Rotating Angle Shear Theories - Unified Theory of Concrete Structures - page 167

τ t are obtained from the global analysis of a structure,
and are considered given values in the design and analysis of a 2-D element. Therefore, two
additional variables must be given before the problem can be solved. Depending on the two
additional given variables, the problem is divided into two types:

In practice, the three stresses σ , σ t ,

Types of problem Given variables Unknown variables

(a) Analysis: σ σ t τ t ρ ρ t f f t σ d ε ε t γ t ε r ε d α r

(b) Design: σ σ t τ t f f t ρ ρ t σ d ε ε t γ t ε r ε d α r

The problems in the analysis and design of 2-D elements boil down to finding the most
efficient way to solve the nine equations. To demonstrate the methodology of the solution
process, the nine equations will first be applied to the analysis problem in Sections 5.3.3 and
5.3.4. This will be followed by the application of the nine equations to the design problem in
Section 5.3.5.

5.3.3 Solution Algorithm
In order to analyze the stresses and strains in the 2-D element shown in Figure 5.3, the
five variables given are the three externally applied stresses σ , σ t and τ t and the two steel

percentages ρ t . The nine equations (5.67)-(5.75) will then be used to solve the remaining

nine unknown variables, including the two steel stresses ( f

ρ and

and f t ), one concrete stress ( ε d ),

the five strains ( ε , ε t , γ t , ε r and ε d ), plus the angle α r . The problem is summarized below:

Given five variables: ρ t (see Figure 5.3(a))

Find nine unknown variables: f , f t ,

σ , σ t , τ t , ρ ,

σ d , ε , ε t , γ t , ε r , ε d , α r

From the three equations (5.67)-(5.69), the stresses in the longitudinal steel f , transverse
steel f t , and concrete struts

σ d , can be expressed as follows:

σ + τ t tan α r

f = (5.77)

ρ

σ t + τ t cot α r

f t = (5.78)

ρ t

− τ t

σ d = (5.79)

sin α r cos α r

Substituting the stress-strain relationships from Equations (5.73), (5.74) and (5.75) into
Equations (5.77), (5.78) and (5.79), respectively, we obtain the strains for the longitudinal
steel

ε , transverse steel ε t , and concrete element, ε d :

σ + τ t tan α r

ε = (5.80)

ρ E s

σ t + τ t cot α r

ε t = (5.81)

ρ t E s

1
E c

τ t

( − ε d ) = (5.82)

α r cos α r

sin