tugas analisis struktur ii
DESCRIPTION
ANALISIS STURKTURTRANSCRIPT
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ANALISIS STRUKTUR II
JURUSAN TEKNIK SIPILSEKOLAH TINGGI TEKNOLOGI NASIONAL
TUGAS
ANALISIS STRUKTUR II
Disusun oleh :
ARHAD HARTADI
( 110011010 )
JURUSAN TEKNIK SIPIL SEKOLAH TINGGI TEKNOLOGI NASIONAL
YOGYAKARTA TAHUN 2012
SEKOLAH TINGGI TEKNOLOGI NASIONAL
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LEMBAR PENGESAHAN
TUGAS ANALISIS STRUKTUR II
Tugas Analisis Struktur ini dibuat guna melengkapi tugas dan syarat
untuk Yudisium pada Jurusan Teknik Sipil
Sekolah Tinggi Teknologi Nasional Yogyakarta
Dikerjakan Oleh :
ARHAD HARTADI
( 110011010 )
Yogyakarta, Juni 2012
Disetujui Asisten
( Retnowati Setioningsih, ST, MT ) ( Adi Siswaya)
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LEMBAR PENGESAHAN
TUGAS ANALISIS STRUKTUR II
ARHAD HARTADI (110011010)
a) Asisten
No Keterangan Pembuat Tugas Asisten
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LEMBAR PENGESAHAN
TUGAS ANALISIS STRUKTUR II
ARHAD HARTADI (110011010)
b) Penanggungjawab Tugas
No Keterangan Pembuat Tugas Dosen
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KATA PENGANTAR
Sujud syukur kepada Allah SWT, sehingga penulis dapat menyelesaikan Tugas
Analisis Strutur II guna melengkapi tugas dan syarat untuk mengikuti Yudisium pada Jurusan
Teknik Sipil Sekolah Tinggi Teknologi Nasional Yogyakarta.
Penulis menyadari bahwa pengerjaan Tugas Analisis Struktur II ini masih jauh dari
sempurna, ini atas keterbatasan waktu, pengetahuan, dan pengalaman. Tersusunnya tugas ini
tidak lepas dari dorongan dan bantuan dari beberapa pihak, oleh karena itu penulis ucapkan
terima kasih kepada :
1. Ibu Retnowati Setioningsih, ST, MT , selaku Dosen Analisis Struktur II
2. Adi Siswaya, selaku Asisten Tugas Analisis Struktur II ini
3. Keluargaku yang telah memberikan dorongan moril dan materiil
4. Teman-teman Sipil 11 ( Made, Yoga, Ibnu, Panggih, Saeful, Agus, Ony, Rio, Dance,
Ran, Alvin )
5. Dan semua pihak yang secara langsung maupun tidak langsung telah membantu
terselesaikannya Tugas Analisis Struktur II ini
Semoga segala apa yang diberikan merupakan amal kebaikan yang dapat memberikan
kemanfaatan. Kritik dan saran tentunya dapat mendekatkan Tugas Analisis Struktur II ini
pada kesempurnaan.
Akhir kata, semoga Tugas Analisis Struktur II ini dapat diambil manfaat bagi
pembaca dan semua pihak yang berkepentingan.
Yogyakarta, Juni 2012
Penyusun
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II. PENGERJAAN SOAL 1
Hitunglah Vertical Deflection di joint B :
Penyelesaian :
1. Kontrol Stabilitas geometri
2� = � + �
2. 12� = 21 + 3
24 = 24���������������
2. Reaksi perletakan.
a) �� = 0
��� + ��� − 100 − 200 − 300 − 400 − 100 = 0
��� + ��� = 1100��
b) �� = 0,
karena tidak ada beban horizontal maka ��� = 0
c) Σ� = 0
� �� = 0
���. 30 − 100.5 − 200.10 − 300.15 − 400.20 − 100.25 = 0
30��� − 500 − 2000 − 4500 − 8000 − 2500 = 0
30��� = 17500
��� = 583,33��↑�
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� Σ�� = 0 ���. 30 − 100.5 − 400.10 − 300.15 − 200.20 − 100.25 = 0 30��� − 500 − 4000 − 4500 − 4000 − 2500 = 0 30��� = 15500 ��� = 516,67 �� ↑�
d) Kontrol.
