Chapter 4, Problem 102P

The steel beam ABCD shown is simply supported at C as shown and supported at B and D by shoulder steel bolts, each having a diameter of 8 mm. The lengths of BE and DF are 50 mm and 65 mm, respectively. The beam has a second area moment of 21(10³) mm⁴. Prior to loading, the members are stress-free. A force of 2 kN is then applied at point A. Using procedure 2 of Sec. 4–10, determine the stresses in the bolts and the deflections of points A, B, and D.

Problem 4–102

To determine

The stresses in the bolts.

The deflection at point A.

The deflection at point B.

The deflection at point D.

Answer to Problem 102P

The stress in the bolt $BE$ is $83.3\text{MPa}$, and the stress in the bolt $DF$ is $-3.8\text{MPa}$.

The deflection at point A is $-0.167\text{mm}$.

The deflection at point B is $-0.0201\text{mm}$.

The deflection at point D is $-0.00118\text{mm}$.

Explanation of Solution

The Figure (1) shows the free body diagram of the steel beam ABCD.

Figure (1)

Here, the applied load at point $A$ is $F$, reaction force at point $B$ is $F_{BE}$, reaction force at point $C$ is $R_C$,reaction force at point $D$ is $F_{DE}$ distance between $AB$ is $d_1$, distance between $AC$ is $d_2$ and distance between $AD$ is $d_3$.

Refer to procedure 2 from Sec. 4–10.

Write the expression for the net force in the beam

    $R_C+F_{BE}-F_{DF}=2000$       (I)

Write the expression for the net moment about point $D$ in the beam.

    $Fl-F_{BE}a-R_{C}b=0$     (II)

Here, the length of the beam is $l$, length between point $BD$ is $a$ and the length between the point $CD$ is $b$.

The Figure (2) shows the beam at section (1).

Figure (2)

Write the expression for the bending moment at section (1).

    $M=-Fx+F_{BE}{\langle x-d_1\rangle }^1+R_C{\langle x-d_2\rangle }^1$                        (III)

Here, the bending moment at section (1) is $M$ and the distance from the left end is $x$.

Write the expression for the bending moment in terms of elastic equation.

    $EI\frac{d^{2}y}{d{x}^2}=M$                                                                                             (IV)

Here, the modulus of elasticity is $E$, the deflection in the beam is $y$ and the second area moment of beam is $I$

Substitute $-Fx+F_{BE}{\langle x-d_1\rangle }^1+R_C{\langle x-d_2\rangle }^1$ for $M$ in Equation (IV).

    $EI\frac{d^{2}y}{d{x}^2}=-Fx+F_{BE}{\langle x-d_1\rangle }^1+R_C{\langle x-d_2\rangle }^1$

Integrate the above expression.

    $EI\frac{dy}{dx}=-F\frac{x^2}{2}+\frac{F_{BE}}{2}{\langle x-d_1\rangle }^2+\frac{R_C}{2}{\langle x-d_2\rangle }^2+C_1$

Further integrate the above expression.

    $EIy=-F\frac{x^3}{6}+\frac{F_{BE}}{6}{\langle x-d_1\rangle }^3+\frac{R_C}{6}{\langle x-d_2\rangle }^3+C_{1}x+C_2$        (V)

Write the expression for the area of the bolt.

    $A=\frac{\pi }{4}{d}^2$                                                                                                (VI)

Here, the area of the bolt is $A$, the diameter of the bolt is $d$.

Write the expression for elongationin the steel boltat point B.

    $y_B=-\frac{F_{BE}{h}_1}{AE}$               (VII)

Here, the elongation is $y_B$ and the height of the bolt is $h_1$.

Write the expression for elongation in the steel bolt at point D

    $y_D=-\frac{F_{DF}{h}_2}{AE}$          (VIII)

Here, the elongation is $y_D$ and the height of the bolt is $h_2$.

Applying Boundary conditions.

At $x=d_1$; $y=y_B$.

Substitute $d_1$ for $x$ and $-\frac{F_{BE}h}{AE}$ for $y$ in Equation (V).

