STRUKTUR RANGKA

Dit : * Hitung Reaksi Perletakkan * Hitung gaya gaya Batang Penyelesaian : MA = o-RB(7)+2,6(7)+3.6(3.5)+2.6(0) =0-7RB + 18,2 + 12,6 + 0 = 0-7RB =-30.8RB = = 4,4 TonMB = 0RA(7) - 2,6(7) -3,6(3,5) 2,6(0) = 07RA 18,2 12,6 0 = 07RA =30,8RA = = 4,4 TonCONTROL = AKSI = REAKSI4,4t + 4,4t - 2,6 + 3,6 + 2,6 = 08,8 -8,8 = 0..okH = 0H1 + H2 + H3 = H1,6 + 2,6 + 1,6 = H5,8 = H.H = 5,8Menghitung gaya batang dari titik buhul AJOINT A

GAMBAR JOINT A Dari ABD diperoleh panjang AD = 7,615mFy = 04,4 ton + S2 sin = 0S2 sin = - 4,4 Ton (BATANG S1= BATANG TEKAN) S2 = - = = - 11,195 ton ( BATANG TEKAN )Sin = = = 0,393Cos = = = 0,919S3 = S2.Cos S3 = -11,195 x 0,919S3 = -10,288ton ( BATANG TEKAN)JOINT C

GAMBAR JOINT CDari CDF diperoleh panjang CF = 7,615mFy = 0-4,4 ton + S6 sin = 0S6 sin = 4,4 Ton (BATANG S5 = BATANG TARIK) S6 = = = 11,195 ton ( BATANG TARIK )Sin = = = 0,393Cos = = = 0,919S7= S2.Cos + H1(1,6)S7 = 11,195 x 0,919 + 1,6S7= 11,888ton ( BATANG TARIK)

JOINT E

GAMBAR JOINT EDari EFH diperoleh panjang EH = 7,615mFy = 04,4 ton + S10 sin = 0S10 sin = - 4,4 Ton (BATANG S9 = BATANG TEKAN) S10 = = = -11,195 ton ( BATANG TEKAN )Sin = = = 0,393Cos = = = 0,919S11= S2.Cos + H2(1,6)S11 = -11,195 x 0,919 + 2,6S11= -7.688 ( BATANG TEKAN)JOINT G

Dari GII diperoleh panjang GI = 4,031mFy = 0-4,4 ton + S13 sin = 0S13 sin = 4,4 Ton)S13 = = = 8,870 ton ( BATANG TARIK )Sin = = = 0,496Cos = = = 0,868S14= S2.Cos + H3(1,6)S14 = 8,870 x 0,868 + 1,6S14 = 9,299 ( BATANG TARIK)JOINT I

GAMBAR JOINT IDari GII diperoleh panjang GI = 4,031mFy = 03,6 ton + S13 sin = 0S16sin = - 3,6 Ton ( BATANG TEKAN )S16 = = = -7,258 ton ( BATANG TEKAN )

TABEL GAYA GAYA BATANGNO. BATANG TARIK (ton)TEKAN(ton)

11 = 44,4

2211,195

3310,288

45=84,4

5611,195

6711,888

79=124,4

81011,195

9117.688

1013=178,870

1114=159,299

12163,6

JADI Gaya batang tarik terbesar yaitu = 11,888 ton Gaya Batang tekan terbesar = 11,195 tonCivil engineeringUniversitas Darma Agung Medan Page 2