��� + ��� = 1100 516,67 �� + 583,33 �� = 1100�� 1100 �� = 1100 ��'(�
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3. Gaya – gaya luar dihilangkan dan diganti dengan beban 1 Ton (1000 kg) ke Joint
B dan dihitung reaksi – reaksi perletakan.
a) Reaksi perletakkan
� �� = 0
��� + ��� − 1000 = 0 ��� + ��� = 1000 ��
� �� = 0
karena tidak ada beban horizontal maka ��� = 0
� Σ� = 0
• Σ�� = 0 ���. 30 − 1000.30 = 0 30��� = 30000 ��� = 1000 ��↑�
• Σ�� = 0 ���. 30 = 0 ��� = 0 ��
b) Kontrol.
��� + ��� = 1000 0 + 1000�� = 1000�� 1000 �� = 1000 �� '(�
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4. Menghitung panjang batang diagonal ( r )
� Batang 12, 13, 19, 21
5 m
3 m
r
� = )5* + 3* � = √25 + 9 � = √34 � = 5,83 - � = 583 .-
� Batang 15, 16, 17, 18
5 m
2 m
r
� = )5* + 2* � = √25 + 4 � = √29 � = 5,39 - � = 539 .-
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� Batang 14, 20.
5 m
3 m
r 1 m
� = )5* + 1* � = √25 + 1 � = √26 � = 5,1 - � = 510 .-
5. Menghitung E.A
Diketahui :
/0123 04516 7 = 1,25 × 109 ��/.-*
�;<=>?<@A15 = 15 .-*
�B4=A>015 = 20 .-*
�C>1D<@15 = 25 .-*
� /0123 × �;<=>?<@A15 = 1,25 × 109 ��/.-* . 15 .-*
= 18,75 × 109 ��
= 18,75 × 10* EF�
� /0123 × �B4=A015 = 1,25 × 109 ��/.-* . 20 .-*
= 25 × 109 ��
= 25 × 10* EF�
� /0123 × �C>1D<@15 = 1,25 × 109 ��/.-* . 25 .-*
= 31,25 × 109 ��
= 31,25 × 10* EF�
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6. Tabel Defleksi.
Keterangan : Jadi Defleksi pada joint B = 0 cm
No.
Batang S (ton) ∝ (ton) L (cm) E.A (ton)
HI =J.∝. KL. M
(cm)
1 -0,86112 0 500 18,75 × 10* 0
2 -0,86112 0 500 18,75 × 10* 0
3 -2,87503 0 500 18,75 × 10* 0
4 -2,87503 0 500 18,75 × 10* 0
5 -0,97225 0 500 18,75 × 10* 0
6 -0,97225 0 500 18,75 × 10* 0
7 -0,10000 0 300 25 × 10* 0
8 -0,70001 0 400 25 × 10* 0
9 -0,30000 0 200 25 × 10* 0
10 -0,80001 0 400 25 × 10* 0
11 -0,10000 0 300 25 × 10* 0
12 1,00423 0 583 31,25 × 10* 0
13 -0,35634 0 583 31,25 × 10* 0
14 1,18978 0 510 31,25 × 10* 0
15 1,25655 0 539 31,25 × 10* 0
16 1,83995 0 539 31,25 × 10* 0
17 1,43606 0 539 31,25 × 10* 0
18 1,66043 0 539 31,25 × 10* 0
19 -0,42111 0 583 31,25 × 10* 0
20 1,35976 0 510 31,25 × 10* 0
21 1,13382 0 583 31,25 × 10* 0
JUMLAH 0
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IV. PENGERJAAN SOAL 3
Hitunglah reaksi perletakan dan gambarkan SFD dan BMD dari konstruksi kabel
dibawah ini :
Penyelesaian :
1. MENGHITUNG NILAI H (GAYA HORIZONTAL/ RDH & RCH)
� Σ�� = 0 ��� + ��� + �N� + �O� − 106� − 30 − 15 − 30 = 0 ��� + ��� + �N� + �O� = 135
� Σ�� = 0 �O� − �N� = 0 �O� = �N� = �
� Σ� = 0
