Modul VIII
Momen = jarak x Gaya
atau secara matematik dpt ditulis:
M= d x F.......(6.11)
Momen positif apabila gaya yg diapplikasikan thd titik momen berputar berlawanan
jarum,Gambar 6.1a.
Momen negatif apabila gaya yg diapplikasikan thd titik momen berputar searah jarum
jam,Gambar 6.11b
6.12. Momen sebuah Gaya dititik C.
M c  d  F .......(6.12)
Gambar.6.13
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Pada Gambar 6.2. ,dilihat dari sumbu Z terlihat Mcz menyatakan bahwa komponen Z dari
Mc.
Besarnya moment Mcz adalah:
Dua Dimensi:
M cz  rx Fy  ry Fx ......(6.13)
Tiga Dimensi:
rP
Karena :
C
 rx I  ry J  rz K ....dan
F  Fx I  Fy J  Fz K , thus
M c  rP  F 
.........(6.14)
c
 ( ry Fz  rz Fy ) I  rz Fx  rx Fz J 
r
x
Fy  ry Fx K
Aturan Tangan Kanan:
Untuk memudahkan dan mengingat cross product dua unit vector,untuk order alphabet
positif dan negatif untuk order alphabet negatif,lihat Gambar dibawah:
I  J  J  I  K
J  K  K  J  I
……..(6.15)
K  I  I  K  J
Gambar 6.13.Aturan Tangan Kanan
Contoh Soal:
6.12Sebuah gaya sebesar 5 kips bekerja pada flange sebuah batang H. Tentukan
momen gaya ini thd pusat C penampang lintang dgn cara2:
(a). mengevaluasi lengan gaya
(b). mengevaluasi vektor
product
(c).pertimbangan komponen
gaya dititik P
(d).komponen gaya pada
web dititik Q.
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Jawab:
1. Gambarkan FBD lengkap dgn sumbu2 ref.x dan y
2. Uraikan gaya2 tsb pada sumbu2 ref. x dan y
3. Applikasikan prinsip2 mek.tek.(momen ).
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
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Soal2 Latihan
6.13.Hitunglah nilai minimal gaya2 yg dikerjakan diujung titik A dan B dari batang pada
Gambar dibawah ini,yang equivalen dengan gaya2 yg diberikan dalam gambar tsb.
Gambar soal 10.
6.14.Seorang pengendara bus mengerjakan dua gaya masing2 sebesar 60N,lihat
Gambar.Berapakah besar torsi kopel tsb dan berapakah besar komponen torsi ini yang
bekerja memutar setir.
Gambar soal 11.
Gambar soal 12.
6.15.Gantilah gaya dan kopel ini dengan sebuah sistem gaya yg equivalen pada ujung B
dari batang berbentuk kurva tsb.
6.16.Tiga grup orang mendorong pinggiran meja seperti pada Gambar dibawah.Hitunglah
sebuah gaya tunggal yg besarnya sama dengan jumlah gaya orang2 tsb,dan dimanakah
letak gaya yg diapplikasikan pada pinggiran BC
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PUSAT PENGEMBANGAN BAHAN AJAR-UMB
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Modul VII:
4.Keseimbangan Moment
4.1 PENDAHULUAN
Mata kuliah statika terutama membahas kondisi gaya yang diperlukan dan cukup untuk
mempertahankan kesetimbangan struktur teknik. Bab mengenai kesetimbangan ini
merupakan bagian yang paling penting dalam statika dan harus benar-benar dikuasai.
Kita akan tetap meng-gunakan konsep yang dikembangkan dalam Modul II yang
meliputi gaya, momen, kopel, dan resul-tan pada saat kita menerapkan prinsip-prinsip
kesetimbangan. Prosedur yang akan dikembangkan dalam Modul V merupakan
pengantar lengkap yang sering dipakai dalam menyelesaikan per-soalan-persoalan
mekanika dan bidang teknik lainnya. Pendekatan ini merupakan dasar dari keberhasilan dalam penguasaan statika, dan para mahasiswa diharapkan membaca dan
mempelajari pasal-pasal berikut ini dengan upaya khusus dan perhatian yang istimewa.
Jika suatu benda berada dalam kesetimbangan, maka resultan semua gaya yang
bekerja pada-nya akan menjadi nol. Jadi gaya resultan R dan kopel resultan M adalah
nol, dan kita memper-oleh persamaan kesetimbangan
R = ΣF = 0
M = ΣM = 0 …….. (4-1)
Kedua syarat ini merupakan kondisi yang diperlukan dan cukup untuk kesetimbangan.
Walaupun semua benda fisis memiliki sifat tiga-dimensi, namun banyak
diantaranya dapat diperlakukan sebagai benda dua-dimensi apabila gaya-gaya yang
dikenakan padanya bekerja pada sebuah bidang tunggal atau dapat diproyeksikan pada
sebuah bidang tunggal. Jikapenye-derhanaan ini tidak mungkin dilakukan, mau tidak
mau persoalan tersebut harus diperlakukan sebagai persoalan tiga-dimensi. Kita akan
mengikuti susunan yang digunakan dalam Modul VI dan membahas dalam Bagian A
mengenai kesetimbangan benda yang dikenai sistem gaya dua-dimensi, dan dalam
Bagian B mengenai kesetimbangan benda yang dikenai sistem gaya tiga-dimensi.
4-2 KESETIMBANGAN DALAM DUA DIMENSI
Sebelum menerapkan Persamaan 4-1, kita harus terlebih dahulu mendefinisikan
dengan jelas benda tertentu atau sistem mekanis yang akan dianalisis dan
menggambarkan secara jelas dan lengkap semua gayayang bekerja pada benda
tersebut. Menghilangkan gaya atau mencantumkan gaya yang tidak bekerja pada benda
yang dibahas akan memberikan hasil yang keliru.
