Chapter 9.4, Problem 90P

Use the method of superposition to solve the following problems and assume that the flexural rigidity El of each beam is constant.

9.90 Before the load P was applied, a gap, δ₀ = 0.5 mm, existed between the cantilever beam AC and the support at B. Knowing that E = 200 GPa, determine the magnitude of P for which the deflection at C is 1 mm.

Fig. P9.90

To determine

Find the magnitude of load P for the given condition using superposition method.

Answer to Problem 90P

The magnitude of load P in the beam is $\underset{\_}{P=5.63\text{kN}(\downarrow )}$.

Explanation of Solution

Given information:

The gap at the point B is ${\delta }_0=0.5\text{mm}$.

The modulus of elasticity of the material is $E=200\text{GPa}$.

The size of the square cross section is $a=60\text{mm}$.

The deflection at point C is ${\delta }_C=1\text{mm}$.

Calculation:

Find the moment of inertia of the square cross section (I) using the relation.

$I=\frac{a^4}{12}$

Here, the size of the square cross section is a.

Substitute 60 mm for a.

$\begin{array}{c}I=\frac{{60}^4}{12}\\ =1.08\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^4\\ =1.08\times {10}^{-6}{\text{m}}^4\end{array}$

Consider the portion AB of the beam.

The load P at point C will be converted into a load and moment at point B.

Show the free-body diagram of the superimposed beam AB as in Figure 1.

Loading I:

The downward load P is acting at point B of the beam.

Refer to Case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

$y=-\frac{P{L}^3}{3EI}$

Find the deflection at point B due to load P at point B as follows;

${\left(y_B\right)}_I=\frac{P{L}^3}{3EI}$

Loading II:

The upward reaction $B_y$ is acting at point B of the beam.

Refer to Case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

$y=-\frac{P{L}^3}{3EI}$

Find the deflection at point B due to reaction at point B as follows;

${\left(y_B\right)}_{II}=-\frac{B_y{L}^3}{3EI}$

Loading III:

The clockwise moment is acting at point B of the beam.

Refer to Case 3 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the deflection equation for moment in a cantilever beam as follows;

$y=-\frac{M{L}^2}{2EI}$

Find the deflection at point B due to the moment at point B as follows;

${\left(y_B\right)}_{III}=\frac{Pa{L}^2}{2EI}$

Find the resultant deflection at point B as follows.

$y_B={\left(y_B\right)}_I+{\left(y_B\right)}_{II}+{\left(y_B\right)}_{III}$

Substitute ${\delta }_0$ for $y_B$, $\frac{P{L}^3}{3EI}$ for ${\left(y_B\right)}_I$, $-\frac{B_y{L}^3}{3EI}$ for ${\left(y_B\right)}_{II}$, and $\frac{Pa{L}^2}{2EI}$ for ${\left(y_B\right)}_{III}$.

$\begin{array}{c}{\delta }_0=\frac{P{L}^3}{3EI}-\frac{B_y{L}^3}{3EI}+\frac{Pa{L}^2}{2EI}\\ =\frac{1}{EI}\left(\frac{P{L}^3}{3}-\frac{B_y{L}^3}{3}+\frac{Pa{L}^2}{2}\right)\end{array}$

Substitute 0.5 mm for ${\delta }_0$, 0.5 m for L, 0.2 m for a, 200 GPa for E, and $1.08\times {10}^{-6}{\text{m}}^4$ for I.

$\begin{array}{l}0.5\text{mm}\times \frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}\times 1.08\times {10}^{-6}}\left(\begin{array}{l}\frac{P{\left(0.5\right)}^3}{3}-\frac{B_y{\left(0.5\right)}^3}{3}\\ +\frac{P\left(0.2\right){\left(0.5\right)}^2}{2}\end{array}\right)\\ \frac{0.5}{{10}^3}=\frac{1}{200\times {10}^9\times 1.08\times {10}^{-6}}\left(0.041667P-0.041667B_y+0.025P\right)\\ 108=0.06667P-0.041667B_y\end{array}$ (1)

Consider the beam ABC.

Show the free-body diagram of the superimposed beam ABC as in Figure 2.

Loading IV:

The downward load P is acting at point C of the beam.

Refer to Case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

$y=-\frac{P{L}^3}{3EI}$

Find the deflection at point C due to load P at point C as follows;

${\left(y_C\right)}_{IV}=\frac{P{\left(L+a\right)}^3}{3EI}$

Loading V:

The upward reaction $B_y$ is acting at point B of the beam.

Refer to Case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the slope and deflection equation for concentrated load acting in a cantilever beam as follows;

$\begin{array}{l}\theta =-\frac{P{L}^2}{2EI}\\ y=-\frac{P{L}^3}{3EI}\end{array}$

Find the deflection at point B due to reaction at point B as follows;

${\left(y_B\right)}_V=-\frac{B_y{L}^3}{3EI}$

Find the slope at point B due to reaction at point B as follows;

${\left({\theta }_B\right)}_V=-\frac{B_y{L}^2}{2EI}$

The portion BC remains straight.

Find the deflection at point C due to reaction at point B as follows;

${\left(y_C\right)}_V={\left(y_B\right)}_V+a{\left({\theta }_B\right)}_V$

Substitute $-\frac{B_y{L}^3}{3EI}$ for ${\left(y_B\right)}_V$ and $-\frac{B_y{L}^2}{2EI}$ for ${\left({\theta }_B\right)}_V$.

${\left(y_C\right)}_V=-\frac{B_y{L}^3}{3EI}-a\frac{B_y{L}^2}{2EI}$

Find the resultant deflection at point C as follows.

$y_C={\left(y_C\right)}_{IV}+{\left(y_C\right)}_V$

Substitute ${\delta }_C$ for $y_C$, $\frac{P{\left(L+a\right)}^3}{3EI}$ for ${\left(y_C\right)}_{IV}$ , and $\left(-\frac{B_y{L}^3}{3EI}-a\frac{B_y{L}^2}{2EI}\right)$ for ${\left(y_C\right)}_V$.

$\begin{array}{c}{\delta }_C=\frac{P{\left(L+a\right)}^3}{3EI}-\frac{B_y{L}^3}{3EI}-a\frac{B_y{L}^2}{2EI}\\ =\frac{1}{EI}\left(\frac{P{\left(L+a\right)}^3}{3}-\frac{B_y{L}^3}{3}-a\frac{B_y{L}^2}{2}\right)\end{array}$

Substitute 1 mm for ${\delta }_C$, 0.5 m for L, 0.2 m for a, 200 GPa for E, and $1.08\times {10}^{-6}{\text{m}}^4$ for I.

$\begin{array}{l}1\text{mm}\times \frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}\times 1.08\times {10}^{-6}}\left(\begin{array}{l}\frac{P{\left(0.5+0.2\right)}^3}{3}-\frac{B_y{\left(0.5\right)}^3}{3}\\ -\left(0.2\right)\frac{B_y{\left(0.5\right)}^2}{2}\end{array}\right)\\ 216=0.11433P-0.041667B_y-0.025B_y\\ 216=0.11433P-0.06667B_y\end{array}$ (2)

Solve the Equation (1) and (2).

$\begin{array}{c}{B}_y=6.42\times {10}^3\text{N}\\ P=5.63\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{{\text{10}}^{\text{3}}\text{N}}\\ =5.63\text{kN}(\downarrow )\end{array}$

Therefore, the magnitude of load P in the beam is $\underset{\_}{P=5.63\text{kN}(\downarrow )}$.