TUGAS SEMESTER I MEKANIKA TEKNIK I NAMA : RANNY A. J. K. FANGGIDAE NIM : 2376/TS-ATK/12 NO. ABSEN : 29 AKADEMI TEKNIK KUPANG 2012 Soal Untuk semua soal dihitung dan gambar bidang Momen (M),Lintang (D), Normal (N). Tipe kontruksi seperti gambar di awah ini, data lain di tentuka sendiri (pilih tumpuan pasangan yang satu sendi yang satu Roll ) P = 6 t 1. q = 4 t/m α = 60 a = 3 m b = 2m c= 3 m L = 8 m 2. P1 = 4t P2 = 3 t P3 = 4 t α = 45 q = 6 t/m a’ = 2 m a = 3 m b = 2 m c = 3 m d = 4 m L = 8 m 3. P = 4 t q = 5 t/m h 1 = 2 m H2 = 3 t H2 = 2 t h = 4 m h 2 = 2 m a = 2 m b = 3 m c = 4 m L = 9 m P = 2 t 4. q 2 = 4t/m q 1 =6 t/m h1 = 3 m h =5 m h2 = 2 m a = 5 m b = 2 m c = 4 m d = 3 m e = 2 m L = 16 m q = 2 t/m P2 = 4t P1 = 2 t 5. a = 3 m b = 4 m L = 7 m 6. q = 8 t/m h4 = 2 m P2 = 3 t h1 = 2m h5 = 2 m h2 = 2 m P3 = 2 t H3 = 3 m L = 5 m KUPANG 16, Desember 2012 DOSEN MATA KULIAH MEKANIKA TEKNIK I Ir. CHARISAL A. MANU, Msi Penyelesaian 1. P = 6 t q = 4 t/m α = 60 a = 3 m b = 2m c= 3 m L = 8 m PH  PV = P.Sin α => 6 t . sin 60’ P  6. 0,886 PV  5,196 t  PH = P.Cos α  6 t . cos 60’  6t. 0,5  3 t  Q = q. a  4 t/m . 3m  12 t/m a. Menghitung Reaksi Horisontal ∑ H = 0 PH = 3 t  PH-RBH = 0  PH = RBH  RBH = PH  3 t b. Menghitung Reaksi Vertikal ∑V = 0 Q PV = 5,196 c= 3 m B ½ a + b + c = 6,5 m RAV L = 8 m  ∑MB = 0  RAV.(L) – Q ( ½ a + b + c ) – PV ( c )  RAV = Q ( ½ a + b + c) + PV (c) L  RAV = 12 (6,5) + 5,196 (3) 8  RAV = 78 + 15,588 8  RAV = 93,588 8  RAV = 11,6985  ∑MA = 0 Q PV = 5,196 A ½ a= 1,5 m a + b = 5 m L = 8 m RBV  Q ( ½ a ) - PV ( a+b ) – RBV (L)  RBV = Q ( ½ a ) + PV (a+b) L  RBV = 12 (1,5) + 5,196 (5) 8  RBV = 18 + 25,98 8  RBV = 43,98 8  RBV = 5,4975 Kontrol  ∑V = 0  Q + PV – RAV + RBV = 0  12 + 5,196-11,6985 +5,4975 = 0 c. Menghitung Momen ( M )  MA = RAV. 0 = 0  MC = RAV . (a) - Q ( ½ a )  11.6985. (3) – 12 ( 1,5)  35,0955 – 18 = 17,0955 m.t  MD = RAV (a+b)- Q( ½ a+b )  58,4925 - 42  16,4925 t.m  MB = RAV ( L) – Q (L- ½ a ) - PV (c)  93,588 – 78-15,588  0  Mmax = RAV.x- ½ q (x)²  11,6985x – ½ 8x  11,6985- 4x  x = 11,6985 = 2,924625 4  11,6985 ( 2,92) – 4(2,92) => Mmax  34,15962 – 11,68  22,47962 Qx =½ q.x RAV x d. Mengtitung Lintang (D) o DA = RAV  11,6985 o DC = RAV – Q  11,6985 – 12  -0,3015 o DD1 = DC  -0,3015 o DD2 = DC – PV  -0,3015 – 5,196  -5,4975 o DB1 = DD2  -5,4975 o DB2 = DD2+ RBV  -5,4975 + 5,4975 = 0 e. Menghitung Normal ( N ) NBD = RAH = 3 t f. Menggambar Momen (M),Lintang (D), dan Normal (N) P = 6 t q = 4 t/m α = 60 a = 3 m b = 2m c= 3 m L = 8 m + Gambar Bidang Momen + _ 2. P1 = 4t P2 = 3 t P3 = 4 t α = 45 q = 6 t/m a’ = 2 m a = 3 m b = 2 m c = 3 m d = 4 m L = 8 m  P2V = P2.sin α  3. Sin 45’  3.0,707  2,121 t  P2H = P2.cos α  3.cos 45’  3.0,707  2,121 t  Q = q. d  6 t/m . 