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:

Fx =0: Bx =0: MA =0 =-(2kips)(3ft) - (3kips)(3ft)-(4kips)(5ft) + By(6ft)=0

By=+5.17kips

MB = 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.Fy = +5.17 kips + 3.83 kips - 2 kips - 3kips - 4 kips =0 SAMPLE PROBLEM 10.2A 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.

Wx = +(5500lb)cos250 = +4980lb

Wy = - (5500lb)sin 25 = -2320lb Equilibrium Equations. We take moments about A to eliminate T and R1 from the computation.

MA =0: -(2320 lb)(25in.) (4980lb)(6in.) + R2 (50in.)=0

R2 =+1758lbNow, taking moments about B to eliminate T and R2 from the computation, we write:

MB =0: (2320lb)(25in.)-(4980lb)(6in.)-R1(50in.)=0

R1=562lb.

The value of T Is found by writing: Fx =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 Fy = 562lb+ 1758 Lb - 2320 Lb = 0A check could also have been obtained by computing moments about any point except A or B. SAMPLE PROBLEM 10.3A 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.

Fx =0: Rx =0

Fy =0: Ry -800N-400N-200N=0

Ry=1400N

MA =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. SAMPLE PROBLEM 10.4A 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 = 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=r, we write F=ks=kr

Equilibrium Equations. Summing the inomenl of W and F about O,we write:

Substituting the given data,we obtain:

Solving by trial and error, we find

PROBLEMS10.1 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 ,(b)if h= 8 ft, L = 10ft, and W = 35 Ib.

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 ,(b)if W=100lb, r = 3in., l = 15 in., and =600.

5.STRUKTURTruss (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.

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.

Space truss (3 dimensi) : m = 3j-6.

Disini : m = jumlah batang

j = jumlah joint

5.1 Metoda joint

Pada setiap joint berlaku :

Fx = 0, Fy = 0, Fz = 0

Gambar 5.2.dibawah ini menunjukkan FBD gaya2 yg bekerja pada setiap joint.

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:

Dari pers.(i) didapat: F6-7=-10.206kN

Idem (ii) didapat (F4)x=-3.835kN

Idem (iii)didapat (F4)Y=5.376kN

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:

Joint.1

Joint.3

Joint.4.

Joint.5

Joint.6

Joint.2

Soal latihan

Hitunglah gaya2 dalam batang dari space truss (rangka batang ruang ) seperti pada Gambar disamping ini dengan metoda joint.

PAGE PUSAT PENGEMBANGAN BAHAN AJAR-UMB Dr. Ir. Abdul Hamid M.Eng. STATIKA STRUKTUR 13

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