Rotating Angle Shear Theories - Unified Theory of Concrete Structures - page 172

Below the first yield load stage the steel and concrete are assumed to obey Hooke's law
and the linear behavior of the 2-D element can be predicted by the Mohr compatibility truss
model . The angle α r of 37.54 ◦ is constant from the cracking load up to the first yield load.

With the application of increasing proportional stresses, however, Hooke's law will no
longer be applicable to steel and concrete, and the angle α r will rotate and decrease. The

nonlinear behavior of the 2-D element can then be predicted by the r otating angle softened
truss model given in Section 5.4. This rotating angle α r will reach a final value of 30 ◦ when

both the longitudinal and the transverse steel yield. This plastic load stage at total yield has
been treated in Section 5.1.5, Example 5.1, using the e quilibrium (plasticity) truss model .

5.3.5 Allowable Stress Design of RC 2-D Elements
We will now study the design of reinforcement in the RC 2-D element shown in Figure 5.9.
The five variables given in this case are the three applied stresses

τ t , and the two steel
stresses f and f t . For design purposes, f and f t will be specified as the allowable stresses
f a and f ta . According to the ACI Code Commentary f a and f ta are 138 MPa (20 ksi) for
Grade 40 and 50 steel and 165 MPa (24 ksi) for Grade 60 and higher strength steel.
The nine equations (5.67)-(5.75), will then be used to solve the remaining nine unknown
variables, including the two steel ratios

σ , σ t ,

ρ and ρ t , the one concrete stress σ d , the five strains

ε , ε t , γ t , ε r , ε d , plus the angle α r . The problem is summarized below:

Given five variables: σ , σ t , τ t , f = f a , f t = f ta

Find nine unknown variables: ρ , ρ t , σ d , ε , ε t , γ t , ε r , ε d , α r

Similar to the analysis problem in Sections 5.3.3 and 5.3.4, the solution of the nine equations
starts with expressing the three equilibrium equations (5.21)-(5.23) by the second type of
equations and setting σ r = 0:

σ + τ t tan α r

ρ = (5.85)

f a

σ t + τ t cot α r

ρ t = (5.86)

f ta

− τ t

σ d = (5.87)

sin α r cos α r

From the linear stress-strain relationships of steel (Equations 5.73 and 5.75), we have

f a
E s

ε = (5.88)

f ta
E s

ε t = (5.89)

σ d
E c

ε d = (5.90)

Substituting Equation (5.87) into the stress-strain equation of compression concrete, (5.90),
gives:

1
E c

− τ t

ε d = (5.91)

sin α r cos α r