1 Menghitung Momen, Gaya Lintang, dan Gaya Normal pada Portal dengan
Menggunakan Metode Takabeya
Ukuran Balok (30/50), Kolom (40/40)
Langkah Penyelesaian:
1. Menentukan momen primer a. MFEF = MFJK = -12 1 q l2 = -12 1 (3.679)(52) = -7.664 TM b. MFFE = MFKJ = 12 1 q l2 = 12 1 (3.679)(52) = 7.664 TM c. MFFG = MFKL = -12 1 q l2 = -12 1 (3.616)(4.852) = -7.088 TM d. MFFG = MFKL = 12 1 q l2 = 12 1 (3.616)(4.852) = 7.088 TM
2 e. MFOP = -12 1 q l2 = -12 1 (1.787)( 52) = -3.723 TM f. MFPO = 12 1 q l2 = 12 1 (1.787)( 52) = 3.723 TM g. MFPQ = -12 1 q l2 = -12 1 (1.757)(4.852) = -3.443 TM h. MFQP = 12 1 q l2 = 12 1 (1.757)( 4.852) = 3.443 TM 2. Menentukan jumlah momen primer di titik kumpul
a. E = J = MFEF = MFJK = -7.664 TM b. F = K = MFFE + MFFG = MFKJ + MFKL = 7.664 + (-7.088) = 0.576 TM c. G = L = MFGF = MFLK = 7.088 TM d. O = MFOP = -3.723 TM e. P = MFPO + MFPQ = 3.723 + (-3.443) = 0.280 TM f. Q = MFQP = 3.443 TM
3. Menentukan kekakuan balok dan kolom IB= 12 1 (b)(h3) = 12 1 (30)(503) = 312500 cm4 IC= 12 1 (b)(h3) = 12 1 (40)(403) = 213333 cm4 K = 1000 cm3
a. Kekakuan balok bentang 500 cm Kb’ = IB/500 = 625 cm3 = 0.625 b. Kekakuan balok bentang 485 cm
Kb’’ = IB/485 = 644.33 cm3 = 0.644 c. Kekakuan balok bentang 100 cm
Kb’’’ = IB/100 = 3125 cm3
= 3.125 d. Kekakuan balok bentang 115 cm
Kb’’’’ = IB/115 = 2717.4 cm3
= 2.717 e. Kekakuan kolom tinggi 400 cm
Kc’ = IC/400 = 533.33 cm3
= 0.533 f. Kekakuan kolom tinggi 380 cm
Kc’’ = IC/380 = 561.4 cm3
3 4. Menentukan nilai , , dan m(0)
a. (jumlah kekakuan pada masing-masing titik kumpul)
1) E = 2 x (0.625 + 3.125 + 0.533 + 0.533) = 9.633 2) F = 2 x (0.644 + 0.625 + 0.533 + 0.533) = 4.672 3) G = 2 x (2.717 + 0.644 + 0.533 + 0.533) = 8.857 4) J = 2 x (0.625 + 3.125 + 0.561 + 0.533) = 9.689 5) K = 2 x (0.644 + 0.625 + 0.561 + 0.533) = 4.728 6) L = 2 x (2.717 + 0.644 + 0.561 + 0.533) = 8.913 7) O = 2 x (0.625 + 3.125 + 0.561) = 8.623 8) P = 2 x (0.644 + 0.625 + 0.561) = 3.661 9) Q = 2 x (2.717 + 0.644 + 0.561) = 7.846 b. Menentukan nilai 1) Titik E EF = E EF k = (0.625/9.633) = 0.065 EJ = E EJ k = (0.533/9.633) = 0.055 2) Titik F FE = F FE k = (0.625/4.672) = 0.134 FG = F FG k = (0.644/4.672) = 0.138 FK = F FK k = (0.533/4.672) = 0.114 3) Titik G GF = G GF k = (0.644/8.857) = 0.073 GL = G GL k = (0.533/8.857) = 0.060