Sistem mekanis didefmisikan sebagai suatu benda atau sekumpulan benda yang dapat
dipi-sahkan dari pengaruh benda-benda lain. Sistem demikian dapat merupakan benda
tunggal atau kombinasi benda yang berhubungan. Benda tersebut dapat tegar ataupun
iak-tegar. Sistem dapat juga merupakan suatu massa fluida terdefinisi, baik cair ataupun
gas, atau dapat pula merupakan kombinasi zat cair dan zat padat. Dalam statika,
perhatian kita terutama dipusatkan pada peng-gambaran gaya-gaya yang bekerja pada
benda tegar dalam keadaan diam, meskipun peninjauan juga diberikan pada statika
fluida. Setelah kita mengambil keputusan mengenai benda atau kombinasi benda mana
yang harus dianalisis, maka benda atau kombinasi benda ini akan diperlakukan sebagai
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benda tunggal yang terpisah dari semua benda di sekitarnya. Pemisahan ini dilaku-kan
dengan memakai diagram benda-bebas, yang merupakan suatu penggambaran
diagramatik dari benda atau kombinasi benda yang terpisah yang ditinjau sebagai benda
tunggal, dan menun-jukkan semua gaya yang dikenakan padanya dengan kontak
mekanis dengan benda-benda lain yang seolah-olah telah dihilangkan. Bila terdapat
gaya benda yang cukup besar, seperti tarikan gravitasi atau tarikan magnetik, maka
gaya ini juga harus ditunjukkan pada diagram benda terpisah tersebut. Hanya setelah
diagram demikian digambar secara benar, barulah persamaan kese-timbangan dapat
ditulis. Karena hal di atas sangat penting, maka kita menekan lagi di sini bahwa diagram
benda-bebas merupakan satu-satunya tahafan terpenting dalam penyelesaian persoalan mekanika.
Sebelum kita mencoba menggambarkan diagram benda-bebas, karakteristik mekanis
dari pengenaan gaya harus diketahui. Dalam Modul VI, karakteristik-karakteristik dasar
dari gaya di-gambarkan dengan perhatian utama dipusatkan pada sifat-sifat vektor dari
gaya tersebut. Kita perhatikan bahwa gaya dapat dikenakan dengan cara kontak fisik
langsung dan dengan aksi dari jauh dan bahwa gaya dapat berupa gaya dalam (internal)
atau gaya luar (external) pada benda yang ditinjau. Lebih lanjut dapat kita amati bahwa
pengenaan gaya luar selalu akan disertai dengan gaya reaktif dan bahwa gaya yang
dikenakan dan gaya reaktif ini dapat terpusat ataupun terdistribusi. Sebagai tambahan,
prinsip transmisibilitas telah diperkenalkan, yang memungkin-kan perlakuan gaya
sebagai sebuah vektor geser sepanjang pengaruh luarnya pada benda tegar yang
ditinjau. Kita sekarang akan memakai karakteristik gaya ini dalam mengembangkan
model analitis dari suatu sistem mekanis terpisah di mana persamaan kesetimbangan
akan diterapkan kemudian.
Gambar 4-1 memperlihatkan jenis-jenis penerapan gaya pada sistem mekanis untuk
analisis dua-dimensi. Dalam setiap contoh pada gambar tersebut ditunjukkan gaya yang
dikenakan pada benda yang dipisahkan oleh benda yang dihilangkan. Hukum Newton
ketiga, yang menyatakan eksistensi dari reaksi yang sama dan berlawanan arah dari
setiap aksi, harus diamati secara sek-sama. Gaya yang dikenakan pada benda yang
disebabkan oleh suatu batang sentuh atau batang sokong selalu berada dalam arahan
yang berlawanan dengan gerakan benda yang akan terjadi apabila batang sentuh atau
batang sokong ini dihilangkan.
Dalam contoh 1 diperlihatkan aksi dari kabel lentur, sabuk, tali, atau rantai pada benda
yang mengait benda-benda tersebut. Karena kelenturannya, sebuah tali atau kabel tak
mungkin me-miliki tahanan terhadap lenturan, geseran, atau desakan (compression)
dan oleh karena itu hanya ada suatu gaya tarikan (tension) dalam arah tangensial (garis
singgung) terhadap kabel di titik kaitnya. Gaya yang dikenakan oleh kabel pada benda
yang mengait kabel tersebut selalu mempunyai arah menjauhi benda. Jika tarikan T jauh
lebih besar bila dibandingkan dengan berat kabel, kita dapat menganggap bahwa kabel
tersebut akan membentuk garis lurus. Jika berat kabel tidak dapat diabaikan bila
dibandingkan dengan besar tarikannya, lenturan kabel akan menjadi hal penting, dan
arah dan besar tarikan dalam kabel akan berubah di sepanjang kabel. Di titik kait kabel
juga terjadi suatu gaya.yang tangensial terhadap dirinya sendiri.
Jika terjadi sentuhan antara permukaan licin dari dua buah benda, seperti dalam Contoh
2, maka gaya yang dikenakan oleh salah satu benda terhadap yang lainnya akan tegak
lurus terhadap garis singgung pada titik sentuh kedua permukaan tersebut dan
merupakan gaya desak (compressive). Meskipun pada kenyataannya tak ada
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permukaan yang benar-benar licin, tapi asumsi ini dapat dibenarkan untuk tujuan praktis
dalam banyak hal.
Jika kedua permukaan benda yang bersentuhan tersebut ternyata kasar,
Contoh 3, maka gaya sentuhnya mungkin tidak tegak lurus terhadap garis singgung
sentuh permukaan tetapi dapat diuraikan menjadi komponen tangensial atau friksional F
dan komponen tegak lurus N.
