Soal Soal Cross 1 ( Tidak Bergoyang

Oleh Raynerius Kapu

22 tayangan
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Transkrip Soal Soal Cross 1 ( Tidak Bergoyang

Soal 1

Gambar bidang M D N
Jawab :
Cara 1 , titik D di tinjau seperti keadaan sebenarnya ( jepit – sendi )
+ Menentukan Koefisien distribusi
MBA : MBC =
MBA =

=

:

= = 0,6

MCB : MCD =
MCB =

:

:

= = 0,4

=

=1: =3:2
MBC =

= = 0,4

:

:1=2:3

MCD =

=

= = 0,6

+ menentukan Momen Primer

MoBC = P.a.b2 = 2.4.22 = 0,889
L2
62
MoCB = ‐ P.b.a2 = ‐ 7,2.4.22 = ‐ 1,778
L2
62

MoCD = qL2 = ‐ 3.6 2 = 13.5

+ mendistribusi semua momen tidak seimbang

A

Joint ( simpul )

B

C

MAB

MBA

MBC

MCB

MCD

Koefisien Distribusi

0

‐0.6

‐0.4

‐0.4

‐0.6

Momen Primer

0

0

0.889

‐2.3444
‐1.4554
0.8732
0.5822

0.4366

0.0175

0.0349

0.0007



0.0014

0.2911

‐0.0582

‐0.1164

0.0233

0.0116

‐0.0023

‐0.0047

0.0009

0.0005

‐0.0001

‐0.0002

‐0.0003

‐6.2849

6.2849

0.0000

0.0001

0.0000

0.4548

0.9097

‐0.9097

Syarat Keseimbangan
Momen

‐1.778
13.5
11.7220
‐4.6888
‐7.0332

∑MB = 0.9097+ ‐0.9097 = 0 ( OK )

+ Menentukan koefisien distribusi
MBA =

=

:

= = 0,6

MCB : MCD =
MCB =

:

:

= = 0,3333

=1: =3:2
MBC =

=

:
MCB =

= = 0,4

= 1 :2
= = 0,6666

+ menentukan Momen Primer

MoBC = P.a.b2 = 2.4.22 = ‐ 6,4
L2
62
MoCB = ‐ P.b.a2 = ‐ 7,2.4.22 = ‐ 3,2
L2
62

MoCD =
MoDC = ‐

qL2 =
qL2 = ‐

‐0.0070

∑MC = ‐6.2849+ 6.2849 = 0 ( OK )

Cara 2 , titik D di tinjau sebagai Jepit tidak sempurna

MBA : MBC =

‐0.1746

3.62 = 9
3.62 = ‐ 9

+ mendistribusi semua momen tidak seimbang

Perhitungan di mulai dari titik C dan di induksi ke titik D dan B , dst

A

C

MBA

MBC

0
0.0000

‐0.6
0.0000

‐0.4
0.8890

‐0.3333
‐0.6666
‐1.7780
9.0000
7.2220

‐1
‐9.0000

‐2.4071

0.1887

‐1.2035
0.1258

0.5766

‐0.9610
0.3844

0.1153

‐0.1922
0.0769

0.0231

‐0.0384
0.0154

0.0046

‐0.0077
0.0031

0.0009

‐0.0015
0.0006

0.0002

‐0.0003
0.0001

0.0000

0.0000

‐0.0001
0.0000

0.0629
5.7035
5.7665
‐1.9220
‐3.8439
0.1922
0.9610
1.1532
‐0.3844
‐0.7687
0.0384
0.1922
0.2306
‐0.0769
‐0.1537
0.0077
0.0384
0.0461
‐0.0154
‐0.0307
0.0015
0.0077
0.0092
‐0.0031
‐0.0061
0.0003
0.0015
0.0018
‐0.0006
‐0.0012
0.0001
0.0003
0.0004
‐0.0001
‐0.0002
0.0000
0.0001

‐2.4071
11.4071

0.4547

0.9037

‐0.9037

‐6.2689

0.2883

0.0577

0.0115

0.0023

0.0005

0.0001

MCB

D

MAB

0.0944



B

MCD

‐4.8142

6.2858

* warna merah adalah penjumlahan momen yang bertemu pada suatu titik ( Δ )
*Mx = Δ x koefisien distribusi

