Chapter 10, Problem 10.5.5P

Solve the preceding problem by integrating the differential equation of the deflection curve.

  

To determine

The reactions for beam at all supports.

Answer to Problem 10.5.5P

The reaction at support $A$ is $\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$ .

The reaction at support $B$ is $-\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}\left(\text{downward}\right)$ .

The reaction at support $C$ is $\frac{9\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$

Explanation of Solution

Given information:

The two spans of beam are $L$ and $\frac{L}{3}$ , flexural rigidity is $EI$ , stiffness of spring is $K$ .

Write the Expression for force equilibrium in vertical direction.

  $R_A+R_B+R_C=0$   .........(I)

Here, reaction produced at $A$ is $R_A$ , reaction produced at $B$ is $R_B$ and reaction produced at $C$ is $R_C$ .

Write the expression for moment about point $C$ .

  $\begin{array}{l}{\displaystyle \sum M_C}=0\\ R_A=\frac{1}{4}{R}_B\end{array}$   .........(II)

Here, $M_C$ is moment about point $C$ .

Write the expression for moment about point $A$ .

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ R_C=\frac{-3}{4}{R}_B\end{array}$   .........(III)

Here, $M_A$ is moment about point $A$ .

Write the expression for double order differential equation for the deflection curve when any value between $0$ to $L$ is substituted for $x$ .

  $EI{V}^{″}=R_{A}x+\frac{EI\alpha \left(T_2-T_1\right)}{h}$   .........(IV)

Here, double order differential of deflection curve is $V^{″}$ , depth of the beam is $h$ , coefficient of thermal expansion is $\alpha$ , flexural rigidity is $EI$ , upper surface temperature is $T_1$ and lower surface is $T_2$ .

Write the expression for single order derivative of deflection curve when any value between $0$ to $L$ is substituted for $x$ .

  $EI{V}^{\prime }=\frac{1}{2}{R}_A{x}^2+\frac{EI\alpha \left(T_2-T_1\right)x}{h}+C_1$   .........(V)

Here, length at which deflection has to be calculate is $x$ , single order differential of deflection curve is $V^{\prime }$ and $C_1$ is integral constant.

Write the expression for deflection curve when any value between $0$ to $L$ is substituted for $x$ .

  $EIV=\frac{1}{6}{R}_A{x}^3+\frac{EI\alpha \left(T_2-T_1\right)x^2}{2h}+C_{1}x+C_2$   .........(VI)

Here, deflection at a point $x$ is $V$ and $C_2$ is integral constant.

Write the expression for first boundary condition.

  $V\left(0\right)=0$   .........(VII)

Here, deflection when $0$ is substituted for $x$ in deflection curve is $V\left(0\right)$ .

Write the expression for second boundary condition when $L$ is substituted for $x$ in Equation (V).

  ${\left(V^{\prime }\right)}_{left}={\left(V^{\prime }\right)}_{right}$   .........(VIII)

Here, single order differential of deflection curve from the loads left to spring is ${\left(V^{\prime }\right)}_{left}$ and single order differential of deflection curve from the loads right to spring is ${\left(V^{\prime }\right)}_{right}$ .

Write the expression for third boundary condition when $L$ is substituted for $x$ in Equation (VI).

  ${\left(V^{\prime }\right)}_{left}={\left(V^{\prime }\right)}_{right}$   .........(IX)

Write the expression for fourth boundary condition.

  $V\left(\frac{4}{3}L\right)=0$   .........(X)

Here, deflection when $\frac{4}{3}L$ is substituted for $x$ in deflection curve is $V\left(\frac{4}{3}L\right)$ .

Write the expression for differential equation for the deflection curve when any value between $L$ to $\frac{4}{3}L$ is substituted for $x$ .

  $EI{V}^{″}=R_{A}x+\frac{EI\alpha \left(T_2-T_1\right)}{h}-4R_A\left(x-L\right)$   .........(XI)

Write the expression for single order derivative of deflection curve when any value between $L$ to $\frac{4}{3}L$ is substituted for $x$ .

  $EI{V}^{\prime }=-\frac{3}{2}{R}_A{x}^2+\frac{EI\alpha \left(T_2-T_1\right)}{h}x+4R_{A}Lx+C_3$   .........(XII)

Write the expression for deflection curve when any value between $L$ to $\frac{4}{3}L$ is substituted for $x$ .

