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No. 1



E 8



4



F



C 7 5



1



9



3



A



6



2



10



L =



12



m



Diketahui Rangka Kap, dengan ukuran/parameter sbb: = = = = =



5.5 m Kemiringan = tan a = 12 m o 30 4 m emen asbes gelombang ( G = 50 kg/m2 )



Diminta : -



Besarnya gaya-gaya batang Dimensi batang atas, bawah dan tegak/diagonal Sambungan baut pada 2 (dua) titik kumpul Sketsa dari rangka kap, termasuk detail 2 (dua) titik kumpul



Digunakan : Baja ST 37 sd = 1600 Kg/cm2



13



G



D



B



Jarak Rangka Kap Panjang Bentangan Kemiringan atap Tinggi Rangka Kap Jenis atap



12 11



4 6



= 0.666666667



a



H



H



1. MENGHITUNG PANJANG BATANG E D



F G



C



4 m H



B A



3



J 3



m



I



30



K L



3



3



m



m



3



m



12.00 m



* Kerena bangunan simetris, ditinjau hanya 1/2 batang EK = 4 m EA =



6 cos



AK =



6 cos



=



6.928 m



=



6.018 m



3



AB = BC = AJ



= 30



CD = DE = ¼ = ¼



JK = ½ x



AK = ½ x



x EA x 6.93 = 1.732 m 6.018 = 3.009 m



BJ² = 1.732 ² + 3.009 ² BJ = √ 2.767 = 1.663 m



2 (



1.732



) (



3.009



)



cos



27 =



2.767



CJ² = 1.732 ² + 1.663 ² CJ = √ 2.488 = 1.577 m



2 (



1.732



) (



1.663



)



cos



55.3 =



2.488



DJ² = 1.732 ² + 1.577 ² DJ = √ 2.756 = 1.660 m



2 (



1.732



) (



1.577



)



cos



120 =



2.756



DK² = 1.732 ² + 4.000 ² DK = √ 11.116 = 3.334 m



2 (



1.732



) (



4.000



)



cos



55.3 =



11.116



REKAPITULASI PANJANG BATANG



BATANG AB, BC, CD, DE, EF, FG, GH, HI AJ, JK, KL, LI EK DK, FK DJ, FL CJ, GL BJ, HL



PANJANG ( m ) 1.732 3.009 4.000 3.334 1.660 1.577 1.663



PERHITUNGAN SUDUT < AJB ,



Cos β = β =



1.663 ² + 3.009 ² 1.732 ² 2 ( 1.663 ) ( 3.009 ) 28.2



γ = 180 -



º 28.2 -



27 = 124.8 º



=



0.881



< CBJ ,



= 180 -



< BJC ,



Cos β = β =



64.7



Cos β = β =



64.7 -



γ = 180 < DJK ,



60



Cos β = β =



γ = 180 < KDE ,



Cos β = β =



γ = 180 -



0.428



-4.6



-4.6 º



120 = - 64.7 -



64.7 =



55.2 º =



0.868



=



0.904



º 33.5 91.3



55.2 = 91.3 º - 33.5 =



55.2 º



3.334 ² + 4.000 ² 1.732 ² 2 ( 3.334 ) ( 4.000 ) 35.2



=



º



120 º



=



64.7 -



= 180 -



< DKE ,



60



3.009 ² + 3.334 ² 1.660 ² 2 ( 3.009 ) ( 3.334 ) 33.5



0.430



º



= 180 -



< JKD ,



55.2 =



1.577 ² + 1.660 ² 1.732 ² 2 ( 1.577 ) ( 1.660 ) 64.7



=



º



= 180 -



< CJD ,



= 55.2 º



1.663 ² + 1.577 ² 1.732 ² 2 ( 1.663 ) ( 1.577 )



