Apakah Struktur Gelagar dibawah ini SST tau SSTT??

Struktur Gelagar GERBER ... Contoh Pengujian Gelagar Gerber. STRUKTUR BALOK GERBER ** Balok : S - C...

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Apakah Struktur Gelagar dibawah ini SST tau SSTT??

Apakah Struktur Gelagar dibawah ini SST tau SSTT??

Apakah Struktur Gelagar dibawah ini SST tau SSTT??

Struktur Gelagar GERBER Penemu Struktur Gerber : JGH.Gerber Merupakan Struktur gelagar yang terdiri dari Gelagar anak dan Gelagar Induk.

Dimana, Salah satu/kedua tumpuan Gelagar anak menumpu pada Gelagar Induk. Reaksi tumpuan pada salah satu tumpuan Gelagar anak akan diteruskan Ke Gelagar induk sebagai beban Terpusat luar.

* SATU sendi dengan TIGA tumpuan.

* DUA sendi dengan EMPAT tumpuan.

* DUA sendi dengan EMPAT tumpuan.

Contoh Pengujian Gelagar Gerber

STRUKTUR BALOK GERBER

** Balok : S - C. MC = 0 RS. 5 - P2. 2,5 = 0 RS = 0,5 P2 MS = 0 RC. 5 - P2. 2,5 = 0 RC = 0,5 P2 V = 0 RS + RC = P2 0,5 P2 + 0,5 P2 = P2 P2 = P2  ok

ME = 2,5 RS

MB = 0 RA. 5 + RS. 1,5 - P1. 2,5 = 0 5 RA + 0,5 P2. 1,5 - 2,5 P1 = 0 5 RA + 0,75 P2 - 2,5 P1 = 0 RA = 0,5 P1 - 0,15 P2

MA = 0 RB. 5 - RS. 6,5 - P1. 2,5 = 0 5 RB - 0,5 P2. 6,5 - 2,5 P1 = 0 5 RB - 3,25 P2 - 2,5 P1 = 0 RB = 0,5 P1 + 0,65 P2 RBS = RS = 0,5 P2 RBA = RB - RS = 0,5 P1 + 0,65 P2 - 0,5 P2 = 0,5 P1 + 0,15 P2

MD = RA. 2,5 = (0,5 P1 - 0,15 P2) 2,5 = 1,25 P1 - 0,375 P2 MB = RS. 1,5 = 0,5 P2. 1,5 = 0,75 P2

46) Gambar bidang momen dan gaya lintang. P1 = 400 kg, P2 = 500 kg

Balok : S – C RS = RC = 0,5 P2 = 0,5. 500 = 250 kg ME = RC. 2 = 250. 2 = 500 kgm

Balok : A – B – S. MB = 0 RA. 4 + RS. 2 - P1. 2 = 0 4 RA + 250. 2 - 400. 2 = 0 4 RA + 500 - 800 = 0 4 RA = 300 RA = 75 kg

MA = 0 RB. 4 - RS. 6 - P1. 2 = 0 4 RB - 250. 6 - 400. 2 = 0 4 RB - 1500 - 800 = 0 4 RB - 2300 = 0 RB = 575 kg RBS = RS = 250 kg V = 0 RA + RB = P1 + RS 75 + 575 = 400 + 250 650 = 650  ok

RBA = RB - RBS = 575 - 250 = 325 kg MB = RS. 2 = 250. 2 = 500 kgm

MD = RA. 2 = 75. 2 = 150 kgm

BIDANG GESER

BIDANG MOMEN

47) Gambar bidang momen dan gaya lintang. P = 500 kg, W = 2000 kg/m

Balok : S – C RS = RC = 0,5 P = 0,5. 500 = 250 kg

MD = RC. 2 = 250. 2 = 500 kgm

Balok : A – B – S. MB = 0 RA. 4 + RS. 2 + W. 2. 1 - W. 4. 2 = 0 4 RA + 250. 2 + 2000. 2 - 2000. 8 = 0 4 RA + 500 + 4000 - 16000 = 0 4 RA - 11500 = 0 RA = 2875 kg