  $EIV=-\frac{1}{2}{R}_A{x}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{x}^2+2R_{A}L{x}^2+C_{3}x+C_4$   .........(XIII)

Write the expression for compatibility Equation for spring when $L$ is substituted for $x$ in deflection curve.

  ${\left(V\right)}_{left}={\left(V\right)}_{right}$ = $\frac{R_B}{K}$   .........(XIV)

Substituted $0$ for $x$ and $0$ for $V$ in Equation (VI).

  $\begin{array}{l}EI\cdot 0=\frac{1}{6}{R}_A\times 0+\frac{EI\alpha \left(T_2-T_1\right)\times 0}{2h}+C_1\times 0+C_2\\ C_2=0\end{array}$   .........(XV)

Substitute $L$ for $x$ in Equation (V).

  $EI{V}^{\prime }=\frac{1}{2}{R}_A{L}^2+\frac{EI\alpha \left(T_2-T_1\right)L}{h}+C_1$

Substitute $L$ for $x$ in Equation (XII).

  $EI{V}^{\prime }=-\frac{3}{2}{R}_A{L}^2+\frac{EI\alpha \left(T_2-T_1\right)}{h}L+4R_{A}L\times L+C_3$   .........(XIV)

Substitute $\frac{1}{2}{R}_A{L}^2+\frac{EI\alpha \left(T_2-T_1\right)L}{h}+C_1$ for $EI{V}^{\prime }$ in Equation (XIV).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{2}{R}_A{L}^2+\\ \frac{EI\alpha \left(T_2-T_1\right)L}{h}+C_1\end{array}\right]=\left[\begin{array}{l}-\frac{3}{2}{R}_A{L}^2+\\ \frac{EI\alpha \left(T_2-T_1\right)}{h}L+4R_{A}L\times L+C_3\end{array}\right]\\ C_1=2R_A{L}^2+C_3\end{array}$   .........(XV)

Substitute $L$ for $x$ and $0$ for $C_2$ in Equation (VI).

  $EIV=\frac{1}{6}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)L^2}{2h}+C_{1}L$   .........(XVI)

Substitute $L$ for $x$ in Equation (XIII).

  $EIV=-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+2R_{A}L\times L^2+C_{3}L+C_4$   .........(XVII)

Substitute $-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+2R_{A}L\times L^2+C_{3}L+C_4$ for $EIV$ and $2R_A{L}^2+C_3$ for $C_1$ in Equation (XVI).

  $\begin{array}{c}-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+2R_{A}L\times L^2+C_{3}L+C_4=\frac{1}{6}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)L^2}{2h}\\ +2R_A{L}^2+C_{3}L\end{array}$

  $\begin{array}{c}{C}_4=\left(2+\frac{1}{6}-\frac{3}{2}\right)R_A{L}^3\\ =\frac{2}{3}{R}_A{L}^3\end{array}$   .........(XVIII)

Substitute $\frac{4}{3}L$ for $x$ , $0$ for $EIV$ and $\frac{2}{3}{R}_A{L}^3$ for $C_4$ in Equation (XIII).

  $0=-\frac{1}{2}{R}_A{\left(\frac{4}{3}L\right)}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{\left(\frac{4}{3}L\right)}^2+2R_{A}L{\left(\frac{4}{3}L\right)}^2+C_3\frac{4}{3}L+\frac{2}{3}{R}_A{L}^3$

  $0=-\frac{64}{54}{R}_A{L}^3+\frac{8EI\alpha \left(T_2-T_1\right)L^2}{9h}+\frac{32}{9}{R}_A{L}^3+C_3\left(\frac{4}{3}L\right)+\frac{2}{3}{R}_A{L}^3$

  $C_3=-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]$   .........(XIX)

Substitute $-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]$ for $C_3$ in Equation (XV).

  $C_1=2R_A{L}^2-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]$   .........(XX)

Substitute $2R_A{L}^2-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]$ for $C_1$ and $0$ for $C_2$ and $L$ for $x$ in Equation (VI).