4 4) Titik J JE = J JE k = (0.533/9.689) = 0.055 JK = J JK k = (0.625/9.689) = 0.065 JO = J JO k = (0.561/9.689) = 0.058 5) Titik K KF = K KF k = (0.533/4.728) = 0.113 KJ = K KJ k = (0.625/4.728) = 0.132 KL = K KL k = (0.644/4.728) = 0.136 KP = K KP k = (0.561/4.728) = 0.119 6) Titik L LG = L LG k = (0.533/8.913) = 0.060 LK = L LK k = (0.644/8.913) = 0.072 LQ = L LQ k = (0.561/8.913) = 0.063 7) Titik O OJ = O OJ k = (0.561/8.623) = 0.065 OP = O OP k = (0.625/8.623) = 0.072 8) Titik P PK = P PK k = (0.561/3.661) = 0.153 PO = P PO k = (0.625/3.661) = 0.171 PQ = P PQ k = (0.644/3.661) = 0.176 9) Titik Q QL = Q QL k = (0.561/7.846) = 0.072 QP = Q QP k = (0.644/7.846) = 0.082
5 c. Menentukan nilai m(0) 1) mE(0) = E E = 633 . 9 664 . 7 = 0.796 2) mF(0) = F F = 672 . 4 576 . 0 = -0.123 3) mG(0) = G G = 857 . 8 088 . 7 = -0.800 4) mJ(0) = J J = 689 . 9 664 . 7 = 0.865 5) mK(0) = K K = 728 . 4 576 . 0 = -0.122 6) mL(0) = L L = 913 . 8 088 . 7 = -0.795 7) mO(0) = O O = 623 . 8 723 . 3 = 0.432 8) mP(0) = P P = 661 . 3 280 . 0 = -0.076 9) mQ(0) = Q Q = 846 . 7 443 . 3 = -0.439
6 5. Pemberesan momen-momen parsil m(0)
7 6. Perhitungan Momen Akhir
a. ME
1) MEA = KEA(2 ME + MA) + MFEA = 0.533 (2(0.758)+0)+0 = 0.808 tm 2) MEF = KEF (2 ME + MF) + MFEF = 0.625 (2(0.758)-0.109)-7.664 = -6.784 tm 3) MEJ = KEJ (2 ME + MJ) + MFEJ = 0.533 (2(0.758)+0.809)+0 = 1.240 tm 4) MED = KED (2 ME + MD) + MFED = 3.125 (2(0.758)+0)+0 = 4.736 tm Jumlah = 0 Ok b. MF 1) MFB = KFB(2 MF + MB) + MFFB = 0.533 (2(-0.109)+0)+0 = -0.116 tm 2) MFG = KFG (2 MF + MG) + MFFG = 0.644 (2(-0.109)-0.749)-7.088 = -7.710 tm 3) MFE = KFE (2 MF + ME) + MFFE = 0.625 (2(-0.109)+0.758)+7.664 = 8.002 tm 4) MFK = KFK (2 MF + MK) + MFFK = 0.533 (2(-0.109)-0.073)+0 = -0.175 tm Jumlah = 0 Ok c. MG 1) MGC = KGC(2 MG + MC) + MFGC = 0.533 (2(-0.749)+0)+0 = -0.799 tm 2) MGF = KGF (2 MG + MF) + MFGF = 0.644 (2(-0.749)-0.109)+7.088 = 6.053 tm 3) MGH = KGH (2 MG + MH) + MFGH = 2.717 (2(-0.749)+0)+0 = -4.071tm 4) MGL = KGL (2 MG + ML) + MFGL = 0.533 (2(-0.749)-0.718)+0 = -1.182 tm Jumlah = 0 Ok d. MJ
