Chapter 9, Problem 9.4.10P

-10 Derive the equations of the deflection curve for beam AB with sliding support at A and roller support at B, supporting a distributed load of maximum intensity q₀acting on the right-hand half of the beam (see figure). Also, determine deflection ${\delta }_A$, angle of rotation ${\theta }_B$ , and deflection $\delta c$at the midpoint. Use the fourth-order differential equation of the deflection curve (the load equation).

To determine

To derive: the equations of deflection curve for beam. Also, find deflection ${\delta }_A$ , Angle of rotation ${\theta }_B$ , and deflection ${\delta }_C$ at the midpoint.

Answer to Problem 9.4.10P

Equations of deflection curve:

With guiding support at A,

  $v\left(x\right)=-\frac{q_0{L}^2}{480EI}\left(-20x^2+19L^2\right)\left(0\le x\le L\right)$

With roller support at B,

  $v\left(x\right)=-\frac{q_0}{960L.EI}\left(80x^{4}L-16x^5-120L^2+x^3+40L^3{x}^2-25L^{4}x+41L^5\right)\left(\frac{L}{2}\le x\le L\right)$

Deflection ${\delta }_A$ :

  ${\delta }_A=\frac{19q_0{L}^4}{480EI}$

Angle of rotation:

  ${\theta }_B=-\frac{13q_0{L}^3}{192EI}$

Deflection ${\delta }_C$ :

  ${\delta }_C=\frac{7}{240EI}{q}_0{L}^4$

Explanation of Solution

Given:

  

Load equation:

  $EIv\text{'}\text{'}\text{'}\text{'}=-q=0\left(0\le x\le \frac{L}{2}\right)$

Upon integration

  $EIv\text{'}\text{'}\text{'}=C_1\left(0\le x\le \frac{L}{2}\right)$

Upon integration

  $EIv\text{'}\text{'}=C_{1}x+C_2\left(0\le x\le \frac{L}{2}\right)$

Boundary condition 1:

  $v\text{'}\text{'}\text{'}\left(0\right)=V\left(0\right)$

Therefore,

  $C_1=0$

Boundary condition 2:

  $v\text{'}\text{'}\left(0\right)=M\left(0\right)=\frac{q_0{L}^2}{12}$

Therefore,

  $C_2=\frac{q_0{L}^2}{12}$

  $EIv\text{'}\text{'}=\frac{q_0{L}^2}{12}\left(L\le x\le \frac{L}{2}\right)$

Upon integration

  $EIv\text{'}=\frac{q_0{L}^{2}x}{12}+C_3\left(L\le x\le \frac{L}{2}\right)$

Boundary condition 3:

  $v\text{'}\left(0\right)=0$

Therefore,

  $C_3=0$

  $EIv=\frac{q_0{L}^2{x}^2}{24}+C_4\left(0\le x\le \frac{L}{2}\right)$

Load equation:

  $EIv\text{'}\text{'}\text{'}\text{'}=-2q_0+\frac{2q_{0}x}{L}\left(\frac{L}{2}\le x\le L\right)$

Upon integration

  $EIv\text{'}\text{'}\text{'}=-2q_{0}x+\frac{q_0{x}^2}{L}+C_5\left(\frac{L}{2}\le x\le L\right)$

Upon integration

  $EIv\text{'}\text{'}=-q_0{x}^2+\frac{q_0{x}^3}{3L}+C_{5}x+C_6\left(\frac{L}{2}\le x\le L\right)$

Boundary condition 4:

  $v\text{'}\text{'}\text{'}\left(\frac{L}{2}\right)=V\left(\frac{L}{2}\right)=0$

Therefore,

  $C_5=\frac{3q_{0}L}{4}$

Boundary condition 5:

  $v\text{'}\text{'}\left(\frac{L}{2}\right)=M\left(\frac{L}{2}\right)=\frac{q_0{L}^2}{12}$

Therefore,

  $C_6=\frac{q_0{L}^2}{12}$

  $EIv\text{'}\text{'}=-q_0{x}^2+\frac{q_0{x}^3}{3L}+\frac{3q_{0}Lx}{4}-\frac{q_0{L}^2}{12}$

Upon integration

  $EIv\text{'}=\frac{-q_0{x}^3}{3}+\frac{q_0{x}^4}{12L}+\frac{3q_{0}L{x}^2}{8}-\frac{q_0{L}^{2}x}{12}+C_7$

Boundary condition 6

  $v{\text{'}}_L\left(\frac{L}{2}\right)=v{\text{'}}_R\left(\frac{L}{2}\right)$

Therefore,

  $C_7=\frac{5}{192}q{L}^3$

  $EIv\text{'}=-\frac{q_0{x}^3}{3}+\frac{q_0{x}^4}{12L}+\frac{3q_{0}L{x}^2}{8}+\frac{q_0{L}^{2}x}{12}+\frac{5q_{0}L{x}^3}{192}$

Upon integration

  $EIv=-\frac{q_0{x}^4}{12}+\frac{q_0{x}^5}{60L}+\frac{3q_{0}L{x}^3}{8}+\frac{q_0{L}^2{x}^2}{24}+\frac{5q_0{L}^{3}x}{192}+C_8$

Boundary condition 7:

  $v\left(L\right)=0$

Therefore,

  $C_8=\frac{-41}{960}{q}_0{L}^4$

  $EIv=-\frac{q_0{x}^4}{12}+\frac{q_0{x}^5}{60L}+\frac{3q_{0}L{x}^3}{8}+\frac{q_0{L}^2{x}^2}{24}+\frac{5q_0{L}^{3}x}{192}-\frac{-41}{960}{q}_0{L}^4\left(\frac{L}{2}\le x\le L\right)$

B.C 8:

  $v_L\left(\frac{L}{2}\right)=v_R\left(\frac{L}{2}\right)$

  $\frac{q_0{L}^2{\left(\frac{L}{2}\right)}^2}{24}+C_4=-\frac{q_0{\left(\frac{L}{2}\right)}^4}{12}+\frac{q_0{L}^2{\left(\frac{L}{2}\right)}^2}{24}+\frac{5q_0{L}^3.\frac{L}{2}}{192}-\frac{41}{960}{q}_0{L}^4$

  $C_4=-\frac{19}{480}{q}_0{L}^4$

  $EIv=\frac{q_0{L}^2{x}^2}{24}-\frac{19}{480}{q}_0{L}^4\left(0\le x\le \frac{L}{2}\right)$

  $v\left(x\right)=-\frac{q_0{L}^2}{480EI}\left(-20x^2+19L^2\right)\left(0\le x\le L\right)$ (With guided support at A)

  $v\left(x\right)=-\frac{q_0}{960L.EI}\left(80x^{4}L-16x^5-120L^2+x^3+40L^3{x}^2-25L^{4}x+41L^5\right)\left(\frac{L}{2}\le x\le L\right)$ (With roller support at B)

Thus, we have,

At $x=0$ ,

  ${\delta }_A=\frac{19q_0{L}^4}{480EI}$

And

At $x=L$ ,

  ${\theta }_B=-v\text{'}\left(L\right)=-\frac{13q_0{L}^3}{192EI}$

And

At $x=\frac{L}{2}$ ,

  ${\delta }_C=-v\left(\frac{L}{2}\right)=\frac{7}{240EI}{q}_0{L}^4$