Slide CIV 209 AnalisisStruktur CIV 209 P1 2

S1

Ilmu Dasar Sains

S2

Statika & Mek Bhn

S3

Analisis Struktur

S4

Peranc.Str. Beton

S6

Din.Str.&Rek.Gempa

S7

Peranc.Str. Bangunan Sipil

Kalkulus

You are here

Peranc.Str. Baja

S5

MK MINOR
STRUKTUR & SB
DAYA AIR

Analisis Struktur (4SKS)

Beam &
Frame

Menghitung
defleksi/perpind
ahan titik
Truss/Rangka
Batang

Castigliano Method

W1

Virtual Load Method

W2

Double Integration

W3

Moment Area Method

W4

Conjugate Beam

W5

Castigliano Method

W6

Virtual Load Method

Analisis Struktur
Statis Tak Tentu

Beam &
Frame

Force Method
Slope-Deflection Eq. Method

Moment Distribution Method

Truss/Rangka
Batang

Force Method

W7-9
W11,12
W13-15

W10

Integrity, Professionalism, & Entrepreneurship
Mata Kuliah
Kode
SKS

: Analisis Struktur
: CIV - 209

: 4 SKS

Prinsip Perpindahan Maya
Pertemuan – 1, 2

Integrity, Professionalism, & Entrepreneurship
• 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

Integrity, Professionalism, & Entrepreneurship
• 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 : 9780471355564

Integrity, Professionalism, & Entrepreneurship

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.

Integrity, Professionalism, & Entrepreneurship

• 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.

Integrity, Professionalism, & Entrepreneurship

Integrity, Professionalism, & Entrepreneurship

Integrity, Professionalism, & Entrepreneurship

• 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.
(1)

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 gayagaya luar yang bekerja dihilangkan.

Integrity, Professionalism, & Entrepreneurship
External work-Axial force
F

P

F 

P
x
D

L, E, A

D

dx
F

1
P
U e   F  dx    x  dx  P D
D
2
0
0
D

x

D

(2)
Kerja yang dilakukan oleh gaya luar P

Integrity, Professionalism, & Entrepreneurship
External work-Bending moment
m

M

df

m

q

M

f

m

f

q

U e   m  df  
q

0

q

0

1
 f  df  Mq
q
2

M

(3)

Kerja yang dilakukan oleh momen lentur M

Integrity, Professionalism, & Entrepreneurship
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 �. Mengingat Hukum Hooke : � = ��.
Maka persamaan defleksi dapat dituliskan menjadi:
D

PL
AE

(4)

Subsitusikan persamaan 4 ke dalam persamaan 2, maka didapat
energi regangan yang tersimpan dalam batang :
P 2L
Ui 
2 AE

(5)

Integrity, Professionalism, & Entrepreneurship
Strain Energy – Bending Moment

dq 

M
dx
EI

1
M2
dU i  M  dq 
dx
2
2 EI

Lihat Mekanika Bahan pt.11.
Dari pers.(3)

Energi regangan yang tersimpan pada balok :
M2
M 2L
Ui  
dx 
2
EI
2 EI
0
L

(6)

Integrity, Professionalism, & Entrepreneurship

• Resume
External Work, Ue
Axial Force
Bending Moment

�∆

�

Internal Work, Ui
�2
��
2

��

Integrity, Professionalism, & Entrepreneurship

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.

Integrity, Professionalism, & Entrepreneurship

Castiglia o’s Theore

for Bea

Deflectio

• 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.

U i
U i
or q i 
Di 
Pi
M i

(7)

Integrity, Professionalism, & Entrepreneurship

• Subtitusi persamaan 6 ke persamaan 7 :

D
P


L



M 2 dx
 M  dx
 M

2 EI
P


 EI
0
L

(8)

• Jika sudut rotasi q, yang hendak dicari :



0

 M  dx


M


 EI

q  M
L

0

(9)

Integrity, Professionalism, & Entrepreneurship

Example 1
• Determine the displacement of point B of the
beam shown in the Figure.
• Take E = 200 GPa, I = 500(106) mm4.

Integrity, Professionalism, & Entrepreneurship
Example 1
•

A vertical force P is placed on the beam at B

•

Internal moments : (taken from the right side of the beam)
M  0

x

M  12 x    P  x  0
2

M  6 x2  Px

M
 x
P

since P  0

•

Fro

Castiglia o’s theore



:



M  6 x 2

 6 x2  xdx 15 103 kN  m3
 M  dx


 0,150m
DB   M 

EI
EI
 P  EI 0
0
L

10

Integrity, Professionalism, & Entrepreneurship

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
ouple M’ is pla ed o the ea at B.

Integrity, Professionalism, & Entrepreneurship

For x1

M  0

M 1  3 x1  0

For x2

M 2  M   35  x2   0

M  0

 M 1  3 x1
M 1
0
M 

 M 2  M   35  x2 

M 2
1
M 

Integrity, Professionalism, & Entrepreneurship
Setting M’ = 0, its a tual alue, a d usi g Castiglia o Theore , e ha e :



 M  dx

qB  M 



M
EI


0
L

qB 


5

 3x1 0dx1 
EI

 112 ,5
 9 ,375 10 3 rad
200  60
0


5

0

 35  x2 1dx2
112 ,5 kN  m 2

EI
EI

The negative sign indicates that qB is opposite to the direction of the couple
moment M’.