1) MJE = KJE(2 MJ + ME) + MFJE = 0.533 (2(0.809)+0.758)+0 = 1.267 tm 2) MJK = KJK (2 MJ + MK) + MFJK = 0.625 (2(0.809)-0.073)-7.664 = 5.054 tm 3) MJO = KJO (2 MJ + MO) + MFJO = 0.561 (2(0.809)+0.383)+0 = -6.723 tm 4) MJI = KJI (2 MJ + MI) + MFJI = 3.125 (2(0.809)+0)+0 = 1.123 tm
Jumlah = 0.721 tm MJE = 1.267 – (2JE x 0.721) = 1.267 – (2 (0.055) x 0.721) = 1.187 tm MJK = 5.054 – (2JK x 0.721) = 5.054 – (2 (0.065) x 0.721) = 4.589 tm MJO = -6.723 – (2JO x 0.721) = -6.723 – (2 (0.058) x 0.721) = -6.816 tm MJI = 1.123 – (2JI x 0.721) = 1.123 – (2 (0.323) x 0.721) = 1.040 tm Jumlah = 0 Ok
8 e. MK 1) MKF = KKF(2 MK + MF) + MFKF = 0.533 (2(-0.112)-0.109)+0 = -0.177 tm 2) MKL = KKL (2 MK + ML) + MFKL = 0.644 (2(-0.112)-0.718)-7.088 = 8.029 tm 3) MKJ = KKJ (2 MK + MJ) + MFKJ = 0.625 (2(-0.112)-0.134)+7.664 = -7.695 tm 4) MKP = KKP (2 MK + MP) + MFKP = 0.561 (2(-0.112)-0.057)+0 = -0.158 tm Jumlah = 0 Ok f. ML 1) MLG = KLG(2 ML + MG) + MFLG = 0.533 (2(-0.718)-0.749)+0 = -1.166 tm 2) MLK = KLK (2 ML + MK) + MFLK = 0.644 (2(-0.718)-0.112)+7.088 = 6.090 tm 3) MLM = KLM (2 ML + MM) + MFLM = 2.717 (2(-0.718)+0)+0 = -3.903tm 4) MLQ = KLQ (2 ML + MQ) + MFLQ = 0.561 (2(-0.718)-0.383)+0 = -1.021 tm Jumlah = 0 Ok g. MO 1) MOJ = KOJ(2 MO + MJ) + MFOJ = 0.561 (2(0.383)+0.809)+0 = 0.884 tm 2) MOP = KOP (2 MO + MP) + MFOP = 0.625 (2(0.383)-0.057)-3.723 = 2.395 tm 3) MON = KON (2 MO + MN) + MFON = 3.125 (2(0.383)+0)+0 = -3.280 tm Jumlah = 0 Ok h. MP 1) MPK = KPK(2 MP + MK) + MFPK = 0.561 (2(-0.057)-0.112)+0 = -0.127 tm 2) MPQ = KPQ (2 MP + MQ) + MFPQ = 0.644 (2(-0.057)-0.383)-3.443 = 3.891 tm 3) MPO = KPO (2 MP + MO) + MFPO = 0.625 (2(-0.057)+0.383)+3.723 = -3.764 tm
Jumlah = 0 Ok i. MQ 1) MQL = KQL(2 MQ + ML) + MFQL = 0.561 (2(-0.383)-0.718)+0 = -0.833 tm 2) MQP = KQP (2 MQ + MP) + MFQP = 0.644 (2(-0.383)-0.057)+3.443 = 2.913 tm 3) MQR = KQR (2 MQ + MR) + MFQR = 2.717 (2(-0.383)+0)+0 = -2.080tm Jumlah = 0 Ok
j. MAE = KAE(2 MA + ME) + MFAE = 0.533 (2(0)+0.758)+0 = 0.404 tm k. MBF = KBF(2 MB + MF) + MFBF = 0.533 (2(0)-0.109)+0 = -0.058 tm l. MCG = KCG(2 MC + MG) + MFCG = 0.533 (2(0)-0.749)+0 = -0.400 tm
9
10 7. Menentukan perletakan momen maksimum, menghitung momen maksimum, dan
