Chapter 11 The serviceability limit state -Deflection of R.C.beam
Chapter 11 The serviceability limit state — —Deflection of R.C. beam
11.1 The allowable deflection Beam and slab 1s(2500L L。一span 吊车梁 ]≤(306d4
11.1 The allowable deflection 吊车梁 0 ) 300 1 250 1 [ f ] ( ~ L L0 — span 0 ) 600 1 500 1 [ f ] ( ~ L Beam and slab
11.2 Stiffness and deformation M 1 曲率 的变化反映了各 阶段截面刚度变化的特 点: B、(刚度) M>Mcr后,曲率突增, 列度减小。 临近钢筋屈服时,曲率 Mer 1 增长迅速,刚度再次减 小 Bo Bs 1 M M P h EI B
11.2 Stiffness and deformation M Mu My Mcr B0 Bs 1 Bs 曲率 的变化反映了各 阶段截面刚度变化的特 点: M>Mcr 后,曲率突增, 刚度减小。 临近钢筋屈服时,曲率 增长迅速,刚度再次减 小 1 s c s B M EI M h 0 1 = (刚度)
Pi 9 P2 Mx 弯矩分布图 Bx 刚度分布图
P1 P2 q Mx Bx 弯矩分布图 刚度分布图
11.3 Bending stiffness 1l.3.1 Method of effective moment of inertia(有效惯性矩法) 1)Pre-cracking stage Ec N.A. Xo As h ho-Xo a (n-1)As Transformed section:4=bh+(n-1)4
11.3 Bending stiffness 11.3.1 Method of effective moment of inertia (有效惯性矩法) As b h (n-1)As N.A. x0 a h0-x0 εc εs 1) Pre-cracking stage A bh n A s ( 1) Transformed section: 0
求形心主轴位置: 2=h-x广+a-A-) h+(n-1)A.h X,= 2 bh+(n-1)A. Moment of inertia: L-9x+h-x门+n-A-x)广 B。=EI
( ) ( 1) ( ) 2 1 2 1 0 0 2 0 2 0 bx b h x n A h x s s s bh n A bh n A h x ( 1) ( 1) 2 1 0 2 0 2 0 0 3 0 3 0 0 [ ( ) ] ( 1) ( ) 3 x h x n A h x b I s 0 0 B E I c 求形心主轴位置: Moment of inertia:
己知I,求I, y a ·形心主轴 b y a) I =1,+(b-a)A b) I=I,+(a-b)A c I,=1,+(b+a)A d I,=1+(b-a2)A
a b y y’ 形心主轴 I y I y (b a )A 2 2 ' I y I y b a A 2 ' ( ) I y I y b a A 2 ' ( ) I y I y (a b )A 2 2 ' a) b) c) d) 已知 I y 求 y ' I
2)Post-cracking stage Ec N.A. As ho ho-Xch b nAs
2) Post-cracking stage As b h0 nAs N.A. xcr a h0-xcr εs εc
Equilibrium condition b A.0,=)x0。 2 AE.g,-1bx.E.8. Constitutive law o,=E6 0。=E8。 Compatibility condition or E.8(-1)-Es. 2 0 x2=n4-X) 2 Solve the quadratic equation to get Xer 0 -x+4h+x) 3
Equilibrium condition s s cr c x b A 2 s E s s c E c c s s s cr E c c A E bx 2 1 Constitutive law Compatibility condition or cr s cr c x h x 0 ( 1) 0 cr s c x h cr c c cr s s c bx E x h A E 2 1 ( 1) 0 2 0 1 ( ) 2 cr s cr bx nA h x Solve the quadratic equation to get xcr 2 0 3 ( ) 3 1 cr cr s cr I bx nA h x
B。=EI。 B.=EI Effective moment of inertia 。=(0+1-0s. M M
cr c cr B E I Effective moment of inertia 0 3 0 3 ( ) (1 ) I I M M I M M I cr cr cr ef M M 0 0 B E I c