Rotating Angle Shear Theories - Unified Theory of Concrete Structures - page 183

Solve the longitudinal steel strain ε :

ε r − ε d
− σ d

0 . 00385 + 0 . 0002

10 − 3 /MPa = = 0 . 545 ×

7 . 43

σ − ρ f − ρ p f p

ε = ε r + ε r − ε d
− σ d

Equation 13

10 − 3 = 0 . 00385 + 0 . 545 × (3 . 44 − 0 . 012 f − 0 . 003 f p )

Assume f = E s ε = 200 × 10 3 ε before yielding

10 3 (0

f p = E ps ( ε dec + ε s ) = 200 × . 005 + ε ) before elastic limit

Then ε + 1.308 ε + 0.327 ε = 0.00385 + 0.00188 − 0.00164

0 . 00409

2

ε = = 0 . 00155 < 0 . 00207( ε y )OK

. 635

( ε ps at 0 . 7 f pu ) = 0 . 7(1862) / 200 000 = 0 . 00652

( ε ps at 0.7 f pu ) − ε dec = 0.00652 − 0.005 = 0.00152

ε = 0 . 00155 ≈ 0 . 00152 OK

Solve the transverse steel strain ε t :

σ t − ρ t f t − ρ tp f tp

ε t = ε r + ε r − ε d
− σ d

Equation 14

10 − 3

= . + . × . − .

012 f t −

0 00385 0 545 (1 72 0 0)

Assume f t = 200 × 10 3 ε t before yielding

Then ε t + 1.308 ε t = 0.00385 + 0.000938

.

0 00479

2

ε t = = 0 . 00208 ≈ 0 . 00207 ( ε y )OK

.

308

The tensile strain ε r can now be checked:

Equation 15 ε r = ε + ε t − ε d = 0 . 00155 + 0 . 00208 + 0 . 0002

= 0 . 00383 ≈ 0 . 00385 (assumed) OK

Now that the strains ε d , ε r , ε , and ε t and the stress, σ d , are calculated from the trial-and-

error process, we can determine the angle α r , the shear stress τ t and the shear strain γ t :

ε t − ε d
ε − ε d =

0 . 00208 + 0 . 0002

Equation 16 tan 2 α r =

0002 =

1 . 303

0 . 00155 + 0 .

tan α r = 1 . 1414

78 ◦ 56 ◦ α r = 48 . 2 α r = 97 .

Equation 3 τ t = ( − σ d )sin α r cos α r = (7 . 43)(0 . 752)(0 . 659) = 3 . 68 MPa

γ t
2 =

Equation 6 ( ε r − ε d )sin α r cos α r = (0 . 00383 + 0 . 0002)(0 . 752)(0 . 659)

= 0 . 00200