menentukan perletakan momen minimum (M=0) a. Batang EF RE => MF = 0 RE (5) – ½ (3.679)(52) – 6.784 + 8.002 = 0 5RE – 45.9875 – 6.784 +8.002 = 0 RE = (44.7695 / 5) = 8.954 T Kontrol R = Q RE + RF = q.l 8.954+ 9.441 = (3.679 x 5) 18.395 = 18.395 Ok
1) Posisi momen maksimum
Dari titik E Mmax = RE (x1) – ½ qx12 - MEF = 8.954x1 – 1.8395x12 – 6.784 dx dMmax =0 8.954 -3.679x1 = 0 ==> x1 = (8.954/3.679) = 2.434 m Dari titik F Mmax = RF (x2) – ½ qx22 – MFE = 9.441x2 – 1.8395x22 – 8.002 dx dMmax =0 9.441 -3.679x2 = 0 ==> x2 = (9.441/3.679) = 2.566 m RF => ME = 0 -RF (5) + ½ (3.679)(5 2 ) – 6.784 + 8.002 = 0 -5RF + 45.9875 – 6.784 +8.002 = 0 RF = (47.2055 / 5) = 9.441 T F P = 3.85 T P = 4.2 T 6.784 tm 8.002 tm 5.00 m E q = 3.679 t/m’
11 2) Momen maksimum Dari titik E Mmax = 8.954x1 – 1.8395x12 – 6.784 = (8.954)(2.434) - (1.8395)(2.4342) – 6.784 = 21.794 – 10.898 – 6.784 = 4.112 tm Dari titik F Mmax = 9.441x2 – 1.8395x22 – 8.002 = (9.441)(2.566) - (1.8395)(2.5662) – 8.002 = 24.226 – 12.112 – 8.002 = 4.112 tm
3) Posisi momen minimum (M=0)
Dari titik E M(0) => 8.954x1 – 1.8395x12 – 6.784 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8395 . 1 ( 2 ) 784 . 6 )( 8395 . 1 ( 4 954 . 8 (8.954) 2 x(a,b) = 679 . 3 501 . 5 (8.954) xa = 0.94 m xb = 3.93 m Dari titik F M(0) => 9.441x2 – 1.8395x22 – 8.002 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8395 . 1 ( 2 ) 002 . 8 )( 8395 . 1 ( 4 441 . 9 (9.441) 2 x(a,b) = 679 . 3 500 . 5 (9.441) xa = 1.07 m xb = 4.06 m
12 b. Batang FG RF => MG = 0 RF (4.85) – ½ (3.616)(4.85 2 ) –7.710 + 6.053 = 0 4.85RF – 42.529 – 7.710 + 6.053 = 0 RF = (44.186 / 4.85) = 9.111 T Kontrol R = Q RF + RG = q.l 9.111+ 8.427 = (3.616 x 4.85) 17.538= 17.538 Ok
1) Posisi momen maksimum
Dari titik F Mmax = RF (x2) – ½ qx22 – MFG = 9.111x2 – 1.808x22 – 7.710 dx dMmax =0 9.111 -3.616x2 = 0 ==> x2 = (9.111/3.616) = 2.52 m Dari titik G Mmax = RG (x1) – ½ qx12 – MGF = 8.427x1 – 1.808x12 – 6.053 dx dMmax =0 8.427-3.616x1 = 0 ==> x1 = (8.427/3.616) = 2.33 m G P = 4.2 T P = 4 T 7.710 tm 6.053 tm 4.85 m F q = 3.616 t/m’ RG => MF = 0 -RG (4.85) + ½ (3.616)(4.852) –7.710 + 6.053 = 0 -4.85RG +42.529 – 7.710 +6.053 = 0 RG = (40.872 / 4.85) = 8.427 T
