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Pembahasan by: I Made Gatot Karohika, ST.MT. LEMBAR KERJA MATA KULIAH STATIKA STRUKTUR SEMESTER GENAP 2013/2014

TUGAS/TAKE HOME TEST KE

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BATAS AKHIR PENGUMPULAN TUGAS/TAKE HOME TEST 17 APRIL 2014

PERNYATAAN

Dengan ini saya menyatakan bahwa tugas ini saya kerjakan sendiri, tidak mengcopy atau meniru tugas dari mahasiswa yang mengambil mata kuliah ini. Jika terbukti saya telah mengcopy atau meniru tugas dari mahasiswa lain maka saya sadar bahwa sanksinya adalah mata kuliah yang saya ambil tersebut diatas memperoleh nilai E.

Tanda Tangan :

Nama

: _______________________________

NIM

: ________________________________

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Page 1

Pembahasan by: I Made Gatot Karohika, ST.MT.

1

The boom on a 9500-lb truck is used to unload a pallet of shingles of weight 3500 lb, Determine the reaction at each of the two (a) rear wheels B, (b) front wheels C.

Pembahasan : FBD:

(a)

+ ΣMc = 0: (3.5 kips) [(1.6+1.3+19.5 cos15o )ft ] – 2FB [(1.6+1.3+14)ft]+(9.5 kips)(1.6 ft ) = 0 2FB = 5,4009 kips FB = 2,70 kips

b) +

ΣMB = 0:

(3.5 kips) [(19.5cos15o -14)ft] - (9.5 kips)[ (14 +1.3)ft] +2Fc [(14+1.3+1.6)ft] = 0 2FC = 7,5991 kips Fc = 3,80 kips


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2

Two children are standing on a diving board that weighs 146 lb. Knowing that the weights of the children at C and D are 63 lb and 90 lb, respectively, determine (a) the reaction at A, (b) the reaction at B.

Pembahasan : FBD:

a) +

Σ MB = 0 - Ay (3.6 ft ) - (146 lb)(1.44 ft ) - (63 lb)(3.24 ft ) - (90 lb)(6.24 ft ) = 0 Ay = - 271,10 lb atau Ay = 271,10 lb

b) +

Σ MA = 0 By (3.6 ft)− (146 lb)(5.04 ft)− (63 lb)(6.84 ft)− (90 lb)(9.84 ft) = 0 By = 570,10 lb


Page 3

Pembahasan by: I Made Gatot Karohika, ST.MT. 3

A load of lumber weighing W = 25 kN is being raised by a mobile crane. The weight of the boom ABC and the combined weight of the truck and driver are as shown. Determine the reaction at each of the two (a) front wheels H, (b) rear wheels K.

Pembahasan : FBD :

a) +

Σ MK = 0 (25 kN)(5.4 m) + (3 kN)(3.4 m) − 2 FH (2.5 m) + (50 kN)(0.5 m) = 0 2 FH = 68,080 kN FH = 34,04 kN

b) +

Σ MH = 0 (25 kN)(2.9 m) + (3 kN)(0.9 m)− (50 kN)(2.0 m) + 2 FK (2.5 m) = 0 2 Fk = 9.9200 kN Fk = 4,96 kN


Page 4

Pembahasan by: I Made Gatot Karohika, ST.MT. 4

A hand truck is used to move two barrels, each weighing 80 lb. Neglecting the weight of the hand truck, determine (a) the vertical force P which should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

Pembahasan : FBD :

a1 = (20 in.) sin α – (8 in.) cosα a2 = (32 in.) cosα – (20 in.) sinα b = (64 in.) cosα dari free-body diagram + ΣMB =0 ;

P (b) − W (a2) + W (a1) = 0 ???????.. (1)

+

P − 2w + 2 B = 0 ????????????.(2)

ΣFy = 0 ;

untuk α = 35° a1 = 20 sin 35° - 8 cos35° = 4.9183 in. a2 = 32 cos35 - 20 sin 35° = 14.7413 in. b = 64 cos35° = 52.426 in. (a) Dari persamaan (1) P(52.426 in.) − 80 lb(14.7413 in.) + 80 lb(4.9183 in.) = 0 P = 14,9896 lb www.statikastruktur.wordpress.com

Page 5

Pembahasan by: I Made Gatot Karohika, ST.MT. b) dari persamaan (2) 14.9896 lb – 2 (80 lb) + 2B = 0 B = 72,505 lb

5

To remove a nail, a small block of wood is placed under a crowbar, and a horizontal force P is applied as shown. Knowing that l = 88 mm and P = 130 N, determine the vertical force exerted on the nail and the reaction at B.

Pembahasan : FBD :

From the free-body diagram:


Page 6


From the force triangle: FN = (130 N) tan Ө = (130 N) tan82.726o = 1018.48 N Atau FN = 1,01848 kN Force on nail is therefore

atau

6

The total weight of a wheelbarrow filled with gravel is 120 lb. If the wheelbarrow is held on an 18° incline in the position shown, determine the magnitude and direction of (a) the force exerted by the worker on each handle, (b) the reaction at C.(Hint: The wheel is a twoforce body.)


Page 7

Pembahasan by: I Made Gatot Karohika, ST.MT. Pembahasan : FBD :

Perhatikan pada roda adalah kondisi benda dengan 2 gaya, sehingga gaya C berada sepanjang AC dan tegak lurus ground/lintasan tanjakan

The wheelbarrow adalah kondisi benda 3 gaya. Misalkan titik D adalah perpotongan dari garis aksi 3 gaya yang bekerja pada wheelbarrow (gaya C, gaya berat G dan gaya B)

Perhatikan segitiga DEG DE = EG tan 72° = (8 in.)tan 72° = 24.6215 in. DF = DE − EF = 24.6215 in. − 9 in. = 15.6215 in.

Dari segitiga DFB :


Page 8


From the force triangle: α = φ − 18° = 68.667° − 18° = 50.667° β = 180° − 50.667° − 18° = 111.333°

menggunakan hukum sinus :

B = 39.809 lb,

C = 9.644 lb

(a) perhatikan gaya pada masing-masing handle adalah B/2 : B/2 = 19,90 lb

;

dengan arah = 90o – 50,667o = 39,33o dari arah lintasan tanjakan

(b) reaksi pada C : C = 99.644 lb dengan arah = 90o – 18o = 72o dari arah lintasan tanjakan


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