4 m  24 t/m a. Menghitung reaksi Horisontal  ∑H = 0  RAH –P2H  RAH = P2H  2,121 t b. Menghitung Reaksi Vertikal ∑V = 0 P1 = 4 t PV2 = 2,121 P3 = 4 t Q = 24 t/m B ½ d C = 3 m b + c = 5 m RAV L = 8 m L + a = 10 m ∑MB = 0  - P1 (L + a)+RAV.(L) – P2V (b+c)-P3 (c)+Q ( ½ d)  RAV = P1 (L + a)+P2V (b+c)+P3 (c)-Q ( ½ d) L  RAV = 4 (10) + 2,121 (5)+4(3)-24(2) 8  RAV = 40 + 10,605 +12-48 = 14,605 = 1,82562 8 8 P1 = 4 t PV2 = 2,121 P3 = 4 t Q = 24 t/m a’ = 2 m a = 3 m a+ b =5 m L = 8 m RBV L + ½ d = 10 m ∑MA = 0  -P1.(a’) + P2V (a) P3 (a+b)-RBV (L)+Q( L + ½ d)  RBV = - P1 (a’)+P2V (a)+P3 (a+b)+Q ( L+ ½ d) L  RBV = 4 (2) + 2,121 (3)+4(5)-24(10) 8  RBV = 8 + 6,363 +20-240 8  RBV =258,363 RBV = 32,295375 8 Kontrol ∑V = 0  P1+P2V+P3+Q-RAV-RBV  4+2,121+4+24-1,82562-32,295375 = 0 c. Menghitung Momen ( M)  MA = 0  MC = -P1 .a’  -4.2 =-8  MB = -P1. (a’ +a ) + RAV. a  -4. 5 + 1,82562. 3  -20 + 5,47686  -14,52314  MD = -P1. (a’ +a+b ) + RAV. a  -4.7 + 1,8256.3  -32 + 5,47686  -26,52314  MB = -P1. (a’ +L) + RAV. a – P2V (b+c)-P3 (c)  -4 (10) + 1,82562.3-2,121.5-4.3  -40+5,47686-10,605-12  -57,13  MB’ = - ½ q (d)²  - ½ 6 (4)²  -3.16 = - 48  Mmax = - ½ q (x)²  - ½ 6. 2x  x = 3 = 1,5 2  - 3 (1,5)²  -3. 2,25= 6,75 d. Menghitung Lintang (D)  DA’ = -P1 = -4  DA = DA’+ RAV  -4 + 1,82562 = -2,174  DC1 = DA = - 2,174  DC2 = DA –P2V  -2,174-2,121= -4,295  DD1 = DC2= -4,295  DD2 = DC2 –P3  -4,295-4= -8,295  DB1 =DD2 = -8,295  DB2= DB1+ RBV  -8,295+32,295375 = 24  DB’ = DB2-Q  24-24 =0 e. Menghitung Normal  NAC = RAH  2,121 f. Mengambar Momen ( M), Lintang (D) dan Normal (N) P1 = 4t P2 = 3 t P3 = 4 t α = 45 q = 6 t/m a’ = 2 m a = 3 m b = 2 m c = 3 m d = 4 m L = 8 m 3. P = 4 t q = 5 t/m h 1 = 2 m α 60 H2 = 3 t H1 = 2 t h = 4 m h 2 = 2 m a = 2 m b = 3 m c = 4 m L = 9 m  PV = P.sin 60 PH = P. Cos 60  4.cos 60 = 4.cos 60  4. 0,866 = 4. 0,5  3,464 = 2  Q = q. c  5 t/m . 4 m  20 t P1H H2 a. Menghitung reaksi Horisontal  ∑H = RAH – PH + H2  RAH = H1 + PH – H2 H1  RAH = 2 t + 2 – 3t  RAH = 1 t b. Menghitung Reaksi Vertikal P1V = 3,464 Q = 20 t / m PH = 2 H2 = 3 t H1 = 2 t h = 4 m h 2 = 2 m ½ c B c = 4 RAV L = 9 m ∑MB = 0  RAV.(L) + H1 (h2)+PH (h)- PV(b+c)-Q( ½ c)- H2 (h)  RAV = - H1 (h2)-PH(h)+PV(c)+Q( ½ c) +H2 (h) L  RAV = -2 (2)-2(4) + 3,464 (4)+20(2)+3(4) 9  RAV = - 4-8 + 13,856 +40+12 9  RAV = 53,865 = 5,948 9 PV = 3,464 Q = 20 t/m PH = 2 H2 = 3 t h = 4 m H1 = 2 t h = 4 m h 2 = 2 m A a+b = 5 m RBV a+b+ ½ c = 7 L = 9 m ∑MA = 0  H1(h2) + PH (h)+PV ( b+a)+Q(a+b ½ c)–H2 (h2)- RBV.L  RBV = H1(h2) +PH (h)+PV ( a+b)+Q( a+b+½ c )-H2 (h2) L  RBV = 2(2) + 2(4)+3,464(5)+20(7)- 3(4) 9  RBV = 4 +8+17,32 + 140-12 9  RBV = 157,32 9  RBV = 17,48 Kontrol ∑V => PV + Q – RAV + RBV = 0  3,464 + 20 – 5,948 +17,48  23,464- 23,464 = 0 c. Menghitung Momen ( M ) PV Q a. Batang AD PH • MA = 0 • MC = RAV. (a) + RAH (h2) H2  5,948. 