Chapter 6.5, Problem 56P

6.56 and 6.57 A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

Fig. p6.56

To determine

The average shearing stress in the bolts.

Answer to Problem 56P

The average shearing stress in the bolts is $\underset{\_}{6.73\text{MPa}}$.

Explanation of Solution

Given information:

The diameter of the bolts is $12\text{mm}$.

The longitudinal spacing is $200\text{mm}$.

The beam is subjected to a vertical shear of $\text{4 kN}$

The modulus of elasticity for wood $E_w$ is $10\text{GPa}$.

The modulus of elasticity for steel $E_s$ is $200\text{GPa}$.

Calculation:

Consider the steel is to be the reference material. So modular ratio of steel is $n_s=1$.

Calculate the modular ratio of timber wood $n_w$ as shown below.

$n_w=\frac{E_w}{E_s}$

Here, $E_w$ is the modulus of elasticity of wood and $E_s$ is the modulus of elasticity of steel.

Substitute $10\text{GPa}$ for $E_w$ and $200\text{GPa}$ for $E_s$.

$\begin{array}{c}{n}_w=\frac{10}{200}\\ =0.05\end{array}$

Total depth of the section d is as follows:

$\begin{array}{c}d=90+84+90\\ =264\text{mm}\end{array}$

Calculate the moment of inertia for the symmetric section I as shown below.

$I=\frac{b{d}^3}{12}$

Here, b is the width of the section and d is the depth of the section.

For steel:

$\begin{array}{c}{I}_s=2\times \frac{6\times {264}^3}{12}\\ =18.4\times {10}^6{\text{mm}}^4\end{array}$

For wood:

$\begin{array}{c}{I}_w=\frac{1}{12}\times 140\times \left({264}^3-{84}^3\right)\\ =207.75\times {10}^6{\text{mm}}^4\end{array}$

Calculate the moment of inertia for the transformed section as shown below.

$I=n_s{I}_s+n_w{I}_w$

Substitute 1 for $n_s$, $18.4\times {10}^6{\text{mm}}^4$ for $I_s$, $0.05$ for $n_w$, and $207.75\times {10}^6{\text{mm}}^4$ for $I_w$.

$\begin{array}{c}I=1\times 18.4\times {10}^6+0.05\times 207.75\times {10}^6\\ =28.7875\times {10}^6{\text{mm}}^4\end{array}$

Calculate the first moment of area as shown below.

$Q={\displaystyle \sum A\overline{y}}$ (1)

For wooden section:

$\begin{array}{c}{Q}_w=90\times 40\times \left(42+45\right)\\ =1.0962\times {10}^6{\text{mm}}^3\end{array}$

Calculate the first moment of area for the transformed section Q as shown below.

$Q=n_w{Q}_w$

Substitute $0.05$ for $n_w$ and $1.0962\times {10}^6{\text{mm}}^3$ for $Q_w$.

$\begin{array}{c}Q=0.05\times 1.0962\times {10}^6\\ =54,810{\text{mm}}^3\end{array}$

Calculate the horizontal shear per unit length q as shown below.

$q=\frac{VQ}{I}$ (2)

Here V is the vertical shear.

Substitute $4\text{kN}$ for V, $54,810{\text{mm}}^3$ for Q, and $28.7875\times {10}^6{\text{mm}}^4$ for I in Equation (2).

$\begin{array}{c}q=\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 54,810{\text{mm}}^3}{28.7875\times {10}^6{\text{mm}}^4}\\ =\frac{219.24\times {10}^6}{28.7875\times {10}^6}\\ =7.62\text{N}/\text{mm}\end{array}$

Calculate the force acting on the bolt $F_{\text{bolt}}$ as shown below.

$F_{\text{bolt}}=qs$

Here, s is the longitudinal spacing.

Substitute $7.62\text{N}/\text{mm}$ for q and $200\text{mm}$ for s.

$\begin{array}{c}q=7.62\times 200\\ =1,524\text{N}\end{array}$

Calculate the area of bolt $A_{\text{bolt}}$ as shown below.

$A_{\text{bolt}}=\frac{\pi d_{\text{bolt}}^2}{4}$

Here, $d_{\text{bolt}}$ is the diameter of the bolt.

Substitute $12\text{mm}$ for $d_{\text{bolt}}$.

$\begin{array}{c}{A}_{\text{bolt}}=\frac{\pi \times {12}^2}{4}\\ =113.097{\text{mm}}^2\end{array}$

The bolt is subjected to double shear.

Calculate the shearing stress of the bolt ${\tau }_{\text{bolt}}$ as shown below.

${\tau }_{\text{bolt}}=\frac{F_{\text{bolt}}}{2A_{\text{bolt}}}$

Substitute $1,524\text{N}$ for $F_{\text{bolt}}$ and $113.097{\text{mm}}^2$ for $A_{\text{bolt}}$.

$\begin{array}{c}{\tau }_{\text{bolt}}=\frac{1,524}{2\times 113.097}\\ =6.73\text{N}/{\text{mm}}^2\times \frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\\ =6.73\text{MPa}\end{array}$

Therefore, the average shearing stress in the bolts is $\underset{\_}{6.73\text{MPa}}$.

To determine

The shearing stress at the center of the cross section.

Answer to Problem 56P

The shearing stress at the center of the cross section is $\underset{\_}{1.515\text{MPa}}$.

Explanation of Solution

Given information:

The diameter of the bolts is $12\text{mm}$.

The longitudinal spacing is $200\text{mm}$.

The beam is subjected to a vertical shear of $\text{4 kN}$

The modulus of elasticity for wood $E_w$ is $10\text{GPa}$.

The modulus of elasticity for steel $E_s$ is $200\text{GPa}$.

Calculation:

Refer to part (a).

Moment of inertia for the transformed section $I=28.7875\times {10}^6{\text{mm}}^4$.

Calculate the first moment of area as shown below.

For the two steel plates:

$\begin{array}{c}{Q}_s=2\times 6\times \left(90+42\right)\times \left(90-42\right)\\ =76,032{\text{mm}}^3\end{array}$

Calculate the first moment of area along the neutral axis for the transformed section as shown below.

$Q=Q_w+Q_s$

Substitute $54,810{\text{mm}}^3$ for $Q_w$ and $76,032{\text{mm}}^3$ for $Q_s$.

$\begin{array}{c}Q=54,810+76,032\\ =130,842{\text{mm}}^3\end{array}$

Calculate the horizontal shear per unit length as shown below.

Substitute $4\text{kN}$ for V, $130,842{\text{mm}}^3$ for Q, and $28.7875\times {10}^6{\text{mm}}^4$ for I in Equation (2).

$\begin{array}{c}q=\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 130,842{\text{mm}}^3}{28.7875\times {10}^6{\text{mm}}^4}\\ =\frac{523.368\times {10}^6}{28.7875\times {10}^6}\\ =18.18\text{N}/\text{mm}\end{array}$

Calculate the shearing stress as shown below.

$\tau =\frac{q}{2t}$

Substitute $18.18\text{N}/\text{mm}$ for q and $6\text{mm}$ for t.

$\begin{array}{c}\tau =\frac{18.18}{2\times 6}\\ =1.515\text{N}/{\text{mm}}^2\times \frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\\ =1.515\text{MPa}\end{array}$

Therefore, the shearing stress at the center of the cross section is $\underset{\_}{1.515\text{MPa}}$.