Ilmu Dasar Sains. Kalkulus. Statika & Mek Bhn. Analisis Struktur. Peranc.Str. Beton. Peranc.Str. Baja. Peranc.Str. Bangunan Sipil

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Ilmu Dasar Sains

S2

Statika & Mek Bhn

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Analisis Struktur

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Peranc.Str. Beton

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Din.Str.&Rek.Gempa

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Peranc.Str. Bangunan Sipil

Kalkulus

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Peranc.Str. Baja

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MK MINOR STRUKTUR & SB DAYA AIR

Analisis Struktur (4SKS)

Menghitung defleksi/perpind ahan titik

Beam & Frame

Truss/Rangka Batang

Analisis Struktur Statis Tak Tentu

Beam & Frame

Castigliano Method

W1

Virtual Load Method

W2

Double Integration

W3

Moment Area Method

W4

Conjugate Beam

W5

Castigliano Method Virtual Load Method Force Method Slope-Deflection Eq. Method Moment Distribution Method

Truss/Rangka Batang

W6

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.


• 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.




• 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

D

x

D

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

(2) Kerja yang dilakukan oleh gaya luar P

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

M df

m

M

q

f

m

f

q q

q

U e   m  df   0

0

M

1  f  df  Mq q 2

(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 𝜎. Mengingat Hukum Hooke : 𝜎 = 𝐸𝜖. Maka persamaan defleksi dapat dituliskan menjadi: PL (4) D AE Subsitusikan persamaan 4 ke dalam persamaan 2, maka didapat energi regangan yang tersimpan dalam batang : P2L 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 : L

M2 M 2L Ui   dx  2 EI 2 EI 0

(6)


• Resume External Work, Ue Axial Force Bending Moment

1 𝑃∆ 2 1 𝑀𝜃 2

Internal Work, Ui 𝑃2 𝐿 2𝐴𝐸 𝑀2 𝐿 2𝐸𝐼


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.


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.

U i U i Di  or q i  Pi M i

(7)


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

L

 0

L

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



(8)

• Jika sudut rotasi q, yang hendak dicari : L



 M  dx    M   EI

q  M 0

(9)


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 x 2  Px

M  x P since P  0

•

M  6 x 2

From Castigliano’s theorem :

 M DB   M   P 0 L





 6 x 2  x dx 15 103 kN  m3  dx    0,150m  EI EI  EI 0 10


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.


For x1 M  0

M 1  3 x1  0

 M 1  3 x1 M 1 0 M 

For x2 M  0

M 2  M   35  x2   0

 M 2  M   35  x2 

M 2 1 M 

Integrity, Professionalism, & Entrepreneurship Setting M’ = 0, its actual value, and using Castigliano Theorem, we have : L

 M  dx qB  M      M EI   0



qB 

5

 0

 3x1 0dx1  EI

5

 0

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

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

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


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.


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


• Applying Castigliano’s Theorem. Setting P = 20 kN : L

 M D Cv  M   P 0



 dx    EI

4

 0

34 x  4 x 0,5x dx 2

1

1

EI

1

1

4



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

 0

18 x2 0,5 x2 dx2 EI


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


Gambar 2.2


1. ∆ = Virtual loading

𝑢 . 𝑑𝐿

(1)

Real displacement

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.

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 : 1. 𝜃 =

loading DimanaVirtual :

𝑀′ 𝜃 𝑢𝜃 dL

𝑢𝜃 . 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.


• Kerja Maya Pada Balok/Frame 1  D   udL

Virtual Loads Real Displ.

um

L

1 D 

 0

mM dx EI

M dL  dq  dx EI

(3)


• Kerja Maya Pada Balok/Frame 1q 

 uq dL

u  mq

L

1q 

 0

mq M dx EI

Virtual Loads Real Displ.

M dL  dq  dx EI

(4)


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


L

1 kN  D B 

 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


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


Real Moment, M

Virtual Moment, mq


• The slope at B, is thus given by : L

1q B 

 0

mq M dx  EI

5

 0

0 3x1  dx EI

1

5

 0

1 35  x2  dx EI

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

2



 112 ,5 kN  m 2 EI


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



L

1q C 

 0

mq M dx  EI

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 qC      0 ,00875 rad EI EI EI 200  15


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

Ilmu Dasar Sains. Kalkulus. Statika & Mek Bhn. Analisis Struktur. Peranc.Str. Beton. Peranc.Str. Baja. Peranc.Str. Bangunan Sipil

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