RBS = RS + Q = 250 + 2000. 2 = 4250 kg RBA = RB - RBS = 9375 - 4250 = 5125 kg

MA = 0 RB. 4 - RS. 6 - W. 4. 2 - W. 2. 5 = 0 4 RB - 250. 6 - 2000. 8 - 2000. 10 = 0 4 RB - 1500 - 16000 - 20000 = 0 4 RB - 37500 = 0 RB = 9375 kg V = 0 RA + RB = Q + RS 2875 + 9375 = 2000. 6 + 250 2250 = 12250  ok

X2

MX = RA. X - 0,5 W. = 2875 X - 0,5. 2000 X2

dMX  2875  2000X dX dMX 0 dX 2000 X = 2875 X = 1,44 m

M maks = 2875. 1,44 - 1000. 1,442 = 4140 - 2074 = 2066 kgm

MB = RS. 2 + W. 2. 1 = 250. 2 + 2000. 2 = 500 + 4000 = 4500 kgm

BIDANG GESER

BIDANG MOMEN

48) Gambar bidang momen dan gaya lintang. P = 2000 kg, W = 2500 kg/m

Balok : S – C. RS = RC = 0,5 P + 0,5 W. 4 = 0,5. 2000 + 0,5. 2500. 4 = 1000 + 5000 = 6000 kg

ME = RC. 2 - W. 2. 1 = 6000. 2 - 2500. 2 = 12000 - 5000 = 7000 kgm

Balok : A – B – S MB = 0 RA. 6 + RS. 2 + W. 2. 1 - W. 6. 3 - P. 3 = 0 6 RA + 6000. 2 + 2500. 2 - 2500. 18 - 2000. 3 = 0 6 RA + 12000 + 5000 - 45000 - 6000 = 0 6 RA - 34000 = 0 RA = 5667 kg MA = 0 RB. 6 - RS. 8 - W. 6. 3 - W. 2. 7 - P. 3 = 0 6 RB - 6000. 8 - 2500. 18 - 2500. 14 - 2000. 3 = 0 6 RB - 48000 - 45000 - 35000 - 6000 = 0 6 RB - 134000 = 0 RB = 22333 kg

V = 0 RA + RB = Q + RS + P 5667 + 22333 = 2500. 8 + 6000 + 2000 28000 = 28000  ok RBS = RS + Q = 6000 + 2500. 2 = 11000 kg RBA = RB - RBS = 22333 - 11000 = 11333 kg

MX = RA. X - 0,5 W. X2 = 5667 X - 0,5. 2500 X2

dMX  5667  2500X dX dMX 0 dX 2500 X = 5667 X = 2,27 m M maks = 5667. 2,27 - 1250. 2,272 = 12864 - 6441 = 6423 kgm

MB = RS. 2 + W. 2. 1 = 6000. 2 + 2500. 2 = 12000 + 5000 = 17000 kgm

MD = RA. 3 - W. 3. 4,5 = 5667.3 - 2500. 4,5 = 17001 - 11250 = 5751 kgm

BIDANG GESER

BIDANG MOMEN

49) Gambar bidang momen dan gaya lintang. P1 = 600 kg, P2 = 1000 kg, P3 = 800 kg

S1

Balok : A – S1

RA = RS1 = 0,5 P1 = 0,5. 600 = 300 kg ME = RA. 1,5 = 300. 1,5 = 450 kgm

Balok : S2 – D MD = 0 RS2. 3 - P3. 2 = 0 3 RS2 - 800. 2 = 0 RS2 = 533 kg MS2 = 0 RD. 3 - P3. 1 = 0 3 RD - 800. 1 = 0 3 RD - 800 = 0 RD = 267 kg

V = 0 RS2 + RD = P3 533 + 267 = 800 800 = 800  ok MG = RD. 2 = 267. 2 = 534 kgm

Balok : S1 – B – C – S2 MC = 0 RB. 5 + RS2. 1 - RS1. 6 - P2. 3 = 0 5 RB + 533. 1 - 300. 6 - 1000. 3 = 0 5 RB + 533 - 1800 - 3000 = 0 5 RB - 4267 = 0 RB = 853 kg