MCD

‐1.9220
1.9220
‐0.3844
0.3844
‐0.0769
0.0769
‐0.0154
0.0154
‐0.0031
0.0031
‐0.0006
0.0006
‐0.0001
0.0001
0.0000

Soal 2

Gambar bidang M D N
Jawab :
+ Menentukan Koefisien distribusi
MBA : MBC : MBD : =
MBA =

=

:

:
MBC =

=

:

:

= 0.5 : 0,75 : 2 = 2 : 3 : 8

=

MBD =

=

+ menentukan Momen Primer

MoBA = ‐ qL2 = ‐ 3.6 2 = ‐ 13.5

MoBC =

PL =

2.4 = 1.5

+ mendistribusi semua momen tidak seimbang

MBD

D
MDB

‐0.2308
1.5




1.8462

‐0.6154
0
‐12
7.3846

2.7692

1.3846

‐11.6539

7.3847

4.2693

1.3847

Joint ( simpul )

MBA
Koefisien Distribusi
Momen Primer


Syarat Keseimbangan
Momen

‐0.1538
‐13.5

B
MBC

∑MB = ‐11.6539 + 7.3847 + 4.2693 = 0 ( OK )

Soal 3

Gambar bidang M D N
Jawab :
+ Menentukan Koefisien distribusi
MAB : MAC =
MAB =

=

:
=

=

:

=

MAC =

: =8:4
=

=

+ menentukan Momen Primer

MoAB =
MoBA = ‐

qL2 =
qL2 = ‐

6,48.52 = 13,5
6,48.52 = ‐ 13,5

MoKantilever = ‐ P.L = ‐ 3,6.1,5 = ‐ 5,4

+ mendistribusi semua momen tidak seimbang
Joint ( simpul )

Koefisien Distribusi

C
MCA

kantilever

0



Syarat Keseimbangan
Momen

MAC

B
MAB

‐0.33333 ‐0.66667

MBA
0

‐5.4

0

13.5

‐13.5

‐1.35



8.1
‐2.7

‐5.4

‐2.7

‐1.35

‐5.4

‐2.7

8.1

‐16.2

Momen Primer



A

∑MA = 8.1 + ‐5.4 + ‐2.7 = 0 ( OK )

Soal 4

Gambar bidang M D N
Jawab :
+ Menentukan Koefisien distribusi
MAB : MAD =
MAB =

:

=

=

,

=2:1

MAD =

MBA : MBC : MBE : =
MBA =

:

,

= 0,4

:

:
MBC =

=

=

:

:

,
,

,

= 0,3

= 2 : 1,5 : 1,5
MBE =

,
,

,

= 0,3

+ menentukan Momen Primer

MoAB =
MoBA = ‐

MoBC =

qL2 =

3.42 = 4

qL2 = ‐

PL =

3.42 = ‐ 4

2.4 = 1.5

MoAD = ‐ PL = ‐ 2.4 = ‐ 1
MoDA = PL = 2.4 = 1

+ mendistribusi semua momen tidak seimbang

Soal 5

Gambar bidang M D N
Jawab :
+ Menentukan Koefisien distribusi
MBA : MBC : MBE : =
MBA =

=

MCB : MCD : MCF : =
MCB=

:

=

:

=

MBC =

:

:
MCD=

:

:

= 0,75 : 2 : 0,75 = 3 : 8 : 3

=

=

:

MBE =

:

=

=

= 2 : 1,5 : 1 = 4 : 3 : 2
MCF=

=

+ menentukan Momen Primer

MoBA= ‐

PL ‐

MoBC =

qL2 =

MoCB = ‐

MoCD=

qL2 = ‐

2.42 = 1,833

qL2 = ‐

PL =

3.4 ‐

2.42 = ‐ 0,833

4.4 = 3

2.42 = ‐ 4,5

MoKantilever = q.L2 = 2.12 = 1

+ mendistribusi semua momen tidak seimbang

+ Menghitung reaksi masing masing bentang

∑MB = 0
RA.4 ‐q.2. – P.2 +3,805 = 0
RA.4 ‐2.2. – 3.2 +3,805 = 0
RA =

,

= 1,548 Ton

∑MB = 0
‐ RB1.4 + q.2.

