1. II MadeMade GatotGatot KarohikaKarohika, ST. MT., ST. MT.
Mechanical EngineeringMechanical Engineering
UdayanaUdayana UniversityUniversity

2. Contents
Introduction
Definition of a Truss
Simple Trusses
Analysis of Trusses by the Method
of Joints
Joints Under Special Loading
Trusses Made of Several Simple
Trusses
Sample Problem 6.3
Analysis of Frames
Frames Which Cease to be Rigid
When Detached From Their
6 - 2
Joints Under Special Loading
Conditions
Space Trusses
Sample Problem 6.1
Analysis of Trusses by the Method
of Sections
When Detached From Their
Supports
Sample Problem 6.4
Machines

3. Introduction
• Untuk keseimbangan struktur yang terdiri dari beberapa bagian
batang yang bersambungan, maka gaya dalam (internal forces)
seperti halnya gaya luar harus diperhitungkan juga.
• Dalam interaksi antara bagian yang terhubung, Newton’s 3rd
Law menyatakan bahwa gaya aksi dan reaksi antara benda
dalam keadaan kontak mempunyai besar yang sama, garis aksi
yang sama, dan berlawanan arah.
• Tiga kategori dari struktur teknik:
a) Portal (Frames): terdiri dari paling kurang satu batang
6 - 3
a) Portal (Frames): terdiri dari paling kurang satu batang
dengan pelbagai gaya (multi-force member), i.e., batang
yang mengalami tiga atau lebih gaya yang umumnya tidak
tidak searah sumbu batang.
b) Rangka batang(Trusses): dibentuk dari batang dengan
dua gaya [two-force members], i.e., batang lurus dengan
ujung-ujung berhubungan, batang mengalami dua gaya
sama besar dan berlawanan yang searah dengan sumbu
batang
c) Machines: struktur yang terdiri dari bagian-bagian yang
bergerak dirancang untuk menyalurkan dan mnengubah
gaya-gaya.

4. 2 - 4

5. Definisi Truss [ rangka batang ]
• Truss terdiri dari bagian berbentuk lurus dan terhubung
pada sambungan. Tidak ada bagian yang menembus
sambungan.
• Sebagaian besar struktur dibentuk dari beberapa truss
yang dihubungkan bersama membentuk kerangka ruang
(space framework). Masing-masing truss menumpu
beban yang beraksi pada bidangnya sehingga dapat
diperlakukan sebagai struktur.
6 - 5
• Sambungan Baut atau las diasumsikan disambung dengan
memakai pin (pasak). Gaya yang beraksi pada ujung-
ujung bagian tereduksi menjadi gaya tunggal dan tidak
ada kopel. Hanya dipandang sebagai bagian dua gaya
(two-force members).
diperlakukan sebagai struktur.
• Ketika gaya cenderung untuk menarik bagian batang,
dikatakan mengalami tegang (tension) [T]. Ketika gaya
cenderung untuk menekan batang, batang dalam
keadaan tekan (compression) [C].

6. Definition of a Truss
6 - 6
Bagian-bagian truss berbentuk batang (slender) dan
hanya bisa mendukung beban ringan dalam arah lateral.
Beban harus diterapkan pada sambungannya bukan
langsung pada bagian-bagiannya.

7. Definition of a Truss
6 - 7

8. 2 - 8

9. Simple Trusses/ Rangka batang sederhana
• Truss tegar (rigid truss) tidak akan
ambruk collapse karena aplikasi
beban.
• A simple truss disusun dengan
menambahkan berturutan dua bagian
dan satu sambungan pada segitiga
dasar (basic) truss.
6 - 9
dasar (basic) truss.
• In a simple truss, m = 2n - 3
dimana m adalah jumlah total
members and n adalahjumlah
sambungan.

10. Analysis Truss dengan Methode Sambungan
• Uraikan truss dan buat freebody diagram untuk
masing-masing bagian dan pin.
• Dua gaya yang ada pada masing-masing bagian (satu
gaya pada masing-masing ujung) adalah besarnya
sama,garis aksi yang sama, dan berlawanan arah.
• Gaya yang ditimbulkan oleh suatu bagian pada pin
atau sambungan pada ujungnya harus berarah
6 - 10
atau sambungan pada ujungnya harus berarah
sepanjang bagian itu dan harus sama dan berlawanan.
• Kondisi keseimbangan pada pins memberikan 2n
persamaan untuk 2n besaran yang tidak diketahui.
untuk simple truss, 2n = m + 3. dapat dipecahkan
untuk m member forces and 3 reaction forces pada
tumpuan.
• Kondisi keseimbangan untuk keseluruhan truss
memberikan 3 persamaan tambahan yang tidak
bebas dengan persamaan pin.