  $EIV=\left[\begin{array}{l}\frac{1}{6}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)L^2}{2h}+\\ \left[2R_A{L}^2-\left[\begin{array}{l}\frac{82}{27}{R}_A{L}^3+\\ \frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\end{array}\right]\left[\frac{3}{4L}\right]\right]L\end{array}\right]$   .........(XXI)

Substitute $-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]$ for $C_3$ , $\frac{2}{3}{R}_A{L}^3$ for $C_4$ and $L$ for $x$ in Equation (XVII).

  $\begin{array}{l}EIV=-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+\\ 2R_{A}L\times L^2+-\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\left[\frac{3}{4L}\right]L+\frac{2}{3}{R}_A{L}^3\end{array}$   .........(XXII)

Substitute $\left[\frac{1}{6}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)L^2}{2h}+\left[2R_A{L}^2-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]\right]L\right]\frac{1}{EI}$ for ${\left(V\right)}_{left}$ in Equation (XIV).

  $\left[\begin{array}{l}\frac{1}{6}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)L^2}{2h}+\\ \left[2R_A{L}^2-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]\right]L\end{array}\right]\frac{1}{EI}=\frac{R_B}{K}$   .........(XXIII)

Substitute $\left[-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+2R_{A}L\times L^2+-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]L+\frac{2}{3}{R}_A{L}^3\right]\frac{1}{EI}$ for ${\left(V\right)}_{right}$ in Equation (XIV).

  $\begin{array}{l}-\frac{1}{2}{R}_A{L}^3+\frac{EI\alpha \left(T_2-T_1\right)}{2h}{L}^2+2R_{A}L\times L^2+\\ \left[-\left[\frac{82}{27}{R}_A{L}^3+\frac{8}{9}\frac{EI\alpha \left(T_2-T_1\right)L^2}{h}\right]\left[\frac{3}{4L}\right]L\right]+\frac{2}{3}{R}_A{L}^3=\frac{R_B}{K}\end{array}$   .........(XXIV)

Solve Equation (XXIII) and Equation (XXIV).

  $R_B=\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}\left(\text{downward}\right)$   .........(XXV)

  $R_A=\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$   .........(XXVI)

Substitute $\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$ for $R_A$ and $\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}\left(\text{downward}\right)$ for $R_B$ considering sense of direction in Equation (I).

  $\begin{array}{l}\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}-\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}+R_C=0\\ R_C=\frac{9\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)\end{array}$   .........(XXVII)

Conclusion:

The reaction at support $A$ is $\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$ .

The reaction at support $B$ is $\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}\left(\text{downward}\right)$ .

The reaction at support $C$ is $\frac{9\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\left(\text{upward}\right)$ .

To determine

The reactions at all the supports when value of $K$ limits to $\infty$ .

Answer to Problem 10.5.5P

The reaction at support $A$ is $\frac{3EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)$ .

The reaction at support $B$ is $\frac{6EI\alpha \left(T_1-T_2\right)}{Lh}\left(\text{downward}\right)$ .

The reaction at support $C$ is $\frac{9EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)$ .

Explanation of Solution

Take limit of $K$ upto $\infty$ in Equation (XXVI).

  $\begin{array}{c}{R}_A=\underset{K\to \infty }{\lim }\frac{3\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\\ =\frac{3EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)\end{array}$ ...... (XXVIII)

Take limit of $K$ upto $\infty$ in Equation (XXV).

  $\begin{array}{c}{R}_B=\underset{K\to \infty }{\lim }\frac{\alpha \left(T_1-T_2\right)L^{2}6EIK}{h\left(36EI+L^{3}K\right)}\\ =\frac{6EI\alpha \left(T_1-T_2\right)}{Lh}\left(\text{downward}\right)\end{array}$ ...... (XXIX)

Take limit of $K$ upto $\infty$ in Equation (XXVII).

  $\begin{array}{c}{R}_C=\underset{K\to \infty }{\lim }\frac{9\alpha \left(T_1-T_2\right)L^{2}EIK}{2h\left(36EI+L^{3}K\right)}\\ =\frac{9EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)\end{array}$ ...... (XXX)

Conclusion:

The reaction at support $A$ is $\frac{3EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)$ .

The reaction ay support $B$ is $\frac{6EI\alpha \left(T_1-T_2\right)}{Lh}\left(\text{downward}\right)$ .

The reaction at support $C$ is $\frac{9EI\alpha \left(T_1-T_2\right)}{2Lh}\left(\text{upward}\right)$ .