
or th
e co
unte
rwei
ght a
t B, w
hich
has
a m
ass
of 4
kg.
If α
= 20
°, w
hat i
s th
e m
ass
of th
e pa
ckag
e at
A?
21
Exam
ple
F &
M4
#2
F&M
22
Exam
ple
F &
M4
#3
F&M G
iven
20
Con
side
r FB
D o
f
00
0
020
0
EEC
x
x
AEC
MrC
C
FEF
C
α=
°
==
=
°
23
0co
s20
0
0
00
(1)
xx
yy
y
FEF
C
EF
FW
C
CW
=−
°−=
=
=−
−=
=−
Exam
ple
F &
M4
#4
Con
side
r FB
D o
f
0
BDC
M
F&M
=
°+
°=
°⋅
=−
°
°⋅
0
(0.1
m)s
in20
(4 N
)(0
.1 m
)cos
100
0.4s
in20
Nm
0.1c
os10
mS
ubst
itute
(1)
0.4s
in20
Nm
D
y
yM
gC
gC
gW
24
==
°°
==
°0.1c
os10
m0.
4sin
201.
3892
kg
0.1c
os10
Mas
s o
f is
1.3
9 kg
Ans
gW
mg
m
mA
7
Exam
ple
Hib
bele
r Ex
6-15
#1
Det
erm
ine
the
horiz
onta
l and
ver
tical
com
pone
nts
of
forc
e w
hich
the
pin
at C
exer
ts o
n m
embe
r ABCD
of th
e fra
me
show
n
F&M
fram
e sh
own.
25
FBD
Exam
ple
Hib
bele
r Ex
6-15
#2
=+
Who
le fr
ame
in e
quilib
rium
0M
F&M
=+
−+
==
=→
+
−
==
0
(98
1 N
)(2 m
) (2
.8 m
)0
700.
71 N
0
070
071
N
A
x
x x xx
M
DD F A
DA
FBD
26
=
=↑
+
−
=
=
70
0.71
N
0 981
N0
981
N
x y y yA F A A
FBD
Exam
ple
Hib
bele
r Ex
6-15
#3
App
ly e
quilib
rium
equ
atio
ns
for m
embe
r CEF
F&M
=+
−−
°=
=−
=→
+
0
(98
1 N
)(2
m)
(si
n45
)(1.6
m)
017
34.2
N
0
C
B
B
xM
FF
F
27
−−
°=
=−
=−
=↑
+
−
°−=
=−
=−
cos4
5
012
26.3
N1.
23 k
NA
ns
0 sin
4598
1 N
0
245.
26 N
245
xB
x
y
yB
yCF
C
F
CF
CN
Ans
Exam
ple
Hib
bele
r Ex
6-17
#1
The
sm
ooth
dis
k sh
own
is p
inne
d at
Dan
d ha
s a
wei
ght o
f 20
lb.
Neg
lect
ing
the
wei
ghts
ofth
eot
herm
embe
rde
term
ine
the
F&M N
egle
ctin
g th
e w
eigh
ts o
f the
oth
er m
embe
r, de
term
ine
the
horiz
onta
l and
ver
tical
com
pone
nts
of re
actio
n of
pin
s B
and D.