• Σ�� = 0 ���. 35 + �O�. 35 − �O�. 9 + �N�. 17 − 105 + 10 + 15 + 20 + 25 + 30� − 30.5 − 1517,5� − 30.30 = 0 35��� + 35�O� − 9� + 17� − 1050 − 150 − 262,5 − 900 = 0 35��� + 35�O� + 8� = 2362,5 … 1�
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• Σ�� = 0 ���. 35 + �N�. 35 + �O�. 9 − �N�. 17 − 105 + 10 + 15 + 20 + 25 + 30� − 30.5 − 1517,5� − 30.30 = 0 35��� + 35�N� + 9� − 17� − 1050 − 150 − 262,5 − 900 = 0 35��� + 35�N� − 8� = 2362,5 … 2�
• Σ�/Q1@1@ = 0 ���. 15 + �O�. 15 − �O�. 4 − 10.5 − 10.10 − 30.10 = 0 15��� + 15�O� − 4� − 50 − 100 − 300 = 0 15��� + 15�O� − 4� = 450 … 3�
� Eliminasi persamaan 1 dan 3
35��� + 35�O� + 8� = 2362,515��� + 15�O� − 4� = 450 R15
35R 525��� + �O�� + 120� = 35437,5525��� + �O�� − 140� = 15750
260 � = 19687,5
� = 75,72 �F�
� �O� = �N� = � 75,72 = 75,72 = � S�T� ∶ VWX = YZ, Y[ \ V]X = YZ, Y[ \
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2. MENGHITUNG NILAI Q
� ^_= `9
a9
=15.835
= 3,43 -
� //b = 3,43 + 4 //b = 7,43 -
∑d2 = 6d
2 = 3d
� ∑�/bQ1@1@ = 0
3d. 15 − d. 5 − d. 10 − �. 7,43 = 0 45d − 5d − 10d − 7,4375,72� = 0 30d = 562,5996 e = fg, YZ \
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3. MENGHITUNG NILAI RAV, RBV, RCV & RDV
a) Menghitung nilai RAV & RBV
� Σ�� = 0 ��� + ��� + 6d − 10.6 − 30 − 15 − 30 = 0 ��� + ��� + 618,75� − 60 − 30 − 15 − 30 = 0 ��� + ��� − 22,5 = 0 ��� + ��� = 22,5
� Σ�� = 0 (����� ��T�� �T� h�h�� ℎF��jF���k, -��� VMX = l
� ∑Mx = 0
• ∑MA = 0 ���. 35 + 18,755 + 10 + 15 + 20 + 25 + 30� − 105 + 10 + 15 + +20 + 25 + 30� − 30.15 − 1517,5� − 30.30 = 0 35��� + 18,75105� − 10105� − 150 − 262,5 − 900 = 0 35��� + 1968,75 − 1050 − 150 − 262,5 − 900 = 0 35��� = 393,75 VIp = ff, [Z \
• ∑MB = 0 ���. 35 + 18,755 + 10 + 15 + 20 + 25 + 30� − 105 + 10 + 15 + +20 + 25 + 30� − 30.15 − 1517,5� − 30.30 = 0 35��� + 18,75105� − 10105� − 150 − 262,5 − 900 = 0 35��� + 1968,75 − 1050 − 150 − 262,5 − 900 = 0 35��� = 393,75 VMp = ff, [Z \
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� Kontrol
��� + ��� = 22,5 11,25 + 11,25 = 22,5 22,5 = 22,5 '(�
S�T� ∶ VMp = ff, [Z \ VIp = ff, [Z \
b) Menghitung nilai RCV & RDV
� Subtitusi nilai RBV ke persamaan 1
35��� + 35�O� + 8� = 2362,5 3511,25� + 35�O� + 875,72� = 2362,5 35�O� + 393,75 + 605,76 = 2362,5 35�O� = 2362,5 − 999,51 VWp = rg, st \
� Subtitusi nilai RAV ke persamaan 2
35��� + 35�N� − 8� = 2362,5 3511, ,25� + 35�N� − 875,72� = 2362,5
35�N� + 393,75 − 605,76 = 2362,5 35�N� = 2362,5 − 212,01 V]p = Yr, Zu \
� S�T� ∶ VWp = rg, st \ V]p = Yr, Zu \
c) Kontrol
��� + ��� + �N� + �O� = 135 11,25 + 11,25 + 73,56 + 38,94 = 135 135 = 135 F��
