Analisa Portal Dengan Perhitungan Meto Hardy Cross
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Transcript of Analisa Portal Dengan Perhitungan Meto Hardy Cross
q= 1 t/m'
C
3EI
DMCD = 0,688 Tm
D2EI 2EIDC 5m MCA
MDC = 0,688 Tm
0 d
A4m
BMAC DA MBD
1) Menghitung Momen a. Momen Tumpuan dan Momen lapangan > Momen Primer MCD = + 1/12. q.l^2 = 1.3330 MDC = - 1/12. q.l ^2 = -1.3330 > Angka Kekakuan KAC = 2EI/5 = 0,4 KCD = 2EI/4 = 0,75 KDB = 2EI/5 = 0,4 > Faktor distribusi Q CA = 0,4 / (0,75+0,4) = Q CD = 0,75 / (0,75+0,4) = Q DC = 0,75 / (0,75+0,4) = Q DB = 0,4 / (0,75+0,4) = > Perataan Momen
0 d A
B
0 d
b. Besaran dan letak M maksimum MX = = RA.x -1/2 qx2-MCD2
>>> RA = 1/2 q l = 4/2 = 2
2.x - 1/2 (1)(x) -0,688 2 = 2x - 1/2 x -0,688 dmx/dx = 0 >>> x = 2 m
(berada di tengah-tengah
0.348 0.652 0.652 0.348
2 M max = RA.x -1/2 qx -MCD 2 (2) (2) - 1/2 (1) (2) - 0,688 4 - 2 - 0,688 =
1.312 Tm
Letak M=0 M0
C0.348 0 -0.464 0 -0.201 0 -0.021 0 -0.002 0 0.000 -0.688 0.652 1.333 -0.869 0.576 -0.376 0.061 -0.040 0.007 -0.004 0.001 0.000 0.688 0.652 -1.333 -0.435 1.152 -0.188 0.122 -0.020 0.013 -0.002 0.001 0.000 -0.688
D0.348 0 0 0.615 0 0.0654 0 0.007 0 0.001 0 0.688M0
RA.x -1/2 qx2-MCD = 0 2 2x - 1/2x -0,688 =0 1/2 x -2x + 0,688 = 0 Gunakan rumus ABC: X untuk M = 0 adalah: x= 22
b s b 4ac 2a
2 2 2 4(0.5)( 0,688 ) = 2(0,5) dan2
2 2 2 4(0.5)(0,688 ) = RA.x -1/2 qx -MCD 2( 0,5) RA(2 . 3,620) - (1/2 (1) (0,620)2 )- 0,688 = 0 Tm (2
-0.344 A
B
0.344
= (pahami caranya dengan memahami rumus xl yg saya gunakan)
0 TM
=
2) Menghitung Gaya Lintang a) Batang A-CTm
C0,688 Tm
0 d
CMDB
MAC MCA MCD MDC MDB MBD
= = = = = =
0,344 Tm 0,688 Tm 0,688 Tm 0,688 Tm 0,688 Tm 0,344 Tm
(-) (-) (+) (-) (+) (+)
-
DA = (-MAC/h)-(MCA/h) = (-0,344/5)-(0,688/5) = DC = (+MAC/h)+(MCA/h) = (+0,344/5)+(0,688/5) = b) Batang C-D
-0.206 T 0.206 T
A DC = RC + (MCD/L)-(MDC/L) = 2 + (0,688/4)-(0,688/4) = DD = RD - (MCD/L)+(MDC/L) = 2 - (0,688/4)+(0,688/4) = c) Batang D-B DD = (MDB/h)+(MBD/h) = (0,688/5)+(0,344/5) = DD = (-MDB/h)-(MBD/h) = (-0,688/5)-(0,344/5) = 3) Gaya Normal NAC = DCD = 2 Ton (tekan) NBD = DDC = 2 Ton (tekan) NCD = DCA = 0,206 Ton ( tarik) 2T 2T2 ton
+
0 d
0.206 T-
-0.206 T
0,206 ton
-
)( 0,688 )
3.620 meter
)(0,688 )
0.380 meterRA.x -1/2 qx2-MCD (2 . 0,380) - (1/2 (1) (0,380)2) - 0,688 = 0 Tm
0 TM
0,688 Tm
0,688 Tm
D0,688 Tm
+1,312 Tm
Gambar Bidang Momen
+
0,344 Tm
B
+ 0,344 Tm
+
-
+
Gambar Bidang gaya lintang
0,206 ton
-
Gambar Bidang gaya Normal
