Chapter 2, Problem 2.4.9P

Repeat Problem 2.4-8, but assume that the bar is made of aluminum alloy and that BC is prismatic. Assume that P = 20 kim. L = 3 ft.t = 314 in., b1 2m.b 2.Sin.andElO.400ksi.

  

To determine

The reaction force at point A and C and displacement at point B.

Answer to Problem 2.4.9P

The reaction force at point A is $-15.55\text{kips}$ .

The reaction force at point C is $-4.45\text{kips}$ .

The displacement at point B is $1.54\times {10}^{-2}\text{in}$ .

Explanation of Solution

Given information:

Load is $200\text{kips}$ , length of the bar is $3\text{ft}$ , thickness is $1/4\text{in}$ , width at point C and B is $2\text{in,}$and the width at point A is $2.5\text{in}$and modulus of elasticity is $10400\text{ksi}$ .

The below figure shows the free body diagram of the bar.

  

Figure-(1)

Here, the reaction force at point A is $R_A$ , load at point D is $P,$and reaction force at point C is $R_C$ .

Write the expression for the thickness at section D.

  $b_D=\frac{b_1+b_2}{2}$…… (I)

Here, the width of the bar at point A is $b_2$and the width of the bar at point B is $b_1$ .

Write the expression for the force balance in horizontal direction.

  $R_A+R_C=20\text{kips}$…… (II)

Write the expression for the total deflection.

  ${\delta }_{AD}+{\delta }_{DB}+{\delta }_{BC}=0$…… (III)

Here, the deflection of the section AD is ${\delta }_{AD}$ , deflection of the section DB is ${\delta }_{DB}$and deflection of the section BC is ${\delta }_{BC}$

Write the expression for the deflection for section AB.

  $\delta =\frac{PL}{AE}$…… (IV)

Here, the cross-section area is $A$ , length is $L$ , load is $P,$and modulus of elasticity is $E$ .

Write the expression for the cross-section area of the bar AD.

  $A_{AD}=\frac{t\left(b_2-b_D\right)}{\ln \left(\frac{b_2}{b_D}\right)}$…… (V)

Here, the thickness is $t$ .

Write the expression for the deflection for section AD.

  ${\delta }_{AD}=\frac{R_A{L}_{AD}}{A_{AD}E}$…… (VI)

Substitute $t\left(b_2-b_D\right)/\ln \left(b_2/b_D\right)$for $A_{AD}$in Equation (V).

  ${\delta }_{AD}=\frac{R_A{L}_{AD}}{\left(t\left(b_2-b_D\right)/\ln \left(b_2/b_D\right)\right)E}$

  ${\delta }_{AD}=\frac{R_A{L}_{AD}}{t\left(b_2-b_D\right)E}\ln \left(b_2/b_D\right)$ …… (VII)

Write the expression for the cross-section area of the bar DB.

  $A_{DB}=\frac{t\left(b_D-b_1\right)}{\ln \left(\frac{b_D}{b_1}\right)}$ …… (VIII)

Here, the width of the bar at point B is $b_1$and the thickness is $t$ .

Substitute $t\left(b_D-b_1\right)/\ln \left(b_D/b_1\right)$for $A_{DB}$in Equation (IV).

  ${\delta }_{DB}=-\frac{R_C{L}_{DB}}{t\left(b_D-b_1\right)E}\ln \left(b_D/b_1\right)$…… (IX)

Write the expression for the deflection for section BC.

  ${\delta }_{BC}=-\frac{R_C{L}_{BC}}{A_{BC}E}$…… (X)

Write the expression for the area of section BC.

  $A_{BC}=t{b}_1$

Substitute $t{b}_1$for $A_{BC}$in Equation (X).

  ${\delta }_{BC}=-\frac{R_C{L}_{BC}}{\left(t{b}_1\right)E}$…… (XI)

Substitute $\frac{R_A{L}_{AD}}{t\left(b_2-b_D\right)E}\ln \left(b_2/b_D\right)$for ${\delta }_{AD}$ , $-\frac{R_C{L}_{BC}}{\left(t{b}_1\right)E}$for ${\delta }_{BC}$and $-\frac{R_C{L}_{DB}}{t\left(b_D-b_1\right)E}\ln \left(b_D/b_1\right)$for ${\delta }_{DB}$in Equation (III).

