Beam: 2f, 1.75f NA------ 26.P 1.75f Section Strain Stress Simplified stress Equivalent elastic distribution stress distribution When 26 Cracks appear M=bhf(0.333+0.25)h=0.292fbh2 2 strength resist bending M cr W Mg=1.75f b 6
Beam: b NA h Section Strain Stress Simplified stress distribution Equivalent elastic stress distribution t , p 2 c c f t 2 f t , p 2 t f Mcr 3 h 4 h t 1.75 f t 1.75 f When t , p 2 Cracks appear 2 (0.333 0.25) 0.292 2 1 Mcr bhf t h f tbh Equivalent elastic strength resist bending t cr cr tm f bh M W M f 1.75 6 2
∫ =1.75 y-the section modulus plastic coefficient 截面抵抗矩塑性发展系数
1.75 t tm m f f －the section modulus plastic coefficient 截面抵抗矩塑性发展系数
10.5 Analytical methods for calculating the magnitude of crack width 10.5.1 Bond-slippage method 最早的方法，以后的方法在 其基础上补充、修正 Example: R.C.bar under uniaxial tension Stage 1:near critical cracking N→N 尚未开裂 0。=f x=0
10. 5 Analytical methods for calculating the magnitude of crack width 10. 5.1 Bond-slippage method 最早的方法，以后的方法在 其基础上补充、修正 Example: R.C. bar under uniaxial tension Stage 1: near critical cracking s c c t f s t f n 0 N N cr 尚未开裂
e/ Ner ① f ① ① T t= 0
Ncr Ncr c s t ft f n 0 ① ① ① ①
Stage 2:Appear of the first crack Ci at the weakest section Cracking section:o.=0 0,= A Within a distance of Lmin from Ci T≠0
Stage 2: Appear of the first crack C1 at the weakest section Cracking section: Within a distance of Lmin from C1 0 c s cr s A N 0