28
FBD
8
Exam
ple
Hib
bele
r Ex
6-17
#2
=
Equ
ilibriu
m o
f ent
ire fr
ame
0M
F&M
=
−+
==
=
−
=
0
(20
lb)(3
ft)
(3.5
ft)
017
.143
lb
0
0 17
143
lb
A
x
x x xx
M
CC F A
CA
29
=
=
−
=
=
17
.143
lb
0 20 lb
0
20 lb
x y y yA F A A
Exam
ple
Hib
bele
r Ex
6-17
#3
F&M
30
Exam
ple
Hib
bele
r Ex
6-17
#4
Equ
ilibriu
m o
f mem
ber
0
AB
M
F&M
=
−+
=
=
=
−
=
0
()(6
ft)
(3 ft
)0
40 lb
0
0
B
yD
D x xx
M
AN
N F AB
31
==
=
−
+=
=
17
.143
17.1
lbA
ns
0
0
20 lb
Ans
x y yd
y
yB F AN
B
B
Exam
ple
Hib
bele
r Ex
6-17
#5
Equ
ilibriu
m o
f dis
k D
F&M
=
→+
=
=↑
+
−
−=
−−
=
0 0A
ns
0 20 lb
0
40 lb
20 lb
0
x x y Dy
y
F D F ND
D
32
=20
lb
Ans
y
yD
9
Exam
ple
Hib
bele
r Ex
6-17
#6
F&M
33
Exam
ple
Hib
bele
r Ex
6-13
#1
Dra
w th
e fre
e-bo
dy d
iagr
ams
of th
e bu
cket
and
the
verti
cal b
oom
of
the
back
hoe
sho
wn
in th
e ph
oto.
The
buc
ket a
nd it
s co
nten
ts
hi
htW
Nl
tth
iht
fth
b
F&M ha
ve a
wei
ght W
. N
egle
ct th
e w
eigh
t of t
he m
embe
rs.
34
Exam
ple
Hib
bele
r Ex
6-13
#2
F&M
35
Exam
ple
F &
M3
#1
For P
= 15
0 N
squ
eeze
on
the
hand
les
of th
e pl
iers
, de
term
ine
the
forc
e F
appl
ied
by e
ach
jaw
.
F&M
36
10
F &
M E
xam
ple3
#2
F&M
37=
Due
to h
oriz
onta
l sym
met
ry o
f foc
es a
ctin
g on
on ja
ws
and
hand
les:
0
xC
Uni
t in
mm
Exam
ple
F &
M3
#3
F&M
=
−=
=FBD
of t
he u
pper
han
dle
0
(0.0
3 m
)(0
.18
m)
0
6(1
)FB
D o
f the
upp
er ja
w
C
y
y
M
BP
BP
Uni
t in
mm
38
=
−=
==
=0
(0.0
6 m
)(0
.02
m)
03
18
2.7
kN
Ans
A
y
y
M
BF
FB
P
F
Exam
ple
Hib
bele
r SI E
x 6-
13 #
1
Dis
asse
mbl
ing
the
stru
ctur
e.
F&M
39
Exam
ple
Hib
bele
r SI E
x 6-
13 #
2
F&M
40
11
Exam
ple
Mer
iam
Ex
47 #
1
Dis
asse
mbl
ing
the
stru
ctur
e.
F&M
41
Exam
ple
Mer
iam
Ex
47 #
2
F&M
42
Exam
ple
Mer
iam
Ex
47 #
3
F&M
43
Exam
ple
Hib
bele
r Ex
6-20
#1
Dis
asse
mbl
ing
the
stru
ctur
e.
F&M
44
12
Exam
ple
Hib
bele
r Ex
6-20
#2
F&M
45
Exam
ple
Hib
bele
r Ex
6-20
#3
F&M
FB
D o
f lev
er
0(1
i)
(8lb
)(4i
)0
32lb
ABG
MF
F
46
=+
−=
→=
=↑
+°−
°=→
=
=
→+
°+°−
=→
=
0 (1
in)
(8 lb
)(4 in
)0
32 lb
FBD
of p
in
0 si
n60
sin
600
0 co
s60
cos6
00
32
lb
BEA
EA
yED
EG
ED
EG
xED
EG
EA
ED
MF
F
E
FF
FF
F
FF
FF
F
Exam
ple
Hib
bele
r Ex
6-20
#4
F&M
=
+
−
°=
=
FBD
of a
rm
0
(6 in
) +
cos3
0(3
in)
013
.9 lb
C SED
S
DC
M FF
F
47
Exam
ple
Bed
ford
6.1
14 #
1
Dis
asse
mbl
ing
the
stru
ctur
e.
F&M
48
13
Trus
s D
efin
ition
A
fram
ewor
k co
mpo
sed
of
bars
join
ed a
t the
ir en
ds b
y sm
ooth
pins
tofo
rma
rigid
Trus
s
smoo
th p
ins
to fo
rm a
rigi
d st
ruct
ure.
