Chapter 10, Problem 10.4.11P

The continuous frame ABCD has a pin support at B: roller supports at A_,C, and D; and rigid corner connections at B and C (see figure). Members AB, BC, and CD each have flexural rigidity EL Moment M₀acts counterclockwise at B and clockwise at C. Note: Disregard axial deformations in member A Band consider only the effects of bending.

1. Find all reactions of the frame.

Find joint rotations $\theta$

al A, B. C, and D.

Repeat parts (a) and (b) if both moments M₀are counter clockwise.

  

To determine

All the reactions of the frame.

Answer to Problem 10.4.11P

The reactions are : $H_A=-H_D=\frac{3M_0}{2L},H_B=V_B=V_C=0.$

Explanation of Solution

Given Information:

The following figure is given with all relevant information,

Calculation:

Consider the following free body diagram,

Take equilibrium of horizontal forces as,

$\begin{array}{l}{\displaystyle \sum F_H}=0\\ \Rightarrow H_A+H_B+H_D=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of vertical forces as,

$\begin{array}{l}{\displaystyle \sum F_V}=0\\ \Rightarrow V_B+V_C=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of moments about B as,

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ \Rightarrow M_0-H_{A}L/2-M_0+V_{C}L-H_{D}L/2=0\mathrm{....}\text{(3)}\end{array}$

The bending moment at distance x from A along AB in part AB is given by,

$M_{AB}=H_{A}x$

Use second order deflection differential equation,

$\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{AB}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_A{x}_0)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{AB}=\frac{1}{EI}(H_A{x}^2/2+C_1)$

Integrate above equation to get rotation as,

$v={\delta }_{AB}=\frac{1}{EI}(H_A{x}^3/6+C_{1}x+C_2)$

The bending moment at distance x from B along BC in part BC is given by,

$M_{BC}=H_{A}L/2-M_0+V_{B}x$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{BC}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{A}L/2-M_0+V_{B}x)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{BC}=\frac{1}{EI}(H_{A}Lx/2-M_{0}x+V_B{x}^2/2+C_3)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_{A}L{x}^2/4-M_0{x}^2/2+V_B{x}^3/6+C_{3}x+C_4)$

The bending moment at distance x from C along CD in part CD is given by,

$M_{CD}=H_{D}x-H_{D}L/2$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{CD}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{D}x-H_{D}L/2)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{CD}=\frac{1}{EI}(H_D{x}^2/2-H_{D}Lx/2+C_5)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_D{x}^3/6-H_{D}L{x}^2/4+C_{5}x+C_6)$

The constraint equations are,

$\begin{array}{l}{\delta }_{AB}(0)=0\mathrm{...}\text{(4)}\\ {\theta }_{AB}(L/2)={\theta }_{BC}\text{(0) }\mathrm{...}\text{(5)}\\ {\delta }_{BC}(0)=0\mathrm{...}\text{(6)}\\ {\delta }_{BC}(L)=0\mathrm{..}\text{(7)}\\ {\theta }_{BC}(L)={\theta }_{CD}\text{(0) }\mathrm{..}\text{(8)}\\ {\delta }_{CD}(L/2)=0\mathrm{..}\text{(9)}\\ {\delta }_{AB}(L/2)=0\mathrm{...}\text{(10)}\end{array}$

Solve equations (1-10) to get integration constants and reactions.

So the reactions are $H_A=-H_D=\frac{3M_0}{2L},H_B=V_B=V_C=0.$

Conclusion:

Therefore, the reactions are: $H_A=-H_D=\frac{3M_0}{2L},H_B=V_B=V_C=0.$

To determine

Rotations at A, B, C and D.