    $\begin{array}{c}EI\left(-\frac{F_{BE}h}{AE}\right)=-F\frac{d_1^3}{6}+\frac{F_{BE}}{6}{\langle d_1-d_1\rangle }^3+0+C_1{d}_1+C_2\\ =-F\frac{d_1^3}{6}+C_1{d}_1+C_2\end{array}$             (IX)

At $x=d_2$; $y=0$.

Substitute $d_2$ for $x$ and $0$ for $y$ in Equation (V).

    $\begin{array}{c}EI\left(0\right)=-F\frac{x^3}{6}+\frac{F_{BE}}{6}{\langle d_2-d_1\rangle }^3+0+C_1{d}_2+C_2\\ 0=-F\frac{x^3}{6}+\frac{F_{BE}}{6}{\langle d_2-d_1\rangle }^3+0+C_1{d}_2+C_2\end{array}$           (X)

At $x=d_3$; $y=y_D$.

Substitute $d_3$ for $x$ and $-\frac{F_{DF}h}{AE}$ for $y$ in Equation (V).

    $\begin{array}{c}EI\left(-\frac{F_{DF}h}{AE}\right)=-F\frac{d_3^3}{6}+\frac{F_{BE}}{6}{\langle d_3-d_1\rangle }^3+\frac{R_C}{6}{\langle d_3-d_2\rangle }^3+C_1{d}_3+C_2\\ -\frac{F_{DF}hI}{A}=-F\frac{d_3^3}{6}+\frac{F_{BE}}{6}{\langle d_3-d_1\rangle }^3+\frac{R_C}{6}{\langle d_3-d_2\rangle }^3+C_1{d}_3+C_2\end{array}$     (XI)

Write the expression for the normal stress in section BE.

    ${\sigma }_{BE}=\frac{F_{BE}}{A}$                                                                                                (XII)

Here, the normal stress is ${\sigma }_{BE}$.

Write the expression for the normal stress in section DF.

    ${\sigma }_{DF}=\frac{F_{DF}}{A}$                                                                                               (XIII)

Here, the normal stress is ${\sigma }_{DF}$.

Conclusion:

Refer to Table A-5 “Physical Constants of Materials”, obtain the properties of modulus of elasticity for steel as $207\times {10}^3\text{MPa}$.

Substitute $8\text{mm}$ for $d$ in Equation (VI).

    $\begin{array}{c}A=\frac{\pi }{4}{\left(8\text{mm}\right)}^2\\ =50.265{\text{mm}}^{\text{2}}\\ \approx 50.27{\text{mm}}^{\text{2}}\end{array}$

Substitute $21\times {10}^3{\text{mm}}^{\text{4}}$ for $I$, $207\times {10}^3\text{MPa}$ for $E$, $50\text{mm}$ for $h_1$, $50.27{\text{mm}}^{\text{2}}$ for $A$, $2\text{kN}$ for $F$ and $75\text{mm}$ for $d_1$ in Equation (IX).

    $\begin{array}{l}\left[\begin{array}{l}\left(21\times {10}^3{\text{mm}}^{\text{4}}\right)\left(207\times {10}^3\text{MPa}\right)\times \\ \left(-\frac{F_{BE}\left(50\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)}\right)\end{array}\right]=\left[\begin{array}{l}-\left(2\text{kN}\right)\frac{{\left(75\text{mm}\right)}^3}{6}+\\ C_1\left(75\text{mm}\right)+C_2\end{array}\right]\\ 0.100904F_{BE}=\left[\begin{array}{l}-\left(2\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)70312.5+\\ C_1\left(75\text{mm}\right)+C_2\end{array}\right]\\ \left(20.89\times {10}^3\right)F_{BE}+75C_1+C_2=\left(140.6\times {10}^6\right)\end{array}$    (XIV)

Substitute $21\times {10}^3{\text{mm}}^{\text{4}}$ for $I$, $207\times {10}^3\text{MPa}$ for $E$, $2\text{kN}$ for $F$ and $150\text{mm}$ for $d_2$ in Equation (X).