Contoh 4 melukiskan sejumlah bentuk tumpuan mekanis yang secara efektif dapat
menghi-langkan gaya gesekan tangensial, dan di sini reaksi bersih yang terjadi adalah
tegak lurus terhadap permukaan tumpuan.
Contoh 5, memperlihatkan aksi suatu batang pengarah licin (smooth guide) pada benda
yang ditumpunya. Hambatan yang sejajar batang pengarah ini ternyata tidak ada.
Contoh 6 melukiskan aksi sebuah perletakan sendi. Perletakan jenis ini dapat menahan
gaya dalam segala arah yang tegak lurus terhadap sumbu jepitnya. Kita biasanya
menggambarkan aksi ini dalam dua komponen persegi panjang. Arahan yang benar dari
komponen-komponen ini dalam persoalan yang sesungguhnya akan bergantung pada
bagaimana batang tersebut dibebani.
Didefinisikan bahwa:
Bila sebuah benda tegar ada dalam keseimbangan static,maka:
“Gaya Resultante dan kopel,keduanya harus =0”
Atau dapat dituliskan dalam persamaan matematik:
 F  0, dan
 M  0......(4.2)
C
Untuk tiga dimensi ,setiap gaya dan momen mempunyai tiga komponen,sehingga
pers.(4.2) diatas dapat ditulis:
F
F
F
Dan:
X
0
Y
0
Z
0
M
M
M
…..(4.3)
CX
0
CY
 0 ….(4.4)
CZ
0
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Contoh2 soal
1.Sebuah gaya 5kN dan sebuah kopel 4kN-m diapplikasikan pada sebuah batang
berbentuk garpu seperti terlihat dalam Gambar.Hitunglah gaya reaksi pada jepit A dan
pena B bila sudut θ=36.87˚.
Jawab:
FBD
i.Gambarkan FBD lengkap dengan sumbu referensi sumbu x dan y
ii.Hitunglah gaya2 aksi dan reaksi
Ambil jumlah moment dititik A untuk mengeliminer gaya2 reaksi yg tidak diketahui.
Jadi:
M
AZ
  N B cos 53.13 (0.09)  N B sin 53.13 (0.250  0.120)  4  5(0.600)  0....i.
Keseimbangan gaya didapat dari:
F
F
X
 N B cos 53.13  AX  0......ii.
Y
 N B sin 53.13  AY  5  0.....iii .
Dari (i) didapat:NB [-0.6(0.09) + 0.8(0.13)] + 4-3=0
NB[-0.05 + 0.10] + 1 =0
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NB  
1
 20kN
0.05
Substitusikan kedlm pers.(ii),maka:
 20(0.60)  AX  0  AX  12kN
Dari ..(iii ) :
 20(0.80)  AY  5  0  AY  21kN
2.Sebuah batang disandarkan diantara pena A dan B,sedangkan ujung lainnya berada
dilantai tanpa gesekan.
Hitunglah semua gaya F dan sudut θ
Jawab:
Step 1
Gambarkan FBD
Step 2.
Ambil sumbu Y // batang dan sumbu X berpangkal di C.
Step 3.
Karena pena dan lantai sangat licin,gaya reaksi adalah normal terhadap permukaan
lantai.
Step 4.
Ambil jumlah momen dititik C,maka:
M
CZ
  N A ( L / 3)  N B (2L / 3)  F sin   L  0......i
Step 5.
Menyamakan jumlah gaya2 pada arah sumbu X dab Y,yaitu:
F
F
X
 N C sin 30   N A  N B  F sin   0....ii.
Y
 N C cos 30   F cos   0.....iii .
Step. 6.
Memecahkan 3 persamaan dengan 3 buah bilangan tak diketahui,NA,NB,dan NC.
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Dari pers.(iii):
NC  
F cos 
 1.1597 F cos 
cos 30
Substitusikan kedalam pers.(ii),maka:
( F
cos 
) sin 30   N A  N B  F sin   0....iv.
cos 30
Dari pers.( i) didapat:
 N A  L / 3  N B  2 L / 3  F sin   L  0
 N A  2 N B  3F sin   0........(v)
Jumlah pers.(iv) x 3 + pers.(v) adalah:
cos 
) sin 30   N A  N B  F sin   0.....  3
cos 30
...........................  N A  2 N B  3F sin   0
( F
cos 
) sin 30   2 N A  N B  0

cos 30
2 N A  N B  3F cos  tan 30   3F cos   0.5774  1.7321F cos  ....(vi)
(3F
Pers.(v) x 2 + pers.(vi):
 2 N A  4 N B  6 F sin
 2 N A  N B  1.7321F cos 
3N B  6F sin   1.7321F cos
 N A  2 N B  3F sin 
 2 F (2 sin   0.576 cos  )  3F sin 
 F sin   1.15F cos 
NC tidak mungkin =0,karena gaya normal harus tegak lurus permukaan ,maka
hendaklah:
cos  0


yang akan dipenuhi untuk 90    270
Untuk NB positif atau nol,maka
2 sin   0.57 cos   0
sin 
0.57

 0.285
cos 
2
  (15.9  180)  165 ..atau..344 
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Untuk NA positif atau nol,maka:
sin   1.15 cos   0
sin 
 1.15    49..atau..
cos 
  180  49  131...atau....360  40  311
3.Sebuah kunci inggris pembuka baut seperti yang terlihat dalam Gambar.
Hitunglah besar momen diujung A bila sudut α=75˚ dan besar gaya F=80N
Hitunglah besar sudut α sedemikian rupa sehingga besar momen diujung A mencapai
nilai maximum.
FBD:
Gambar soal 3
Jawab:
i.Gambarkan sketsa diagram bebasnya(FBD),dan tentukan posisi sumbu x dan y yg
strategis
ii.Uraikanlah gaya F terhadap sumbu2nya.
