T 1) 2) ( ) T. T 4 T. R T. T U A doi / THE ANALYSIS ON STATIC CHARACTERISTICS OF CURVED T-BEAMS IN CONS

Size: px

Start display at page:

Download "T 1) 2) ( ) T. T 4 T. R T. T U A doi / THE ANALYSIS ON STATIC CHARACTERISTICS OF CURVED T-BEAMS IN CONS"

读毓通
4 years ago
Views:

1 T T. T 4 T. T. T U doi 1.652/ THE NLYSIS ON STTIC CHCTEISTICS OF CUVED T-BEMS IN CONSIDETION OF SELF-EQUILIBIUM 1 GN Yanan 2 SHI Feiting School of Civil Engineering and rchitecture, Yancheng Institute of Technology, Yancheng 22451, Jiangsu, China bstract In consideration of the shear lag effects and the shear deformation, a new warping displacement mode of curved T-beams is chosen to satisfy the axial self-equilibrium condition for the shear lag warping stress, and an accurate approach is proposed to analyze the static characteristics of curved T-beams widely used in engineering. The energy-variational principle is applied to establish the governing differential equations and the corresponding natural boundary conditions, and thus the closed-form solutions of the generalized displacements are obtained. The variations of the shear lag coefficients and the stress in the curved T-beams against the span-width ratio and the type of loading are discussed, and the role played by the self-equilibrium condition is analyzed. Key words curved T-beams, self-equilibrium condition, static characteristics, energy-variational principle T [1-2] T [3-4]. T T. [5-8] [9-12] KJB gyn-12@163.com,. T., 215, 371: Gan Yanan, Shi Feiting. The analysis on static characteristics of curved T-beams in consideration of self-equilibrium. Mechanics in Engineering, 215, 371: 79-85

2 T T. T T, L wx θx, φx, Ux, T ux [1,11,13] ux = ρ T y, zux = ρ M s zφ s yux φ s y = 1 b y2 b 2, y b T, x, z, y T T y, z M s b T ρ b ρ. σ sx = E ux x τ sy = G ux y, M s = E ρ = G ρ Ms zφ s y U x 2 dφ s y zux 3 dy σ sx d = M s = 4h 1 tb/3 h 1 T y t T T E, G. M sy, I sy = M sy = σ sy zd = EI sy U 4 /ρzm s zφ s yd. M y = EI y φ θ EI sy U 5 [ σ za = Ez φ θ + E M s zφy U ]6 ρ, I y = [/ρz] 2 d. V y = 1 σ 2 za 2 E + τ sy 2 G { 1 l E 2 ddx = [I s U 2 + 2I sy φ θ U + I y φ θ 2 ] + Gk sy U }dx 2 7 I s = k sy = ρ [/ρm s zφ s y] 2 d dφ s y 2d z dy T V 1 3 T V 1 = V y l M 2 k1 dx l kg φ w 2dx x

3 1 T 81 M k1 = θ + φ/ 8, k, J k. V 2 V 2 = l q z w + m x θdx Q z w + M y φ + M y U + T x θ l 9 M y φx, M y, q z, m x, Q z. V = V 1 + V δv = T [14] EI y φ 2 φ EI y + θ + EI sy U kgφ w = 11 EI y + φ + θ EI y 2 θ+ EI sy U + m x = 12 EI sy φ EI sy θ + EI s U Gk sy U = 13 kgφ w q z = 14 [EI y φ θ ] + EI sy U l M y δφ = 15 [EI s U + EI sy φ θ ] l M y δ U = 16 [ θ + φ ] l T x δθ = 17 [ kgφ w Q z ] l δw = 18 x Ux 11 φ + 1/ 2 φ + θ 3 + 1/θ + [1/ ]kgφ w = 19 Ux 13 φ 3 + B 1 φ + B 2 θ 4 + B 3 θ + B 4 θ + B 5 m x = 2 B 1 = m 1 = EI y, m 2 = / 2 m 3 = + EI y /, m 5 = EI y / 2, m 7 = EI s, m 8 = Gk sy m 4 = EI sy m 6 = EI sy / m 3 m 8 m m, B 2 = m 2m 7 3 3m 7 m m 3m 7 B 3 = m 7m 1 m m 2 m 8 4 m 2 4 m 1m 7 + m 2 m 7 2 B 4 = m 1 m 8 m m 3m 7, B m 8 5 = m m 3m φx θ 6 3 B 1 B 2 B θ 4 B 2 B 1 B 3 2 B 4 2 θ + B 2 B 4 2 B 2 θ + B 5 2 B 2 m x B 1 2 B 2 q z = r 1,2 = ±α 1 + β 1 i, r 3,4 = ±α 2 + β 2 i r 5,6 = ±α 2 + β 2 i η 1 = α 1 + β 1 i, η 2 = α 2 + β 2 i, θx = c 1 chη 1 x + c 2 shη 1 x + c 3 chη 2 x + c 4 xchη 2 x+ c 5 shη 2 x + c 6 xshη 2 x + c 7 x 2 + c 8 x + c 9 22 θx φx, wx θx. T Ux Ux : θ3 3 θ + EI2 sy I y I s I sy U 4 + Gk sy I y 2 + J k I s 2 I sy U G2 J k k sy 2 EI sy U = 23

