Section 11.3.6.4 Determination of β and θ

ε_x is originally calculated using the flexural steel calculated due to only the flexural/axial demand and without shear and torsion tension included.

This value of ε_x will be used to calculate θ and β used throughout the shear and torsion calculations. Additionally the longitudinal flexural/axial steel design will be performed including the shear and torsion tension, using the calculated value of ε_x, θ, and β to determine the shear and torsion tension. A longitudinal steel design is also performed including shear and torsion tension to limit ε_x to a maximum of the calculated value.

ε_x is limited -0.2×10^-3 and 3.0×10^-3.

The angle of inclination θ is calculated using equation 11-12:

θ = 29 + 7,000ε_x

The value of β is calculated using Equation 11-11:

β = \frac{0.40}{(1 + 1, 500 ε_{x})} \frac{1, 300}{(1, 000 + s_{z e})}

Initially it is assumed that no shear reinforcement required to calculate crack spacing parameter S_ze as per equation 11-10.

S_{z e} = \frac{35 s_{z}}{15 + a_{g}}

If f’_c exceeds 70 Mpa, the a_g shall be taken as "0 and from 60 to 70 Mpa, a_g shall be linearly reduced to zero".

The crack spacing parameter, s_z, is calculated as s_z = dv = max (0.9d, 0.72h)

A single layer of reinforcement is assumed.

If it is determined that shear reinforcement is required, than the crack spacing parameter S_ze will considered as 300 mm as per section 11.3.6.4 and the design restarted.