3.1 Web Shear Element
As stated before, the web shear element can be treated as a membrane element subjected to pure shear force (1 − α)V. After diagonal cracking, the concrete is separated by diagonal cracks into a series of concrete struts as shown in Fig. 6a. The shear force produces a set of biaxial stress field constituted with the compressive principal stress f_{
2
} and the tensile principal stress f_{
1
} with inclination angle of θ as shown in Fig. 6. The reinforcing bars are assumed to be a link member transmitting axial force only. The behavior of such a membrane element is well explained by either Collins’ MCFT (Vecchio and Collins 1986) or Hsu’s STM (Hsu 1993).
The equilibrium conditions, which relate the concrete stresses and the stirrup stress to the applied shear stress, can be expressed in terms of average stresses. These relationships are derived from the Mohr’s circle shown in Fig. 6. These are
$$ f_{2} = (1  \alpha )v\left( {\tan \theta + \cot \theta } \right)  f_{1} $$
(8)
$$ f_{t} = (1  \alpha )v\tan \theta  f_{1} $$
(9)
$$ f_{x} = (1  \alpha )v\cot \theta  f_{1} $$
(10)
where v is an average shear stress defined by V/b_{
w
}z, f_{2} is positive in compression, and f_{1} is positive in tension. From the vertical force equilibrium the transverse concrete stress f_{
t
} of Eq. (9) must be balanced by the stirrup stress:
$$ \rho_{v} f_{v} = \left( { 1  \alpha } \right)v { \tan } \theta  f_{ 1} ,{\text{ for }}\rho_{\text{v}} > 0 $$
(11a)
$$ 0 = \left( { 1  \alpha } \right)v { \tan } \theta  f_{ 1} ,{\text{ for }}\rho_{\text{v}} = 0 $$
(11b)
In the same manner, the longitudinal concrete stress f_{
x
} of Eq. (10) is balanced by the chords and the horizontal web steels, if any. When the horizontal web steels are not provided, it is transferred to the top and bottom chord. The equilibrium of the resultants on the right face is shown by the force polygon in Fig. 6b, and the longitudinal compressive resultant N_{
b
} produced by the f_{
x
} is
$$ N_{b} = f_{x} b_{w} z = (1  \alpha )V\cot \theta  f_{1} b_{w} z $$
(12)
The acting point of this resultant will be z/2 from the center of the bottom chord. The average stress and average strain of the concrete in each principal direction (axis 1–2 in Fig. 6a) is assumed to obey the material laws developed by Vecchio and Collins (Vecchio and Collins 1986), as summarized in Figs. 7a and 7b. Thus, after diagonal cracking with taking f_{
cr
} = 0.33 \( \sqrt {f_{c}^{'} } \) (MPa), for principal tensile direction;
$$ \varepsilon_{1} = 0.002\left( {\frac{{0.33\sqrt {f_{c}^{'} } }}{{f_{1} }}  1} \right)^{2} $$
(13)
For principal compressive direction with taking ɛ_{
co
} = 0.002;
$$ \varepsilon_{2} = 0.002\left( {1  \sqrt {1  (0.8 + 170\varepsilon_{1} )f_{2} /f_{c}^{'} } } \right) $$
(14)
In beams having vertical stirrups, the transverse average strain ɛ_{
t
} can be approximately evaluated using CEB/FIP MC90 (1990) the equation for tension stiffening effect, which is shown in Fig. 7c. Replacing f_{
s
} by f_{
v
} of Eq. (11), then ɛ_{
sm
} corresponds to ɛ_{
t
}. Thus the following relationship is obtained:
$$ \varepsilon_{t} = \frac{1}{{E_{s} \rho_{v} }}\left[ {(1  \alpha )v\tan \theta  \left( {f_{1} + 0.132\sqrt {f_{c}^{'} } } \right)} \right] $$
(15)
From the compatibility condition satisfying Mohr’s strain circle shown in Fig. 7d, the average shear strain γ_{
w
}of the web shear element is expressed in terms of ɛ_{1}, ɛ_{2}, and ɛ_{
t
} as follows:
$$ \gamma_{w} = (\varepsilon_{1} + \varepsilon_{2} )\sin 2\theta $$
(16a)
$$ = 2(\varepsilon_{t} + \varepsilon_{2} )\tan \theta $$
(16b)
Substituting Eqs. (13), (14) and (15) for ɛ_{1}, ɛ_{2} and ɛ_{
t
} respectively in Eq. (16), the shear strain γ_{
w
} of the web element is eventually expressed in terms of f_{1}, f_{2} and θ.
3.2 Redistribution of Resultants in a Section
As previously stated, the vertical component of the axial force in the inclined top chord is the shear carried by the arch action, and it is designated by αV. This shear force will be combined with flexural compression resultant C, eventually leading to form an inclined thrust line (arch). For the extension of the present approach to a simple strutand tie model, it may be technically required that the shear force αV is accompanied by the longitudinal resultant N_{
a
}, and whose acting point is temporarily denoted by z_{
a
} as shown in Fig. 8. That is,
$$ N_{a} = \alpha V\cot \theta $$
(17)
It has been known that significant longitudinal forces are caused by each base action. By such induction of the longitudinal forces at a cross section, the internal couple C_{
o
} and T_{
o
} calculated from the conventional beam theory due to the bending moment is redistributed to meet the equilibrium condition over the section. Figure 9 shows the forces N_{
a
} and N_{
b
} acting on the points z_{
a