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4. MENGHITUNG MOMEN
� �� = 11,2535� + 18,755 + 10 + 15 + 20 + 25 + 30� − 105 + 10 +
15 + 20 + 25 + 30� − 30.5 − 1517,5� − 30.30
= 393,75 + 1968,75 − 1050 − 150 − 262,5 − 900
= 0E
� �� = 11,2535� + 18,755 + 10 + 15 + 20 + 25 + 30� − 105 + 10 +
15 + 20 + 25 + 30� − 30.5 − 1517,5� − 30.30
= 393,75 + 1968,75 − 1050 − 150 − 262,5 − 900
= 0E
� ��0>=> = 11,25 − 152,5�
= 56,25 − 37,5
= 18,75E
� �v0>=> = 11,25.10 + 18,75�. 5 − 10.5 − 30.5
= 112,5 + 93,75 − 50 − 150
= 6,25E
� ��0>=> = 11,25.15 + 18,75�5 + 10� − 1055 + 10� − 30.10
= 168,75 + 281,25 − 150 − 300
= 0E
� �w01@ = 11,25.15 + 18,75�5 + 10� − 1055 + 10� − 30.10
= 168,75 + 281,25 − 150 − 300
= 0E
� ��01@ = 11,25.10 + 18,75�. 5 − 10.5 − 30.5
= 112,5 + 93,75 − 50 − 150
= 6,25E
� �(01@ = 11,25 − 152,5�
= 56,25 − 37,5
= 18,75E
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5. MENGHITUNG MX ( MOMEN MAKSIMAL )
� � `
• � ` = 11,25 − 3 . `*
= 11,25 − a* *
• O → 11,25 − 3 = 0
=−11,25−3
= 3,75-
• � ` = 11,25 � − a* *
= 11,253,75� −323,75*�
= 21,09E�
� � *
• � * = 11,255 + � + 18,75 − 10 − 152,5 + � − 3 . `*
= 56,25 + 11,25 + 18,75 − 10 − 37,5 − 15 −32 *
= 56,25 − 37,5 + 11,25 + 18,75 − 10 − 15 −32 *
= 18,75 + 5 −32 *
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• O → 5 − 3 = 0
=−5−3
= 1,67-
• � * = 18,75 + 5 − a* *
= 18,75 + 51,67� −321,67*�
= 22,92E�
� � a
• � a = 11,2515 + � + 18,75 + 5 + + 10 + � − 10 + 5 +
+10 + � − 3010 + � − 3 .12
= 168,75 + 11,25 + 18,753 + 15� − 103 − 15� − 300
−30 −32 *
= 168,75 + 11,25 + 56,25 + 281,25 − 30 − 150 − 300
−30 − a* *
= 168,75 + 281,25 − 150 − 300 + 11,25 + 56,25 − 30
−30 − a* *
= 7,5 −32 *
• O → 7,5 − 3 = 0
=−7,5−3
= 2,5-
• � a = 7,5 − a* *
= 7,52,5� −322,5*�
= 9,38E�
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� � y
• � y = 11,255 + � + 18,75 − 10 − 152,5 + � − 3 . `*
= 56,25 + 11,25 + 18,75 − 10 − 37,5 − 15 −32 *
= 56,25 − 37,5 + 11,25 + 18,75 − 10 − 15 −32 *
= 18,75 + 5 −32 *
• O → 5 − 3 = 0
=−5−3
= 1,67-
• � y = 18,75 + 5 − a* *
= 18,75 + 51,67� −321,67*�
= 22,92E�
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� � 9
• � 9 = 11,25 − 3 . `*
= 11,25 − a* *
• O → 11,25 − 3 = 0
=−11,25−3
= 3,75-
• � 9 = 11,25 − a* *
= 11,253,75� − a*3,75*�
= 21,09E�
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III. PENGERJAAN SOAL 2
Hitunglah reaksi dan gambarkan SFD, NFD & BMD dari kontruksi pelengkung sendi
berikut ini :
Penyelesaian:
1. MENGHITUNG PANJANG (L’) UNTUK MENENTUKAN JARAK S KE TITIK A
DAN B DENGAN RUMUS PERSAMAAN PARABOLA.