    $\begin{array}{l}0=\left[\begin{array}{l}-\left(2\text{kN}\right)\frac{{\left(150\text{mm}\right)}^3}{6}+\\ \frac{F_{BE}}{6}{\langle \left(150\text{mm}\right)-\left(75\text{mm}\right)\rangle }^3\\ +0+C_1\left(150\text{mm}\right)+C_2\end{array}\right]\\ 0=\left[\begin{array}{l}-\left(2\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\times 562500+\\ F_{BE}\left(70312.5\right)+150C_1+C_2\end{array}\right]\\ \left[\begin{array}{l}\left(70.31\times {10}^3\right)F_{BE}+\\ 150C_1+C_2\end{array}\right]=\left(1.125\times {10}^9\right)\end{array}$                                                (XV)

Substitute $21\times {10}^3{\text{mm}}^{\text{4}}$ for $I$, $207\times {10}^3\text{MPa}$ for $E$, $65\text{mm}$ for $h_2$, $50.27{\text{mm}}^{\text{2}}$ for $A$, $2\text{kN}$ for $F$ and $225\text{mm}$ for $d_3$ in Equation (XI).

    $\begin{array}{l}\left[\begin{array}{l}-\left(21\times {10}^3{\text{mm}}^{\text{4}}\right)\left(207\times {10}^3\text{MPa}\right)\times \\ \left(-\frac{F_{DF}\left(65\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)}\right)\end{array}\right]=\left[\begin{array}{l}-\left(2\text{kN}\right)\frac{{\left(225\text{mm}\right)}^3}{6}+\\ \frac{F_{BE}}{6}{\langle 225\text{mm}-75\text{mm}\rangle }^3\\ +\frac{R_C}{6}{\langle \left(225\text{mm}\right)-\left(150\text{mm}\right)\rangle }^3\\ +C_1\left(225\text{mm}\right)+C_2\end{array}\right]\\ 27.15\times {10}^3{F}_{DF}=\left[\begin{array}{l}-\left(2\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\frac{{\left(225\right)}^3}{6}\\ +\frac{F_{BE}}{6}{\langle 150\rangle }^3\\ +\frac{R_C}{6}{\langle 75\rangle }^3+C_1\left(225\right)+C_2\end{array}\right]\\ \left[\begin{array}{l}\left(70.31\times {10}^3\right)R_C+\left(562.5\times {10}^3\right)\\ -\left(27.15\times {10}^3\right)F_{DF}+225C_1+C_2\end{array}\right]=\left(3.797\times {10}^9\right)\end{array}$ (XVI)

Write Equation (I), Equation (II), Equation (XIV), Equation (XV), and Equation (XVI) in matrix form.

    $\left[\begin{array}{ccccc}1& 1& -1& 0& 0\\ 1& 2& 0& 0& 0\\ 0& \left(20.89\times {10}^3\right)& 0& 75& 1\\ 0& \left(70.31\times {10}^3\right)& 0& 150& 1\\ \left(70.31\times {10}^3\right)& \left(562.5\times {10}^3\right)& \left(-27.15\times {10}^3\right)& 225& 1\end{array}\right]\left\{\begin{array}{c}{R}_C\\ F_{BE}\\ F_{DF}\\ C_1\\ C_2\end{array}\right\}=\left\{\begin{array}{c}\left(2\times {10}^3\right)\\ \left(6\times {10}^3\right)\\ \left(140.6\times {10}^6\right)\\ \left(1.125\times {10}^9\right)\\ \left(3.797\times {10}^9\right)\end{array}\right\}$

Solve the above matrix Equation to obtain reactions as follows.

    $R_C=-2378\text{N}$

    $F_{BE}=4189\text{N}$

    $F_{DF}=-189.2\text{N}$

Solve the above matrix to obtain the constants.

    ${\text{C}}_1=\left(1.036\times {10}^7\right)\text{N}\cdot {\text{mm}}^2$

    $C_2=\left(-7.243\times {10}^8\right)\text{N}\cdot {\text{mm}}^2$

Substitute $4189\text{N}$ for $F_{BE}$ and $50.27{\text{mm}}^{\text{2}}$ for $A$ in Equation (XII).

