Modul X
SAMPLE PROBLEM 10.1
Three loads arc applied to a steel plate as shown. The plate is supported by a roller at
A and by a pin at B. Determine the reaction at A and B.
Solution. A free-body diagram of the plate is drawn. The reaction at A is vertical and is
denoted by A. The reaction at B is represented by components Bx and By . Each
component is assumed to act in the direction shown.
Equilibrium Equations. We write the following three equations and solve for the
reaction indicated:
ΣFx =0:
Bx =0:
ΣMA =0
=-(2kips)(3ft) - (3kips)(3ft)-(4kips)(5ft) + By(6ft)=0
By=+5.17kips
ΣMB = 0:
-A(6ft)+(2kips)(5ft) + (3 kips)(3ft) + (4kips)(1 ft) = 0
A =+3.83kips.
Check.The results are checked by adding the vertical components of all the external
forces.
ΣFy = +5.17 kips + 3.83 kips - 2 kips - 3kips - 4 kips =0
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Dr. Ir. Abdul Hamid M.Eng.
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SAMPLE PROBLEM 10.2
A loading car is at rest on a track forming an angle of 250 with the vertical. The gross
weight of the car and its load is 5500 Ib, and it is applied at a point 30 in. from the track,
hallway between the two axles. The car is held by a cable attached 24 in. from the track.
Determine the tension in the cable and the reaction at each pair of wheels.
Solution. A free-body diagram of the car is drawn. The reaction at each wheel is
perpendicular to the track, and the tension force T is parallel to the track. For
convenience, we choose the x axis parallel to the track and the y axis perpendicular to
the track. The 5500-lb weight is then resolved into x and y components.
W x = +(5500lb)cos250 = +4980lb
Wy = - (5500lb)sin 25° = -2320lb
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Dr. Ir. Abdul Hamid M.Eng.
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Equilibrium Equations. We take moments about A to eliminate
T and R1 from the
computation.
ΣMA =0:
-(2320 lb)(25in.) – (4980lb)(6in.) + R2 (50in.)=0
R2 =+1758lb
Now, taking moments about B to eliminate T and R2 from the computation, we write:
ΣMB =0: (2320lb)(25in.)-(4980lb)(6in.)-R1(50in.)=0
R1=562lb.
The value of T Is found by writing:
ΣFx =0:
4950 lb – T=0
T = +4930 lb
The computed values of the reactions are shown in the adjacent sketch. Check. The
computations are verified by writing
ΣFy = 562lb+ 1758 Lb - 2320 Lb = 0
A check could also have been obtained by computing moments about any point except
A or B.
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Dr. Ir. Abdul Hamid M.Eng.
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SAMPLE PROBLEM 10.3
A cantilever beam is loaded as shown.The beam is fixed at the left end and free at the
right end. Determine the reaction at the fixed end.
Solution. The portion of the beam which is embedded in the wall is subjected to a
large number of forces. These forces ,however, are equivalenl to a force of components
Rx and Ry and a couple M.
Equilibrium Equations.
ΣFx =0:
Rx =0
ΣFy =0:
Ry -800N-400N-200N=0
Ry=1400N
ΣMA =0:
=-( 800N)(1.5m)- (400N)(4m) –(200N)(6m)+M =0
M= +4000N. m.
The reaction at the fixed end consists of a vertical upward force of 1400N and of a
4000N.m counterclockwise couple.
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Dr. Ir. Abdul Hamid M.Eng.
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SAMPLE PROBLEM 10.4
A 400-lb weight is attached to the lever AO as shown. The constant of the spring BC is
k = 250 lb /in., and the spring is unstretched when θ = 0. Determine the position or
positions of equilibrium.
Solution, Force Exerted by Spring, Denoting by s the deflection of the spring from
its undeformed position, and noting that s=rθ, we write
F=ks=krθ
Equilibrium Equations. Summing the inomenl of W and F about O,we write:
M O  0 :
Wl sin   r (kr )  0
sin  
kr 2

Wl
Substituting the given data,we obtain:
(250lb / in.)(3in.) 2
sin  

(400lb )(8in.)
 0.703
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Solving by trial and error, we find   0,....  80.3
PROBLEMS
10.1 KoowLtii; thai the inatinltude of the vertical force F is 400 N, determine \a\ llic
leiision In ihe cable CD, \b] Ihe reaclion al B.
10.2 The ladder AB, of length L and weight W , can be raised by the cable BC.
Determine the tension T required to raise end B just off the floor (a)in the term of W and
θ,(b)if h= 8 ft, L = 10ft, and W = 35 Ib.