13 2) Momen maksimum Dari titik F Mmax = 9.111x2 – 1.808x22 – 7.710 = (9.111)(2.52) - (1.808)(2.522) – 7.710 = 22.96 – 11.482 – 7.710 = 3.768 tm Dari titik G Mmax = 8.427x1 – 1.808x12 – 6.053 = (8.427)(2.33) - (1.808)(2.332) – 6.053 = 19.635 – 9.815 – 6.053 = 3.767 tm
3) Posisi momen minimum (M=0)
Dari titik F M(0) => 9.111x2 – 1.808x22 – 7.710 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 808 . 1 ( 2 710 . 7 )( 808 . 1 ( 4 111 . 9 (9.111) 2 x(a,b) = 616 . 3 22 . 5 (9.111) xa = 1.08 m xb = 3.96 m Dari titik G M(0) => 8.427x1 – 1.808x12 – 6.053 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 808 . 1 ( 2 ) 053 . 6 )( 808 . 1 ( 4 427 . 8 (8.427) 2 x(a,b) = 616 . 3 22 . 5 (8.427) xa = 3.77 m xb = 0.89 m
14 c. Batang JK RJ => MK = 0 RJ (5) – ½ (3.679)(52) – 6.816 + 8.029 = 0 5RJ – 45.9875 – 6.816 +8.029 = 0 RJ = (44.7745 / 5) = 8.955 T Kontrol R = Q RJ + RK = q.l 8.955+ 9.440 = (3.679 x 5) 18.395= 18.395 Ok
1) Posisi momen maksimum
Dari titik J Mmax = RJ (x1) – ½ qx12 – MJK = 8.955x1 – 1.8395x12 – 6.816 dx dMmax =0 8.955 -3.679x1 = 0 ==> x1 = (8.955/3.679) = 2.434 m Dari titik K Mmax = RK (x2) – ½ qx22 – MKJ = 9.440x2 – 1.8395x22 – 8.029 dx dMmax =0 9.440-3.679x2 = 0 ==> x2 = (9.440/3.679) = 2.566 m RK => MJ = 0 -RK (5) + ½ (3.679)(5 2 ) – 6.816 + 8.029 = 0 -5RK + 45.9875 – 6.816 +8.029 = 0 RK = (47.201 / 5) = 9.440 T K P = 3.85 T P = 4.2 T 6.816 tm 8.029 tm 5.00 m J q = 3.679 t/m’
15 2) Momen maksimum Dari titik J Mmax = 8.955x1 – 1.8395x12 – 6.816 = (8.955)(2.434) - (1.8395)(2.4342) – 6.816 = 21.796 – 10.898 – 6.816 = 4.082 tm Dari titik K Mmax = 9.440x2 – 1.8395x22 – 8.029 = (9.440)(2.566) - (1.8395)(2.5662) – 8.029 = 24.223 – 12.112 – 8.029 = 4.082 tm
4) Posisi momen minimum (M=0)
Dari titik J M(0) => 8.955x1 – 1.8395x12 – 6.816 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8395 . 1 ( 2 ) 816 . 6 )( 8395 . 1 ( 4 955 . 8 (8.955) 2 x(a,b) = 679 . 3 481 . 5 (8.955) xa = 0.94 m xb = 3.92 m Dari titik K M(0) => 9.440x2 – 1.8395x22 – 8.029 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8395 . 1 ( 2 ) 029 . 8 )( 8395 . 1 ( 4 440 . 9 (9.440) 2 x(a,b) = 679 . 3 481 . 5 (9.440) xa = 1.08 m xb = 4.06 m