11. Joints Under Special Loading Conditions
• Gaya pada bagian yang berlawanan berpotongan
pada dua garis lurus pada sambungan harus sama
• Gaya pada dua bagian yang berlawanan harus
sama. Jika beban P dihubungkan dengan
bagian ketiga, gaya pada bagian ketiga harus
sama dengan dengan beban P ( termasuk jika
beban nol).
• Gaya pada dua bagian terhubung pada
sambungan adalah sama jika bagian itu segaris
6 - 11
sambungan adalah sama jika bagian itu segaris
(aligned) dan nol begitu sebaliknya.
• Dengan mengetahui sambungan dalam kondisi
pembebanan khusus akan menyederhanakan
analisa truss .

12. Space Trusses
• An elementary space truss consists of 6 members
connected at 4 joints to form a tetrahedron.
• A simple space truss is formed and can be
extended when 3 new members and 1 joint are
added at the same time.
• In a simple space truss, m = 3n - 6 where m is the
6 - 12
• Equilibrium for the entire truss provides 6
additional equations which are not independent of
the joint equations.
• In a simple space truss, m = 3n - 6 where m is the
number of members and n is the number of joints.
• Conditions of equilibrium for the joints provide 3n
equations. For a simple truss, 3n = m + 6 and the
equations can be solved for m member forces and
6 support reactions.

13. 2 - 13

14. Sample Problem 6.1
SOLUTION:
• Based on a free-body diagram of the
entire truss, solve the 3 equilibrium
equations for the reactions at E and C.
• Joint A is subjected to only two unknown
member forces. Determine these from the
joint equilibrium requirements.
6 - 14
Using the method of joints, determine
the force in each member of the truss.
joint equilibrium requirements.
• In succession, determine unknown
member forces at joints D, B, and E from
joint equilibrium requirements.
• All member forces and support reactions
are known at joint C. However, the joint
equilibrium requirements may be applied
to check the results.

15. Sample Problem 6.1
SOLUTION:
• Based on a free-body diagram of the entire truss,
solve the 3 equilibrium equations for the reactions
at E and C.
( )( ) ( )( ) ( )ft6ft12lb1000ft24lb2000
0
E
MC
−+=
=∑
6 - 15
↑= lb000,10E
∑ == xx CF 0 0=xC
∑ ++−== yy CF lb10,000lb1000-lb20000
↓= lb7000yC

16. Sample Problem 6.1
• Joint A is subjected to only two unknown
member forces. Determine these from the
joint equilibrium requirements.
6 - 16
joint equilibrium requirements.
534
lb2000 ADAB FF
==
CF
TF
AD
AB
lb2500
lb1500
=
=
• There are now only two unknown member
forces at joint D.
( ) DADE
DADB
FF
FF
5
32=
=
CF
TF
DE
DB
lb3000
lb2500
=
=

17. Sample Problem 6.1
• There are now only two unknown member
forces at joint B. Assume both are in tension.
( )
lb3750
250010000 5
4
5
4
−=
−−−==∑
BE
BEy
F
FF
CFBE lb3750=
( ) ( )3750250015000 5
3
5
3 −−−==∑ BCx FF
6 - 17
lb5250
55
+=BCF TFBC lb5250=
• There is one unknown member force at joint
E. Assume the member is in tension.
( )
lb8750
375030000 5
3
5
3
−=
++==∑
EC
ECx
F
FF
CFEC lb8750=

18. Sample Problem 6.1
• All member forces and support reactions are
known at joint C. However, the joint equilibrium
requirements may be applied to check the results.
( ) ( )
( ) ( )checks087507000
checks087505250
5
4
5
3
=+−=
=+−=
∑
∑
y
x
F
F
6 - 18

19. Analysis of Trusses by the Method of Sections
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagram for the left side.
6 - 19
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.

20. Trusses Made of Several Simple Trusses
• Compound trusses are statically
determinant, rigid, and completely
constrained.
32 −= nm
• Truss contains a redundant member
and is statically indeterminate.
32 −> nm
6 - 20
32 −> nm
• Necessary but insufficient condition
for a compound truss to be statically
determinant, rigid, and completely
constrained,
nrm 2=+
non-rigid rigid
32 −< nm
• Additional reaction forces may be
necessary for a rigid truss.
42 −< nm

21. Sample Problem 6.3
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.
• Pass a section through members FH,
GH, and GI and take the right-hand
section as a free body.
6 - 21
Determine the force in members FH,
GH, and GI.
section as a free body.
• Apply the conditions for static
equilibrium to determine the desired
member forces.

