Let's solve the problem **step-by-step**: --- # 1. Understanding the Problem - **Composite beam:** 3 boards, bolted together. - **Bolt diameter:** 10 mm. - **Allowable shear stress per bolt:** \( \tau_B \) - **Shear force applied:** \( V \) - **Spacing between bolts:** \( s \) - **Goal:** (a) Find the maximum spacing \( s_{max} \) so that bolts can safely transfer the shear force \( V \). (b) Suggest ways to double the bolt spacing, using equations. --- ## (a) **Determine Maximum Bolt Spacing \( s_{max} \)** ### **Step 1: Shear Flow Calculation** The **shear flow** \( q \) (force per unit length transferred between boards) is: \[ q = \frac{VQ}{I} \] Where: - \( V \) = Shear force on beam - \( Q \) = First moment of area about neutral axis for the portion above (or below) the bolt line - \( I \) = Second moment of area (moment of inertia) of the entire cross section --- ### **Step 2: Shear Force per Bolt** - Shear force per bolt = \( q \cdot s \) (where \( s \) = spacing between bolts) --- ### **Step 3: Shear Strength of One Bolt** - **Shear area of one bolt:** \( A_{bolt} = \frac{\pi}{4} d^2 \) (For \( d = 10\, \text{mm} \)) - **Maximum allowable shear force per bolt:** \( F_{bolt,max} = \tau_B \cdot A_{bolt} \) --- ### **Step 4: Equate Bolt Capacity to Shear Flow** Set maximum force per bolt equal to the force resisted by shear flow over spacing \( s \): \[ q \cdot s_{max} = F_{bolt,max} \] \[ s_{max} = \frac{F_{bolt,max}}{q} \] --- ### **Step 5: Final Expression for \( s_{max} \)** Substitute values: \[ s_{max} = \frac{\tau_B \cdot A_{bolt}}{q} \] \[ s_{max} = \frac{\tau_B \cdot \frac{\pi}{4}d^2}{\frac{VQ}{I}} \] \[ s_{max} = \frac{\tau_B \cdot \frac{\pi}{4}d^2 \cdot I}{VQ} \] --- ### **Summary Formula** \[ \boxed{ s_{max} = \frac{\tau_B \cdot \frac{\pi}{4}d^2 \cdot I}{VQ} } \] **Where:** - \( \tau_B \) = Allowable shear stress in bolt - \( d \) = Bolt diameter - \( I \) = Moment of inertia of full cross section - \( Q \) = First moment of area of section above or below bolt line - \( V \) = Applied shear force --- ## (b) **Suggest Two Ways to Double Bolt Spacing** To **double the bolt spacing** (\( 2s_{max} \)), you must either: 1. **Increase the bolt's shear capacity** 2. **Decrease the required shear flow** ### **Option 1: Increase Bolt Shear Capacity** - Increase \( d \) (bolt diameter): \( A_{bolt} \propto d^2 \) - Use bolts with higher allowable shear stress (\( \tau_B \)) **Equation:** \[ s_{max,new} = \frac{\tau_{B,new} \cdot \frac{\pi}{4}d_{new}^2 \cdot I}{VQ} \] To **double** \( s_{max} \): \[ \frac{\tau_{B,new} \cdot d_{new}^2}{\tau_B \cdot d^2} = 2 \] So, either double \( \tau_B \), or increase \( d \) by \( \sqrt{2} \) times. --- ### **Option 2: Reduce Shear Flow \( q \)** - Reduce applied shear force \( V \) (not practical in design) - Increase \( Q \) or \( I \) by redesigning the cross-section (e.g., make the beam deeper or wider) - **Split the shear force among more interface layers** (e.g., use more boards, so each interface carries less force) **Equation:** \[ q = \frac{VQ}{I} \] If you can **reduce \( q \) by half**, spacing \( s \) can double. --- ## **Final Answers** ### **(a) Maximum Bolt Spacing** \[ \boxed{ s_{max} = \frac{\tau_B \cdot \frac{\pi}{4}d^2 \cdot I}{VQ} } \] ### **(b) Two Design Concepts to Double Bolt Spacing** 1. **Use larger or stronger bolts:** Increase \( d \) or \( \tau_B \) so that \( A_{bolt} \) or allowable force per bolt doubles. 2. **Reduce interface shear flow:** Increase \( I \) or reduce \( Q \) (change cross-section), or split the force among more interfaces. --- If you want to see a **numerical example**, let me know!