Diketahui α=75˚ ,F=80N
Besar Moment dititik O adalah:
M O   dy  Fx  dx  Fy .......i.
dx  OC  sin 53.13  BO  0.7999  0.03  2.4cm
dy  AC  AB  BC  AB  cos 53.13  BO 
 25cm  1.8cm  26.8cm
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Fx  F cos(90  75)  80 N  0.9659  77.2741N
Fy  F sin( 90  75)  80 N  0.2588  20.7055 N
 M O  26.8cm  77.2741N  2.4cm  20.7055 N 
 2070.9460 N  cm  49.6932 N  cm  2021.2528 N  cm
20.21N  m.
Besar sudut α akan mencapai nilai maximum bila:
dM O
 0.....ii.
d
M O   dy  Fx  dx  Fy
 26.8  80 sin   2.4  80 cos  
 2144 sin   192 cos 
dM O
 2144 cos   192 sin   0
d
sin 
2144

 11.166
cos 
192
   84.88     90  84.84  5.12 
Soal2 Latihan:
4.1 Sebuah kerekan dipasang pada batang besi ABC bentuk huruf L, lihat Gambar.
Sebuah gaya tarik sebesar 200 lb diapplikasikan thd ujung bebas kabel tersebut. Hitung
gaya tarik dalam kabel, gaya-gaya reaksi dipena A, jepit B dan roller C.
Gambar soal 4.1
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Gambar soal 4.2
Dr. Ir. Abdul Hamid M.Eng.
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4.2.Sebuah kunci pas dipasang pada mur (baut) sebuah ban mobil seberat massa 20 kg,
lihat Gambar. Hitunglah besar gaya F sehingga ban dapat berputar melalui ujung
tembok B.
Gambar soal 4.3.
4.4.Sebuah pelat berbentuk empat pesegi bermassa 2000kg diikat dengan sebuah
engsel dititik A,dan ditopang dengan pengikat horizontal dititik E,kemudian diikatkan
oleh dua buah kabel,lihat Gambar.
Gaya beratnya bekerja dititik pusat G.Hitunglah tegangan dalam kabel2 tsb dan gaya2
reaksi dititik A.
4.5.Struktur truss seperti yang terlihat dalam pada Gambar dibawah ini,dibebani oleh
gaya2 horizontal dan vertical.
Hitunglah resultante dari beban tsb dan tentukanlah besar ,arah dan titik applikasinya
pada garis AB.
4.3 Moment of a Couple. Two forces F and having the same magnitude, parallel lines
of action, and opposi sense are said to form a couple (Fig. 4.5). Clearly, the sum the
components of the two forces in any direction is zero. The sum of the moments of the
two forces about a given point
however, is not zero. While the two forces will not translate the body on which they act,
they will tend to make it rotate.
Denoting by rA and rB, respectively, the position vectors of the points of application of F
and -F (Fig. 4.6), we find that the sum of the moments of the two forces about O is
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rA  F  rB  (F )  (rA  rB )  F
Setting (rA  rB )  r , where r is the vector joining the points of application of the two
forces, we conclude that the sum of the moments of F and — F about O is represented
by the vector.
M = r × F …..(4.5)
The vector M is called the moment of the couple; it is a vector perpendicular to the plane
containing the two forces and the magnitude is
M=rFsinθ= Fd …..(4.6)
where d is the perpendicular distance between the lines of action of F and — F. The
sense of M is defined by the right-hand rule. Since the vector r in (4.5) is independent of
the choice of the origin O of the coordinate axes, we note that the same rest would have
been obtained if the moments of F and — F had been computed about a different point
O'. Thus, the moment M of a couple is a free vector which may be applied at any point
(Fig. 4.7).
Fig.4.5
Fig.4.6
Fig 4.7
From the definition of the moment of a couple, it also follows that two couples, one
consisting of the forces F1 and — F1, the other of the forces F2 and — F2 (Fig. 3.33), will
have equal moments if
F1d1= F2d2.....(4.7)
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and if the two couples lie in parallel planes (or in the same plane) and have the same
sense.
4.4. Equivalent Couples. Consider the three couples shown in Fig. 4.8, which are
made to act successively on the same rectangular box. As seen in the preceding
section, the only motion a couple may impart to a rigid body is a rotation. Since each of
the three couples shown has the same moment M (same direction and same magnitude
M = 120 Ib • in.), we may expect the three couples to have the same effect on the box.
Fig. 4.8
As reasonable as this conclusion may appear, we should not accept it hastily. While
intuitive feeling is of great help in the itudy of mechanics, it should not be accepted as a
substitute for logical reasoning. Before stating that two systems (or groups) of forces
have the same effect on a rigid body, we should prove that fact on the basis of the
experimental evidence introduced so far. This evidence consists of the parallelogram law
for the addition of two forces and of the principle of transmissibility . Therefore, we shall
state that two systems of forces are equivalent ('i.e., they have the same effect on a rigid
body) if we transform one of them into the other by means of one or several the
following operations: (1) replacing two forces acting on the same particle by their
resultant; (2) resolving a force into two components; (3) canceling two equal and
opposite forces acting on the same particle; (4) attaching to the same
Fig.8
particle two equal and opposite forces; (5) moving a force al its line of action. Each of
these operations is easily justified on the basis of the parallelogram law or the principle
of transmissibility.
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Let us now prove that two couples having the same moment are equivalent. First, we
shall consider two couples contained in the same plane, and we shall assume that this
plane coincide with the plane of the figure (Fig. 4.8).
