## 3 First-Order Theory of Beams (TBT)
### Exact Solutions  
#### 3.5.1 General Solution  
In this section we present exact solutions to the _linear_ equations of equilibrium of FGM beams without the modified couple stress effect. By setting \(f=0\) in Equations (50)-(52), we obtain  
\[-\frac{d}{dx}\bigg{[}A_{xx}\bigg{(}\frac{du}{dx}\bigg{)}+B_{xx} \frac{d\phi_{x}}{dx}\bigg{]}=f, \tag{53}\] \[-\frac{d}{dx}\bigg{[}S_{xz}\bigg{(}\phi_{x}+\frac{dw}{dx}\bigg{)} \bigg{]}-q=0,\] (54) \[-\frac{d}{dx}\bigg{(}B_{xx}\frac{du}{dx}+D_{xx}\frac{d\phi_{x}}{ dx}\bigg{)}+S_{xz}\bigg{(}\phi_{x}+\frac{dw}{dx}\bigg{)}=0. \tag{55}\]  
Again, we further assume that the beam stiffness coefficients are all constant and \(f=0\).  
Equations (53)-(55), when expressed in terms of the stress resultants \(N_{xx}\), \(N_{xz}\), and \(M_{xx}\) (see Equations (43)-(45) with \(P_{xy}=0\)) take the following form:  
\[\frac{dN_{xx}}{dx}=0,\ \ -\frac{dN_{xz}}{dx}-q=0,\ \ -\frac{dM_{xx}}{dx}+N_{xz}=0 \tag{56}\]  
Substituting for \(N_{xz}\) from the third equation into the second equation, we obtain  
\[\frac{dN_{xx}}{dx}=0,\ \ -\frac{d^{2}M_{xx}}{dx^{2}}-q=0. \tag{57}\]  
Integrating the above equations,  
\[N_{xx}=c_{1},\ \ \frac{dM_{xx}}{dx}=-\int^{x}q(\xi)d\xi+c_{2} \tag{58}\]  
Integrating the second equation, we obtain  
\[M_{xx}=-\int^{x}\int^{\xi}q(\eta)\,d\eta d\xi+c_{2}x+c_{3}\equiv F(x), \tag{59}\]  
Here \(c_{i}\) (\(i=1,2,3\)) denote the constants of integration.  
The left sides of Equations (58) and (59) can be expressed in terms of the displacements \((u,\phi_{x})\) using Equations (48a) and (48b); we have  
\[\left[\begin{array}{cc}A_{xx}&B_{xx}\\ B_{xx}&D_{xx}\end{array}\right]\left\{\begin{array}{cc}\frac{du}{dx}\\ \frac{du}{dx}\end{array}\right\}=\left\{\begin{array}{cc}c_{1}+N_{xx}^{T} \\ F(x)+M_{xx}^{T}\end{array}\right\} \tag{60}\]  
Solving for \(du/dx\) and \(d\phi_{x}/dx\), we obtain  
\[\frac{du}{dx} =\frac{1}{D_{xx}^{*}}[D_{xx}c_{1}-B_{xx}F(x)] \tag{61}\] \[\frac{d\phi_{x}}{dx} =\frac{1}{D_{xx}^{*}}[-B_{xx}c_{1}+A_{xx}F(x)] \tag{62}\]  
where  
\[D_{xx}^{*}=A_{xx}D_{xx}-B_{xx}^{2},\;\;F(x)=-\int^{x}\int^{\xi}q(\eta)\,d\eta \,d\xi+c_{2}x+c_{3} \tag{63}\]  
Integrating Equations (61) and (62), we obtain  
\[u(x) =\frac{D_{xx}}{D_{xx}^{*}}c_{1}\,x+\frac{B_{xx}}{D_{xx}^{*}}\int^ {x}\int^{\xi}\int^{\eta}q(\zeta)\,d\xi\,d\eta\,d\xi\] \[\qquad-\frac{B_{xx}}{D_{xx}^{*}}\bigg{(}c_{2}\frac{x^{2}}{2}+c_{3 }x+c_{4}\bigg{)}, \tag{64}\] \[\phi_{x}(x) =-\frac{B_{xx}}{D_{xx}^{*}}c_{1}\,x-\frac{A_{xx}}{D_{xx}^{*}}\int ^{x}\int^{\xi}\int^{\eta}q(\zeta)\,d\zeta\,d\eta d\xi\] \[\qquad+\frac{A_{xx}}{D_{xx}^{*}}\bigg{(}c_{2}\frac{x^{2}}{2}+c_{3 }x+c_{5}\bigg{)}. \tag{65}\]  
From Equations (56) and (58), we arrive at  
\[N_{xz}=\frac{dM_{xx}}{dx}=-\int^{x}q(\xi)\,d\xi+c_{2} \tag{66}\]  
and using Equation (48d) we obtain  
\[\frac{dw}{dx} =\frac{1}{S_{xz}}\bigg{(}-\int^{x}q(\xi)\,d\xi+c_{2}\bigg{)}-\phi _{x}\] \[=\frac{1}{S_{xz}}\bigg{(}-\int^{x}q(\xi)\,d\xi+c_{2}\bigg{)}+ \frac{B_{xx}}{D_{xx}^{*}}c_{1}\,x\] \[\quad-\frac{A_{xx}}{D_{xx}^{*}}\bigg{(}-\int^{x}\int^{\xi}\int^{ \eta}q(\zeta)\,d\zeta\,d\eta d\xi+c_{2}\frac{x^{2}}{2}+c_{3}x+c_{5}\bigg{)} \tag{67}\]  
or  
\[w(x) =\frac{1}{S_{xz}}\bigg{(}-\int^{x}\int^{\xi}q(\eta)\,d\eta\,d\xi+ c_{2}x\bigg{)}+\frac{B_{xx}}{D_{xx}^{*}}c_{1}\,\frac{x^{2}}{2}\] \[\qquad-\frac{A_{xx}}{D_{xx}^{*}}\bigg{(}-\int^{x}\int^{\xi}\int^{ \eta}\int^{\xi}q(\mu)\,d\mu\,d\zeta\,d\eta d\xi+c_{2}\frac{x^{3}}{6}+c_{3} \frac{x^{2}}{2}+c_{5}x+c_{6}\bigg{)} \tag{68}\]  
The six constants of integration are determined using six boundary conditions, three at each end of the beam. One (and only one) element of the each of three duality pairs at each boundary point must be known (see Equation (43)): \((u,N_{xx})\), \((w,N_{xz})\), and \((\phi_{x},M_{xx})\). We note that, in TBT, \(\phi_{x}\) has replaced \(-dw/dx\) as the primary variable, and it is dual to the bending moment \(M_{xx}\). _One should not specify \(dw/dx\) in place of \(\phi_{x}\) in the TBT_. The stress resultants \((N_{xx},M_{xx},P_{xy},N_{xz})\) can be computed with the help of Equations (48a)-(48d).