T�� ∶ = 40-
z = 6-
ℎ = 10-
� Persamaan parabola
z =4ℎ {b − �
{b*
6 =4.10.40{b − 40�
{′*
6kb = 1600{b − 40�
6kb = 1600{b − 6400
6kb − 1600{b + 6400 = 0
��ℎ����� ∶
� = 6
h = −1600
. = 6400
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� {b`}* =}~±√~�}y1�
*1
= −−1600� ± )−1600�* − 4.6.640002.6
= 1600 ± √102400012
• {b`}* = `����√`�*y���`*
= 217,66 - ��T�� -�-���ℎ� �z�����
• {b`}* = `���}√`�*y���`*
= 49 - -�-���ℎ� �z�����
� ��ℎ����� S���� � → � ∶ 1 2� . 49 = 24,5 - � → � ∶ 40 − 24,5 = 15,5 -
2. MENGHITUNG REAKSI PERLETAKKAN.
� Σ�� = 0 ��� + ��� − 2.40� = 0 ��� + ��� − 80 = 0 ��� + ��� = 80
� Σ�� = 0 ��� − ��� + 3.10� = 0 ��� − ��� + 30 = 0 ��� − ��� = −30 −��� + ��� = 30 ��� − ��� = 30
� �� = 0
• Σ�� = 0 ���. 40 + ���. 6 − 2.40�20 − 3.10�5 = 0 40��� + 6��� − 1600 − 150 = 0 40��� + 6��� − 1750 = 0 40��� + 6��� = 1750 … 1�
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• Σ�� = 0 ���. 40 − ���. 6 − 2.40�20 − 3.6�3 + 3.4�2 = 0 40��� − 6��� − 1600 − 54 + 24 = 0 40��� − 6��� − 1630 = 0 40��� − 6��� = 1630 … 2�
• Σ��Q1@1@ = 0 ���. 15,5 − ���. 4 − 2 . 15,5� 1 2� . 15,5 = 0 15,5��� − 4��� − 240,25 = 0 15,5��� − 6��� = 240,25 … 3�
• Σ��Q>=> = 0 ���. 24,5 − ���. 10 − 3.10�5 − 2 . 24,5� 1 2� . 24,5 = 0 24,5��� − 10��� − 150 − 600,25 = 0 24,5��� − 10��� = 750,25 … 4�
� Elimininasi ( menghitung RAH & RBH)
• Eliminasi persamaan 1 & 3
40��� + 6��� = 175015,5��� − 4��� = 240,05 R× 15,5
× 40 R 620��� + 93��� = 27125620��� − 160��� = 9610
253��� = 17515 VIX = us, [r \
• Eliminasi persamaan 2 & 4
40��� − 6��� = 1630 24,5��� − 10��� = 750,25 R× 24,5
× 40R 980��� − 147��� = 39935980��� − 400��� = 30010
253��� = 9925 VMX = rs, [r \
• Kontrol
��� − ��� = 30 69,23 − 39,23 = 30 30 = 30 F��
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� Subtitusi (menghitung nilai RAV & RBV)
• Subtitusi nilai RBH ke persamaan 1
40��� + 6��� = 1750 40��� + 669,23� = 1750 40��� + 415,38 = 1750 40��� = 1750 − 415,38 40��� = 1334,62 VIp = rr, rY \
• Subtitusi nilai RAH ke persamaan 2
40��� − 6��� = 1630 40��� − 639,23� = 1630 40��� − 235,38 = 1630 40��� = 1630 + 235,38 40��� = 1865,38 VMp = tu, ur \
• Kontrol
��� + ��� = 80 46,63 + 33,37 = 80 80 = 80 F��
3. MENGHITUNG GAYA LINTANG, NORMAL & MOMEN
Persamaan parabola
z = 4ℎ {b − �{b*
z = 4.10. 49 − �49*
z = 40 49 − �2401
z = 1960 − 40 *
2401
� = l, gfu� − l, lfuY�[
![Page 30: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/30.jpg)
��� = zb ��� = 0,816 − 0,0334
� = ��� ��l, gfu − l, lrrt��
� Mencari NFD, SFD dan Momen pada titik C yang berjarak 4 m dr titik A
• z� = 0,816.4 − 0,01674�*