MB = 0 RC. 5 + RS1. 1 - P2. 2 - RS2. 6 = 0 5 RC + 300. 1 - 1000. 2 - 533. 6 = 0 5 RC + 300 - 2000 - 3198 = 0 5 RC - 4898 = 0 RC = 980 kg

V = 0 RB + RC = RS1 + P2 + RS2 853 + 980 = 300 + 1000 + 533 1833 = 1833  ok

RBS1 = RS1 = 300 kg RBC = RB - RBS1 = 853 - 300 = 553 kg

MB = RS1. 1 = 300. 1 = 300 kgm

MC = RS2. 1 = 533. 1 = 533 kgm

RCS2 = RS2 = 533 kg RCB = RC - RCS2 = 980 - 533 = 447 kg

MF = RB. 2 - RS1. 3 = 853. 2 - 300. 3 = 1706 - 900 = 806 kgm

BIDANG GESER

BIDANG MOMEN

50) Gambar bidang momen dan gaya lintang. W1 = 1500 kg/m, W2 = 1800 kg/m, W3 = 2200 kg/m

Balok : A – S1 RA = RS1 = 0,5 W 1. 3 = 0,5. 1500. 3 = 2250 kg

M maks = 1/8 W1. 32 = 1/8. 1500. 9 = 1688 kgm

Balok : S2 – D RS2 = RD = 0,5 W 2. 4 = 0,5. 1800. 4 = 3600 kg M maks = 1/8 W 2. 42 = 1/8. 1800. 16 = 3600 kgm

Balok : S1 – B – C – S2 MC = 0 RB. 5 + RS2. 1 + W 2. 1. 0,5 - RS1. 6 - W 1. 1. 5,5 - W 3. 5. 2,5 = 0 5 RB + 3600. 1 + 1800. 0,5 - 2250. 6 - 1500. 5,5 - 2200. 12,5 = 0 5 RB + 3600 + 900 - 13500 - 8250 - 27500 = 0 5 RB – 44750 = 0 RB = 8950 kg MB = 0 RC. 5 + RS1. 1 + W 1. 1. 0,5 - RS2. 6 - W 2. 1. 5,5 - W 3. 5. 2,5 = 0 5 RC + 2250. 1 + 1500. 0,5 - 3600. 6 - 1800. 5,5 - 2200. 12,5 = 0 5 RC + 2250 + 750 - 21600 - 9900 - 27500 = 0 5 RC - 56000 = 0 RC = 11200 kg

V = 0 RB + RC = RS1 + RS2 + W 1. 1 + W 2. 1 + W 3. 5 8950 + 11200 = 2250 + 3600 + 1500. 1 + 1800. 1 + 2200. 5 20150 = 20150  ok

RBS1 = RS1 + Q = 2250 + 1500. 1 = 3750 kg RBC = RB - RBS1 = 8950 - 3750 = 5200 kg

RCS2 = RS2 + Q = 3600 + 1800. 1 = 5400 kg RCB = RC - RCS2 = 11200 - 5400 = 5800 kg

MX = RB. X - RS1 (1 + X) - W 1. 1 (0,5 + X) - 0,5 W 3. X2 = 8950 X - 2250 (1 + X) - 1500 (0,5 + X) - 0,5. 2200 X2 = 8950 X - 2250 - 2250 X - 750 - 1500 X - 1100 X2 = 5200 X - 3000 - 1100 X2

dMX  5200  2000 x dX dMX 0 dX 2200 X = 5200 X = 2,36 m M maks = 5200. 2,36 - 3000 - 1100. 2,362 = 12272 - 3000 - 6127 = 3145 kgm

MB = RS1. 1 + W1. 1. 0,5 = 2250. 1 + 1500. 0,5 = 2250 + 750 = 3000 kgm MC = RS2. 1 + W2. 1. 0,5 = 3600. 1 + 1800. 0,5 = 3600 + 900 = 4500 kgm