16 d. Batang KL RK => ML = 0 RK (4.85) – ½ (3.616)(4.85 2 ) –7.695 + 6.090 = 0 4.85RK – 42.529 – 7.695 + 6.090 = 0 RK = (44.134 / 4.85) = 9.100 T Kontrol R = Q RK + RL = q.l 9.100+ 8.438 = (3.616 x 4.85) 17.538= 17.538 Ok
1) Posisi momen maksimum
Dari titik K Mmax = RK (x2) – ½ qx22 – MKL = 9.100x2 – 1.808x22 – 7.695 dx dMmax =0 9.100 -3.616x2 = 0 ==> x2 = (9.100/3.616) = 2.52 m Dari titik L Mmax = RL (x1) – ½ qx12 – MLK = 8.438x1 – 1.808x12 – 6.090 dx dMmax =0 8.438-3.616x1 = 0 ==> x1 = (8.438/3.616) = 2.33 m L P = 4.2 T P = 4 T 7.695 tm 6.090 tm 4.85 m K q = 3.616 t/m’ RL => MK = 0 -RL (4.85) + ½ (3.616)(4.852) –7.710 + 6.053 = 0 -4.85RL +42.529 – 7.695 +6.090 = 0 RL = (40.924 / 4.85) = 8.438 T
17 2) Momen maksimum Dari titik K Mmax = 9.100x2 – 1.808x22 – 7.695 = (9.100)(2.52) - (1.808)(2.522) – 7.695 = 22.932 – 11.482 – 7.695 = 3.755 tm Dari titik L Mmax = 8.438x1 – 1.808x12 – 6.090 = (8.438)(2.33) - (1.808)(2.332) – 6.090 = 19.661 – 9.815 – 6.090 = 3.756 tm
3) Posisi momen minimum (M=0)
Dari titik K M(0) => 9.100x2 – 1.808x22 – 7.695 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 808 . 1 ( 2 ) 695 . 7 )( 808 . 1 ( 4 100 . 9 (9.100) 2 x(a,b) = 616 . 3 21 . 5 (9.100) xa = 1.08 m xb = 3.96 m Dari titik L M(0) => 8.438x1 – 1.808x12 – 6.090 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 808 . 1 ( 2 ) 090 . 6 )( 808 . 1 ( 4 438 . 8 (8.438) 2 x(a,b) = 616 . 3 21 . 5 (8.438) xa = 3.77 m xb = 0.89 m
18 e. Batang OP RO => MP = 0 RO (5) – ½ (1.787)(5 2 ) – 3.280 + 3.891 = 0 5RO – 22.3375 – 3.280 +3.891 = 0 RO = (21.7265 / 5) = 4.345 T Kontrol R = Q RO + RP = q.l 4.345+ 4.59 = (1.787 x 5) 8.935= 8.935 Ok
1) Posisi momen maksimum
Dari titik O Mmax = RO (x1) – ½ qx12 – MOP = 4.345x1 – 0.8935x12 – 3.280 dx dMmax =0 4.345 – 1.787x1 = 0 ==> x1 = (4.345/1.787) = 2.431 m Dari titik P Mmax = RP (x2) – ½ qx22 – MPO = 4.59x2 – 0.8935x22 – 3.891 dx dMmax =0 4.59 - 1.787x2 = 0 ==> x2 = (4.59/1.787) = 2.569 m RP => MO = 0 -RP (5) + ½ (1.787)(5 2 ) – 3.280 + 3.891 = 0 -5RP + 22.3375 – 3.280 +3.891 = 0 RP = (22.9485 / 5) = 4.59 T P P = 3.4 T P = 3.7 T 3.280 tm 3.891 tm 5.00 m O q = 1.787 t/m’
19 2) Momen maksimum Dari titik O Mmax = 4.345x1 – 0.8935x12 – 3.280 = (4.345)(2.431) - (0.8935)(2.4312) – 3.280 = 10.563 – 5.280 – 3.280 = 2.003 tm Dari titik P Mmax = 4.59x2 – 0.8935x22 – 3.891 = (4.59)(2.569) - (0.8935)(2.5692) – 3.891 = 11.792 – 5.897 – 3.891 = 2.004 tm
