Chapter 2, Problem 2.4.8P

Bar ABC is fixed at both ends (see figure) and has load P applied at B. Find reactions at A and C and displacement S_Bif P = 200 kN. L = 2 m, t = 20 mm, b, = 100 mm, b₂= 115 mm, and E = 96 GPa.

  

To determine

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

Answer to Problem 2.4.8P

The reaction force at point A is $80\text{kN}$ .

The reaction force at point C is $120\text{kN}$ .

The displacement at point B is $4.666\times {10}^{-4}\text{m}$ .

Explanation of Solution

Given:

Load at point B is $200\text{kN}$ , length of the bar is $2\text{m}$ , thickness is $20\text{mm}$ , width at point C and A is $115\text{mm,}$and the width at point B is $100\text{mm,}$and modulus of elasticity is $96\text{GPa}$ .

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 B is $P$and reaction force at point C is $R_A$ .

Write the expression for the force balance in horizontal direction.

  $R_A+R_C=200\text{kN}$…… (I)

Write the expression for the total deflection.

  ${\delta }_{AB}+{\delta }_{BC}=0$ …… (II)

Here, the deflection of the section AB is ${\delta }_{AB}$and deflection of the section BC is ${\delta }_{BC}$ .

Write the expression for the deflection for section AB.

  $\delta =\frac{PL}{AE}$…… (III)

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 AB.

  $A_{AB}=\frac{t\left(b_2-b_1\right)}{\ln \left(\frac{b_2}{b_1}\right)}$ …… (IV)

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

Write the expression for the deflection for section AB.

  ${\delta }_{AB}=\frac{R_A{L}_{AB}}{A_{AB}E}$…… (V)

Substitute $t\left(b_2-b_1\right)/\ln \left(b_2/b_1\right)$for $A_{AB}$in Equation (V).

  ${\delta }_{AB}=\frac{R_A{L}_{AB}}{\left(t\left(b_2-b_1\right)/\ln \left(b_2/b_1\right)\right)E}$

  ${\delta }_{AB}=\frac{R_A{L}_{AB}}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)$ …… (VI)

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

  $A_{BC}=\frac{t\left(b_1-b_2\right)}{\ln \left(\frac{b_1}{b_2}\right)}$ …… (VII)

Here, the width of the bar at point C is $b_2$ , the width of the bar at point B is $b_1$and the thickness is $t$ .

Write the expression for the deflection for section BC.

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

Substitute $t\left(b_1-b_2\right)/\ln \left(b_{12}/b_2\right)$for $A_{BC}$in Equation (VIII).

  ${\delta }_{BC}=-\frac{R_C{L}_{BC}}{\left(t\left(b_1-b_2\right)/\ln \left(b_1/b_2\right)\right)E}$

  ${\delta }_{BC}=-\frac{R_C{L}_{BC}}{t\left(b_1-b_2\right)E}\ln \left(b_1/b_2\right)$ …… (IX)

Substitute $-\frac{R_C{L}_{BC}}{t\left(b_1-b_2\right)E}\ln \left(b_1/b_2\right)$for ${\delta }_{BC}$and $\frac{R_A{L}_{AB}}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)$for ${\delta }_{AB}$in Equation (II).

  $\begin{array}{l}\frac{R_A{L}_{AB}}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)-\frac{R_C{L}_{BC}}{t\left(b_1-b_2\right)E}\ln \left(b_1/b_2\right)=0\\ \frac{R_A{L}_{AB}}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)=\frac{R_C{L}_{BC}}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)\end{array}$

  $R_A{L}_{AB}=R_C{L}_{BC}$…… (X)

Substitute $3L/5$for $L_{AB}$and $2L/5$for $L_{BC}$in Equation (X).

  $\begin{array}{l}{R}_A\left(3L/5\right)=R_C\left(2L/5\right)\\ R_A=\frac{2}{3}{R}_C\end{array}$

Calculation:

Substitute $\frac{2}{3}{R}_C$for $R_A$in Equation (I).

  $\begin{array}{l}\frac{2}{3}{R}_C+R_C=200\text{kN}\\ \frac{5}{3}{R}_C=200\text{kN}\\ R_C=120\text{kN}\end{array}$

Substitute $120\text{kN}$for $R_C$in Equation (I).

  $\begin{array}{l}{R}_A+120\text{kN}=200\text{kN}\\ R_A=200\text{kN}-120\text{kN}\\ R_A=80\text{kN}\end{array}$

Substitute $3L/5$for $L_{AB}$in Equation (VI).

  ${\delta }_{AB}=\frac{R_A\left(3L/5\right)}{t\left(b_2-b_1\right)E}\ln \left(b_2/b_1\right)$ …… (XI).

Substitute $80\text{kN}$for $R_A$ , $2\text{m}$for $L$ , $20\text{mm}$for $t$ , $96\text{GPa}$for $E$ , $100\text{mm}$for $b_1$and $115\text{mm}$for $b_2$in Equation (XI).

  $\begin{array}{c}{\delta }_{AB}=\frac{\left(80\text{kN}\right)\left(\left(3\times 2\text{m}\right)/5\right)}{\left(20\text{mm}\right)\left(115\text{mm}-100\text{mm}\right)\left(96\text{GPa}\right)}\ln \left(115\text{mm}/100\text{mm}\right)\\ =\frac{\left(80\text{kN}\right)\left(\left(3\times 2\text{m}\right)/5\right)}{\left(20\text{mm}\right)\left(115\text{mm}-100\text{mm}\right)\left(96\text{GPa}\right)}\left(0.13976\right)\\ =\frac{\left(80\text{kN}\right)\left(\left(3\times 2\text{m}\right)/5\right)}{\left(20\text{mm}\right)\left(15\text{mm}\right)\left(96\text{GPa}\right)}\left(0.13976\right)\\ =\frac{\left(96\text{kN}\cdot \text{m}\right)}{\left(300{\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\right)\left(96\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)}\left(0.13976\right)\end{array}$

  $\begin{array}{c}{\delta }_{AB}=\frac{\left(96\text{kN}\cdot \text{m}\right)0.13976}{\left(300{\text{mm}}^2\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\right)\left(96\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)}\\ =4.666\times {10}^{-4}\text{m}\end{array}$

Conclusion:

The reaction force at point A is $80\text{kN}$ .

The reaction force at point C is $120\text{kN}$ .

The displacement at point B is $4.666\times {10}^{-4}\text{m}$ .