BIDANG GESER

BIDANG MOMEN

51) Gambar bidang momen dan gaya lintang. P1 = 500 kg, P2 = 800 kg, W1 = 1500 kg/m, W2 = 1800 kg/m, W3 = 2200 kg/m

Balok : A – S1 MS1 = 0 RA 3 - P1. 1 - W2. 3. 1,5 = 0 3 RA - 500. 1 - 2000. 4,5 = 0 3 RA - 1500 - 9000 = 0 3 RA - 10500 = 0 RA = 3167 kg

MX = RA. X - W 2. X. 0,5 X = 3167 X - 0,5. 2000 X2

dMX  3167  2000 x dX dMX 0 dX 2000 X = 3167 X = 1,58 m

MA = 0 RS1 - P1. 2 - W2. 3. 1,5 = 0 3 RS1 - 500. 1 - 2000. 4,5 = 0 3 RS1 - 500 - 9000 = 0 3 RS1 - 9400 = 0 RS1 = 3333 kg

V = 0 RA + RS1 = P1 + Q 3167 + 3333 = 500 + 2000. 3 6500 = 6500  ok

M maks = 3167. 1,58 - 1000. 1,582 = 5004 - 2496 = 2508 kgm ME = RS1. 1 - W 2. 1. 0,5 = 3333. 1 - 2000. 0,5 = 3333 - 1000 = 2333 kgm

Balok : S2 – D MD = 0 RS2 3 - P1. 2 - W 1. 3. 1,5 = 0 3 RS2 - 500. 2 - 1800. 4,5 = 0 3 RS2 - 1000 - 8100 = 0 3 RS2 - 9100 = 0 RS2 = 3033 kg

MS2 = 0 RD. 3 - P1. 1 - W 1. 3. 1,5 = 0 3 RD - 500. 1 - 1800. 4,5 = 0 3 RD - 500 - 8100 = 0 3 RD - 8600 = 0 RD = 2867 kg

V = 0 RS2 + RD = P1 + Q 3033 + 2867 = 500 + 1800. 3 5900 = 5900  ok

MX = RD. X - W1. X. 0,5 X = 2867 X - 0,5. 1800 X2

dMX  2867  1800 x dX dMX 0 dX 1800 X = 2867 X = 1,59 m

M maks = 2867. 1,59 - 900. 1,582 = 5004 - 2496 = 2283 kgm MG = RS2. 1 - W 1. 1. 0,5 = 3033. 1 - 1800. 0,5 = 3033 - 900 = 2133 kgm

Balok : S1 – B – C – S2 MC = 0 RB. 4 + RS2. 1 + W 3. 1. 0,5 - S1. 5 - W 3. 1. 4,5 - W 3. 4. 2 - P2. 2 = 0 4 RB + 3033. 1 + 2200. 0,5 - 3333. 5 - 2200. 4,5 - 2200. 8 - 800. 2 = 0 4 RB + 3033 + 1100 - 16665 - 9900 - 17600 - 1600 = 0 4 RB - 41632 = 0 RB = 10408 kg MB= 0 RC. 4 + RS1. 1 + W 3. 1. 0,5 - RS2. 5 - W 3. 1. 4,5 - W 3. 4. 2 - P2. 2 = 0 4 RC + 3333. 1 + 2200. 0,5 - 3033. 5 - 2200. 4,5 - 2200. 8 - 800. 2 = 0 4 RC + 3333 + 1100 - 15165 - 9900 - 17600 - 1600 = 0 4 RC - 39832 = 0 RC = 9958 kg

V = 0 RB + RC = RS1 + RS2 + P2 + Q 10408 + 9958 = 3333 + 3033 + 800 + 2200. 6 20366 = 20366  ok

RBS1 = RS1 + Q = 3333 + 2200.1 = 5533 kg

RBC = RB - RBS1 = 10408 - 5533 = 4875 kg

RCS2 = RS2 + Q = 3033 + 2200. 1 = 5233 kg RCB = RC - RCS2 = 9958 - 5233 = 4725 kg