5) Posisi momen minimum (M=0)
Dari titik O M(0) => 4.345x1 – 0.8935x12 – 3.280 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8935 . 0 ( 2 ) 280 . 3 )( 8935 . 0 ( 4 345 . 4 (4.345) 2 x(a,b) = 787 . 1 675 . 2 (4.345) xa = 0.93 m xb = 3.93 m Dari titik P M(0) => 4.59x2 – 0.8935x22 – 3.891 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8935 . 0 ( 2 ) 891 . 3 )( 8935 . 0 ( 4 59 . 4 (4.59) 2 x(a,b) = 787 . 1 676 . 2 (4.59) xa = 1.07 m xb = 4.07 m
20 f. Batang PQ RP => MQ = 0 RP (4.85) – ½ (1.757)(4.85 2 ) –3.764 + 2.913 = 0 4.85RP – 20.6645 – 3.764 + 2.913 = 0 RP = (21.5155 / 4.85) = 4.436 T Kontrol R = Q RP + RQ = q.l 4.435+ 4.085 = (1.757 x 4.85) 8.52= 8.52 Ok
1) Posisi momen maksimum
Dari titik P Mmax = RP (x2) – ½ qx22 – MPQ = 4.436x2 – 0.8785x22 – 3.764 dx dMmax =0 4.436 -1.757x2 = 0 ==> x2 = (4.436/1.757) = 2.525 m Dari titik Q Mmax = RQ (x1) – ½ qx12 – MQP = 4.085x1 – 0.8785x12 – 2.913 dx dMmax =0 4.085-1.757x1 = 0 ==> x1 = (4.085/1.757) = 2.325 m Q P = 3.7 T P = 3.5 T 3.764 tm 2.913 tm 4.85 m P q = 1.757 t/m’ RQ => MP = 0 -RQ (4.85) + ½ (1.757)(4.852) –3.764 + 2.913 = 0 -4.85RQ +20.6645 – 3.764 +2.913 = 0 RQ = (19.8135 / 4.85) = 4.085 T
21 2) Momen maksimum Dari titik P Mmax = 4.436x2 – 0.8785x22 – 3.764 = (4.436)(2.525) - (0.8785)(2.5252) – 3.764 = 11.201 – 5.601 – 3.764 = 1.836 tm Dari titik Q Mmax = 4.085x1 – 0.8785x12 – 2.913 = (4.085)(2.325) - (0.8785)(2.3252) – 2.913 = 9.4976 – 4.7488 – 2.913 = 1.836 tm
3) Posisi momen minimum (M=0)
Dari titik P M(0) => 4.436x2 – 0.8785x22 – 3.764 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8785 . 0 ( 2 ) 764 . 3 )( 8785 . 0 ( 4 436 . 4 (4.436) 2 x(a,b) = 757 . 1 54 . 2 (4.436) xa = 1.08 m xb = 3.97 m Dari titik Q M(0) => 4.085x1 – 0.8785x12 – 2.913 = 0 x(a,b) = a 2 ac 4 b b 2 x(a,b) = ) 8785 . 0 ( 2 ) 913 . 2 )( 8785 . 0 ( 4 085 . 4 (4.085) 2 x(a,b) = 757 . 1 54 . 2 (4.085) xa = 3.77 m xb = 0.88 m
22 8. Menghitung gaya lintang