Tr
usse
s ar
e su
ppor
ted
and
load
ed a
t the
ir jo
ints
.
If
we
negl
ectt
hew
eigh
tsof
the
Riv
er K
wai
Brid
ge
49
If
we
negl
ect t
he w
eigh
ts o
f the
ba
rs, e
ach
bar i
s a
two-
forc
e m
embe
r.
g
Trus
s Jo
ints
Trus
s
50
Trus
s Ty
pe #
1
Trus
s
51
Trus
s Ty
pe #
2
Trus
s
52
14
Trus
s Ty
pe #
3
Trus
s
How
e B
ridge
Tru
ss
K B
ridge
Tru
ss
53
g
Subd
ivid
ed W
arre
n Br
idge
Tru
ssw
ww.
nasa
.gov
Trus
s C
ateg
orie
s
P
lane
Tru
ss –
2D
Trus
s
S
pace
Tru
ss –
3D
54
Plan
e Tr
uss
Rig
id F
ram
es
R
igid
fram
es d
o no
t col
laps
e an
d ha
ve n
eglig
ible
de
form
atio
n.
Trus
s
55
Plan
e Tr
uss
Sim
ple
Trus
ses
Th
e ba
sic
elem
ent i
s a
trian
gle.
R
igid
sim
ple
truss
es a
re b
uilt
by a
ddin
g un
its o
f tw
o en
d-co
nnec
ted
bars
Trus
s
conn
ecte
d ba
rs.
56=
−S
D p
lane
trus
s:
23
mj
15
Plan
e Tr
uss
Stab
ility
>−
23
mj
SI t
russ
with
redu
ndan
t mem
ber(s
).
Trus
s
=−
23
mj
SD
pla
ne tr
uss
<−
23
mj
Trus
s co
llaps
es u
nder
load
.
57
SI w
ith re
dund
ant s
uppo
rts
SI w
ith im
prop
er s
uppo
rts
Plan
e Tr
uss
Solv
ing
Proc
edur
es
M
ain
Ana
lyse
s
Met
hod
of jo
ints
–di
sass
embl
ing
all j
oint
s an
d m
embe
rs
Trus
s
mem
bers
M
etho
ds o
f sec
tions
–se
ctio
ning
trus
s as
nee
ded
H
elpi
ng H
ands
Ze
ro-fo
rce
mem
bers
–id
entif
ying
mem
bers
that
do
not
sup
port
load
s
58
Met
hod
of J
oint
s Pr
oced
ure
Fo
r SD
trus
ses
Is
olat
e th
e tru
ss, d
raw
FB
D a
nd fi
nd th
e su
ppor
t re
actio
ns
Trus
s
reac
tions
Is
olat
e ea
ch jo
int,
draw
FB
D a
nd fi
nd fo
rces
of
mem
bers
on
join
ts
59
D
raw
FB
D o
f the
ent
ire tr
uss
Met
hod
of J
oint
s D
emo
#1
D
eter
min
e th
e ax
ial f
orce
s in
eac
h m
embe
r
Trus
s
60
16
Met
hod
of J
oint
s D
emo
#2
D
eter
min
e th
e re
actio
ns a
t its
sup
ports
.
Trus
s
=+
=
=
−
=−
+=
=
050
0 N
0
500
N
0(2
m)(
500
N)
20
500
N
xx x
Ay
y
FA A
MB
B
61
=+
=
=
−
00
500
Ny
yy
y
FA
B
A
FB
Ds
of m
embe
rs a
nd jo
ints
S
elec
t ord
er o
f joi
nt a
naly
ses
Trus
s
62
Met
hod
of J
oint
s D
emo
#4
D
raw
the
FBD
of a
join
t hav
ing
at le
ast o
ne k
now
n fo
rce
and
at m
ost t
wo
unkn
own
forc
es.
