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SLOPE DEFLECTION

Lereng metode defleksi

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The slope deflection method is a structural analysis method for beams and frames introduced in 1915 by George A. Maney. [ 1 ] The slope deflection method was widely used for more than a decade until the moment distribution method was developed. Metode defleksi lereng adalah analisis struktur metode untuk balok dan frame diperkenalkan pada tahun 1915 oleh George A. Maney. [1] Metode defleksi lereng secara luas digunakan lebih dari satu dekade sampai metode distribusi saat dikembangkan.

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[ edit ] Introduction [ sunting ] Pendahuluan

By forming slope deflection equations and applying joint and shear equilibrium conditions, the rotation angles (or the slope angles) are calculated. Dengan membentuk persamaan lendutan kemiringan dan menerapkan kondisi kesetimbangan geser dan sendi, sudut rotasi (atau sudut kemiringan) dihitung. Substutituting them back into the slope deflection equations, member end moments are readily determined. Substutituting mereka kembali ke lereng persamaan lendutan, anggota saat akhir dapat segera ditentukan.

[ edit ] Slope deflection equations [ sunting ] persamaan defleksi Lereng

The slope deflection equations express the member end moments in terms of rotations angles. kemiringan persamaan defleksi akhir ini mengungkapkan saat-saat anggota dalam hal sudut rotasi. The slope deflection equations of member ab of flexural rigidity E I a b and length L a b are: Persamaan defleksi kemiringan ab anggota E kekakuan lentur aku a, b dan panjang L b adalah:
M_ (ab) = \ frac (ab) EI_ () (L_ 
(ab)) \ left (4 \ theta_a + 2 \ theta_b - 6 \ frac (\ Delta) (L_ (ab)) \
 right)
M_ (ba) = \ frac (ab) EI_ () (L_ (ab)) \ left (2 \ 
theta_a + 4 \ theta_b - 6 \ frac (\ Delta) (L_ (ab)) \ right)
where θ a , θ b are the slope angles of ends a and b respectively, Δ is the relative lateral displacement of ends a and b. di mana θ a, b θ adalah sudut-sudut kemiringan dari ujung a dan b masing-masing, Δ adalah perpindahan lateral relatif berakhir dan b. The absence of cross-sectional area of the member in these equations implies that the slope deflection method neglects the effect of shear and axial deformations. Tidak adanya luas penampang anggota dalam persamaan ini menunjukkan bahwa metode lendutan kemiringan mengabaikan pengaruh deformasi geser dan aksial.
The slope deflection equations can also be written using the stiffness factor kemiringan persamaan lendutan juga dapat ditulis menggunakan faktor kekakuan K = \ frac (ab)
 I_ () (L_ (ab)) and the chord rotation dan rotasi akord \ Psi = \ frac (\ Delta) (L_ (ab)) : :
M_ (ab) = 2EK \ left (2 \ 
theta_a + \ theta_b - 3 \ kanan \ psi)
M_ (ba) = 2EK \ left \ (theta_a + 2 \ theta_b - 3 \ 
kanan \ psi)

[ edit ] Derivation of slope deflection equations [ sunting ] Penurunan persamaan lendutan kemiringan

When a simple beam of length L a b and flexural rigidity E I a b is loaded at each end with clockwise moments M a b and M b a , member end rotations occur in the same direction. Ketika sebuah balok sederhana dengan panjang L a, b dan E kekakuan lentur aku b dibuka setiap akhir dengan momen searah jarum jam M a, b dan M b, mengakhiri anggota rotasi terjadi pada arah yang sama. These rotation angles can be calculated using the unit dummy force method or the moment-area theorem . Sudut rotasi ini dapat dihitung dengan menggunakan metode unit boneka gaya atau daerah-teorema saat .
\ Theta_a - \ frac 
(\ Delta) (L_ (ab)) = \ frac (L_ (ab)) ((3EI_ ab) (ab) M_) - \ frac (L_ 
(ab)) ((6EI_ ab)) M_ (ba)
\ Theta_b - frac \ (\ Delta) (L_ (ab)) = - frac \ (L_ 
(ab)) ((6EI_ ab)) (ab + M_) \ frac (L_ (ab)) ((3EI_ ab)) M_ (ba)
Rearranging these equations, the slope deflection equations are derived. Menata ulang persamaan ini, persamaan lendutan kemiringan adalah berasal.

