S1 Ilmu Dasar Sains Kalkulus - UPJIntegrity, Professionalism, & Entrepreneurship • Kemampuan Akhir...
Transcript of S1 Ilmu Dasar Sains Kalkulus - UPJIntegrity, Professionalism, & Entrepreneurship • Kemampuan Akhir...
MK MINOR STRUKTUR & SB
DAYA AIR
Ilmu Dasar Sains Kalkulus
Statika & Mek Bhn
Analisis Struktur
Peranc.Str. Beton Peranc.Str. Baja
Din.Str.&Rek.Gempa
S1
S2
S3
S4 S5
S7
S6
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Peranc.Str. Bangunan Sipil
Analisis Struktur (4SKS)
Menghitung defleksi/perpind
ahan titik
Analisis Struktur Statis Tak Tentu
Beam & Frame
Castigliano Method
Virtual Load Method
Double Integration
Conjugate Beam
Moment Area Method
Truss/Rangka Batang
Castigliano Method
Virtual Load Method
Beam & Frame
Force Method
Slope-Deflection Eq. Method
Moment Distribution Method
Truss/Rangka Batang
Force Method
W1
W2
W3
W4
W5
W6
W7-9
W10
W11,12
W13-15
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Pertemuan – 1, 2
Prinsip Perpindahan Maya
Mata Kuliah : Analisis Struktur
Kode : CIV - 209
SKS : 4 SKS
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• Kemampuan Akhir yang Diharapkan
• Mahasiswa dapat menjelaskan prinsip kerja dan Energi dalam perhitungan deformasi struktur
• Sub Pokok Bahasan :
• Prinsip Dasar Metode Energi
• Kerja dan Energi
• Prinsip Konservasi Energi
• Virtual work
• Aplikasi kerja maya
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• Text Book :
• Hibbeler, R.C. (2010). Structural Analysis. 8th edition. Prentice Hall. ISBN : 978-0-13-257053-4
• West, H.H., (2002). Fundamentals of Structural Analysis. John Wiley & Sons, Inc. ISBN : 978-0471355564
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Deflections
• Deflections of structures can occur from various sources, such as loads, temperature, fabrication errors, or settlement.
• In design, deflections must be limited in order to provide integrity and stability of roofs, and prevent cracking of attached brittle materials such as concrete, plaster or glass.
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• Before the slope or displacement of a point on a beam or frame is determined, it is often helpful to sketch the deflected shape of the structure when it is loaded in order to partially check the results.
• This deflection diagram represents the elastic curve of points which defines the displaced position of the centroid of the cross section along the members.
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• Kerja Prinsip Konservasi Energi (conservation of energy principle) :
Kerja akibat seluruh gaya luar yang bekerja pada sebuah struktur (external forces) Ue, menyebabkan terjadinya gaya-gaya dalam pada struktur (internal work or strain energy) Ui seiring dengan deformasi yang terjadi pada struktur.
Apabila tegangan yang terjadi tidak melebihi batas elastis material struktur tersebut, elastic strain energy akan mengembalikan bentuk struktur ke tahap awal sebelum terjadinya pembebanan, jika gaya-gaya luar yang bekerja dihilangkan.
(1)
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External work-Axial force
L, E, A
F
dx
P
F
x D
xP
F D
DD
DD
PdxxP
dxFU e2
1
00
Kerja yang dilakukan oleh gaya luar P
(2)
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External work-Bending moment
M
m
f q
fq
M
m
qffq
fqq
MdM
dmU e2
1
00
Kerja yang dilakukan oleh momen lentur M
m
df
(3)
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Strain Energy – Axial force
Gaya P yang bekerja pada sebuah Bar seperti yang terlihat pada Gambar, dikonversikan menjadi strain energy yang menyebabkan pertambahan panjang pada batang sebesar ∆ dan timbulnya tegangan 𝜎. Mengingat Hukum Hooke : 𝜎 = 𝐸𝜖.
Maka persamaan defleksi dapat dituliskan menjadi:
Subsitusikan persamaan 4 ke dalam persamaan 2, maka didapat energi regangan yang tersimpan dalam batang :
AE
PLD (4)
AE
LPU i
2
2
(5)
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Strain Energy – Bending Moment
Energi regangan yang tersimpan pada balok :
(6)
dxEI
Md q Lihat Mekanika Bahan pt.11.
dxEI
MdMdU i
22
1 2
q
EI
LMdx
EI
MU
L
i22
2
0
2
Dari pers.(3)
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External Work, Ue Internal Work, Ui
Axial Force 1
2𝑃∆
𝑃2𝐿
2𝐴𝐸
Bending Moment 1
2𝑀𝜃
𝑀2𝐿
2𝐸𝐼
• Resume
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Castigliano Theorem • Italian engineer Alberto Castigliano (1847 – 1884) developed
a method of determining deflection of structures by strain energy method.
• His Theorem of the Derivatives of Internal Work of Deformation extended its application to the calculation of relative rotations and displacements between points in the structure and to the study of beams in flexure.
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Castigliano’s Theorem for Beam Deflection • For linearly elastic structures, the partial derivative of the
strain energy with respect to an applied force (or couple) is equal to the displacement (or rotation) of the force (or couple) along its line of action.
i
ii
i
ii
M
U
P
U
D qor (7)
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• Subtitusi persamaan 6 ke persamaan 7 :
• Jika sudut rotasi q, yang hendak dicari :
D
LL
EI
dx
P
MM
EI
dxM
P00
2
2(8)
L
EI
dx
M
MM
0
q (9)
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Example 1
• Determine the displacement of point B of the beam shown in the Figure.
