# Analysis of Equivalent Flexural Stiffness of Steel-Concrete Composite Beams in Frame Structures
## 5 Simplified Design Method
### Simplified Method for Calculating Rotation Constraint Stiffness  
The stiffness of the rotation constraints at the beam ends can be directly reflected by the bending moment distribution on the composite beam. As a rule, strong rotation constraints at the beam ends lead to large negative moment regions on the composite frame beam. Therefore, the rotation constraint stiffness can be represented by the length of the negative moment region by identifying the distribution trend of the internal force. For a composite beam with a uniform flexural stiffness distribution and no concrete slab cracking, the reaction force at the beam support and the distributions of the bending moments, rotations, and vertical deflections on the composite beam can be derived using the theoretical model described in Section 2. These are expressed as follows:  
\[F_{1}=\frac{1}{2}ql+\frac{k_{1}\theta_{1}+k_{2}\theta_{2}}{l}, \tag{19}\]  
Figure 12: Comparison between the calculation results of the proposed formula and the FE analysis. ((**a**) Load – deflection curve of 6-m-span frame at middle span in 1 st story; (**b**) Load –deflection curve of 6-m-span frame at side span in 4 th story; (**c**) Comparison of deflection results between proposed formula and FE models). \[M(x)=F_{1}x-k_{1}\theta_{1}-\frac{1}{2}qx^{2}, \tag{20}\]  
\[EI^{+}\theta(x)=-\int M(x)dx=-\frac{1}{2}F_{1}x^{2}+k_{1}\theta_{1}x+\frac{1}{6}qx ^{3}+C_{1},\text{ and} \tag{21}\]  
\[EI^{+}w(x)=\int\theta(x)dx=-\frac{1}{6}F_{1}x^{3}+\frac{1}{2}k_{1}\theta_{1}x^ {2}+\frac{1}{24}qx^{4}+C_{1}x+C_{2}, \tag{22}\]  
where the definitions of the notations are consistent with those specified in Section 2. For frame beams with constant sectional stiffness, the boundary condition functions at the beam ends can be expressed as  
\[\left\{ \begin{array}{l}\theta(0)=\theta_{1}\\ \theta(l)=\theta_{2}\\ w(0)=0\\ \theta(l)=0\end{array} \right.. \tag{23}\]  
By substituting the boundary condition functions into Equations (21) and (22), \(\theta_{1}\) and \(\theta_{2}\) can be expressed as  
\[\left\{ \begin{array}{l}\theta_{1}=\frac{q^{3}\left(6 EI^{+}+k_{2}l\right)}{1 2\left[12\left(EI^{+}\right)^{2}+4 EI^{+}(k_{1}+k_{2})l+k_{1}k_{2}l^{2}\right] }\\ \theta_{1}=-\frac{q^{3}\left(6 EI^{+}+k_{1}l\right)}{12\left[12\left(EI^{+} \right)^{2}+4 EI^{+}(k_{1}+k_{2})l+k_{1}k_{2}l^{2}\right]}\end{array} \right.. \tag{24}\]  
Subsequently, the internal force distribution functions listed in Equations (20)-(23) can be solved. To build the relationship between the rotation constraint stiffness and the bending moment distribution, two additional parameters \(\alpha_{cr}\) and \(l_{cr}\) are introduced to describe the length of a negative moment region, which are defined as follows:  
\[\left\{ \begin{array}{l}\alpha_{cr}=\frac{l_{cr1}+l_{cr2}}{l}\\ l_{cr}=l_{cr1}+l_{cr2}=\alpha_{cr}l\end{array} \right., \tag{25}\]  
where \(l_{cr1}\) and \(l_{cr2}\) are the lengths of the negative moment regions located on the composite beam at the left and right ends, respectively; \(l_{cr}\) is the total length of the negative moment region; and \(l\) is the length of the composite beam span. Based on the definitions of \(l_{cr1}\) and \(l_{cr2}\), the following relationship should be satisfied:  
\[M(l_{cr1})=M(l-l_{cr2})=0 \tag{26}\]  
Thus, the rotation restraint stiffness can be expressed as  
\[\left\{ \begin{array}{l}k_{1}=\frac{12 EI^{+}l_{cr1}(l-l_{cr2})}{l(l^{2}-2 ll_{ cr2}-4 ll_{cr1}+6l_{cr1}l_{cr2})}\\ k_{2}=\frac{12 EI^{+}l(-l_{cr1})l_{cr2}}{l(l^{2}-2 ll_{cr1}-4 ll_{cr2}+6l_{cr1}l_{ cr2})}\end{array}