[ edit ] Equilibrium conditions [ sunting ] kondisi Equilibrium

[ edit ] Joint equilibrium [ sunting ] ekuilibrium Bersama

Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments ie be in equilibrium. kondisi kesetimbangan Bersama menyiratkan bahwa setiap sendi dengan derajat kebebasan seharusnya tidak seimbang yaitu saat-saat berada dalam ekuilibrium. Therefore, Oleh karena itu,
\ Sigma \ left (M ^ (f) M_ + (anggota) \ right) = \ 
Sigma M_ (bersama)
Here, M m e m b e r are the member end moments, M f are the fixed end moments , and M j o i n t are the external moments directly applied at the joint. Di sini, M m e m b e r adalah saat-saat akhir anggota, M f adalah saat-saat akhir tetap , dan M j o n t i adalah saat-saat eksternal secara langsung diterapkan pada sendi.

[ edit ] Shear equilibrium [ sunting ] ekuilibrium Shear

When there are chord roations due to sidesway in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account. Bila ada akord roations karena sidesway dalam bingkai, kondisi kesetimbangan tambahan, yaitu kondisi kesetimbangan geser harus diperhitungkan.

[ edit ] Example [ sunting ] Contoh


Example Contoh
The statically indeterminate beam shown in the figure is to be analysed. Balok statis tak tentu ditunjukkan pada gambar adalah untuk dianalisis.
  • Members AB, BC, CD have the same length Anggota AB, BC, CD memiliki panjang yang sama L = 10 \ m . .
  • Flexural rigidities are EI, 2EI, EI respectively. Kekakuan lentur adalah EI, 2EI, EI masing.
  • Concentrated load of magnitude Konsentrat beban besar P = 10 \ kN acts at a distance bekerja pada jarak a =
 3 \ m from the support A. dari dukungan A.
  • Uniform load of intensity Uniform beban intensitas q = 1 kN \ / m acts on BC. bertindak atas SM.
  • Member CD is loaded at its midspan with a concentrated load of magnitude CD Anggota dimuat di midspan dengan beban terkonsentrasi dari besarnya P = 10 \ kN . .
In the following calcuations, clockwise moments and rotations are positive. Dalam calcuations berikut, saat rotasi searah jarum jam dan positif.

[ edit ] Degrees of freedom [ sunting Derajat] kebebasan

Rotation angles θ A , θ B , θ C , θ D of joints A, B, C, D respectively are taken as the unknowns. Rotasi sudut θ A, B θ, θ C, θ D sendi A, B, C, D masing-masing diambil sebagai tidak diketahui. There are no chord rotations due to other causes including support settlement. Ada rotasi chord tidak karena penyebab lain, termasuk penyelesaian dukungan.

[ edit ] Fixed end moments [ sunting Tetap saat akhir]

Fixed end moments are: saat akhir tetap:
M _ (AB) ^ f = - frac \ (P ab ^ 2) (L ^ 2) = - \ frac 
(10 \ times 3 \ times 7 ^ 2) (10 ^ 2) = -14,7 mathrm \ (\, kN \, m)
M _ (BA) ^ f = \ frac (P 2 ^ (b) L ^ 2) = \ frac (10 \
 times 3 ^ 2 \ times 7) (10 ^ 2) = 6,3 \ mathrm (\, \ kN , m)
M _ (SM) ^ f = - \ frac (kw ^ 2) (12) = - \ frac (1 \ 
times 10 ^ 2) (12) = - 8,333 \ mathrm (\, \ kN, m)
M _ (CB) ^ f = \ frac (kw ^ 2) (12) = \ frac (1 \ 
times 10 ^ 2) (12) = 8,333 \ mathrm (\, \ kN, m)
M _ (CD) ^ f = - frac \ (PL) (8) = - \ frac (10 \ 
times 10) (8) = -12,5 \ mathrm (\, \ kN, m)
M _ (DC) ^ f = \ frac (PL) (8) = \ frac (10 \ times 
10) (8) = 12,5 \ mathrm (\, \ kN, m)