Answer to Problem 10.4.11P

Rotations at A, B, C and D are ${\theta }_A=\frac{-M_{0}L}{16EI},{\theta }_B=\frac{M_{0}L}{8EI},{\theta }_C=-{\theta }_B,{\theta }_D=-{\theta }_A.$

Explanation of Solution

Given Information:

The following figure is given with all relevant information,

Calculation:

Consider the following free body diagram,

Take equilibrium of horizontal forces as,

$\begin{array}{l}{\displaystyle \sum F_H}=0\\ \Rightarrow H_A+H_B+H_D=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of vertical forces as,

$\begin{array}{l}{\displaystyle \sum F_V}=0\\ \Rightarrow V_B+V_C=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of moments about B as,

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ \Rightarrow M_0-H_{A}L/2-M_0+V_{C}L-H_{D}L/2=0\mathrm{....}\text{(3)}\end{array}$

The bending moment at distance x from A along AB in part AB is given by,

$M_{AB}=H_{A}x$

Use second order deflection differential equation,

$\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{AB}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_A{x}_0)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{AB}=\frac{1}{EI}(H_A{x}^2/2+C_1)$

Integrate above equation to get rotation as,

$v={\delta }_{AB}=\frac{1}{EI}(H_A{x}^3/6+C_{1}x+C_2)$

The bending moment at distance x from B along BC in part BC is given by,

$M_{BC}=H_{A}L/2-M_0+V_{B}x$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{BC}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{A}L/2-M_0+V_{B}x)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{BC}=\frac{1}{EI}(H_{A}Lx/2-M_{0}x+V_B{x}^2/2+C_3)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_{A}L{x}^2/4-M_0{x}^2/2+V_B{x}^3/6+C_{3}x+C_4)$

The bending moment at distance x from C along CD in part CD is given by,

$M_{CD}=H_{D}x-H_{D}L/2$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{CD}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{D}x-H_{D}L/2)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{CD}=\frac{1}{EI}(H_D{x}^2/2-H_{D}Lx/2+C_5)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_D{x}^3/6-H_{D}L{x}^2/4+C_{5}x+C_6)$

The constraint equations are,

$\begin{array}{l}{\delta }_{AB}(0)=0\mathrm{...}\text{(4)}\\ {\theta }_{AB}(L/2)={\theta }_{BC}\text{(0) }\mathrm{...}\text{(5)}\\ {\delta }_{BC}(0)=0\mathrm{...}\text{(6)}\\ {\delta }_{BC}(L)=0\mathrm{..}\text{(7)}\\ {\theta }_{BC}(L)={\theta }_{CD}\text{(0) }\mathrm{..}\text{(8)}\\ {\delta }_{CD}(L/2)=0\mathrm{..}\text{(9)}\\ {\delta }_{AB}(L/2)=0\mathrm{...}\text{(10)}\end{array}$

Solve equations (1-10) to get integration constants and reactions.

So the reactions are $H_A=-H_D=\frac{3M_0}{2L},H_B=V_B=V_C=0.$

Substitute reactions and integration constants in expression of rotation to get,

$\begin{array}{l}{\theta }_{AB}(0)={\theta }_A=\frac{-M_{0}L}{16EI},\\ {\theta }_{AB}(L/2)={\theta }_B=\frac{M_{0}L}{8EI},\\ {\theta }_{BC}(L)={\theta }_C=-\frac{M_{0}L}{8EI},\\ {\theta }_{CD}(L/2)={\theta }_D=\frac{M_{0}L}{16EI}.\end{array}$

Conclusion:

Therefore, rotations at A, B, C and D are ${\theta }_A=\frac{-M_{0}L}{16EI},{\theta }_B=\frac{M_{0}L}{8EI},{\theta }_C=-{\theta }_B,{\theta }_D=-{\theta }_A.$

To determine

Reactions and rotations at A, B, C and D.