• Take E = 200 GPa, I = 500(106) mm4.
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Example 1 • A vertical force P is placed on the beam at B
• Internal moments : (taken from the right side of the beam)
• From Castigliano’s theorem :
PxxM
xPx
xM
M
26
02
12
0
26 0 since xMP
xP
M
m,
EI
mkN
EI
dxxx
EI
dx
P
MM
L
B 150010156 3310
0
2
0
D
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Example 2 • Determine the slope at point B of the beam shown in Figure.
• Take E = 200 GPa, I = 60(106) mm4.
• Since the the slope at B is to be determined, an external couple M’ is placed on the beam at B.
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1
53 053 0
For
0
3 03 0
For
2
2222
2
1
1111
1
M
M
xMMxMMM
x
M
M
xMxMM
x
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rad10375960200
5112
mkN 511215303
3
25
0
22
5
0
11
0
,,
EI
,
EI
dxx
EI
dxx
EI
dx
M
MM
B
L
B
q
q
Setting M’ = 0, its actual value, and using Castigliano Theorem, we have : The negative sign indicates that qB is opposite to the direction of the couple moment M’.
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Example 3 • Determine the vertical displacement of point C of the beam.
• Take E = 200 GPa, I = 150(106) mm4.
External Force P. A vertical force P is applied at point C. Later this force will be set equal to a fixed value of 20 kN. Internal Moments M. In this case two x coordinates are needed for the integration, since the load is discontinuous at C.
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22
22
22
2
11
2111
11
11
1
50
508
0508 0
For
50
45024
02
85024 0
For
x,P
M
xP,M
xP,MM
x
x,P
M
xxP,M
Mx
xxP,M
x
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• Applying Castigliano’s Theorem. Setting P = 20 kN :
mm 14,2 m 01420150200
7426
mkN7426mkN192mkN7234
501850434
333
4
0
222
4
0
112
11
0
D
,,
EI
,
EIEI
,
EI
dxx,x
EI
dxx,xx
EI
dx
P
MM
L
Cv
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• 9.47
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Soal Latihan (Chapter IX, Hibbeler)
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• Prinsip perpindahan maya (virtual work) Prinsip ini dikembangkan oleh John Bernoulli pada tahun
1717 dan lebih dikenal dengan nama Unit Load Method. General Statement : • If we take a deformable structure of any shape
or size and apply a series of external loads P to it, it will cause internal loads u at points throughout the structure.
• It is necessary that the external and internal loads be related by the equations of equilibrium.
• As a consequence of these loadings, external displacements D will occur at the P loads and internal displacements d will occur at each point of internal load u.
Gambar 2.1
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Gambar 2.2
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1. ∆ = 𝑢 . 𝑑𝐿
Dimana :
𝑃′ = 1 = beban maya luar yang bekerja searah dengan ∆
∆ = perpindahan yang disebabkan oleh beban nyata
u = beban dalam maya yang bekerja dalam arah dL
dL = deformasi dalam benda yang disebabkan oleh beban nyata.
(1)
Virtual loading Real displacement
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Dengan cara yang sama, apabila kita ingin menentukan besar sudut rotasi pada lokasi tertentu dari sebuah benda, kita dapat mengaplikasikan beban momen maya M’ sebesar 1 satuan, lalu mengintegrasikannya dengan persamaan rotasi akibat beban momen nyata, sehingga :
1. 𝜃 = 𝑢𝜃 . 𝑑𝐿
Dimana :
𝑀′ = 1 = beban maya luar yang bekerja searah dengan ∆
𝜃 = perpindahan rotasi yang disebabkan oleh beban nyata
𝑢𝜃 = kerja dalam maya yang bekerja dalam arah dL
dL = deformasi dalam benda yang disebabkan oleh beban nyata.
(2)
Virtual loading Real displacement
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• Kerja Maya Pada Balok/Frame
D
L
dxEI
mM
0
1 (3)
dxEI
MddL q
D udL1
Virtual Loads
Real Displ.
mu
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• Kerja Maya Pada Balok/Frame
L
dxEI
Mm
0
1 qq (4)
dxEI
MddL q
dLuqq1
Virtual Loads
Real Displ.
qmu
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Example 4
• Determine the displacement of point B of the beam shown in the Figure.
• Take E = 200 GPa, I = 500(106) mm4.
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• The vertical displacement of point B is obtained by placing a virtual unit load of 1 kN at B.
Real Moment, M Virtual Moment, mq
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mm 150 m 1500500200
00015
mkN 0001561kN 1
3210
0
2
0
D
D
,.
EI
.
EI
dxxxdx
EI
mM
B
L
B
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Example 5 • Determine the slope at point B of the beam shown in Figure.
• Take E = 200 GPa, I = 60(106) mm4.
• The slope at B is determined by placing a virtual unit couple of 1 kN.m at B.
• Calculate virtual momen mq and real moment M
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Real Moment, M Virtual Moment, mq
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• The slope at B, is thus given by :
rad103759rad 009375060200
5112
5112531301
3
2
5
0
22
5
0
11
0
,,,
mkNEI
,dx
EI
xdx
EI
xdx
EI
Mm
B
L
B
q
q q
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Example 7 • Determine the tangential
rotation at point C of the frame shown in figure.
• Take E = 200 GPa,
I = 15(106) mm4
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rad 00875015200
2526mkN 2526152511
5715211
2
2
0
2
3
0
11
0
,,
EI
,
EIEI
,
EI
dx,
EI
dxx,dx
EI
Mm
C
L
C
q
q q
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Soal Latihan (Chapter IX, Hibbeler)