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Dr. Ir. Abdul Hamid M.Eng.
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10.3 A block of weight W is to be supported by the winch shown. . Determine the
required magnitude of the force P (a) in terms of W, r, l,and θ,(b)if W=100lb, r = 3in., l =
15 in., and θ =600.
5.STRUKTUR
Truss (Rangka Batang)
Truss adalah sebuah kerangka tunggal (rigid framework) dari batang dua gaya yang
dihubungkan oleh joint,seperti yg terlihat pada Gambar 5.1. dibawah ini.
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Gambar 5.1.Beberapa Kerangka Tunggal Truss.
Kriteria truss
1. semua batang adalah lurus
2. center lines (sumbu) semua batang bertemu disatu titik pada joint (titik simpul)
3. gaya-gaya luar yang bekerja pada struktur tersebut harus diapplikasikan pada
titik simpul.
Planar truss (2 dimensi) : m = 2j-3.
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Space truss (3 dimensi) : m = 3j-6.
Disini : m = jumlah batang
j = jumlah joint
5.1 Metoda joint
Pada setiap joint berlaku :
ΣFx = 0, ΣFy = 0, ΣFz = 0
Gambar 5.2.dibawah ini menunjukkan FBD gaya2 yg bekerja pada setiap joint.
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Contoh
1.Hitunglah gaya-gaya dalam setiap batang dari truss seperti Gambar dibawah ini
dengan menggunakan metoda joint.
Jawab:
i.Gambarkanlah FBD dengan sumbu2 referensinya x dan y,
ii. Uraikanlah gaya2 aksi dan reaksinya // sumbu x dan y
Assumsikan bahwa gaya F6-7 adalah gaya tarik dan ambiljoint 4 untuk penjumlahan
moment.Persamaan keseimbangannya adalah:
M
4z
 10(4)  4(4)  ( F67 sin 36.87  )  8
 (5 sin 30  )  8  (5 cos 30  )  3  0.......i.
F
F
X
  F67 cos 36.87   ( F4 ) X  5 cos 30   0......ii.
Y
  F67 sin 36.87   ( F4 ) Y  4  10  5 sin 30   0....iii .
Dari pers.(i) didapat: F6-7=-10.206kN
Idem (ii) didapat (F4)x=-3.835kN
Idem (iii)didapat (F4)Y=5.376kN
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Step.iii.
Pertimbangkan keseimbangan gaya pada setiap joint,untuk itu gambarkanlah truss tsb
lengkap dengan data2 yg diperoleh dari hitungan diatas.
Untuk perhitungan yg terbaik
adalah dengan mempertimbangkan joint yg
menghubungkan batang yg tersedikit untuk dihitung terlebih dahulu. Jadi urutannya
adalah:
(1)  (3)  (4)  (5)  (6)  (2)....atau....
(3)  (1)  (4)  (5)  (6)  (2)
Joint.1
F
F
X
 F1 2  5 cos 30   0
Y
  F16  5 sin 30   0
F1 2  4.330kN
F16  2.50kN
Joint.3
F
F
X
  F2 3  0
Y
  F3 4  0
F2 3  F3 4  0
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Joint.4.
F
X
  F45  F2 4 cos 36.87   3.835  0
  F45  0.8 F2 4  3.835  0
F
Y
 F3 4  F2 4 sin 36.87   5.376  0
 0  0.6 F2 4  5.376  0
 F2 4  8.960kN.......F45  3.333kN
Joint.5
F
X
 F4 5  F56  0
F5 6  3.333kN
F
Y
 F2 5  4  0
F2 5  4kN
Joint.6
F
X
 F56  F2 6 cos 36.87   10.206 cos 36.87   0
 3.333  0.8 F2 6  8.165  0
F
Y
 F16  F2 6 sin 36.87   10.206 sin 36.87   0
 2.50  0.6 F2 6  6.124  0
F2 6  14.373kN
Joint.2
F
X
 F23  F23 cos 36.87   F26 cos 36.87   F1 2  0
 0  (8.960)0.8  (14.373)0.8  4.330  0
F
Y
 ( F26  F2 4 ) sin 36.87   F25  10.0  0
 (14.373  8.960)0.6  4  10.0  0
Soal latihan
Hitunglah gaya2 dalam batang dari space truss (rangka batang ruang ) seperti pada
Gambar disamping ini dengan metoda joint.
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Dr. Ir. Abdul Hamid M.Eng.
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