MX = RC. X - RS2 (1 + X) - W 3. 1 (0,5 + X) - W 3. X. 0,5 X = 9958 X - 3033 (1 + X) - 2200 (0,5 + X) - 0,5. 2200 X2 = 9958 X - 3033 - 3033 X - 1100 - 2200 X - 1100 X2 = 4725 X - 4133 - 1100 X2

dMX  4725  2200 x dX dMX 0 dX

2200 X = 4725 X = 2,15 m > 2 m

 tidak mungkin

MX = RB. X - RS1 (1 + X) - W 3. 1 (0,5 + X) - W 3. X. 0,5 X = 10408 X - 3333 (1 + X) - 2200 (0,5 + X) - 0,5. 2200 X2 = 10408 X - 3333 - 3333 X - 1100 - 2200 X - 1100 X2 = 4875 X - 4433 - 1100 X2

dMX  4875  2200 x dX dMX 0 dX

2200 X = 4875 X = 2,22 m > 2 m  tidak mungkin

M maks = MF = RB. 2 - RS1. 3 - W 3. 1. 2,5 - W 3. 2. 1 = 10408. 2 - 3333. 3 - 2200. 2,5 - 2200. 2 = 20816 - 9999 - 5500 - 4400 = 917 kgm

MB = RS1. 1 + W3. 1. 0,5 = 3333. 1 + 2200. 0,5 = 3333 + 1100 = 4433 kgm MC = RS2. 1 + W 3. 1. 0,5 = 3033. 1 + 2200. 0,5 = 3033 + 1100 = 4133 kgm

BIDANG GESER

BIDANG MOMEN

52) Gambar bidang momen dan gaya lintang. P1 = 1000 kg, P2 = 1250 kg, P3 = 1500 kg

Balok : S1 – S2. RS1 = RS2 = 0,5. P2 = 0,5. 1250 = 625 kg

Balok : A – B - S1

V = 0 RA + RB = P1 + RS1 594 + 1031 = 1000 + 625 1625 = 1625  ok

RBS1 = RS1 = 625 kg

MF = RS1. 2 = 625. 2 = 1250 kgm

MB = 0 RA. 4 + RS1. 1 - P1. 3 = 0 4 RA + 625. 1 - 1000. 3 = 0 4 RA + 625 - 3000 = 0 4 RA - 2375 = 0 RA = 594 kg

RBA = RB - RBS1 = 1031 - 625 = 406 kg MB = RS1. 1 = 625. 1 = 625 kgm

MA = 0 RB. 4 - RS1. 5 - P1. 1 = 0 4 RB - 625. 5 - 1000. 1 = 0 4 RB - 3125 - 1000 = 0 4 RB - 4125 = 0 RB = 1031 kg

ME = RA. 1 = 594. 1 = 594 kgm

Balok : S2 – C - D MD = 0 RC. 4 - RS2. 5 - P3. 3 = 0 4 RC - 625. 5 - 1500. 3 = 0 4 RC - 3125 - 4500 = 0 4 RC - 7625 = 0 RC = 1906 kg MC = 0 RD. 4 + RS2. 1 - P3. 1 = 0 4 RD + 625. 1 - 1500. 1 = 0 4 RD + 625 - 1500 = 0 4 RD - 875 = 0 RD = 219 kg

V = 0 RC + RD = RS2 + P3 1906 + 219 = 625 + 1500 2125 = 2125  ok

RCS2 = RS2 = 625 kg

RCD = RC – RCS2 = 1906 – 625 = 1281 kg

MC = RS2. 1 = 625. 1 = 625 kgm

MG = RD. 3 = 219. 3 = 657 kgm

BIDANG GESER

BIDANG MOMEN

53) Gambar bidang momen dan gaya lintang. W1 = 1500 kg/m, W2 = 1800 kg/m, W3 = 2100 kg/m