γ = 180 < JDC ,



124.8



º 35.2 -



55.2 = 89.6 º



Sin a



=



0.500



Cos a



=



0.866



Tan a



=



0.577



1



A



a 2



2. MENGHITUNG BEBAN - BEBAN Untuk Perhitungan Gording Beban Yang Perlu Diperhitungkan Adalah : a. Beban Atap b. Beban Angin c. Beban Tak Terduga d. Beban Gording Uraian : a. Beban Atap L' m L' m



Q Qx



Qy



a



Jarak Antar Gording (a) = AC/3 = 3.464 m Jarak Kap (b) = 5.5 m Berat Genteng dengan reng dan usuk (G) = 50 Kg/m2 Jadi : (sesuai PMI 1970) Q = axG Q = 3.464 x = 173.210 Kg/m 50 = 86.605 Kg/m ~ Qx = Q.Sin ao Qy = Q.Cos ao = 150.004 Kg/m ~ Mx = 1/8 Qy b2 = 567.204 Kg.m My = 1/8 Qx b2 = 327.475 Kg.m



b. Beban Angin P



P P



P



2 = 100 Kg/m (sesuai PMI 1970) = 0,02a - 0,4 = 0.200 = -0.4 (belakang angin untuk semua a) = c1 x a x P = 0.200 x 3.464 x 100 = 69.284 Kg/m



Tekanan Angin (P) Koefisien Angin (c1)



P



c2 W



a ~



Mx My



= 1/8 W b2 = 0



= 261.980 Kg.m



c. Beban Tak Terduga Px



P Py



P = 125 Kg ~ Px = P.Sin a Py = P.Cos a ~ Mx = 1/4 Py b My = 1/4 Px b



= = = =



62.500 108.253 148.848 85.938



Kg Kg Kg.m Kg.m



a d. Beban Gording Diketahui berat gording dalam perbandingan normal (q) = 10 Kg/m - 50 Kg/m Direncanakan berat gording (q) =



qx



q



12 ~



qy



~



qx qy Mx My



kg/m = = = =



q.Sin a q.Cos a 1/8 qy b2 1/8 qx b2



= = = =



6.000 10.392 39.296 22.688



Kg/m Kg/m Kg.m Kg.m



a Kombinasi Pembebanan : Keadaan 1



: Beban Atap + Beban Gording + Tekanan Angin Mx = 567.204 + 39.296 + 261.980 = 868.480 Kg.m My = 327.475 + 22.688 + 0.000 = 350.163 Kg.m



Keadaan 2



: Beban Atap + Beban Gording + Beban Tak Terduga Mx = 567.204 + 39.296 + 148.848 = 755.348 Kg.m My = 327.475 + 22.688 + 85.938 = 436.1 Kg.m



Ambil Maksimalnya : Mx = 755.348 Kg.m My = 436.100 Kg.m 3. MENGONTROL PROFIL GORDING Mengontrol Tegangan : Mx Kontrol : Wx +



My Wy < s



Dimana



s s =



leleh



1.5 2400 s = = 1600 1.5 Wy diambil = 1/8 Wx Mx + Wx



My (1/8)WX



Mx + 8 My Wx



< s



Misalkan : Mx + 8 My Wx maka :



Wx perlu =



Mx



Kg/cm2



< s



= s +



8 My s



=



755.348 +



( 8 ) . 436.100 x 100 = 265 Cm3 1600



Wx perlu = 265.259 Cm3 Diambil Profil kanal12 dengan data sbb: Wx Wy Ix Iy q



( Untuk Bj A37)



tw =



= 449 cm3 = 103 cm3 = 4490 cm4 = 771 cm4 = 13.4 Kg/m



Mx My + < s Wx Wy 755.35 436.100 ( + ) . 100 < 449 103



0.9 cm



Kontrol :



591.63



s




0.141 > 0.141



2340 123.6



9



0.769



oke!!



ayfiktif ay1 x sd x 0.299 x 0.769 x 367.890 Kg/cm2 P = 2F



~



=



18.932



1600




0.121 > 0.121



2050 123.6



18



0.736



oke!!



ay1 x sd x x 0.736 x Kg/cm2



ayfiktif 0.266 313.24



= = =



=



=



1600



16.586