z� = 3 -
• � = ��. ��0,816 − 0,0334 . 4� � = 34,31°
• Σ�� = 0 ��� − 2.4� − � = 0 46,63 − 8 − � = 0 � = 38,63 E ↓�
• Σ�� = 0 ��� + 3.3� − � = 0 39,23 + 9 − � = 0 � = 48,23 E ←�
• ��N = �.F�� − ����� = 38,63 cos34,31� − 48,23 sin34,31� = 4,7 T
• ��N = ����� + �.F�� = 38,63 sin34,31� + 48,23 cos34,31� = 61,61 T
![Page 31: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/31.jpg)
• �N = ���. 4 − ���. 3 − 2.4�2 − 3.3�1,5
= 46,63�4 − 39,23�3 − 8.2 − 91,5�
= 39,33Tm
� Mencari NFD, SFD dan Momen pada titik D yang berjarak 8 m dr titik A
• zC = 0,816. 8 − 0,01678�*
zC = 5,5-
• � = ��.��0,816 − 0,0334. 8�
� = 28,76°
• Σ�� = 0
��� − 2.8� − � = 0
46,63 − 16 − � = 0
� = 30,63E↓�
• Σ�� = 0
��� + 3. 5,5� − � = 0
39,23 + 16,5 − � = 0
� = 55,73E←�
• ��O = �.F�� − �����
= 30,63 cos28,76� − 55,73 sin28,76�
= 0,038T
• ��O = ����� + �.F��
= 30,63 sin28,76� + 55,73 cos28,76�
= 63,6T
• �O = ���. 8 − ���. 5,5 − 2.8�4 − 3. 5,5� 1 2� . 5,5
= 46,63�8 − 39,23�5,5 − 16. 4 − 3. 5,5� 1 2� . 5,5
= 47,9E-
![Page 32: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/32.jpg)
� Mencari NFD, SFD dan Momen pada titik E yang berjarak 12 m dr titik A
• z� = 0,816.12 − 0,016712�*
z� = 7,4-
• � = ��.��0,816 − 0,0334. 12�
� = 22,5°
• Σ�� = 0
��� − 2.12� − � = 0
46,63 − 24 − � = 0
� = 22,63E↓�
• Σ�� = 0
��� + 3. 7,4� − � = 0
39,23 + 3. 7,4� − � = 0
� = 61,43E←�
• ��/ = �.F�� − �����
= 22,63 cos22,5� − 61,43 sin22,5�
= −2,6E
• ��/ = ����� + �.F��
= 22,63 sin22,5� + 61,43 cos22,5�
= 65,41E
• �/ = ���. 12 − ���. 7,4 − 2.12�6 − 3. 7,4� 1 2� . 7,4
= 46,63�12 − 39,23�7,4 − 24�6 − 3. 7,4� 1 2� . 7,4
= 43,12E-
![Page 33: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/33.jpg)
� Mencari NFD, SFD dan Momen pada titik F yang berjarak 16 m dr titik A
• z� = 0,816. 16 − 0,016716�*
z� = 8,8-
• � = ��.��0,816 − 0,0334. 16�
� = 15,73°
• Σ�� = 0
��� − 2.16� − � = 0
46,63 − 32 − � = 0
� = 14,63E↓�
• Σ�� = 0
��� + 3. 8,8� − � = 0
39,23 + 3. 8,8� − � = 0
� = 65,63E←�
• ��� = �.F�� − �����
= 14,63 cos15,73� − 65,63 sin15,73�
= −3,7E
• ��� = ����� + �.F��
= 14,63 sin15,73� + 65,63 cos15,73�
= 67,13E
• �� = ���. 16 − ���. 8,8 − 2.16�8 − 3. 8,8� 1 2� . 8,8
= 46,63�16 − 39,23�8,8 − 32�8 − 3. 8,8� 1 2� . 8,8
= 28,7E-
![Page 34: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/34.jpg)
� Mencari NFD, SFD dan Momen pada titik G yang berjarak 20 m dr titik A
• z� = 0,816. 20 − 0,016720�*
z� = 9,64-
• � = ��.��0,816 − 0,0334. 20�
� = 8,42°
• Σ�� = 0
��� − 2.20� − � = 0
46,63 − 40 − � = 0
� = 6,63E↓�
• Σ�� = 0
��� + 3. 9,64� − � = 0
39,23 + 3. 9,64� − � = 0
� = 68,15E←�
• ��v = �.F�� − �����
= 6,63 cos8,42� − 68,15 sin8,42�
= −3,42E
• ��v = ����� + �.F��
= 6,63 sin8,42� + 68,15 cos8,42�
= 68,38E
• �v = ���. 20 − ���. 9,64 − 2.20�10 − 3. 9,64� 1 2� . 9,64
= 46,63�20 − 39,23�9,64 − 40�10 − 3. 9,64� 1 2� . 9,64
= 15,03 E-
![Page 35: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/35.jpg)
� Mencari NFD, SFD dan Momen pada titik H yang berjarak 24 m dr titik A
• z; = 0,816. 24 − 0,016724�*
z; = 9,96-
• � = ��.��0,816 − 0,0334. 24�
� = 0,825°
• Σ�� = 0
��� − 2.24� − � = 0
46,63 − 48 − � = 0
� = −1,37E↓�
• Σ�� = 0
��� + 3. 9,96� − � = 0
39,23 + 3. 9,96� − � = 0
� = 69,11E←�
• ��� = �.F�� − �����
= −1,37 cos0,825� − 69,11 sin0,825�
= −2,36E
• ��� = ����� + �.F��
= −1,37 sin0,825� + 69,11 cos0,825�
= 69,08E
• �� = ���. 24 − ���. 9,96 − 2.24�12 − 3. 9,96� 1 2� . 9,96
= 46,63�24 − 39,23�9,96 − 48�12 − 3. 9,96� 1 2� . 9,96
= 3,59 E-
![Page 36: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/36.jpg)
� Mencari NFD, SFD dan Momen pada titik S kiri yang berjarak 24,5 m dari titik A
• z�0>=>� = 0,816. 24,5 − 0,016724,5�*
z�0>=>� = 10-
• � = ��.��0,816 − 0,0334. 24,5�
� = 0°
• Σ�� = 0
��� − 2.24,5� − � = 0
46,63 − 49 − � = 0
� = −2,37E↓�
• Σ�� = 0
��� + 3. 10� − � = 0
39,23 + 3. 10� − � = 0
� = 69,23E←�
• ���0>=> = �.F�� − �����
= −2,37 cos0� − 69,23 sin0�
= −2,37E
• ���0>=> = ����� + �.F��
= −2,37 sin0� + 69,23 cos0�
= 69,23E
• ��0>=> = ���. 24,5 − ���. 10 − 2. 24,5� 1 2� . 24,5 − 3. 10�5
= 46,63�24,5 − 39,23�10 − 2. 24,5� 1 2� . 24,5 − 30. 5
= −0,115 E-
![Page 37: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/37.jpg)
� Mencari NFD, SFD dan Momen pada titik S kanan yang berjarak 15,5 m dari titik
B
• {b − { = 49 − 40 = 9 -
• z� 01@� = 0,816. 24,5 − 0,016715,5 + 9�*
z� 01@� = 10 -
• � = ��. ��0,816 − 0,0334 . 15,5 + 9� � = 0°
• Σ�� = 0 ��� − 2. 15,5� − � = 0 33,37 − 2. 15,5� − � = 0 � = 2,37 E ↓�
• Σ�� = 0 � − ��� = 0 � = 69,23 E →�
• ���01@ = −�.F�� + ����� = −2,37 cos0� + 69,23 sin0� = −2,37 E
• ���01@ = −����� − �.F�� = −2,37 sin0� − 69,23 cos0� = −69,23 E
• ��01@ = ���. 15,5 − ���. 4 − 2. 15,5� 1 2� . 15,5
= 33,3715,5� − 69,23�4 − 2.15,5� 1 2� . 15,5 = 0,065 E-
![Page 38: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/38.jpg)
� Mencari NFD, SFD dan Momen pada titik I yang berjarak 28 m dr titik A
• {b − { = 49 − 40 = 9 -
• z7 + 6 = 0,816. 12 + 9� − 0,016712 + 9�*
z7 + 6 = 9,77 z7 = 3,77 -
• � = ��. ��0,816 − 0,0334 . 21� � = 6,54°
• Σ�� = 0 ��� − 2. 12� − � = 0 33,37 − 2. 12� − � = 0 � = 9,37 E ↓�
• Σ�� = 0 � − ��� = 0 � = 69,23 E →�
