Modul XI
The Method of Joints
SAMPLE PROBLEM 6.1
Using the method of joints, determine the force in each member oft truss shown.
Solution. A free-body diagram of the entire truss is drawn; exten forces acting on this
free body consist of the applied loads and I reactions at C and E.
Equilibrium of Entire Truss.
+ ΣMC = 0:
(2000 lb)(24 ft) + (1000 lb)(12 ft) - E(6 ft) = 0
E = +10,000 Ib
Fx  0
F y  0
Cx=0
-2000 Ib - 1000 Ib + 10,000 Ib + Cy = 0
Cy = -7000lb
Joint A. This joint is subjected to only two unknown forces, namely the forces exerted by
members AB and AD. A force triangle is used to determine FAB and FAD. We note that
member AB pulls on the joint and thus is in tension and that member AD pushes on the
joint and thus is in compression. The magnitudes of the two forces are obtained from the
proportion
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
1
2000lb FAB FAD


4
3
5
FAB  1500lb..(T ); FAD  2500lb..(C )
FAB = 1500 Ib T F AD = 2500 Ib C
Joint D.
Since the force exerted by member AD has been determined, only two
unknown forces are now involved at this joint. Again,a force triangle is used to determine
the unknown forces in menbers DB and DE.
FDB =FDA,
FDB =25001b(T)
FDE =2(3/5)FDA
FDE = 3000 lb (C)
Joint B. Since more than three forces act at this joint, we determine the two unknown
forces FBc and FBE by solving the equilibrium equations ΣFx = 0. and ΣFy = 0. We
arbitrarily assume that both unknown forces act away from the joint, i.e., that the
members are in tension. The positive value obtained for FBC indicates that our assumption was correct; member BC is in tension. The negative value of FBE indicates that
our assumption was wrong; member BE is in compression.
ΣFy = 0.
ΣFx = 0:
-1000 - 4/5(2500)-4/5FBE = 0
FBE = -3750 Ib
FBE = 3750 Ib (C)
FBC -1500 – 3/5(2500) – 3/5(3750) = 0
FBC = 5250 Ib (T )
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
2
Joint E. The unknown force FEC is assumed to act away from the joint.
components, we write
ΣFx = 0:
3/5FEC + 3000 + 3/5(3750) = 0
FEC = -8750 Ib atau
FEC = 8750 Ib (C )
Summing x
Summing y components, we obtain a check of our computations:
ΣFy = 10,000 – 4/5(3750) – 4/5(8750)
= 10,000 - 3000 - 7000 = 0 (checks)
Joint C. Using the computed values of FCB and FCE, we may determine the reactions Cx
and Cy by considering the equilibrium of this joint. Since these reactions have already
been determined from the equilibrium of the entire truss, we will obtain two checks of our
computations. We may also merely use the computed values of all forces acting on the
joint (forces in members and reactions) and check that the joint is in equilibrium:
ΣFx = -5250 + 3/5(8750) = -5250 + 5250 = 0
ΣFy = -7000 + 4/5(8750) = -7000 + 7000 = 0
(checks)
(checks)
PROBLEMS
12.1 through 12.3 Using the method of joints, determine thd force in each member of
the truss shown. State whether each membaf is in tension or compression.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
3
5.2. Metoda Seksi
Yaitu memotong seksi-seksi yang kita inginkan.
Gambar.5.3.Pemotongan yang Diinginkan
Potongan a-a
Pada Gambar 5.3 ini hanya batang FCD,FCH dan FHI saja yg tidak diketahui.
Gambar.5.3.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
4
Potongan b-b
Dalam pemotongan bagian2 tsb dapat juga dilakukan seperti potongan b-b,untuk
mengetahui gaya dalam batang CI,misalnya.
Dapat juga dengan cara seperti pada Gambar 5.4.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
5
Umpama kita ingin tahu gaya dalam batang CM,maka dilakukan pemotongan melalui
a-a,seperti pada Gambar 5.5,dimana FY satu2nya gaya reaksi yg diketahui.Jadi hanya
ada 3 persamaan keseimbangan yg diketahui dengan empat bilangan yg tidak
diketahui yaitu FCD,FCM,FHM dan FGH.Hal itu tidak mungkin dapat dipecahkan oleh tiga
persamaan tsb.
Kita coba lagi pemotongan melalui b-b,seperti pada Gambar.5.6.Dan hal tsb tidak
mungkin lagi dapat dipecahkan. Sehingga kita harus ber-kali2 melakukan pemotongan
agar gaya dalam batang yg diinginkan dapat dihitung.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
6
Contoh.
1.Tentukanlah gaya-gaya dalam batang CF dan FG dari truss pada Gambar ini.
Jawab.
i.Gambar dahulu FBD dari seluruh struktur.
Penjumlahan gaya2 dan moment terhadap titik A menghasilkan :
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
7
M
AB
 4 BY  10(4)  10(8)  20(4)  20(8)  0
4 BY  360  0
F
F
X
 AX  10  10  0
Y
 AY  BY  20  20  0
 BY  90kN...... AY  50kN...... AX  20kN
Selanjutnya dilakukan pemotongan melalui a-a.
Potongan b-b.
Harga FCF dapat dicari dari persamaan tunggal dengan menjumlahkan gaya2 pada arah
X.Jadi:
F
X
  FCF cos 45  10  0
FCF  14.14kN
Selanjutnya dilakukan pemotongan melalui b-b.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
8
Untuk potongan b-b,ambil bagian sebelah kanan saja, karena hanya ada satu gaya
saja yg bekerja. Harga FFG didapat dengan persamaan keseimbangan tunggal dengan
menjumlahkan moment terhadap joint J.
M
JZ
 4 FGF  20(4)  0
 FGF  20kN.
12.2 Hitunglah tegangan dalam batang 4-8 dan 7-8 dari truss, Gambar ini, dengan
metoda seksi.
Jawab.
Untuk keseimbangan truss :
F
F
M
X
 ( F1 ) X  ( F5 ) X  0
Y
 ( F1 ) Y  ( F5 ) Y  4  4  0
1Z