Answer to Problem 10.4.11P

Reactions and rotations at A, B, C and D are $\begin{array}{l}{H}_A=H_D=V_B=-V_C=\frac{M_0}{L},H_B=\frac{-2M_0}{L}\\ {\theta }_A=\frac{-M_{0}L}{24EI},{\theta }_B=\frac{M_{0}L}{12EI},{\theta }_C={\theta }_B,{\theta }_D={\theta }_A.\end{array}$

Explanation of Solution

Given Information:

The following figure is given with all relevant information,

Calculation:

Consider the following free body diagram,

Take equilibrium of horizontal forces as,

$\begin{array}{l}{\displaystyle \sum F_H}=0\\ \Rightarrow H_A+H_B+H_D=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of vertical forces as,

$\begin{array}{l}{\displaystyle \sum F_V}=0\\ \Rightarrow V_B+V_C=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of moments about B as,

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ \Rightarrow M_0-H_{A}L/2+M_0+V_{C}L-H_{D}L/2=0\mathrm{....}\text{(3)}\end{array}$

The bending moment at distance x from A along AB in part AB is given by,

$M_{AB}=H_{A}x$

Use second order deflection differential equation,

$\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{AB}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_A{x}_0)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{AB}=\frac{1}{EI}(H_A{x}^2/2+C_1)$

Integrate above equation to get rotation as,

$v={\delta }_{AB}=\frac{1}{EI}(H_A{x}^3/6+C_{1}x+C_2)$

The bending moment at distance x from B along BC in part BC is given by,

$M_{BC}=H_{A}L/2-M_0+V_{B}x$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{BC}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{A}L/2-M_0+V_{B}x)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{BC}=\frac{1}{EI}(H_{A}Lx/2-M_{0}x+V_B{x}^2/2+C_3)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_{A}L{x}^2/4-M_0{x}^2/2+V_B{x}^3/6+C_{3}x+C_4)$

The bending moment at distance x from C along CD in part CD is given by,

$M_{CD}=H_{D}x-H_{D}L/2$

Use second order deflection differential equation,

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}=\frac{M_{CD}}{EI}\\ \Rightarrow \frac{d^{2}v}{d{x}^2}=\frac{1}{EI}(H_{D}x-H_{D}L/2)\end{array}$

Integrate above equation to get rotation as,

$\frac{dv}{dx}={\theta }_{CD}=\frac{1}{EI}(H_D{x}^2/2-H_{D}Lx/2+C_5)$

Integrate above equation to get rotation as,

$v={\delta }_{BC}=\frac{1}{EI}(H_D{x}^3/6-H_{D}L{x}^2/4+C_{5}x+C_6)$

The constraint equations are,

$\begin{array}{l}{\delta }_{AB}(0)=0\mathrm{...}\text{(4)}\\ {\theta }_{AB}(L/2)={\theta }_{BC}\text{(0) }\mathrm{...}\text{(5)}\\ {\delta }_{BC}(0)=0\mathrm{...}\text{(6)}\\ {\delta }_{BC}(L)=0\mathrm{..}\text{(7)}\\ {\theta }_{BC}(L)={\theta }_{CD}\text{(0) }\mathrm{..}\text{(8)}\\ {\delta }_{CD}(L/2)=0\mathrm{..}\text{(9)}\\ {\delta }_{AB}(L/2)=0\mathrm{...}\text{(10)}\end{array}$

Solve equations (1-10) to get integration constants and reactions.

So the reactions are $H_A=H_D=V_B=-V_C=\frac{M_0}{L},H_B=\frac{-2M_0}{L}$

Substitute reactions and integration constants in expression of rotation to get,

$\begin{array}{l}{\theta }_{AB}(0)={\theta }_A=\frac{-M_{0}L}{24EI},\\ {\theta }_{AB}(L/2)={\theta }_B=\frac{M_{0}L}{12EI},\\ {\theta }_{BC}(L)={\theta }_C=\frac{M_{0}L}{12EI},\\ {\theta }_{CD}(L/2)={\theta }_D=\frac{-M_{0}L}{24EI}.\end{array}$

Conclusion:

Therefore, reactions and rotations at A, B, C and D are

$\begin{array}{l}{H}_A=H_D=V_B=-V_C=\frac{M_0}{L},H_B=\frac{-2M_0}{L}\\ {\theta }_A=\frac{-M_{0}L}{24EI},{\theta }_B=\frac{M_{0}L}{12EI},{\theta }_C={\theta }_B,{\theta }_D={\theta }_A.\end{array}$