Orie
ntx
andy
axes
such
that
the
forc
eson
the
FBD
Trus
s
O
rient
xan
d y
axes
suc
h th
at th
e fo
rces
on
the
FBD
ca
n be
eas
ily re
solv
ed in
to th
eir x
and y
com
pone
nts.
S
olve
for t
hem
.
At j
oint
A
63
=−
=
=
=−
=
=
050
0 N
0
500
N
050
0 N
0
500
N
x yFAB
AB
FAC
AC
Met
hod
of J
oint
s D
emo
#5
C
ontin
ue w
ith th
e re
st o
f joi
nts
usin
g th
e sa
me
proc
edur
e.
Trus
s
=°+
=
=
−At j
oint
0si
n45
500
N0
707.
1 N
y
B
FBC
BC
Ai
lfi
b
64
= = =
Axi
al fo
rces
in m
embe
rs a
rem
embe
r:
500
N
: 50
0 N
: 70
7 N
Ans
ABAB
TACAC
TBCBC
C
17
Met
hod
of J
oint
s Ti
ps &
Hin
ts
P
roce
dura
l tip
s to
avo
id c
onfu
sion
A
ssig
nte
nsio
non
join
tsan
dm
embe
rsin
the
orig
inal
FBD
Trus
s
A
ssig
n te
nsio
n on
join
ts a
nd m
embe
rs in
the
orig
inal
FBD
C
heck
for t
heir
corre
ct s
ense
s la
ter
+
= te
nsio
n, p
ullin
g on
joi
nts
and
mem
bers
−
= co
mpr
essi
on, p
ushi
ng o
n th
e jo
ints
& m
embe
rs
65
Met
hod
of J
oint
sW
WW
Inte
ract
ive
Trus
s
B
ridge
Des
igne
r @ J
ohns
Hop
kins
Uni
vers
ity
Sm
all p
acka
ge
http
://w
ww
.jhu.
edu/
~virt
lab/
brid
ge/tr
uss.
htm
FR
AME3
DD
@ D
uke
Uni
vers
ity
ht
tp://
fram
e3dd
.sou
rcef
orge
.net
/
D
octo
r Fra
me
ht
tp://
ww
w.d
rsof
twar
e-ho
me.
com
/inde
x.ht
ml
66
Exam
ple
Met
hod
of J
oint
s 1
#1
Det
erm
ine
max
imum
axi
al fo
rces
in e
ach
mem
ber w
hen
0 ≤
θ≤
90°.
Trus
s
67
Exam
ple
Met
hod
of J
oint
s 1
#2
Trus
s
θθ
θ
=
+
==
−
=
−
−=
0
(10
kN)s
in0
10si
n k
N
0
(4 m
)(8
m)(1
0 kN
)cos
0
x
x x
A
y
F
A A
M
C
68
θ
θθ
=
=
+
−=
=−
20
cos
kN
0 (10
kN)c
os0
10co
s k
N
y
y
yy
y
C
F
AC
A
Whi
ch jo
ints
sho
uld
be c
hose
n?A
nd in
wha
t ord
er?
18
Exam
ple
Met
hod
of J
oint
s 1
#3
α−
=
1A
t joi
nt
,ta
n(3
/4)
0A
F
Trus
s
αθ
θ αθ
θθ
=
−
==
=
+
−=
0 sin
(10
kN)c
os(1
0 kN
)0
(50
cos
)/3
kN
0
cos
(10
kN)s
in0
10si
n(4
0co
s)/
3kN
y xF ABAB
F AC
AB
AC
69
θθ
=−
10si
n(4
0co
s)/
3 kN
AC Ex
ampl
e M
etho
d of
Joi
nts
1 #4
=
At j
oint
0y
C
F
Trus
s
θ θ
θθ
+=
=−
=
−
==
−
20co
s k
N0
20co
s k
N
0
010
sin
(40
cos
)/3
kN
y xBCBC F CD
AC
CD
70
Exam
ple
Met
hod
of J
oint
s 1
#5
α−
=1
At j
oint
,
tan
(3/4
)D
Trus
s
αθ
θ
=
−
==
0 sin
(10
kN)c
os0
(50
cos
)/3
kN
yF BDBD
71
Whe
n yo
u ge
t use
d to
the
proc
edur
e,
you
can
draw
onl
y FB
Ds
of jo
ints
.