2 + P.2 +3,805 = 0

‐ RB1.4 + 2.2.

2 + 3.2 +3,805 = 0

,

RB1 =

= 5,452 Ton

∑MC = 0
RB2.4 ‐ q.2.

2 ‐ 3,110 + 1,059 = 0

RB2.4 ‐ 2.2.

2 ‐ 3,110 + 1,059 = 0

,

RB2 =

= 3,5128 Ton

∑MB = 0
‐ RC1.4 + q.2.

‐ 3,110 + 1,059 = 0

‐ RC1.4 + 2.2.

‐ 3,110 + 1,059 = 0

RB1 =

,

= 0,4872 Ton

∑MD = 0
RC2.4 – P.2 ‐ 1,635 + 1 = 0
RC2.4 – 4.2 ‐ 1,635 + 1 = 0
,

RC2 =

= 2,158 Ton

∑MC = 0
‐ RD.4 + P.2 ‐ 1,635 + 1 = 0
‐ RD.4 + 4.2 ‐ 1,635 + 1 = 0
RD =

,

= 1,842 Ton

RD2 = q.L = 2.1 = 2 Ton

∑ME = 0
HB.4 ‐ 0,695 = 0
HB =

,

= 0,628 Ton

∑MB = 0
‐HE.4 ‐ 0,695 = 0
,

HE = ‐

= ‐ 0,628 Ton (

)

∑MF = 0
HC.4 + 0,288 + 0,576 = 0
,

HC = ‐

= ‐ 0,216Ton (

∑MC = 0
‐ HF.4 + 0,288 + 0,576 = 0
HF =

,

= 0,216 Ton

+ menggambar Diagram Momen , Lintang dan Normal

)

Diagram Momen
X = 2 = RA.2 = 1,549.2 = 3,098 TM
X = 4 = RA.4 – P.2 – q.2.
= 1,549.4 – 3.2 – 2.2.

Diagram Lintang
Dx = 0 = RA = 1,549 Ton
Dx = 2 = RA ‐ P = 1,549 – 3 = 1,451 Ton
Dx = 4 = RA ‐ P = 1,549 – 3 – 2.2 + RB = 0 Ton

= 3,805

Diagram Momen
M MAX terjadi di XMAX
0=

[ M + RB2X – q.X2 ]

0 = [RB2 – q.X ]
Xmax =

,

= 1,757 M

M+MAX = M + RB2.1,757 – q.1,7572
= ‐3,11+ 3,513.1,757 – 2.1,7572
= 0,25Ton
Momen di X= 2
= M + RB2.2– q.22
= ‐3,11 + 3,513.2 ‐ 2.22
= 0,085 Ton Meter

Diagram Lintang
Dx = 0 = RB2 = 3,513Ton
Dx = 2 = RB2 – q.2 = 3,513 – 2.2 = 0,487 Ton
Dx = 2 = RB2 – q.2 + RC1 = 3,513 – 2.2 + 0,487 = 0 Ton

Diagram Momen
X =2
X =4

= M + RC2.2 = ‐1,635 + 2,159.2 = 2,683 TM
= M + RC2.4 – P.2
= ‐1,635 + 2,159.4 – 4.2
= 2,683 TM

Diagram Momen
X =0

= q.1. = 2.1. = 1

Diagram Momen Superposisi

Diagram Lintang
Dx = 0 = RC2 = 2,159 Ton
Dx = 2 = RC2 ‐ P = 2,159 – 4 = 1,841 Ton
Dx = 2 = RC2 ‐ P + RD1 = 2,159 – 4 + 1,841 = 0 Ton

Diagram Lintang
Dx = 0 = RD2 = 2 Ton
Dx = 1 = RC2 – q.1 = 2 – 2.1 = 0 Ton

Judul: Soal Soal Cross 1 ( Tidak Bergoyang

Oleh: Raynerius Kapu


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