} and 0.5z from the reinforcement level in addition to the internal force couple C_{
o
} − ΔC and T_{
o
} + ΔT. To satisfy the equilibrium condition in longitudinal direction, the change in the compression resultant ΔC and that in steel tension ΔT are seen to be
$$ \Updelta C = \frac{1}{{z_{o} }}(N_{a} z_{a} + 0.5N_{b} z) $$
(18a)
$$ \Updelta T = \frac{1}{{z_{o} }}\left[ {N_{a} (z_{o}  z_{a} ) + N_{b} (z_{o}  0.5z)} \right] $$
(18b)
The redistributed compression resultant C_{
o
} − ΔC and the longitudinal force due to the arch action N_{
a
} should be combined to form a final compression resultant C in a section. Hence, as shown in Fig. 9, the final acting point z will be
$$z = \frac{(C_{o}  \Updelta C)z_{o} + N_{a} z_{a}} {C_{o}  \Updelta C + N_{a} } $$
(19)
The lever arm length z above is established on the basis of the force equilibrium conditions at a section. Whereas the z defined by Eq. (7) is mathematically derived on the basis of the assumption of constant α throughout the shear span. These values of z must be identical to each other. By equating these two equations with substituting Eqs. (12), (17) and (18) and the approximation of z_{
a
} ≅ z, the inclined angle θ is expressed in terms of α and f_{1}:
$$ \cot \theta_{x} = \frac{{\frac{{M_{x} }}{{z_{o} }}\left( {\frac{1}{{(R_{x} )^{\alpha } }}  1} \right) + f_{1} b_{w} z_{o} (R_{x} )^{\alpha } \left( {1.5  0.5(R_{x} )^{\alpha } } \right)}}{{V_{x} \left( {\alpha \left\{ {1  (R_{x} )^{\alpha } } \right\} + (1  \alpha )\left\{ {1.5  0.5(R_{x} )^{\alpha } } \right\}} \right)}} $$
(20)
where R_{
x
} = M_{
x
}/M_{max}, that is the sectional moment ratio with respect to the maximum moment of a beam.
From the moment equilibrium at a section with z_{
a
} ≅ z, the final compression force C in the top chord and the tension force T in the bottom chord should be
$$ C = (C_{o}  \Updelta C)\frac{{z_{o} }}{z} + N_{a} = \frac{M}{z}  \frac{{N_{b} }}{2} $$
(21a)
$$ T = T_{o} + \Updelta T + (C_{o}  \Updelta C)\frac{{z_{o}  z}}{z} = \frac{M}{z} + \frac{{N_{b} }}{2} $$
(21b)
Figure 10 shows the final configuration of the resultants acting at a section. From the figure, it can be realized that the equilibriums in both the shear and the moment are simultaneously satisfied at the section. Also, the slope of the thrust in the top chord is V_{
a
}/C, which is equivalent to dz/dx in Fig. 2c. As the section approaches toward the support, the slope dz/dx becomes steeper because C becomes smaller owing to the smaller sectional moment. This is the primary reason for which a tiedarch action is formed in cracked reinforced concrete beams.
3.3 Relative Displacements of Top and Bottom Chord
As described before, the relative displacements of the top and bottom chords u_{
m
} and u_{
n
} can be evaluated by Eq. (3). Utilizing Eqs. (2) and (7) together with Fig. 4, dC/dx and dT/dx at each section are simply expressed in terms of α and V_{
x
}:
$$ \frac{dC}{dx} = \frac{dT}{dx} = \frac{{(1  \alpha )V_{x} }}{{z_{o} (R_{x} )^{\alpha } }} $$
(22)
For the evaluation of the axial stiffness of the top and bottom chord, some approximations are required because the actual stress distributions are relatively complex. It may be reasonable estimation that the effective depth of the compression chord is equal to the depth of the rectangular compression stress block and remains constant along the span as sketched in Fig. 11. With this approximation and the substitution of Eq. (22), Eq. (3a) becomes
$$ u_{m} = \int\limits_{a}^{x} {\frac{1}{{E_{c} A_{tc} }}\left( {\int_{a}^{x} {\frac{{(1  \alpha )V_{x} }}{{z_{o} (R_{x} )^{\alpha } }}dx} } \right)dx} = \frac{Va}{{z_{o} }}\frac{x}{{E_{c} A_{tc} (2  \alpha )}}\left[ {1  \left( {R_{x} } \right)^{1  \alpha } } \right] $$
(23a)
where E_{
c
} is the concrete elastic modulus, and A_{
tc
} is the effective area of the top chord estimated based on the effective depth defined above. The axial deformation of the tension chord due to dT/dx, which corresponds to the relative displacement u_{
n
}, is approximately evaluated using CEB/FIP MC90 the tension stiffening effect expression (Fig. 7c). In applying this formula to a beam, the height of the effective tensile tie is normally assumed to be about 2.5(h − d) as shown in Fig. 11 (CEBFIP 1990). Hence, Eq. (3b) can be rewritten as follow:
$$ u_{n} = \frac{Va}{{z_{o} }}\frac{x}{{E_{s} A_{s} (2  \alpha )}}\left[ {1  \left( {R_{x} } \right)^{1  \alpha } } \right]  x\left( {\frac{{0.13\sqrt {f_{ck} } }}{{E_{s} \rho_{eff} }}} \right) \ge 0 $$
(23b)
where \( \rho_{eff} = \frac{{A_{s} }}{b \cdot 2.5(h  d)} \) and x is distance from the support.
3.4 Solution Algorithm
All of the relationships required to determine the value of α have been discussed above. For a specific section in a beam with vertical stirrups at a given load, a suitable iterative procedure is as follows:

Step1: Assume a value of α

Step2: Choose a value of ɛ_{1}, then calculate f_{1} from Eq. (13)

Step3: Calculate z from Eq. (7), and θ from Eq. (20)

Step4: Calculate f_{2} from Eq. (8), and f_{
v
} from Eq. (11)

Step5: Calculate ɛ_{2} from Eq. (14), and ɛ_{
t
} from Eq. (15)

Step6: Calculate γ_{
w
}from Eq. (16a) and γ_{
w
} from Eq. (16b)
Then, check that γ_{
w
}from Eq. (16a) = γ_{
w
} from Eq. (16b) or not.
If γ_{
w
}from Eq. (16a) ≠ γ_{
w
} from Eq. (16b), return to Step2.
If γ_{
w
}from Eq. (16a) ≅ γ_{
w
} from Eq. (16b), go to Step7.

Step7: Calculate u_{
m
}from Eq. (23a), and u_{
n
} from Eq. (23b)

Step8: Check that γ_{
w
} = (u_{
m
} + u_{
n
})/z or not.
If γ_{
w
} ≠ (u_{
m
} + u_{
n
})/z, return to Step1 and repeat
If γ_{
w
} ≅ (u_{
m
} + u_{
n
})/z, take the last assumed α value.
With the help of a spread sheet calculator, the procedure above is easily performed. As a tip for fast iteration, it is recommended to use the inverse value of the shear spantodepth ratio of the beam considered as an initial value of α in Step1.