Balok : S1 – S2

1800

RS1 = RS2 = 0,5. W 2. 4 = 0,5. 1800. 4 = 3600 kg M maks = 1/8. W. 42 = 1/8. 1800. 16 = 3600 kgm

Balok : A – B -S1

V = 0 RA + RB = RS1 + W1. 5 1913 + 9187 = 3600 + 1500. 5 11100 = 11100  ok RBS1 = RS1 + Q = 3600 + 1500. 1 = 5100 kg

MB = 0 RA. 4 + RS1. 1 + W 1. 1. 0,5 - W 1. 4. 2 = 0 4 RA + 3600. 1 + 1500. 0,5 - 1500. 8 = 0 4 RA + 3600 + 750 - 12000 = 0 4 RA - 7650 = 0 RA = 1913 kg MA = 0 RB. 4 - RS1. 5 - W 1. 4. 2 - W 1. 1. 4,5 = 0 4 RB - 3600. 5 - 1500. 8 - 1500. 4,5 = 0 4 RB - 18000 - 12000 - 6750 = 0 4 RB - 36750 = 0 RB = 9187 kg RBA = RB - RBS1 = 9187 - 5100 = 4087 kg

Balok : A – B -S1

V = 0 RA + RB = RS1 + W1. 5 1913 + 9187 = 3600 + 1500. 5 11100 = 11100  ok

RBS1 = RS1 + Q = 3600 + 1500. 1 = 5100 kg RBA = RB - RBS1 = 9187 - 5100 = 4087 kg

MX = RA. X - 0,5 W1. X2 = 1913 X - 0,5. 1500 X2 dMX  1913 1500X dX dMX 0 dX 1500 X = 1913 X = 1,28 m

M maks = 1913. 1,28 - 750. 1,282 = 2449 - 1229 = 1220 kgm MB = RS1. 1 = 3600. 1 = 3600 kgm

Balok : S2 – C – D MD = 0 RC. 4 - RS2. 5 - W 3. 1. 4,5 - W 3. 4. 2 = 0 4 RC - 3600. 5 - 2100. 4,5 - 2100. 8 = 0 4 RC - 18000 - 9450 - 16800 = 0 4 RC - 44250 = 0 RC = 11063 kg MC = 0 RD. 4 + RS2. 1 + W 3. 1. 0,5 - W 3. 4. 2 = 0 4 RD + 3600. 1 + 2100. 0,5 - 2100. 8 = 0 4 RD + 3600 + 1050 - 16800 = 0 4 RD - 12150 = 0 RD = 3037 kg V = 0 RC + RD = RS2 + W 3. 5 11063 + 3037 = 3600 + 2100, 5 14100 = 14100  ok

RCS2 = RS2 + Q = 3600 + 2100. 1 = 5700 kg RCD = RC - RCS2 = 11063 - 5700 = 5363 kg

Balok : S2 – C – D RCS2 = RS2 + Q = 3600 + 2100. 1 = 5700 kg RCD = RC - RCS2 = 11063 - 5700 = 5363 kg

MX = RD. X - 0,5 W 3. X2 = 3037 X - 0,5. 2100 X2

dMX  3037  2100X dX dMX 0 dX 2100 X = 3037 X = 1,45 m

M maks = 3037. 1,45 - 1050. 1,452 = 4404 - 2208 = 2196 kgm MC = RS2. 1 + W 3. 1. 0,5 = 3600. 1 + 0,5. 2100 = 3600 + 1050 = 4650 kgm

BIDANG MOMEN

BIDANG MOMEN

54) Gambar bidang momen dan gaya lintang. P = 1500 kg, W = 1800 kg/m

Balok : S1 – S2

MS2 = 0 RS1. 4 - W. 1,5. 3,25 - P. 1 = 0 4 RS1 - 1800. 4,875 - 1500. 1 = 0 4 RS1 - 8775 - 1500 = 0 4 RS1 - 10275 = 0 RS1 = 2569 kg MS1 = 0 RS2. 4 - W. 1,5. 0,75 - P. 3 = 0 4 RS2 - 1800. 1,125 - 1500. 3 = 0 4 RS2 - 2025 - 4500 = 0 4 RS1 - 6525 = 0 RS2 = 1631 kg