2 + 2 . 2  11,896 + 4 =15,896 • MD = RAV (a+b) + RAH.(h) – H1(h1)  5,948.5 + 2. 4 – 2.2  29,74+8-4  33,74 b. Batang DE Q • MD2 = RBV .(c) – Q ( ½ c )  69,92- 40  29,92 c ½ c RBV c. Batang BE • ME = 0 E H2 • MB = 0 B RBV PV d. Menghitung Lintang (D) • PVD = PV.c /h  3,464 . 5/4  4,33 • PHD = PH.h/c = 2 .4/5  1,6 • H1D = H1. h/c = 2 . 4/5  1,6 • RAHD = RAH. h/c  1. 4/5  0,8 • RAVD = RAV. h/c  5,948 . 4/5  4,7584 • PVN = PV. h/c  3,464. 4/5  2,77 • PHN = PH. c/h  2. 5/4  2,5 • H1N = H1. c/h  2. 5/4  2,5 • RAHN = RAH. c/h  1. 5/4  1,25 • RAVN = RAV. c/h  5,948. 5/4  7,435  DA = RAVD + RAHD  5,948 + 1  6,948  DC1 = DA  6,948  DC2 = DA – H1D  6,968 – 1,6  5,368  DD = DC  5,368 Batang DE  DD = RAV – PVD = 5,948 – 4,33 = 1,618  DE = DD- Q = 1,618 – 20 = - 18,382 Batang EB  DE = H2 = 3  DB = 0 e. Menghitung Normal Batang AD  NAC = - RAVN + RAHN  - 7,435 + 1,25  - 6,185  NCD = -NAC + H1N  - 6,185 + 2,5  - 3,685 Batang AD  NDE = - PHN  - 2,5 Batang DE  NBE = - RBV  - 17, 48 P = 4 t q = 5 t/m h 1 = 2 m α 60 H2 = 3 t H1 = 2 t h = 4 m h 2 = 2 m a = 2 m b = 3 m c = 4 m L = 9 m 4. q 2 = 4t/m q 1 =6 t/m h1 = 3 m h =5 m h2 = 2 m a = 5 m b = 2 m c = 4 m d = 3 m e = 2 m L = 16 m • Q1 = q1 . a  6 . 5 = 30 t/m • Q2 = q2 . c  4 . 4 = 16 t/m a. Menghitung Reaksi Horizontal  RAH = 0 b. Menghitung Reaksi Vertikal Q2 = 16 t/m P = 2 t Q1 =30 t/m h =5 m e= 2 m ½ c + d + e = 7 m L – ½ a = 13,5 m RAV L = 16 m  ∑MB = 0  RAV . L - Q1 . ( L – ½ a ) – Q2 ( ½ c + d + e ) – P ( e )  RAV = Q1 . ( L – ½ a ) + Q2 ( ½ c + d + e ) + P ( e ) L  RAV = 30 . ( 13,5 ) + 16 (7) + 2 ( 2 ) 16  RAV = 405 + 112 + 4 16  RAV = 521 16  RAV = 32,5625 Q2 = 16 t/m P = 2 t Q1 =30 t/m h =5 m ½ a = 2,5 m a + b + ½ c = 9 m L – e = 14 m L = 16 m RBV  ∑MA = 0  Q1. ( ½ a ) + Q2. ( a + b + ½ c) + P. (L-e) – RBV. L  RBV = Q1. ( ½ a ) + Q2. ( a + b + ½ c) + P. (L-e) L  RBV = 30 (2,5) + 16(9)+2(14) 16  RBV = 75 + 144 + 28 16  RBV = 247 16  RBV = 15,4375 ∑V => RAV + RBV – Q1 + Q2 + P = 0 Q1  32,5625 + 15,4375 – 30 + 16 + 2  48 - 48 = 0 c. Menghitung Momen ( M )  MA = RAV.0 = 0  MC = RAV.a – Q1 ( ½ a ) ½ x  MX = RAV.x - ½ q1.x² RAV x  MD = RAV. (a+ ½ b)- Q1 ( ½ a + ½ b)  ME = RAV. (a+b)- Q1( ½ a+b)-Q2( ½ c+d)  ME = RBV . (c+d) – P1 .c  MF = RBV. d  MB = 0  MX2 = RAV. ( a + ½ b + x ) – ½ q. x ² d. Menghitung lintang  DA = RAV. a/h  DC1 = RAV. a/h – Q1 a/h  DC 2= RAV- Q1  DD = DC2  DE = RAV-Q1-Q2  DB= - RBV. ( c+d)/ h  DF1 = DB  DF2 = DB + P. ( c + d ) / h  DE = DF2 e. Menghitung normal  NAC = - RAV. h/a + Q1. h/a  NBF = RBV. h/ (c + d )  NFE = RBV. h/ (c+d) – P. h /(c+d)  q = 2 t/m P2 = 4t P1 = 2 t 5. a = 3 m b = 4 m L = 7 m