a. DED = q . l – P7 = (1.56)(1) – 3.85 = - 2.29 T b. DEF = REF – P7 = 8.954 – 3.85 = 5.104 T c. DFE = RFE – P8 = 9.441 – 4.2 = 5.241 T d. DFG = RFG – P8 = 9.111 – 4.2 = 4.911 T e. DGF = RGF – P9 = 8.427 – 4 = 4.427 T f. DGH = q.l – P9 = (1.794)(1.15) – 4 = - 1.94 T g. DJI = q.l – P4 = (1.56)(1) – 3.85 = - 2.29 T h. DJK = RJK – P4 = 8.955 – 3.85 = 5.105 T i. DKJ = RKJ – P5 = 9.440 – 4.2 = 5.240 T j. DKL = RKL – P5 = 9.100 – 4.2 = 4.900 T k. DLK = RLK – P6 = 8.438 – 4 = 4.438 T l. DLM = q.l – P6 = (1.794)(1.15) – 4 = - 1.94 T m. DON = q.l – P1 = (0.758)(1) – 3.4 = - 2.64 T n. DOP = ROP – P1 = 4.345 – 3.4 = 0.945 T o. DPO = RPO – P2 = 4.59 – 3.7 = 0.890 T p. DPQ = RPQ – P2 = 4.436 – 3.7 = 0.736 T q. DQP = RQP – P3 = 4.085 – 3.5 = 0.585 T r. DQR = q.l – P3=(0.871)(1.15)– 3.5 = - 2.50 T
9. Menghitung gaya normal a. Balok 1) NEF = 4 ) M M M M ( ) M M M (MEA AE EJ JE FB BF FK KF = 4 0.177 -0.175 -0.058 116 . 0 1.187 1.240 0.404 0.808 = 0.78 ton (tarik) 2) NFG = 4 ) M M M M ( ) M M M (M - FB BF FK KF GC CG GL LG = 4 166 . 1 182 . 1 400 . 0 799 . 0 177 . 0 175 . 0 058 . 0 0.116 - = 1.02 ton (tarik) 3) NJK = 4 ) M M ( ) M (MJE EJ KF FK + 8 . 3 ) M M ( ) M (MJO OJ KP PK
23 = 4 0.175 177 . 0 240 . 1 187 . 1 8 . 3 0.127 158 . 0 844 . 0 040 . 1 = 0.94 ton (tarik) 4) NKL = 4 ) M M ( ) M (M - KF FK LG GL 8 . 3 ) M M ( ) M (M - KP PK LQ QL = 4 182 . 1 166 . 1 175 . 0 0.177 - 8 . 3 833 . 0 021 . 1 127 . 0 0.158 - = 1.24 ton (tarik) 5) NOP = 8 . 3 ) M M ( ) M (MOJ JO PK KP = 8 . 3 0.158 127 . 0 040 . 1 844 . 0 = 0.42 T (tarik) 6) NPQ = 8 . 3 ) M M ( ) M (M - PK KP QL LQ = 8 . 3 021 . 1 833 . 0 158 . 0 0.127 - = 0.56 T (tarik) b. Kolom
1) NEA = REF + RED – P7 = 8.954 + 1.560 – 3.85 = 6.664 T (tarik) 2) NFB = RFE + RFG – P8 = 9.441 + 9.111 – 4.2 = 14.352 T (tarik) 3) NGC = RGF + RGH – P9 = 8.427 + 1.794 – 4 = 6.221 T (tarik) 4) NJE = RJI + RJK – P4 = 1.560 + 8.955 – 3.85 = 6.665 T (tarik) 5) NKF = RKJ + RKL – P5 = 9.440 + 9.100 – 4.2 = 14.34 T (tarik) 6) NLG = RLK + RLM – P6 = 8.438 + 1.794 – 4 = 6.232 T (tarik) 7) NOJ = ROP + RON – P1 = 4.345 + 0.758 – 3.4 = 1.703 T (tarik) 8) NPK = RPO + RPQ – P2 = 4.59 + 4.436 – 3.7 = 5.326 T (tarik) 9) NQL = RQP + RQR – P3 = 4.085 + 0.871 – 3.5 = 1.456 T (tarik)
24 10.Gambar bidang momen, gaya lintang, dan gaya normal
25 2) Gambar bidang gaya lintang
26 3) Gambar bidang gaya normal
