Chapter 8, Problem 20P

To determine

Find the slope and the vertical deflection of the cantilever beam at the points B and C.

Find the minimum value of moment of inertia, if the deflection of point C is 0.4 in.

Answer to Problem 20P

The slope at point C is $\underset{\_}{{\theta }_C=\frac{720}{EI}}$.

The deflection at point C is $\underset{\_}{{\delta }_C=\frac{P{L}^3}{3EI}+\frac{w{L}^4}{8EI}}$.

The slope at point B is $\underset{\_}{{\theta }_B=\frac{576}{EI}}$.

The deflection at point B is $\underset{\_}{{\delta }_B=\frac{1,998}{EI}}$.

The required minimum value of I is $\underset{\_}{6,531.8\text{in}{.}^4}$.

Explanation of Solution

Given information:

The value of E is $4,000\text{kips/in}{.}^2$ and length of the beam is $12\text{ft}$.

Deflection at point C is 0.4 in.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Find the deflection at C of the cantilever beam as follows:

For Q system, remove all external loads and consider a dummy load of 1 kips at the point C of the beam.

Sketch the P-system and Q-system for deflection at C as shown in Figure 1.

From P-system,

The moment at a distance of x from C is,

$\begin{array}{c}{M}_P=Px+wx\frac{x}{2}\\ =Px+\frac{w{x}^2}{2}\end{array}$

From Q-system,

$\begin{array}{c}{M}_Q=1\left(x\right)\\ =x\end{array}$

Find the deflection at point C $\left({\delta }_{\text{C}}\right)$ as follows:

$\begin{array}{c}\left(1\text{kips}\right)\left({\delta }_{\text{C}}\right)={\displaystyle \underset{0}{\overset{L}{\int }}{M}_Q{M}_P\frac{dx}{EI}}\\ {\delta }_{\text{C}}={\displaystyle \underset{0}{\overset{L}{\int }}\left(x\right)\left(Px+\frac{w{x}^2}{2}\right)\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{L}{\int }}\left(P{x}^2+\frac{w{x}^3}{2}\right)\frac{dx}{EI}}\\ =\frac{1}{EI}{\left[\frac{P{x}^3}{3}+\frac{w{x}^4}{8}\right]}_0^L\end{array}$

$\begin{array}{c}{\delta }_{\text{C}}=\frac{1}{EI}\left(\frac{P{L}^3}{3}+\frac{w{L}^4}{8}\right)\\ =\frac{P{L}^3}{3EI}+\frac{w{L}^4}{8EI}\end{array}$

Therefore, the deflection at point C is $\underset{\_}{{\delta }_C=\frac{P{L}^3}{3EI}+\frac{w{L}^4}{8EI}}$.

Find the minimum value of I, if the value of ${\delta }_C$ is $0.4\text{in}\text{.}$

Substitute 6 kips for P, $1\text{kip/ft}$ for w, $0.4\text{in}\text{.}$ for ${\delta }_C$, and 12 ft for L.

$\begin{array}{l}0.4\text{in}\text{.}=\frac{\left(6\text{kips}\right){\left(12\text{ft}\right)}^3}{3\left(4,000\text{kips/in}{\text{.}}^{\text{2}}\right)I}+\frac{\left(1\text{kip/ft}\right){\left(12\text{ft}\right)}^4}{8\left(4,000\text{kips/in}{\text{.}}^{\text{2}}\right)I}\\ 0.4\text{in}\text{.}=\frac{\left(6\text{kips}\right){\left(12\text{ft}\times \frac{12\text{in}.}{1\text{ft}}\right)}^3}{3\left(4,000\text{kips/in}{\text{.}}^{\text{2}}\right)I}+\frac{\left(1\text{kip/ft}\times \frac{1\text{ft}}{12\text{in}.}\right){\left(12\text{ft}\times \frac{12\text{in}.}{1\text{ft}}\right)}^4}{8\left(4,000\text{kips/in}{\text{.}}^{\text{2}}\right)I}\\ 0.4=\frac{1,492.992}{I}+\frac{1,119.744}{I}\\ I_{\min }=6,531.8\text{in}{.}^4\end{array}$

Therefore, the required minimum value of I is $\underset{\_}{6,531.8\text{in}{.}^4}$.

Find the slope at C of the cantilever beam as follows:

For Q system, remove all external loads and consider a dummy moment of $1\text{kip}\cdot \text{ft}$ at the point C of the beam.

Sketch the P-system and Q-system for slope at C as shown in Figure 2.

From P-system,

The moment at a distance of x from C is,

$\begin{array}{c}{M}_P=Px+wx\frac{x}{2}\\ =Px+\frac{w{x}^2}{2}\end{array}$

From Q-system,

$M_Q=1$

Find the slope at point C $\left({\theta }_{\text{C}}\right)$ as follows:

$\begin{array}{c}{\theta }_{\text{C}}={\displaystyle \underset{0}{\overset{L}{\int }}{M}_Q{M}_P\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{L}{\int }}\left(1\right)\left(Px+\frac{w{x}^2}{2}\right)\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{L}{\int }}\left(Px+\frac{w{x}^2}{2}\right)\frac{dx}{EI}}\\ =\frac{1}{EI}{\left[\frac{P{x}^2}{2}+\frac{w{x}^3}{6}\right]}_0^L\end{array}$

${\theta }_C=\frac{P{L}^2}{2EI}+\frac{w{L}^3}{6EI}$

Substitute 6 kips for P, $1\text{kip/ft}$ for w, $0.4\text{in}\text{.}$ for ${\delta }_C$, and 12 ft for L.