[ edit ] Slope deflection equations [ sunting ] persamaan defleksi Lereng

The slope deflection equations are constructed as follows: kemiringan persamaan defleksi dibangun ini adalah sebagai berikut:
M_ (AB) = \ 
frac (EI) (L) \ left (4 \ theta_A + 2 \ theta_B \ right) = 0.4EI \ 
theta_A + 0.2EI \ theta_B
M_ (BA) = \ frac (EI) (L) \ left (2 \ theta_A + 4 \ 
theta_B \ right) = 0.2EI \ theta_A + 0.4EI \ theta_B
M_ (SM) = \ frac (2EI) (L) \ left (4 \ theta_B + 2 \ 
theta_C \ right) = 0.8EI \ theta_B + 0.4EI \ theta_C
M_ (CB) = \ frac (2EI) (L) \ left (2 \ theta_B + 4 \ 
theta_C \ right) = 0.4EI \ theta_B + 0.8EI \ theta_C
M_ (CD) = \ frac (EI) (L) \ left (4 \ theta_C \ right)
 = 0.4EI \ theta_C
M_ (DC) = \ frac (EI) (L) \ left (2 \ theta_C \ right)
 = 0.2EI \ theta_C

[ edit ] Joint equilibrium equations [ sunting ] Bersama persamaan kesetimbangan

Joints A, B, C should suffice the equilibrium condition. Sendi A, B, C harus mencukupi kondisi ekuilibrium. Therefore Karena itu
\ Sigma M_A = M_ 
(AB) + M_ (AB) ^ f = 0.4EI \ theta_A + 0.2EI \ theta_B - 14,7 = 0
\ Sigma M_B = M_ (BA) + M_ (BA) ^ f + M_ (SM) + M_ 
(SM) ^ f = 0.2EI \ theta_A + 1.2EI \ theta_B + 0.4EI \ theta_C - 2,033 =
 0
\ Sigma M_C = M_ (CB) + M_ (CB) ^ f + M_ (CD) + M_ 
(CD) ^ f = 0.4EI \ theta_B + 1.2EI \ theta_C - 4,167 = 0

[ edit ] Rotation angles [ sunting ] Rotasi sudut

The rotation angles are calculated from simultaneous equations above. Sudut rotasi dihitung dari persamaan simultan di atas.
\ Theta_A = \ frac (40,219)) (KTI
\ Theta_B = \ frac (-6,937) () KTI
\ Theta_C = \ frac (5,785)) (KTI

[ edit ] Member end moments [ sunting ] Anggota saat akhir

Substitution of these values back into the slope deflection equations yields the member end moments (in kNm): Pergantian nilai-nilai ini kembali ke lereng persamaan defleksi hasil akhir saat-saat anggota (dalam KNM):
M_ (AB) = 0,4 \ times 40,219 + 0,2 
kali \ \ left (-6,937 \ kanan) - 14,7 = 0
M_ (BA) = 0,2 \ times 40,219 + 0,4 kali \ \ left 
(-6,937 \ right) + 6,3 = 11,57
M_ (SM) = 0,8 \ times \ left (-6,937 \ right) + 0,4 \ 
times 5,785-8,333 = -11,57
M_ (CB) = 0,4 \ times \ left (-6,937 \ right) + 0,8 \ 
times 5,785 + 8,333 = 10,19
M_ (CD) = 0,4 \ times 5,785-12,5 = -10,19
M_ (DC) = 0,2 \ times 5,785 + 12,5 = 13,66

[ edit ] Notes [ sunting ] Catatan

  1. ^ Maney, George A. (1915). Studies in Engineering . ^ Maney George, A. (1915) Rekayasa. Studi di. Minneapolis: University of Minnesota Minneapolis: University of Minnesota  

[ edit ] References [ sunting ] Referensi

[ edit ] See also [ sunting ] Lihat pula

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