V = 0 RS1 + RS2 = Q + P 2569 + 1631 = 1800. 1,5 + 1500 4200 = 4200  ok MX = RS1. X - 0,5 W. X2 = 2569 X - 0,5. 1800 X2

dMX  2569  1800X dX dMX 0 dX 1800 X = 2569 X = 1,43 m

M maks = 2569. 1,43 - 900. 1,432 = 3674 - 1840 = 1833 kgm MF = RS1. 1,5 - 0,5 W. 1,52 = 2569. 1,5 - 0,5. 1800. 2,25 = 3854 - 2025 = 1829 kgm MG = RS2. 1 = 1631. 1 = 1631 kgm

Balok : A – B – S1 V = 0 RA + RB = RS1 + Q1 + Q2 1833 + 6136 = 2569 + 1800. 1 + 1800. 2 7969 = 7969  ok

RBS1 = RS1 + Q = 2569 + 1800. 1 = 4369 kg RBA = RB - RBS1 = 6136 - 4369 = 1767 kg MA = 0 RB. 4 - RS1. 5 - W. 1. 4,5 - W. 2. 1 = 0 4 RB - 2569. 5 - 1800. 4,5 - 1800. 2 = 0 4 RB - 12845 - 8100 - 3600 = 0 4 RB - 24545 = 0 RB = 6136 kg MB = 0 RA. 4 + RS1. 1 + W. 1. 0,5 - W. 2. 3 = 0 4 RA + 2569. 1 + 1800. 0,5 - 1800. 6 = 0 4 RA + 2569 + 900 - 10800 = 0 4 RA - 7331 = 0 RA = 1833 kg

ME = RA. 2 - Q. 1 = 1833. 2 - 1800. 2 = 3666 - 3600 = 66 kgm MB = RS1. 1 + Q. 0,5 = 2569. 1 + 1800. 0,5 = 2569 + 900 = 3469 kgm

Balok : A – B – S1 MX = RA. X - 0,5 W. X2 = 1833 X - 0,5. 1800 X2

dMX  1833 1800X dX dMX 0 dX

1800 X = 1833 X = 1,02 m M maks = 1833. 1,02 - 900. 1,022 = 1870 - 936 = 934 kgm

Balok : S2 – C – D

MD = 0 RC. 4 - RS2. 5 - W. 3. 3,5 - P. 1 = 0 4 RC - 1631. 5 - 1800. 10,5 - 1500. 1 = 0 4 RC - 8155 - 18900 - 1500 = 0 4 RC - 28555 = 0 RC = 7139 kg MC = 0 RD. 4 + RS2. 1 + W. 1. 0,5 - W. 2. 1 - P. 3 = 0 4 RD + 1631. 1 + 1800. 0,5 - 1800. 2 - 1500. 3 = 0 4 RD + 1631 + 900 - 3600 - 4500 = 0 4 RD - 5569 = 0 RD = 1392 kg

V = 0 RC + RD = Q + P + RS2 7139 + 1392 = 1800. 3 + 1500 + 1631 8531 = 8531  ok

RCS2 = RS2 + Q = 1631 + 1800. 1 = 3431 kg RCD = RC - RCS2 = 7139 - 3431 = 3708 kg

Balok : S2 – C – D

MX = RC. X - RS2 (1 + X) - W. 1 (0,5 + X) - 0,5 W. X2 = 7139 X - 1631 (1 + X) - 1800 (0,5 + X) - 0,5. 1800 X2 = 7139 X - 1631 - 1631 X - 900 - 1800 X - 900 X2 = 3708 X - 2531 - 900 X2

dMX  3708  1800X dX dMX 0 dX

1800 X = 3708 X = 2,06 m > 2 m  tidak mungkin

MC = RS2. 1 + W. 1. 0,5 = 1631. 1 + 1800. 0,5 = 1631 + 900 = 2531 kgm MH = RD. 2 - P. 1 = 1392. 2 - 1500. 1 = 2784 - 1500 = 1284 kgm MI = RD. 1 = 1392. 1 = 1392 kgm