  $\frac{R_A{L}_{AD}}{t\left(b_2-b_D\right)E}\ln \left(b_2/b_D\right)-\frac{R_C{L}_{DB}}{t\left(b_D-b_1\right)E}\ln \left(b_D/b_1\right)-\frac{R_C{L}_{BC}}{\left(t{b}_1\right)E}=0$…… (XII)

Substitute $L/4$for $L_{AD}$ , $L/4$for $L_{DB}$and $L/2$for $L_{BC}$in Equation (XII).

  $\frac{R_A\left(L/4\right)}{t\left(b_2-b_D\right)E}\ln \left(b_2/b_D\right)-\frac{R_C\left(L/4\right)}{t\left(b_D-b_1\right)E}\ln \left(b_D/b_1\right)-\frac{R_C\left(L/2\right)}{\left(t{b}_1\right)E}=0$

  $\frac{R_A}{4\left(b_2-b_D\right)}\ln \left(b_2/b_D\right)-\frac{R_C}{4\left(b_D-b_1\right)}\ln \left(b_D/b_1\right)-\frac{R_C}{2b_1}=0$ …… (XIII)

Calculation:

Substitute $2\text{in}$for $b_1$and $2.5\text{in}$for $b_2$in Equation (I).

  $\begin{array}{c}{b}_D=\frac{2\text{in}+2.5\text{in}}{2}\\ =\frac{4.5\text{in}}{2}\\ =2.25\text{in}\end{array}$

Substitute $2\text{in}$for $b_1$ , $2.25\mathrm{in}$for $b_D$and $2.5\text{in}$for $b_2$in Equation (XIII).

  $\begin{array}{l}\frac{R_A}{4\left(2.5\text{in}-2.25\text{in}\right)}\ln \left(b_2/2.25\text{in}\right)-\frac{R_C}{4\left(2.25\text{in}-2\text{in}\right)}\ln \left(2.25\text{in}/2\text{in}\right)-\frac{R_C}{2\text{in}}=0\\ \frac{R_A}{1\text{in}}\ln \left(b_2/2.25\text{in}\right)-\frac{R_C}{1\text{in}}\ln \left(2.25\text{in}/2\text{in}\right)-\frac{R_C}{2\text{in}}=0\end{array}$

  $R_A=3.491R_C$…… (XIV)

Substitute $3.491R_C$for $R_A$in Equation (II).

  $\begin{array}{l}3.491R_C+R_C=20\text{kips}\\ 4.491R_C=20\text{kips}\\ R_C=4.45\text{kips}\end{array}$

Substitute $4.45\text{kips}$for $R_C$in Equation (XIV).

  $\begin{array}{c}{R}_A=3.491\left(4.45\text{kips}\right)\\ =15.55\text{kips}\end{array}$

Substitute $L/2$for $L_{BC}$in Equation (XI).

  ${\delta }_{BC}=-\frac{R_C\left(L/2\right)}{\left(t{b}_1\right)E}$ …… (XV).

Substitute $4.45\text{kips}$for $R_C$ , $3\text{ft}$for $L$ , $1/4\text{in}$for $t$ , $10400\text{ksi}$for $E$and $2\text{in}$for $b_1$in Equation (XV).

  $\begin{array}{c}{\delta }_{BC}=-\frac{\left(4.45\text{kips}\right)\left(3\text{ft}/2\right)}{\left(\left(\frac{1}{4}\text{in}\right)\left(2\text{in}\right)\right)\left(10400\text{ksi}\right)}\\ =-\frac{\left(4.45\text{kips}\right)\left(\frac{1000\text{Psi}}{1\text{kips}/{\text{in}}^2}\right)\left(3\text{ft}/2\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(\left(\frac{1}{4}\text{in}\right)\left(2\text{in}\right)\right)\left(10400\text{ksi}\right)\left(\frac{1000\text{Psi}}{1\text{ksi}}\right)}\\ =1.54\times {10}^{-2}\text{in}\end{array}$

Conclusion:

The reaction force at point A is $-15.55\text{kips}$ .

The reaction force at point C is $-4.45\text{kips}$ .

The displacement at point B is $1.54\times {10}^{-2}\text{in}$ .