Integrity, Professionalism, & Entrepreneurship

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.

Integrity, Professionalism, & Entrepreneurship

For x1
M  0

x 
 24  0 ,5 P x1  8 x1  1   M 1  0
2

 M 1  24  0 ,5 P x1  4 x12
M 1
 0 ,5 x1
P

For x2

M  0

 M 2  8  0 ,5 P x2  0

 M 2  8  0 ,5 P x2
M 2
 0 ,5 x2
P

Integrity, Professionalism, & Entrepreneurship

• Applyi g Castiglia o’s Theore . Setti g P = 20 kN :



 M
D Cv  M 
 P
0
L

 dx


 EI


4

0

34 x  4 x 0,5x dx
2

1

1

EI

1

1



234 ,7 kN  m 3 192kN  m 3 426 ,7 kN  m 3



EI
EI
EI



426 ,7
 0 ,0142 m  14,2 mm
200  150


4

0

18 x2 0,5 x2 dx2
EI

Integrity, Professionalism, & Entrepreneurship

Soal Latihan (Chapter IX, Hibbeler)

•
•
•
•
•
•
•

9.47
9.49
9.53
9.56
9.61
9.63
9.66

•
•
•
•
•
•
•

9.69
9.72
9.74
9.80
9.83
9.85
9.87

•
•
•
•
•

9.89
9.92
9.94
9.96
9.98

Integrity, Professionalism, & Entrepreneurship
• 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

Integrity, Professionalism, & Entrepreneurship

Gambar 2.2

Integrity, Professionalism, & Entrepreneurship

Virtual loading

.∆ =

�.�

(1)
Real displacement

Dimana :
�′
= 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.

Integrity, Professionalism, & Entrepreneurship
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 :
.� =
loading
DimanaVirtual
:
′

�
��
dL

�� . �

(2)
Real displacement

= 1 = beban maya luar yang bekerja searah dengan ∆
= perpindahan rotasi yang disebabkan oleh beban nyata
= kerja dalam maya yang bekerja dalam arah dL
= deformasi dalam benda yang disebabkan oleh beban nyata.

Integrity, Professionalism, & Entrepreneurship

• Kerja Maya Pada Balok/Frame

1  D   udL
um
1 D 


L

0

mM
dx
EI

Virtual
Loads
Real
Displ.

M
dL  dq 
dx
EI

(3)

Integrity, Professionalism, & Entrepreneurship

• Kerja Maya Pada Balok/Frame
1q 

 uq dL

u  mq
1q 


L

0

mq M
EI

Virtual
Loads
Real
Displ.

M
dL  dq 
dx
EI

dx

(4)

Integrity, Professionalism, & Entrepreneurship

Example 4
• Determine the displacement of point B of the
beam shown in the Figure.
• Take E = 200 GPa, I = 500(106) mm4.

Integrity, Professionalism, & Entrepreneurship
• The vertical displacement of point B is obtained by placing a
virtual unit load of 1 kN at B.
Virtual Moment, mq

Real Moment, M

Integrity, Professionalism, & Entrepreneurship

1 kN  D B 


L

0

mM
dx 
EI



10

0

 1x 6 x 2 dx  15.000 kN 2  m 3
EI

15.000
DB 
 0 ,150 m  150 mm
200  500

EI

Integrity, Professionalism, & Entrepreneurship

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

Integrity, Professionalism, & Entrepreneurship

Real Moment, M

Virtual Moment, mq

Integrity, Professionalism, & Entrepreneurship

• The slope at B, is thus given by :
1q B 


L

mq M
EI

dx 


5

0 3x1  dx
EI

1


5

1 35  x2  dx
EI

 112 ,5
 0 ,009375 rad  9 ,375 10 3 rad
qB 
200  60
0

0

0

2



 112 ,5
kN  m 2
EI

Integrity, Professionalism, & Entrepreneurship

Example 7
• Determine the tangential
rotation at point C of the
frame shown in figure.
• Take E = 200 GPa,
I = 15(106) mm4

Integrity, Professionalism, & Entrepreneurship

Integrity, Professionalism, & Entrepreneurship

1q C 


L

0

mq M
EI

dx 


3

0

 1 2,5 x1 dx1  17 ,5dx2


2

EI

0

EI

11,25 15 26,25 kN  m 2
26,25



 0 ,00875 rad
qC 
200  15
EI
EI
EI

Integrity, Professionalism, & Entrepreneurship

Soal Latihan (Chapter IX, Hibbeler)

•
•
•
•
•
•
•

9.46
9.48
9.50
9.51
9.52
9.54
9.55

•
•
•
•
•
•
•

9.57
9.58
9.59
9.60
9.62
9.64
9.65

•
•
•
•
•
•
•

9.67
9.68
9.70
9.71
9.73
9.75
9.78

•
•
•
•
•
•
•

9.79
9.81
9.82
9.84
9.86
9.88
9.90

•
•
•
•

9.91
9.93
9.95
9.97