BIDANG GESER

BIDANG MOMEN

55) Gambar bidang momen dan gaya lintang. P = 1200 kg, W1 = 1600 kg/m, W2 = 1200 kg/m, W3 = 2200 kg/m

Balok : S1 – S2 RS1 = RS2 = 0,5. 1200. 3,5 = 2100 kg 1 1200. 3,52 8 = 1838 kgm

M maks =

Balok : A – B – S1

MB = 0 RA. 4 + RS1. 1 + W 1. 1. 0,5 - W 1. 4. 2 - P. 2 = 0 4 RA + 2100. 1 + 1600. 0,5 - 1600. 8 - 1200. 2 = 0 4 RA + 2100 + 800 - 12800 - 2400 = 0 4 RA - 12300 = 0 RA = 3075 kg

MA = 0 RB. 4 - RS1. 5 - W 1. 5. 2,5 - P. 2 = 0 4 RB - 2100. 5 - 1600. 12,5 - 1200. 2 = 0 4 RB - 10500 - 20000 - 2400 = 0 4 RB - 32900 = 0 RB = 8225 kg

V = 0 RA + RB = RS1 + Q + P 3075 + 8225 = 2100 + 1600. 5 + 1200 11300 = 11300  ok

RBS1 = RS1 + Q = 2100 + 1600. 1 = 3700 kg

RBA = RB - RBS1 = 8225 - 3700 = 4525 kg

Balok : A – B – S1

MX = RA. X - 0,5 W 1. X2 = 3075 X - 0,5. 1600 X2

dMX  3075  1600X dX dMX 0 dX 1600 X = 3075 X = 1,92 m

M maks = 3075. 1,92 - 800. 1,922 = 5904 - 2949 = 2955 kgm ME = RA. 2 - W 1. 2. 1 = 3075. 2 - 1600. 2 = 6150 - 3200 = 2950 kgm

MB = RS1. 1 + W 1. 1. 0,5 = 2100. 1 + 1600. 0,5 = 2100 + 800 = 2900 kgm

Balok : S2 – C – D MD = 0 RC. 4 - RS2. 5,5 - W 3. 5,5. 2,75 - P. 2 = 0 4 RC - 2100. 5,5 - 2200. 15,125 - 1200. 2 = 0 4 RC - 11550 - 33275 - 2400 = 0 4 RC - 47225 = 0 RC = 11806 kg MC = 0 RD. 4 + RS2. 1,5 + W 3. 1,5. 0,75 - W 3. 4. 2 - P. 2 = 0 4 RD + 2100. 1,5 + 2200. 1,125 - 2200. 8 - 1200. 2 = 0 4 RD + 3150 + 2475 - 17600 - 2400 = 0 4 RD - 14375 = 0 RD = 3594 kg V = 0 RC + RD = RS2 + P + W 3. 5,5 11806 + 3594 = 2100 + 1200 + 2200. 5,5 15400 = 15400  ok

RCS2 = RS2 + W 3. 1,5 = 2100 + 2200. 1,5 = 5400 kg RCD = RC - RCS2 = 11806 - 5400 = 6406 kg

Balok : S2 – C – D MX = RD. X - 0,5 W 3. X2 = 3594 X – 0,5. 2200 X2 dMX  3594  2200X dX dMX 0 dX

2200 X = 3594 X = 1,63 m

M maks = 3594. 1,63 - 1100. 1,632 = 5858 - 2923 = 2935 kgm MC = RS2. 1,5 + W 3. 1,5. 0,75 = 2100. 1,5 + 2200. 1,125 = 3150 + 2475 = 5625 kgm

MF = RD. 2 - W 3. 2. 1 = 3594. 2 - 2200. 2 = 7188 - 4400 = 2788 kgm

BIDANG GESER

BIDANG MOMEN

Nahhh, kalau soalnya seperti ini, gimana ngerjakannya ya…????