The first couple coiw of the forces F1- and - Fl of magnitude F1 and at a distance d1 from
each other (Fig. .8a), and the second couple of the for F2 and - F2, of magnitude F2 and
at a distance d2 from another (Fig. 4.8b). Since the two couples have the same moment
M perpendicular to the plane of the figure, they must have the same sense (assumed
here counterclockwise) and the relation
F1d1
= F2d2 …..(4.8)
must be satisfied. To prove that they are equivalent, we will show that the first couple
may be transformed into the second by means of the operations listed above.
Denoting by A, B, C, D the points of intersection of the lines action of the two couples,
we first slide the forces F1 and –F1 . until they are attached, respectively, at A and B, as
shown in Fig.4.8d. The force F1 is then resolved into a component P ah line AB and a
component Q along AC (Fig. 4.8c); similarly, F1 force - F1 is resolved into - P along AB
and -Q along ED. The forces P and - P have the same magnitude, same line of action
and opposite sense; they may be moved along their common line of action until they are
applied at the same point and then canceled. Thus the couple formed by = F1, and –F1
reduces to a couple consisting of Q and – Q.
We shall now show that the forces Q and - Q are respective equal to the forces — F2
and F2. The moment of the couple formed by Q and — Q may be obtained by
computing the moment of Q about B; similarly, the moment of the couple formed by F1
and - F1 is the moment of Fa about B. But, by Varignon's :'ieorem, the moment of Fj is
equal to the sum of the moments of its components P and Q. Since the moment of P
about B is zero, :ae moment of the couple formed by Q and — Q must be equal to the
moment of the couple formed by F1 and - F1. We write:
Qd2 = F1d1 = F2d2
and
Q = F2
Kalau perlu:
Thus the forces Q and — Q are respectively equal to the forces
-F2 and F2, and the couple of Fig. 4.8a is equivalent to the couple of Fig. 4.8d.
Next we shall consider two couples contained in parallel planes P-^ and P2 and prove
that they are equivalent if they have the same moment. In view of the foregoing we may
assume that the couples consist of forces of the same magnitude F acting along parallel
lines (Fig. 3.36a and d). We propose to show that the couple contained in plane P1 may
be transformed into the couple contained in plane P2 by means of the standard
operations listed above.
Let us consider the two planes defined respectively by the lines ofaction of Fl and -F2,
and of -F-, and F2 (Fig. 3.36fo). At a uiiit on their line of intersection we attach two forces
F3 and
-F3, respectively equal to Fl and — F1. The couple formed by 1 md — F3 may be
replaced by a couple consisting of F3 and
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Dr. Ir. Abdul Hamid M.Eng.
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-F2 ^Fig, 3.36c), since both couples have clearly the same moment and are contained in
the same plane. Similarly, the couple formed by — F: and F3 may be replaced by a
couple consisting of
-F3 and F2. Canceling the two equal and opposite forces F3 and
-F3, we obtain the desired couple in plane P2 (Fig. 3.36d). luis, we conclude that two
couples having the same moment M .re equivalent, whether they are contained in the
same plane or iii parallel planes.
The property we have just established is very important for the jrrect understanding of
the mechanics of rigid bodies. It indices that, when a couple acts on a rigid body, it does
not matter •here the two forces forming the couple act, or what magnitude ,.nd direction
they have. The only thing which counts is the ".merit of the couple (magnitude and
direction). Couples with :liesame moment will have the same effect on the rigid body.
3.13. Addition of Couples. Consider two intersecting planes l\ and P2 and two couples
acting respectively in P: and P2. We may, without any loss of generality, assume that the
couple in?, consists of two forces Ft and — Fj perpendicular to the line
of intersection of the two planes and acting respectively at A -a B (Fig. 3.37«). Similarly,
we assume that the couple in P2 com. of two forces F2 and — F2 perpendicular to AB
and acting rr spectively at A and B. It is clear that the resultant R of Fj amlF and the
resultant — R of — Ft and — F2 form a couple. Denolr by r the vector joining B to A, and
recalling the definition of tt moment of a couple (Sec. 3.11), we express the moment M of
tli resulting couple as follows:
M = r x R = r x (F1 + F2)
and, by Varignon's theorem,
But the first term in the expression obtained represents the in; ment Mj of the couple in l\,
and the second term the momer M2 of the couple in P2. We have
M = Mt + M2 (3..?
and we conclude that the sum of two couples of moments Mj a' M2 is a couple of
moment M equal to the vector sum of M, a: M2 (Fig. 3.37fo).
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3.14. Couples May Be Represented by Vectors
As we saw in Sec. 3.12, couples which have the same momer whether they act in the
same plane or in parallel planes, as equivalent. There is therefore no need to draw the
actual fore: forming a given couple in order to define its effect on a rign body (Fig.
3.38a). It is sufficient to draw an arrow equal i; magnitude and direction to the moment M
of the couple (R 3.38b). On the other hand, we saw in Sec. 3.13 that the sum'. two
couples is itself a couple and that the moment M of tt. resultant couple may be obtained
by forming the vector suni' the moments M.^ and M2 of the given couples. Thus, couplt
obey the law of addition of vectors, and the arrow used in Fit 3.38£> to represent the
couple defined in Fig. 3.38« may truly Iv considered a vector.
The vector representing a couple is called a couple vector. Note that, in Fig. 3.38, a
colored arrow is used to distinguish the couple vector, which represents the couple itself,
from the vector representing the moment of the couple, and that the symbol ^ is added to
avoid any confusion with vectors representing forces. A couple vector, like the moment
of a couple, is a free vector. Its point of application, therefore, may be chosen at the
origin of the system of coordinates, if so desired (Fig. 3.38c). Furthermore, the couple
vector M may be resolved into component vectors Mx, M , and M^, directed along the
axes of coordinates (Fig. 3.38d) and representing couples acting, respectively, in the yz,
zx, and it/ planes.