 
 

 ( F5 ) Y 4(10 cot 30  )  4 2(10 cot 30  )  4 3(10 cot 30  )  0
 ( F5 ) Y  5kips......( F1 ) Y  3kips.......( F5 ) X  ( F1 ) X kips
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
9
Untuk menentukan (F1)x dan (F5)x maka potonglah melalui
a-a,yaitu:
Kita dapatkan (F1)x langsung dengan penjumlahan moment terhadap joint 3.Jadi:
M
3Z
 ( F1 ) X (20)  3(20 cot 30  )  0
( F1 ) X  3 cot 30   5.196kips
( F5 ) X  ( F1 ) X  5.196kips
Untuk menentukan gaya2 dalam batang 4-8 dan 7-8,potonglah bidang b-b.
Pilihlah joint 4 untuk penjumlahan moment guna mengeliminer F4-3 dan F4-8.
Persamaan keseimbangannya adalah:
M
4Z
  F7 8 sin 60  (10 cot 60  )  F7 8 cos 60  (10)
 5(10 cot 30  )  (5.196)(10)  0
F
F
X
  F7 8 cos 60   F4 8  F43 cos 30   5.196  0
Y
 F3 4 sin 30   F7 8 sin 60   5  4  0
 F7 8  3.464kips......F4 3  8.00kips.....F48  0kips
Problem
12-1Determine the force in members BD and CD of the truss shown.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
10
Fig.Problem 12-1 and Problem 12-2
12-2 Determine the force in members DF and DG of the truss shown.
12-3Determine the force in members DF and DE of the truss shown.
Fig.Problem 12-3 ,Problem 12-4 and Problem 12-5.
12-4 Determine the force in members CD and CE of the truss shown.
12-5 Determine the force in members EF and DF of the truss shown.
12-6
Determine the force in member CD of the Fink roof truss shown.
Fig. P 12-6
12-7 Determine the force in members CE, CD, and BD of them shown.
12-8 Determine the force in members EG, EF, and DFof the shown.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
11
Fig. P 12-6 and 12-6
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
12