• ��w = −�.F�� + ����� = −9,37 cos6,54� + 69,23 sin6,54� = −1,42 E
• ��w = −����� − �.F�� = −9,37 sin6,54� − 69,23 cos6,54� = −69,85 E
• �w = ���. 12 − ���. 3,77 − 2. 12�6 = 33,37�12 − 69,23�3,77 − 24. 6 = −4,56 E-
![Page 39: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/39.jpg)
� Mencari NFD, SFD dan Momen pada titik J yang berjarak 32 m dr titik A
• {b − { = 49 − 40 = 9 -
• z7 + 6 = 0,8168 + 9� − 0,01678 + 9�*
z7 + 6 = 9,05 z7 = 3,05 -
• � = ��. ��0,816 − 0,0334 . 17� � = 13,94°
• Σ�� = 0 ��� − 2. 8� − � = 0 33,37 − 16 − � = 0 � = 17,37 E ↓�
• Σ�� = 0 � − ��� = 0 � = 69,23 E →�
• ��� = −�.F�� + ����� = −17,37 cos13,94� + 69,23 sin13,94� = −0,18 E
• ��� = −����� − �.F�� = −17,37 sin13,94� − 69,23 cos13,94� = −71,38 E
• �� = ���. 8 − ���. 3,05 − 2. 8�4 = 33,37�8 − 69,23�3,05 − 16. 4 = −8,19 E-
![Page 40: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/40.jpg)
� Mencari NFD, SFD dan Momen pada titik K yang berjarak 36 m dr titik A
• {b − { = 49 − 40 = 9 -
• z7 + 6 = 0,8164 + 9� − 0,01674 + 9�*
z7 + 6 = 7,8 z7 = 1,8 -
• � = ��. ��0,816 − 0,0334 . 13� � = 20,9°
• Σ�� = 0 ��� − 2.4� − � = 0 33,37 − 8 − � = 0 � = 25,37 E ↓�
• Σ�� = 0 � − ��� = 0 � = 69,23 E →�
• ��( = −�.F�� + ����� = −25,37 cos20,9� + 69,23 sin20,9� = 1 E
• ��( = −����� − �.F�� = −25,37 sin20,9� − 69,23 cos20,9� = −73,72 E
• �( = ���. 4 − ���. 1,8 − 2. 4�2 = 33,37�4 − 69,23�1,8 − 8. 2 = −7,13 E-
![Page 41: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/41.jpg)
� Mencari NFD, SFD dan Momen di tumpuan A
• � = ��.��0,816 − 0,0334. 0�
� = 39,21°
• Σ�� = 0
��� − � = 0
� = 46,63E↓�
• Σ�� = 0
��� − � = 0
� = 39,23E←�
• ��� = �.F�� − �����
= 46,63 cos39,21� − 39,23 sin39,21�
= 11,33E
• ��� = ����� + �.F��
= 46,63 sin39,21� + 39,23 cos39,21�
= 59,87E
• �� = 0E-
![Page 42: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/42.jpg)
� Mencari NFD, SFD dan Momen di tumpuan B
• � = ��.��0,816 − 0,0334. 9�
� = 27,27°
• Σ�� = 0
��� − � = 0
� = 33,37E↓�
• Σ�� = 0
� − ��� = 0
� = 69,23E→�
• ��� = −�.F�� + �����
= −33,37 cos27,27� + 69,23 sin27,27�
= 2,06E
• ��� = −����� − �.F��
= −33,37 sin27,27� − 69,23 cos27,27�
= −76,83E
• �� = 0E-
4. MENGHITUNG MOMEN MAKSIMAL (MX)
� ^by}^
= `�,`_
0,18 = 4 −
1,18 = 4
′ = 3,4-
![Page 43: Tugas Analisis Struktur II](https://reader034.vdocuments.net/reader034/viewer/2022050805/55cf9724550346d0338feaec/html5/thumbnails/43.jpg)
� − 9 = b + 4 − 9 = 3,4 + 4 − 9 = 7,4 = 16,4 -
� z7 + 6 = 0,8167,4 + 9� − 0,01677,4 + 9�* z7 + 6 = 8,9 z7 = 8,9
� � = ���7,4� − ���2,9� − 2. 7,4� 1 2� . 7,4
= 33,377,4� − 69,232,9� − 2. 7,4� 1 2� . 7,4 = −8,6 E-