    $\begin{array}{c}{\sigma }_{BE}=\frac{4189\text{N}}{50.27{\text{mm}}^{\text{2}}}\\ =83.3\text{MPa}\end{array}$

Thus, the stress in the bolt $BE$ is $83.3\text{MPa}$.

Substitute $-189.2\text{N}$ for $F_{DF}$ and $50.27{\text{mm}}^{\text{2}}$ for $A$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_{DF}=\frac{\left(-189.2\text{N}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)}\\ =-\text{3}\text{.8}\text{MPa}\end{array}$

Thus, the stress in the bolt $DF$ is $-\text{3}\text{.8}\text{MPa}$.

Calculate deflection at point $A$.

Substitute $21\times {10}^3{\text{mm}}^{\text{4}}$ for $I$, $207\times {10}^3\text{MPa}$ for $E$, $0\text{mm}$ for $x$, $4189\text{N}$ for $F_{BE}$, $2\text{kN}$ for $F$, $75\text{mm}$ for $d_1$, $150\text{mm}$ for $d_2$,$-2378\text{N}$ for $R_C$, $\left(1.036\times {10}^7\right)\text{N}\cdot {\text{mm}}^2$ for ${\text{C}}_1$ and $\left(-7.243\times {10}^8\right)\text{N}\cdot {\text{mm}}^2$ for $C_2$ in Equation (V).

    $\begin{array}{l}\left(207\times {10}^3\text{MPa}\right)\left(21\times {10}^3{\text{mm}}^{\text{4}}\right)y_A=\left[\begin{array}{l}-\left(2\text{kN}\right)\frac{{\left(0\text{mm}\right)}^3}{6}+0+0+\\ \left(1.036\times {10}^7\text{N}\cdot {\text{mm}}^2\right)\left(0\text{mm}\right)\\ +\left(-7.243\times {10}^8\text{N}\cdot {\text{mm}}^2\right)\end{array}\right]\\ y_A=\frac{-7.243\times {10}^8\text{N}\cdot {\text{mm}}^2}{\left(207\times {10}^3\right)\left(21\times {10}^3{\text{mm}}^{\text{4}}\right)}\\ y_A=-0.167\text{mm}\end{array}$

Thus, the deflection at point $A$ is $-0.167\text{mm}$.

Calculate deflection at point $B$.

Substitute $207\times {10}^3\text{MPa}$ for $E$, $4189\text{N}$ for $F_{BE}$, $50.27{\text{mm}}^{\text{2}}$ for $A$ and $50\text{mm}$ for $h_1$ in Equation (VII).

    $\begin{array}{c}{y}_B=-\frac{\left(4189\text{N}\right)\left(50\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)}\\ =-\frac{\left(4189\text{N}\right)\left(50\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)\left(\frac{1\text{N}/{\text{mm}}^2}{1\text{MPa}}\right)}\\ =-\frac{209450\text{mm}}{10405890}\\ =-0.0201\text{mm}\end{array}$

Thus, the deflection at point $C$ is $-0.0201\text{mm}$.

Calculate deflection at point $D$.

Substitute $207\times {10}^3\text{MPa}$ for $E$, $-189.2\text{N}$ for $F_{DF}$, $50.27{\text{mm}}^{\text{2}}$ for $A$ and $65\text{mm}$ for $h_2$ in Equation (VIII).

    $\begin{array}{c}{y}_D=-\frac{\left(-189.2\text{N}\right)\left(65\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)}\\ =-\frac{\left(-189.2\text{N}\right)\left(65\text{mm}\right)}{\left(50.27{\text{mm}}^{\text{2}}\right)\left(207\times {10}^3\text{MPa}\right)\left(\frac{1\text{N}/{\text{mm}}^2}{1\text{MPa}}\right)}\\ =-\frac{12298\text{mm}}{10405890}\\ =-0.00118\text{mm}\end{array}$

Thus, the deflection at point $D$ is $-0.00118\text{mm}$.