Reazzjonijiet għat-tagħbijiet fuq gerder sempliċi

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Contoh Soal.

Hitung Reaksi tumpuan, gaya lintang (Bid.D), bidang momen (Bid M) dan gaya Normal ( Bid N) serta gambarnya, gelagar sederhana, dengan muatan di bawah ini :

PV

45

^o

PH

Q1 Q2 Q3

Penyelesaian:

P PV = P sin∝ = 2√2 x 1

2√2 = 2 ton PH = P cos∝ = 2√2 x 1

2√2 = 2 ton.

Q1 = 1/2ql = ½. 1. 3 = 1,5 ton Q2 = q.l = 2x 2 = 4 ton Q3 = qxl = 2 x 4 = 8 ton Menghitung Reaksi Tumpuan.

R_AV

∑

^{MB = 0}

R_AV . 10 – Q1.8- PV.4 – Q2 . 1 + Q3 . 2 = 0 10 R_AV – 1,5 x 8 – 2 x 4 – 4 x 1 + 8 x 2 = 0 10 R_AV - 8 = 0

R_AV = 8/10 = 0,8 ton ( ) R_BV

∑

^MA^{= 0}

-R_BV . 10 + Q1 x 2 + PV x 6 + Q2 x 9 + Q3 x 12 = 0 - 10 ^RBV + 1,5 x 2 + 2 x 6 + 4 x 9 + 8 x 12 = 0 - 10 R_BV + 147 = 0

P = 2√2T

C D

A

q1 = 1 T/m’

q2 = 2 T/m’

E B F

4 m 2 m

2 m 3 m 3m

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R_BV = 147/10 = 14,7 ton ( ) Kontrol

∑

^KV^{= 0}

R_AV + _R_BV – Q1 – PV – Q2 – Q3 = 0 0,8 + 14,7 – 1,5 – 2 – 4 – 8 = 0

Menghitung R_AH

∑

^KH^{= 0}

R_AH - PH = 0

R_AH - 2 = 0 R_AH = 2 ton ( ) Menghitung Gaya Lintang (Bid D)

DA = R_AV = 0,8 ton.

DC = DA – Q1 = 0,8 – 1,5 = - 0,7 ton DD = DC – PV = - 0,7 – 2 = - 2,7 ton DE = DD = - 2,7 ton

DB kr = DE – Q2 = - 2, 7 – 4 = - 6,7 ton DB kn = DB kr + ^RBV = - 6,7 + 14,7 = 8 ton DF = DB kn – Q3 = 8 – 8 = 0 ton.

Menghitung Bidang Momen ( Bid. M ).

MA = R_AV. x 0 = 0 tm

MC = R_AV x 3 – Q1 x 1 = 0,8 x3 – 1,5 x 1 = 0, 9 tm MD = R_AV x 6 – Q1 x 4 = 0,8 x 6 – 1,5 x 4 = - 1,2 tm.

ME = ^RAV x 8 – Q1 x 6 – PV x 2 = 0,8 x 8 – 1,5 x 6 – 2 x 2 = - 8,2 tm

MB = R_AV x 10 – Q1 x 8 – PV x 4 – Q2 x 1 = 0,8 x 10 – 1,5 x 8 – 2 x 4 – 4 x 1 = - 16 tm.

MF = ^RAV x 14 – Q1 x 12 – PV x 8 – Q2 x 5 + ^RBV x 4 – Q3 x 2 = 0,8 x 14 – 1,5 x 12 – 2 x 8

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– 4 x 5 + 14.7 x 4 – 8 x 2 = 0 tm.

Menghitung Gaya Normal ( Bid N) NA = R_AH = 2 ton

NC = NA = 2 ton

ND = NC – PH = 2 – 2 = 0 ton