$\begin{array}{c}{\theta }_C=\frac{\left(6\right){\left(12\right)}^2}{2EI}+\frac{\left(1\right){\left(12\right)}^3}{6EI}\\ =\frac{432}{EI}+\frac{288}{EI}\\ =\frac{720}{EI}\end{array}$

Therefore, the slope at point C is $\underset{\_}{{\theta }_C=\frac{720}{EI}}$.

Find the deflection at B of the cantilever beam as follows:

For Q system, remove all external loads and consider a dummy load of 1 kips at the point C of the beam.

Sketch the P-system and Q-system for deflection at B as shown in Figure 3.

From P-system,

The moment at a distance of x from B is,

$\begin{array}{c}{M}_P=w\left(6+x\right)\left(\frac{6+x}{2}\right)+P\left(6+x\right)\\ =\frac{w}{2}\left(36+12x+x^2\right)+6P+Px\\ =18w+6wx+\frac{w{x}^2}{2}+6P+Px\end{array}$

From Q-system,

$\begin{array}{c}{M}_Q=1\left(x\right)\\ =x\end{array}$

Find the deflection at point B $\left({\delta }_{\text{B}}\right)$ as follows:

$\begin{array}{c}\left(1\text{kips}\right)\left({\delta }_{\text{B}}\right)={\displaystyle \underset{0}{\overset{L}{\int }}{M}_Q{M}_P\frac{dx}{EI}}\\ {\delta }_{\text{B}}={\displaystyle \underset{0}{\overset{6}{\int }}\left(x\right)\left(18w+6wx+\frac{w{x}^2}{2}+6P+Px\right)\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{6}{\int }}\left(18wx+6w{x}^2+\frac{w{x}^3}{2}+6Px+P{x}^2\right)\frac{dx}{EI}}\\ =\frac{1}{EI}{\left[\frac{18w{x}^2}{2}+\frac{6w{x}^3}{3}+\frac{w{x}^4}{8}+\frac{6P{x}^2}{2}+\frac{P{x}^3}{3}\right]}_0^6\end{array}$

$\begin{array}{c}{\delta }_{\text{B}}=\frac{1}{EI}\left(\frac{18w{\left(6\right)}^2}{2}+\frac{6w{\left(6\right)}^3}{3}+\frac{w{\left(6\right)}^4}{8}+\frac{6P{\left(6\right)}^2}{2}+\frac{P{\left(6\right)}^3}{3}\right)\\ =\frac{1}{EI}\left(324w+432w+162w+108P+72P\right)\\ =\frac{1}{EI}\left(324\left(1\right)+432\left(1\right)+162\left(1\right)+108\left(6\right)+72\left(6\right)\right)\\ =\frac{1,998}{EI}\end{array}$

Therefore, the deflection at point B is $\underset{\_}{{\delta }_B=\frac{1,998}{EI}}$.

Find the slope at B of the cantilever beam as follows:

For Q system, remove all external loads and consider a dummy moment of $1\text{kip}\cdot \text{ft}$ at the point B of the beam.

Sketch the Q-system for slope at B as shown in Figure 4.

From P-system,

The moment at a distance of x from B is,

$\begin{array}{c}{M}_P=w\left(6+x\right)\left(\frac{6+x}{2}\right)+P\left(6+x\right)\\ =\frac{w}{2}\left(36+12x+x^2\right)+6P+Px\\ =18w+6wx+\frac{w{x}^2}{2}+6P+Px\end{array}$

From Q-system,

$M_Q=1$

Find the slope at point B $\left({\theta }_{\text{B}}\right)$ as follows:

$\begin{array}{c}{\theta }_{\text{B}}={\displaystyle \underset{0}{\overset{L}{\int }}{M}_Q{M}_P\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{6}{\int }}\left(1\right)\left(18w+6wx+\frac{w{x}^2}{2}+6P+Px\right)\frac{dx}{EI}}\\ ={\displaystyle \underset{0}{\overset{6}{\int }}\left(18w+6wx+\frac{w{x}^2}{2}+6P+Px\right)\frac{dx}{EI}}\\ =\frac{1}{EI}{\left[18wx+\frac{6w{x}^2}{2}+\frac{w{x}^3}{6}+6Px+\frac{P{x}^2}{2}\right]}_0^6\end{array}$

$\begin{array}{c}{\theta }_B=\frac{1}{EI}\left[18w\left(6\right)+\frac{6w{\left(6\right)}^2}{2}+\frac{w{\left(6\right)}^3}{6}+6P\left(6\right)+\frac{P{\left(6\right)}^2}{2}\right]\\ =\frac{1}{EI}\left(108w+108w+36w+36P+18P\right)\\ =\frac{1}{EI}\left(108\left(1\right)+108\left(1\right)+36\left(1\right)+36\left(6\right)+18\left(6\right)\right)\\ =\frac{576}{EI}\end{array}$

Therefore, the slope at point B is $\underset{\_}{{\theta }_B=\frac{576}{EI}}$.