3.15. Resolution of a Given Force into a Force at O and a Couple. Consider a force
F acting on a rigid body at a point A defined by the position vector r (Fig. 3.39a).
Suppose that for some reason we would rather have the force act at point 0, We know
that we can move F along its line of action (principle of transmissibility); but we cannot
move it to a point O away from the original line of action without modifying the action of F
on the rigid body.
Fig. 3.39
We may, however, attach two forces at point O, one equal to F and the other equal to —
F, without modifying the action of the original force on the rigid body (Fig. 3.39k). As a
result of this transformation, a force F is now applied at O; the other two forces form a
couple of moment M0 = r X F. Thus, any force F acting on a rigid body may be moved to
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Dr. Ir. Abdul Hamid M.Eng.
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an arbitrary point O, prodded that a couple is added, of moment equal to the moment ufF
about O. The couple tends to impart to the rigid body the ^ame motion of rotation about
O that the force F tended to oroduce before it was transferred to O. The couple is repre>cnted by a couple vector M0 perpendicular to the plane con-laining r and F. Since M0 is
a free vector, it may be applied anywhere; for convenience, however, the couple vector
is usually attached at 0, together with F, and the combination obtained is referred to as &
force-couple system (Fig. 3.39c).
If the force F had been moved from A to a different point ( (Fig. 3.40a and c), the
moment MQ, — r' x F of F about U should have been computed, and a new force-couple
systffi consisting of F and of the couple vector M0/, would have bed attached at O'. The
relation existing between the moments off about O and O' is obtained by writing
MQ, = r' X F = (r + s) x F = r x F + s X F
Mff = M0 + s X F (3.5.
where s is the vector joining O' to O. Thus, the moment M0- off about O' is obtained by
adding to the moment Mo of F about( the vector product s x F representing the moment
about O'a the force F applied at O.
This result could also have been established by observing tk in order to transfer to O' the
force-couple system attached at! (Fig. 3AOb and c), the couple vector M0 may be freely
moved! O'; to move the force F from O to O', however, it is necessary! add to F a couple
vector s x F representing the moment aboi O' of the force F applied at O. Thus, the
couple vector M0- mib be the sum of M0 and s X F.
As noted above, the force-couple system obtained by transfei ring a force F from a point
A to a point O consists of F and of. couple vector M0 = r X F perpendicular to F.
Conversely, an force-couple system consisting of a force F and of a coup!: vector Mo
which are mutually perpendicular may be replaced b a single equivalent force. This is
done by moving the force Fii the plane perpendicular to M0 until its moment about 0 k
comes equal to the couple vector M0 to be eliminated.
SAMPLE PROBLEM 4.6
Two couples act on a block as shown. Replace these two couples with a single
equivalent couple.
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Solution. Each of the given couples is represented by a couple vector which is
perpendicular to the plane of the couple and of a magnitude equal to the moment of the
couple. The sense of each vector is obtained by applying the right-hand rule, and for
convenience both couple vectors are attached at the origin.
The single couple equivalent to the two given couples will be represented by the
resultant of the two couple vectors. The magnitude of the resultant couple vector M is
obtained from the law of cosines.
M2 = (24)2 + (30)2 - (2) (24) (30) cos 120°
M = 46.9 Ib • in.
The angle φy that the resultant couple vector forms with the vertical is obtained from the
law of sines.
sin  y
30lb.in

sin 120 
46.9lb.in
φy=33.60
The angle that the couple vector forms with the x axis is
φx = 90° - 33.6° = 56.4°
and the angle it forms with the z axis is φz = 90°. Thus the resultant couple vector M is
defined by
M = 46.9 Ib • in.
φx = 56.4° φy = 33.6°
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φz = 90°.
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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The. single couple equivalent to the two original couples is a couple of -magnitude M
=46.9 Ib • in., acting in a plane parallel to the z axis and forming an angle of 33.6° with
the horizontal plane. This couple may be formed in many ways, for example, by two
7.82-lb forces acting at corners C and D in the directions shown, or by two 9.77-Ib forces
parallel to the z axis and applied at E and F as shown.
SAMPLE PROBLEM 4.7
Replace the couple and force shown by an equivalent single applied to the
lever.Determine the distance from the shaft to the point of application of this equivalent
force.
Solution. First the given force and couple are replaced by an equivalent force-couple
system at O.
We move the F =- (400 N)j to O and at the same time add a
couple of moment Mo equal to the moment about O of the force in its
original position.
M0 = OB x F = [(0.150 m)i + (0.260 m)j] × (-400N)j
= -(60N-m)k
This couple is added to the couple of moment -(24 N • m)k formed by the two 200-N
forces, and a couple of moment -(84N-m)k is obtained. This last couple may be
eliminated by applying F at a point C chosen in such a way that
-(84N-m)k = OC× F
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= [(OC) cos 60°i + (OC) sin 60°j] × (-400N)j
= -(OC)cos60°(400N)k
We conclude that
(OC) cos 60° = 0.210 m = 210 mm
OC = 420 mm
Alternate Solution. Since the effect of a couple does not depend on its location, the
couple of moment -(24 N • m)k may be moved to B;we thus obtain a force-couple system
at B. The couple may now be eliminated by applying F at a point C chosen in such a way
that
-(24N-m)k = BC x F
= -(BC)cos60°(400N)k
We conclude that
(BC) cos 60° = 0.060 m = 60 mm BC - 120 mm
OC = OB + BC = 300 mm + 120 mm OC = 420mm
PROBLEM 4.7
4.1 The two couples shown are applied to a 120- by 160-mm plate, Knowing that P1 = P2
= 150 N and Q1 = Q2 = 200 N, prove that their sum is zero (a) by adding their moments,
(b) by combining P1 and Q1 into their resultant R1 ,combining P2 and Q2 into their
resultant R2,and then showing that R1 and R2 are equal and opposite and have the
same line of action.