22. Sample Problem 6.3
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.
6 - 22
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )
↑=
++−==
↑=
+−−
−−−==
∑
∑
kN5.12
kN200
kN5.7
m25kN1m25kN1m20
kN6m15kN6m10kN6m50
A
ALF
L
L
M
y
A

23. Sample Problem 6.3
• Pass a section through members FH, GH, and GI
and take the right-hand section as a free body.
• Apply the conditions for static equilibrium to
6 - 23
( )( ) ( )( ) ( )
kN13.13
0m33.5m5kN1m10kN7.50
0
+=
=−−
=∑
GI
GI
H
F
F
M
• Apply the conditions for static equilibrium to
determine the desired member forces.
TFGI kN13.13=

24. Sample Problem 6.3
( )( ) ( )( ) ( )( )
( )( )
kN82.13
0m8cos
m5kN1m10kN1m15kN7.5
0
07.285333.0
m15
m8
tan
−=
=+
−−
=
°====
∑
FH
FH
G
F
F
M
GL
FG
α
αα
CFFH kN82.13=
6 - 24
( )
( )( ) ( )( ) ( )( )
kN371.1
0m10cosm5kN1m10kN1
0
15.439375.0
m8
m5
tan
3
2
−=
=++
=
°====
∑
GH
GH
L
F
F
M
HI
GI
β
ββ
CFGH kN371.1=

25. Analysis of Frames
• Frames and machines are structures with at least one
multiforce member. Frames are designed to support loads
and are usually stationary. Machines contain moving parts
and are designed to transmit and modify forces.
• A free body diagram of the complete frame is used to
determine the external forces acting on the frame.
• Internal forces are determined by dismembering the frame
6 - 25
• Internal forces are determined by dismembering the frame
and creating free-body diagrams for each component.
• Forces between connected components are equal, have the
same line of action, and opposite sense.
• Forces on two force members have known lines of action
but unknown magnitude and sense.
• Forces on multiforce members have unknown magnitude
and line of action. They must be represented with two
unknown components.

26. Frames Which Cease To Be Rigid When Detached From Their Supports
• Some frames may collapse if removed from
their supports. Such frames can not be treated
as rigid bodies.
• A free-body diagram of the complete frame
indicates four unknown force components which
can not be determined from the three equilibrium
6 - 26
can not be determined from the three equilibrium
conditions.
• The frame must be considered as two distinct, but
related, rigid bodies.
• With equal and opposite reactions at the contact
point between members, the two free-body
diagrams indicate 6 unknown force components.
• Equilibrium requirements for the two rigid
bodies yield 6 independent equations.

27. Sample Problem 6.4
SOLUTION:
• Create a free-body diagram for the
complete frame and solve for the support
reactions.
• Define a free-body diagram for member
BCD. The force exerted by the link DE
has a known line of action but unknown
6 - 27
Members ACE and BCD are
connected by a pin at C and by the
link DE. For the loading shown,
determine the force in link DE and the
components of the force exerted at C
on member BCD.
has a known line of action but unknown
magnitude. It is determined by summing
moments about C.
• With the force on the link DE known, the
sum of forces in the x and y directions
may be used to find the force
components at C.
• With member ACE as a free-body,
check the solution by summing
moments about A.

28. Sample Problem 6.4
SOLUTION:
• Create a free-body diagram for the complete frame
and solve for the support reactions.
N4800 −==∑ yy AF ↑= N480yA
( )( ) ( )mm160mm100N4800 BM A +−==∑
6 - 28
→= N300B
xx ABF +==∑ 0
←=
−=
N300Ax
NAx 300
°== −
07.28tan 150
801
α
Note:

29. Sample Problem 6.4
• Define a free-body diagram for member
BCD. The force exerted by the link DE has a
known line of action but unknown
magnitude. It is determined by summing
moments about C.
( )( ) ( )( ) ( )( )
NF
mmN480mm0N300mmFM
DE
DEC
561
1008250sin0
−=
++==∑ α
CFDE N561=
6 - 29
NFDE 561
CFDE N561=
• Sum of forces in the x and y directions may be used to find the force
components at C.
( ) N300cosN5610
N300cos0
+−−=
+−==∑
α
α
x
DExx
C
FCF
N795−=xC
( ) N480sinN5610
N480sin0
−−−=
−−==∑
α
α
y
DEyy
C
FCF
N216=yC

30. Sample Problem 6.4
• With member ACE as a free-body, check
the solution by summing moments about A.
6 - 30
( )( ) ( )( ) ( )
( )( ) ( )( ) ( )( ) 0mm220795mm100sin561mm300cos561
mm220mm100sinmm300cos
=−−−+−=
−+=∑
αα
αα xDEDEA CFFM
(checks)

31. Machines
• Machines are structures designed to transmit
and modify forces. Their main purpose is to
transform input forces into output forces.
• Given the magnitude of P, determine the
magnitude of Q.
• Create a free-body diagram of the complete
machine, including the reaction that the wire
6 - 31
machine, including the reaction that the wire
exerts.
• The machine is a nonrigid structure. Use
one of the components as a free-body.
• Taking moments about A,
P
b
a
QbQaPM A =−==∑ 0