Exam
ple
Met
hod
of J
oint
s 1
#6
θ=
=
Mag
nitu
de o
f for
ces
in m
embe
rs a
re50
cos
3 k
N10
i40
3kN
AB
BD
AC
CD
Trus
s
θθ
θ=
=−
=−
10si
n40
cos
3 k
N20
cos
kN
AC
CD
BC
72
19
Exam
ple
Met
hod
of J
oint
s 1
#7
θθ
=≤
≤°
Max
imum
mag
nitu
de o
f w
hen
0.Th
us, i
n th
e ra
nge
090
,F
dFd
Trus
s
θ==
==
= =°
max
max
max
max
max
16.7
kN
13
.3 k
N
20 k
N
all a
t 0
Ans
AB
BD
TAC
CD
CBC
C
73
Zero
-For
ce M
embe
r ide
ntifi
catio
n #1
Th
ey s
uppo
rt no
load
.
They
are
use
d to
incr
ease
sta
bilit
y of
the
truss
dur
ing
cons
truct
ion
Trus
s
cons
truct
ion.
Th
ey p
rovi
de s
uppo
rt if
the
appl
ied
load
cha
nges
.
74
=→
=
=
→=
3 1
2
00
0y xF
F
FF
F
Zero
-For
ce M
embe
r ide
ntifi
catio
n #2
Trus
s
′
=→
=
=
→=
1 2
00
00
x xFF
FF
75
′
=→
=
=
→=
3
4
12
0 0x xF
FF
FF
F
Zero
-For
ce M
embe
r Rem
ovin
g
Trus
s
76
=→
=
=
→=
1 2
00
00
y xFF
FF
20
Exam
ple
Zero
-For
ce M
embe
r 1 #
1
Trus
s
77
Exam
ple
Zero
-For
ce M
embe
r 1 #
2
Trus
s
78
Exam
ple
Zero
-For
ce M
embe
r 1 #
3
Trus
s
79
Exam
ple
Zero
-For
ce M
embe
r 2 #
1
Trus
s
80
21
Exam
ple
Zero
-For
ce M
embe
r 2 #
2
Trus
s
How
e B
ridge
Tru
ss
K B
ridge
Tru
ss
Zero
-forc
e m
embe
rs in
st
anda
rd tr
usse
s
81Su
bdiv
ided
War
ren
Brid
ge T
russ
Met
hod
of S
ectio
ns B
asic
s #1
Im
agin
ary
cut t
hrou
gh th
e co
nsid
ered
regi
on
If th
e w
hole
bod
y is
in e
quili
briu
m, i
ts p
arts
mus
t als
o be
Imag
inar
y S
ectio
nsTr
uss
in e
quili
briu
m.
Bui
lt-in
Sup
port
82
Met
hod
of S
ectio
ns B
asic
s #2
Fi
nd re
sulta
nt in
tern
al lo
ad b
y m
etho
d of
sec
tion
In
tern
al lo
ads
are
norm
al fo
rce N
, she
ar fo
rce V
, tor
sion
T
Trus
sIm
agin
ary
Sec
tions
and
bend
ing
mom
ent M
83
Met
hod
of S
ectio
ns C
once
pt
If a
body
is in
equ
ilibr
ium
, any
par
t of
the
body
isal
soin
equi
libriu
m.
Trus
s We
can
draw
an
imag
inar
y se
ctio
n th
roug
h th
e bo
dy.
of th
e bo
dy is
als
o in
equ
ilibr
ium
.
84
If bo
dies
are
two-
forc
e m
embe
rs,
ther
e ar
e on
ly n
orm
al in
tern
al
forc
es.