4 θ3 + 3 θ + EI2 sy + EI y I s 2 I sy U + Gk sy + EI y 2 EI sy U = U 4 + Gk syi y 2 + EI 2 sy I y I s EI 2 sy I y I s 2 U + Gk sy I y EI 2 sy I y I s 2 U = r 1,2 = ±α 1 + β 1 i, r 3,4 = ±α 2 + β 2 i Ux Ux = F 1 shη 1 x + F 2 chη 1 x+ F 3 shη 2 x + F 4 chη 2 x + F 5 26 θx, φx, wx, Ux 11 14, θx, φx, wx Ux. θx = c 1 chη 1 x + c 2 shη 1 x + c 3 chη 2 x + c 4 xchη 2 x+ c 5 shη 2 x + c 6 xshη 2 x + 2 EI y m x EI y + EI y 3 q z 27 φx = c 1 D 1 shη 1 x + c 2 D 1 chη 1 x + c 3 D 32 shη 2 x+ c 4 D 4 xshη 2 x + D 31 chη 2 x + c 5 D 32 chη 2 x+ c 6 D 4 xchη 2 x + D 31 shη 2 x+ k 2 J k + k 2 E 7 2 q z x 28 F 1 = 2 η 3 1m 2 m 4 + m 6 m 3 m 5 m 4 η 1 η 2 1 m 6m 4 + m 7 m 3 + m 8 m 3 D 1 = m 6η 1 F 1 + m 5 2 m 2 η 2 1 m 3 η 1 D 4 = m 5 2 m 2 η 2 2 m 3 η 2 F 31 = m 3m η 2 2m 2 m 4 D 4 m 3 m 4 η 2 m 8 m 3 + η 2 2 m 6m 4 + m 7 m 3 F 32 = 2 η 3 2m 2 m 4 + η 2 m 6 m 3 m 5 m 4 m 8 m 3 + η 2 2 m 6m 4 + m 7 m 3 D 32 = F 32m 6 η 2 + m 5 2 m 2 η 2 2 m 3 η 2 D 31 = F 31m 6 η m 2 η 2 D 4 m 3 m 3 η 2 Ux = c 1 F 1 shη 1 x + c 2 F 1 chη 1 x + c 3 F 32 shη 2 x+ c 4 F 31 chη 2 x + c 5 F 32 chη 2 x + c 6 F 31 shη 2 x 29 D 1 D 1 D 32 wx = c 1 chη 1 x + c 2 shη 1 x + c 3 chη 2 x+ η 1 η 1 η 2 D4 c 4 xchη 2 x + D 31η 2 D 4 shη 2 x + η 2 η 2 2 D 32 D4 c 5 shη 2 x + c 6 xshη 2 x+ η 2 η 2 D 31 η 2 D 4 chη 2 x + E 7 x + E 8 η kg + 2 q z x T 1 T wx l =, θx l = U x l =, φ x l = 31 T p k l k1 l k2 k U k1 l k1 = U k2, U k1 l k1 = U k2 φ k11 l k1 = φ k21, θ k1l k1 = θ k2 θ k1 l k1 = θ k2, U k1l k1 = U k2 U k1 l k1 = U k2 φ k1 l k1 φ k2 = p k /kg 2 T φ =, U =, θ = w =, U l =, φl = w l θ l + φl =, φ l θl = T,

5 1 T 83 E = 35 GPa, G = 15 GPa, t w =.15 m, t =.11 m, b = 2.85 m, h = 1 m, = 5 m. p z = 1.57 MN q z = N/m. T. NSYS 2 T T NSYS Extrude T x, y, z z, y T. 4 T L = 8 m 1 T x = L/2, L = 12 m, L/2b = 2.1 T /m /MPa /MPa /MPa /MPa /% /MPa T x = L/2, L = 12 m, L/2b = 2.1 T /m /MPa /MPa /MPa /MPa /% /MPa

6 T T 3 T T T. T. 3 T x = L/2, L = 8 m, L/2b = 1.4 T /m /MPa /MPa /MPa /MPa /% /MPa T x = L/2, L = 12 m, b = 2.35 m, L/2b = 2.6 T /m /MPa /MPa /MPa /MPa /% /MPa T x = L/2, L = 12 m, t w =.25 m, L/2b = 2.1 T /m /MPa /MPa /MPa /MPa /% /MPa

7 1 T 85 6%, 14%. T T. 4 5 T L = 12 m T. T T. 6 T L = 12 m 7 T 8% 7 T L = 8 m 1,. T., 24, 233: ,.., 212, 343: ,,.., 213, 356: Song QG, Sordelis C. Shear lag analysis of T, I and box beams. Structural Engineering, 199, 1165: ,.., 29, 261: ,,.., 29, 586: Sanguanmanasak J, Chaisomphob T, Yamaguchi E. Stress concentration due to shear lag in continuous box girders. Engineering Structures, 27, 297: ,,.., 21, 2712: Kang YJ, Yoo CH. Thin-walled curved beams. I: formulation of nonlinear equations. J Engrg Mech, SCE, 1994, 121: ,.., 22, 231: ,.., 22, 194: ,.., 24, 215: ,.., 28, 256: ,1981 :

,!! #! > 1? = 4!! > = 5 4? 2 Α Α!.= = 54? Β. : 2>7 2 1 Χ! # % % ( ) +,. /0, , ) 7. 2

,!! #! > 1? = 4!! > = 5 4? 2 Α Α!.= = 54? Β. : 2>7 2 1 Χ! # % % ( ) +,. /0, , ) 7. 2 ! # %!% # ( % ) + %, ). ) % %(/ / %/!! # %!! 0 1 234 5 6 2 7 8 )9!2: 5; 1? = 4!! > = 5 4? 2 Α 7 72 1 Α!.= = 54?2 72 1 Β. : 2>7 2 1 Χ! # % % ( ) +,.