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4.2 A couple formed by two 975-N forces is applied to the pulley assembly shown.
Determine an equivalent couple which is formed by (a) vertical forces acting at A and C,
(b) the smallest possible forces acting at B and D, (c) the smallest possible forces which
can be attached to the assembly.
4.3 Four l-in.-diameter pegs are attached to a board as shown. Two strings are passed
around the pegs and pulled with forces of magnitude P = 20 Ib and Q = 35 Ib.
Determine the resultant couple acting on the board.
4.4 A multiple-drilling machine is used to drill simultaneously six hople in the steel plate
shown. Each drill exerts a clockwise couple of magnitude 40 Ib • in. on the plate.
Determine an equivalent couple formed by the smallest possible forces acting (a) at A
and C, (b) at A and D, (c) on the plate.
3.53 The axles and drive shaft of an automobile are acted upon by by three couples
shown. Replace these three couples by a single equivalent couple.
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Dr. Ir. Abdul Hamid M.Eng.
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Modul VIII:
4.9 Equivalent Systems of Forces. We have seen in the preceding section that any
system of forces acting on a rigid body may be reduced to a force-couple system at a
given point 0. This equivalent force-couple system characterizes completely the effect of
the given system on the rigid body. Two i of forces are equivalent, therefore, if they may
be reduced same force-couple system at a given point O.
We state: Two systems of forces Fl5 F2, F3, etc., and F’1, F’2, F’3,ete., are equivalent if,
and only if, the sums of the forces and the sums of the moments about a given point O
of the forces of the two systems are, respectively,, equal. Expressed mathematically,
the necessary and sufficient conditions for the two systems of forces to be equivalent are
ΣF = ΣF'
and
Σ M0 = ΣM0’ (8.1)
Note that, to prove that two systems of forces are equivalent t second of the relations
(3.57) needs to be established with respt; to only one point O. It will hold, however, with
respect to the point if the two systems are equivalent.
Resolving the forces and moments in (8.1) into their rectangular components, we may
express the necessary and sufficient conditions for the equivalence of two systems of
forces acting on a rigid body as follows:
Fx  Fx/ ........Fy  Fy/ ............Fz  Fz/
M x  M x/ .....M y  M y/ .........M z  M z/
……..(8.2)
These equations have a simple physical significance. They express that two systems of
forces are equivalent if they tend impart to the rigid body (1) the same translation in the
x, y,z directions, respectively, and (2) the same rotation about the x,y and z axes,
respectively.
4.10. Equipollent Systems of Vectors. When the systems of vectors satisfy Eqs. (8.1)
or (8.2), i.e., when the resultants and their moment resultants about an arbitrary point
are respectively equal, the two systems are said to be equipollent. The result established
in the preceding section may be restated as follows: if two systems of forces acting on a
rigid body are equipollent, they are also equivalent.
It is important to note that this statement does not apply any system of vectors. Consider
for example a system of force acting on a set of independent particles which do not form
a rigid body. A different system of forces acting on the same particle may happen to be
equipollent to the first one, i.e., it may have the same resultant and the same moment
resultant. Yet, since different forces will now act on the various particles, their effet on
these particles will be different; the two systems of forces, while equipollent, are not
equivalent..
4.11. Further Reduction of a System of Force; We saw in that any given system of
forces acting ot rigid body may be reduced to an equivalent force-couple system at O,
consisting of a force R equal to the sum of the forces of the system, and of a couple
R
vector M O equal to the moment resultant of the system.
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R
When R =0, the force-couple system reduces to the couple vector M O . The given
system of forces may then be reduced to a single couple, called the resultant couple of
the system.
We shall now investigate the conditions under which a given system of forces may be
reduced to a single force. It follows that the force-couple system at O may be replaced
R
by a single force R acting along a new line of action if R and M O are mutually
perpendicular. The systems of forces which may be reduced to a single force, or
R
resultant, are therefore the systems for which the force R and the couple vector M O
are mutually perpendicular. While this condition is generally not satisfied by systems of
forces in space, it will be satisfied by systems consisting (1) of concurrent forces, (2) of
coplanar forces, or (3) of parallel forces. We shall discuss these cases separately.
1.Concurrent forces are applied at the same point and may therefore be added directly
into their resultant R, Thus, they always reduce to a single force.
2.Coplanar forces act in the same plane, which we shall assume here to he the plane of
the figure (Fig. 8.1). The sum R of the forces of the system will also lie in the plane of the
R
figure, while the moment of each force about O, and thus the moment resultant M O ,
will be perpendicular to that plane. The force-couple system at O consists therefore of a
R
force R and a couple vector M O which are mutually perpendicular (Fig. 8.1). They may
be reduced to a single force R by moving R in the plane of the figure until its moment
R
about O becomes equal to M O . The distance from O to the line of action of R is (Fig.
\8.1c).
Fig.8.1
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Fig.8.2
The reduction of a system of force is considerably simplified if the forces are resolved
into angular components. The force-couple system at 0is then characterized by the
components (Fig. 8.2a)
R x  Fx .....R y  Fy
M zR = M OR = M O …(8.3)
To reduce the system to a single force R we shall express the moment of R about O
R
must be equal to M O . Denoting x and y the coordinates of the point of application A of
the resultant, we write
R
xRy - yRx = M O
which represents the equation of the line of action of R. We may also determine directly
R
the x and y intercepts of the line of action of the resultant by noting that M O must be
equal to the moment about O of the y component of R when R attached at B (Fig. 8.2),
and to the moment of its x component when R is attached at C (Fig. 8.2c).