22
Met
hod
of S
ectio
ns In
tern
al F
orce
s
Trus
s
85
Met
hod
of S
ectio
ns P
roce
dure
#1
Trus
s
86
Met
hod
of S
ectio
ns P
roce
dure
#2
Trus
s
87
Met
hod
of S
ectio
ns D
emo
#1
Det
erm
ine
the
axia
l for
ces
in m
embe
rs BD
, CD
and CE
.
Trus
s
88
23
Met
hod
of S
ectio
ns D
emo
#2
D
raw
FB
D o
f ent
ire
stru
ctur
e.
Trus
s
C
hoos
e an
imag
inar
y lin
e to
“cut
” or s
ectio
n th
roug
h th
e m
embe
rs
whe
re fo
rces
are
to b
e de
term
ined
.
89
Met
hod
of S
ectio
ns D
emo
#3
C
onsi
der F
BD
s of
sec
tions
Trus
s
90
Met
hod
of S
ectio
ns D
emo
#4
D
raw
FB
D o
f tha
t par
t of t
he s
ectio
ned
truss
with
the
leas
t num
ber o
f for
ces
on it
.
Trus
s
91
A
pply
equ
ilibr
ium
equ
atio
ns to
the
chos
en F
BD
.
Met
hod
of S
ectio
ns D
emo
#5
=+
0
DM
Trus
s
′
+
°−
=
=
=
°−°=
=
0
()(1
m)s
in45
(10
kN)(1
m)
0
102
kN
0
sin
45(1
0 kN
)sin
450
10kN
D yM
CE
CE
F
CDCD
92
′
=
=
−
−°−
−°=
=−
10
kN 0
cos4
5(1
0 kN
)cos
450
202
kN
x
CD
F
BD
CD
CE
BD
= = =
102
kN
202
kN
10 k
N
A
ns
CE
T
BD
CCD
T
24
Spac
e Tr
usse
sTr
uss
M
embe
rs a
re c
onne
cted
by
join
ts th
at d
o no
t res
ist m
omen
ts.
Th
e ba
sic
elem
ent i
s a
tetra
hedr
on.
R
igid
sim
ple
spac
e tru
sses
can
be
built
by
addi
ng u
nits
of t
hree
end
-con
nect
ed b
ars.
Ana
le
inth
esa
me
aas
2Dtr
sses
93=
−S
D s
pace
trus
s:
36
mj
An
alyz
e in
the
sam
e w
ay a
s 2D
trus
ses.
Exam
ple
Spac
e Tr
usse
s 1
#1
A s
pace
trus
s is
pla
ced
on a
sm
ooth
floo
r. Jo
int A
is s
uppo
rted
by th
e co
rner
whe
re th
e sm
ooth
Trus
s
byt
eco
ee
et
es
oot
wal
ls m
eet,
and
join
t Cre
sts
agai
nst t
he s
moo
th b
ack
wal
l. D
eter
min
e ax
ial f
orce
s in
mem
ber
AC
.
