Fixed Angle Shear Theories - Unified Theory of Concrete Structures - page 258

Stress equilibrium equations

c
1

cos 2

c
2

sin 2

c
12 2sin

σ = σ α 1 + σ α 1 − τ α 1 cos α 1 + ρ f

1

1 sin 2

c

2 cos 2

c c

12 2sin

σ t = σ α 1 + σ α 1 + τ α 1 cos α 1 + ρ t f t

2

1 2 )sin 12 (cos 2

sin 2

τ t = ( σ − σ α 1 cos α 1 + τ α 1 − α 1 )

3

Strain compatibility equations

α 1 − γ 12
2

ε 1 cos 2 ε 2 sin 2

ε = ¯ ¯ α 1 + ¯ 2sin α 1 cos α 1

11

α 1 + γ 12
2

ε 1 sin 2 ε 2 cos 2 ε t = ¯ ¯ α 1 + ¯ 2sin α 1 cos α 1

12

γ t
2 =

α 1 + γ 12
2

(cos 2 sin 2 (¯ ε 1 − ε 2 )sin ¯ α 1 cos α 1 − α 1 )

6 a

Constitutive laws of materials
Concrete in compression

2 ¯

2

¯

ε 2
ζε o

ε 2
ζε o

c
2

f c σ = ζ − ¯ ε 2 /ζ ε o ≤ 1

13 a

1

2

¯

ε 2 /ζ ε o − 1

f c

2

σ = ζ − ε 2 /ζ ε o ≥ ¯ 1

13 b

4 /ζ − 1

5

9

1

8
f c ≤

. 1 | β |

24 ◦

ζ = 0 . √ 1 −

14

+ 400¯ ε 1

2 tan − 1

1 γ 12

β =

15

(¯ ε 1 − ε 2 ) ¯

Concrete in tension

c
1

σ = E c ¯ ε 1 ε 1 ≤ ε cr ¯ and ε cr = 0 . 00008 mm / mm ,

16 a

f cr ε cr
¯

0 . 4

31 f c ( MPa )

1

σ = ¯ ε 1 >ε cr and f cr = 0 .

16 b

ε 1

Concrete in shear

c c

2

σ

1 − σ

12

τ =

ε 2 ) γ 12

17

2(¯ ε 1 − ¯

Mild steel

ε s ≤ ε y

f s = E s ¯ ε s ¯

18 a , 19 a

ε s >ε y

f s =

. −

2 B ) f y +

. + . ε s

(0 91 (0 02 0 25 B ) E s ¯ ¯

18 b , 19 b

f s = f p − E s (¯ ε p − ¯ ε s ) ε s < ¯ ε p ¯

18 c , 19 c

ε y = f y / f y = E s (0 . 93 − 2 B ) f y

18 d , 19 d

f cr
f y

1 . 5

31 f c (MPa) and

1
ρ

B = f cr = 0 . ρ ≥ 0 . 15%

18 e , 19 e