3. Parallel forces have parallel lines of action and may or may not have the same sense.
Assuming here that the forces parallel to the y axis (Fig. 8.3a), we note that their sum R
will also be parallel to the y axis. On the other hand, since moment of a given force must
be perpendicular to that force, the moment about O of each force of the system, and
R
thus the moment resultant M O , will lie in the zx plane. The force-couple system at O
R
consists therefore of a force R and couple vector M O which are mutually perpendicular
(Fig. 8.3). They may be reduced to a single force R (Fig. 8,3c) or, if R = 0, to a single
R
couple of moment M O .
In practice, the force-couple system at O will be characterized by the components
Ry = ΣFy
M xR = M x
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M OR = M z
(8.4)
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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Fig.8.4
The reduction of the system to a single force may be carried out by moving R to a new
R
point of application A(x,0,z) chosen so that the moment of R about O is equal to M O .
We write
R
r x R =MO .
R
(xi + zk) × Ryj = M x i + M zR k
Computing the vector products and equating the coefficients of the corresponding unit
vectors in both members of the equation, we obtain two scalar equations which define
the coordinates of A:
R
-zRy= M x
xRy= M zR
These equations express that the moments of R about the x axes must, respectively, be
R
equal to M x and M zR .
Ih the general case of a system of forces in space, the forces couple system at O
R
consists of a force R and a couple vector M O which are not perpendicular, and neither
of which is zero (Fig.8.4 ). Thus, the system of forces cannot be reduced to a single
couple. The couple vector, however, may be two other couple vectors obtained by
R
resolving M O .
Fig.8.5
An important particular case of the reduction of a system forces to a force-couple system
R
occurs when both the force R and the couple vector M O are equal to zero. The system
of force is said to be equivalent to zero. Such a system has no effect out-rigid body on
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which it acts, and the rigid body is said to be equilibrium. This case will be considered in
detail in the next chapter.
SAMPLE PROBLEM 8.1
A 4.80-m beam is subjected to the forces shown, Reduce the given system of forces to
(a) an equivalent force-couple system at A, (b) an equivalent force-couple system at B,
(c) a single force or resultant.
Note. Since the reactions at the supports are not included in the given system of forces,
the given system will not maintain the beam in equilibrium.
a. Force-couple System at A. The force-couple system at A equivalent to the given
system of forces consists of a force R and a couple M AR defined as follows:
R = ΣF
= (150 N)j -(600 N)j + (100 N)j - (250 N)j = -(600 N)j
M AR = Σ{r × F)
= (1.6i) x (-600j) + (2.8i) x (100j) + (4.8i) x (-250j)
= -(1880N-m)k
The equivalent force-couple system at A is thus
M AR = 1880 N • m (+)
R = 600 N(+)
b. Force-couple System at B. We shall find a force-couple system at B equivalent to the
force-couple system at A determined in part a. The force R is unchanged, but a new
couple M BR must be determined, the moment of which is equal to the moment about B
of the force-couple system determined in part a. Thus, we have
M BR = M AR + BA x R
= -(1880N-m)k + (-4.8 m)i x (-600 N)j
= -(1880 N - m)k + (2880 N • m)k = +(1000 N • m)k
The equivalent force-couple system at B is thus
M BR = 1000 N - m (+)
R = 600 N(+)
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c. Single Force or Resullant. The resultant of the given system of forces is equal to R
and its point of application must be such that the moment of R about A is equal to M AR .
We write
r x R = M AR
xi × (-600 N)j = -(1880N-m)k
-x(600 N)k = -(1880 N • m)k
and conclude that x= -3.13 m. Thus, the single force equivalent to the given system is
defined as
R= 600N,
x= 3.13m
SAMPLE PROBLEM 8.2
Four tugboats are used to bring an ocean liner to its pier. Each tugboat exerts a 5000-lb
force in the direction shown. Determine the equivalent force-couple system at the
foremast O, (b) the point on the hull where a single, more powerful tugboat should push
to produce the same effect as the original four tugboats.
a. Force-couple System at O. Each of the given forces is resolved into components in
the diagram shown (kip units are used), The force-couple system at O equivalent to the
R
given system of force consists of a force R and a couple M O defined as follows:
R = ΣF
= (2.50i - 4.33j) + (3.00i - 4.00j) + (-5.00j) + (3.54i + 3.54j)
= 9.04i - 9.79j
M BR = Σ(r x F)
= (-90i + 50j) x (2.50i - 4.33j)
+ (400i + 70j) x (3.00i - 4.00j)
+ (400i + 70j) x (-5.00j)
+ (300i - 70j) x (3.54i + 3.54j)
= (390 - 125 - 400 - 210 - 2000 + 1062 + 248)k
= -1035k
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
31
The equivalent force-couple system at O is thus
M OR = -(1035 kip • ft)k
R = (9.04 kips)i - (9.79 kips)j
or
R = 13.33 kips θ=47.3°
M OR = 1035 kip • ft(-)
Remark. Since all the forces are contained in the plane of the figure we could have
expected the sum of their moments to be perpendicular to that plane. Note that the
moment of each force component could have been obtained directly from the diagram by
forming the product of its magnitude and its perpendicular distance to O, and assigning
this product a positive or a negative sign, depending upon the sense the moment.
b. Single Tugboat. The force exerted by a single tugboat must be equal to R and its point
R
of application A must be such that the moment of R about O is equal to M O . Observing
that the position vector A is
r = xi + 70j
we write
R
r x R = MO
(xi + 70j) x (9.04i - 9.79j) = - 1035k
-x(9.79)k - 633k
= -1035k
x=41.1ft
PROBLEMS
8.1 A 4-m beam is loaded in the various ways represented in the figure 8.5. Find two
loadings which are equivalent.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
32
Fig.8.5
8.2 A 4-m beam is loaded as shown in Fig.8.5. Determine the loading of Prob.8.1 which
is equivalent to this loading.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
33