94
Exam
ple
Spac
e Tr
usse
s 1
#2
App
lied
load ˆ
ˆˆ
26
kNF
ijk
=−
−−
Trus
s
26
kN
All
smoo
th s
uppo
rtsJo
int
is n
este
d in
a c
orne
rˆ
ˆˆ
kN
Join
t is
pla
ced
on a
floo
rx
yz
Fi
jk
A
AAiAjAk
B= =+
+
95
p ˆ kN
Join
t re
sts
agai
nst a
wal
lˆ
ˆ kN
y yz
BBj
C
CCjCk
= =+
Exam
ple
Spac
e Tr
usse
s 1
#3
=−
++
+−
++
−=
×+
×+
×
ˆˆ
ˆ0
(2)
6)(
1)0
(1)
0(
)(
)(
)0
(2)
xy
yy
zz
FA
iA
BC
jA
Ck
Mr
Br
Cr
F
Trus
s
=×
+×
+×
=
0(
)(
)(
)0
(2)
AAB
AC
AD
Mr
Br
Cr
F Sol
ving
(1) a
nd (2
)ˆ
ˆˆ
96
24
1 k
Nˆ 1 k
N
ˆ 1 k
N
Ai
jk
Bj
Cj
=+
+
= =
25
Exam
ple
Spac
e Tr
usse
s 1
#4
Trus
s
97
Exam
ple
Spac
e Tr
usse
s 1
#5
Trus
s
At j
oint
ˆ
ˆCA
CA
CA
CA
C
FF
nF
i=
=−
98
ˆˆ
ˆˆ
ˆ(
23
)(
0.55
470
0.83
205
)13
ˆˆ
ˆˆ
(2
31
)14
ˆˆ
ˆ(
0.53
452
0.80
178
0.26
726
)
CB
CB
CB
CB
CB
CD
CD
CD
CD
CD
CD
FF
Fn
ik
Fi
k
FF
Fn
ij
k
FF
ij
k
==
−+
=−
+
==
−+
+
=−
++
Exam
ple
Spac
e Tr
usse
s 1
#6
=
+
++
=
0
0CA
CB
CD
y
F
FF
FC
Trus
s
−−
−+
++
+=
=−
ˆ(
0.55
470
0.53
452
)ˆ
(0.8
0178
1)ˆ
(0.8
3205
0.26
726
)0
Thus
,1.
2472
2kN
CA
CB
CD
y
CA
CB
CD
CD
CB
CD
CD
FF
Fi
Fj
FF
k
F
99
= =
=
Thus
,1.
2472
2 kN
0.
4006
2 kN
0.44
444
kN
444
N
Ans
CD
CB
CA
CA
F F F
FT
Trus
ses
Tips
& H
ints
M
embe
rs
All
mem
ber a
re o
f tw
o-fo
rce
mem
ber t
ypes
.
Trus
s
Lo
ok fo
r zer
o-fo
rce
mem
bers
to s
impl
ify p
robl
ems
In
tern
al fo
rces
U
se te
nsio
n as
+ve
and
com
pres
sion
as
−ve
100
V
isua
lize
the
‘flow
’ of f
orce
s/lo
ads
thro
ugh
the
truss
es
Che
ck th
e re
sults
26
Sum
mar
y
Fram
e: A
stru
ctur
e w
hich
is d
esig
ned
to re
mai
n st
atio
nary
an
d su
ppor
t loa
ds.
Mac
hine
:Ast
ruct
ure
whi
chis
desi
gned
tom
ove
and
exer
t
Mac
hine
: A s
truct
ure
whi
ch is
des
igne
d to
mov
e an
d ex
ert
load
s.
Trus
s: A
stru
ctur
e w
hich
is c
ompr
ised
ent
irely
of t
wo-
forc
e m
embe
rs.
Tens
ion:
axia
lfor
ces
atth
een
dsar
edi
rect
edaw
ayfro
m
101
Te
nsio
n: a
xial
forc
es a
t the
end
s ar
e di
rect
ed a
way
from
ea
ch o
ther
.
Com
pres
sion
: axi
al fo
rces
at t
he e
nds
are
dire
cted
tow
ard
each
oth
er
Con
cept
s
Stru
ctur
es a
re s
yste
ms
of in
terc
onne
ctin
g rig
id b
odie
s an
d fa
sten
ing
(pin
s, jo
ints
, etc
.). S
truct
ured
can
be
anal
yzes
by
disa
ssem
blin
gth
esy
stem
into
inte
r-re
late
dFB
Ds
andR
evie
w
disa
ssem
blin
g th
e sy
stem
into
inte
r-re
late
d FB
Ds
and
cons
ider
ing
the
equi
libriu
m o
f ind
ivid
ual m
embe
rs.
Fr
ames
and
mac
hine
sar
e st
ruct
ures
that
are
not
tr
usse
s. F
ram
es s
tay
stat
iona
ry w
hile
mac
hine
s ca
n m
ove.
102
Tr
usse
sar
e st
ruct
ures
whi
ch c
ompr
ise
of o
nly
inte
rcon
nect
ing
2-fo
rce
mem
bers
.