Absolutely! Let’s break down the problem step by step: --- # **Problem Summary** - **Given:** - A beam supported by pins at \(A\), \(B\), and \(C\). - Loads \(P\) act downward at two points. - Pins have **double shear**. - **Maximum average shear stress** allowed: \(70~\text{MPa}\). - **Pin diameter:** \(18~\text{mm}\). - **Find:** - Maximum allowable \(P\) such that shear stress in any pin **does not exceed** \(70~\text{MPa}\). --- ## **Step 1: Understanding Double Shear** In **double shear**, each pin splits the force across **two shear planes** (see diagram below): ``` |---[Shear Plane]---|Pin|---[Shear Plane]---| ``` So, for a force \(F\) on the pin, each shear plane resists \(F/2\). --- ## **Step 2: Shear Stress Formula** The **average shear stress** in a pin is: \[ \tau = \frac{F}{nA} \] Where: - \(\tau\): Average shear stress - \(F\): Total force transmitted by the pin - \(n\): Number of shear planes (for **double shear**, \(n = 2\)) - \(A\): Area of pin cross-section (\(A = \frac{\pi d^2}{4}\)) --- ## **Step 3: Area of Pin Cross-Section** Given \(d = 18~\text{mm} = .018~\text{m}\): \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.018)^2 \approx 2.545 \times 10^{-4}~\text{m}^2 \] --- ## **Step 4: Maximum Allowable Force per Pin** Given \(\tau_{\text{max}} = 70~\text{MPa} = 70 \times 10^6~\text{Pa}\): \[ F_{\text{max}} = nA \tau_{\text{max}} = 2 \times 2.545 \times 10^{-4}~\text{m}^2 \times 70 \times 10^6~\text{Pa} \] \[ F_{\text{max}} = 2 \times 2.545 \times 10^{-4} \times 70 \times 10^6 \] \[ F_{\text{max}} = 2 \times 2.545 \times 70 \times 10^2 \] \[ F_{\text{max}} = 2 \times 2.545 \times 700 \] \[ F_{\text{max}} = 2 \times 17815 \] \[ F_{\text{max}} = 35630~\text{N} \] --- ## **Step 5: Relating Pin Shear Force to Load \(P\)** You need to determine **which pin experiences the highest shear force** due to \(P\). This requires a **free-body diagram** and equilibrium analysis. But usually, in this type of setup, the pin at **A** or **B** will see the largest force, depending on the geometry and the loads applied. ### **Quick Check (Pin at B):** If you sum forces and moments, pin B is likely to experience the full load \(P\) (worst case). So, **maximum \(P\) is when the pin force equals \(F_{\text{max}}\):** \[ P_{\text{max}} = F_{\text{max}} \] --- ## **Step 6: Final Answer with Units and Significant Figures** \[ \boxed{P_{\text{max}} = 3.56 \times 10^4~\text{N}~\text{or}~35.6~\text{kN}} \] --- ## **Summary Table** | Quantity | Value | |----------|-------| | Pin diameter (\(d\)) | \(18~\text{mm} = .018~\text{m}\) | | Max shear stress (\(\tau_{max}\)) | \(70~\text{MPa}\) | | Area (\(A\)) | \(2.55 \times 10^{-4}~\text{m}^2\) | | Max force per pin (\(F_{max}\)) | \(35,600~\text{N}\) | | **Max Load \(P\)** | \(\mathbf{35.6~\text{kN}}\) | --- **If you want to see the full equilibrium calculation (to check exactly which pin is critical), let me know!**

# **Understanding the Beam and Shear Forces** Let’s analyze the problem involving a beam supported by pins and subjected to loads. ## **Problem Summary** - **Beam Supports:** Pins at points \(A\), \(B\), and \(C\). - **Loads:** Downward forces \(P\) applied at two points. - **Double Shear:** Each pin experiences shear across two planes. - **Maximum Shear Stress:** \(70 \, \text{MPa}\). - **Pin Diameter:** \(18 \, \text{mm}\). - **Objective:** Find the maximum allowable load \(P\) such that shear stress does not exceed \(70 \, \text{MPa}\). ## **Step 1: Double Shear Concept** In double shear, each pin divides the load across two shear planes: ``` |---[Shear Plane]---|Pin|---[Shear Plane]---| ``` This means if a pin bears a force \(F\), each shear plane carries \(F/2\). ## **Step 2: Shear Stress Calculation** The average shear stress \(\tau\) is defined as: \[ \tau = \frac{F}{nA} \] Where: - \(F\) = Total force on the pin - \(n\) = Number of shear planes (2 for double shear) - \(A\) = Cross-sectional area of the pin: \(A = \frac{\pi d^2}{4}\) ## **Step 3: Cross-Sectional Area of the Pin** Given the diameter \(d = 18 \, \text{mm} = 0.018 \, \text{m}\): \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.018)^2 \approx 2.545 \times 10^{-4} \, \text{m}^2 \] ## **Step 4: Maximum Allowable Force on Each Pin** Using the maximum shear stress \(\tau_{\text{max}} = 70 \, \text{MPa} = 70 \times 10^6 \, \text{Pa}\): \[ F_{\text{max}} = nA \tau_{\text{max}} = 2 \times 2.545 \times 10^{-4} \, \text{m}^2 \times 70 \times 10^6 \, \text{Pa} \] Calculating: \[ F_{\text{max}} \approx 2 \times 2.545 \times 70 \times 10^2 = 35630 \, \text{N} \] ## **Step 5: Relating Pin Shear Force to Load \(P\)** To find the maximum allowable load \(P\), identify which pin experiences the highest shear force. Typically, this is either pin \(A\) or pin \(B\). Assuming pin \(B\) bears the full load in a worst-case scenario: \[ P_{\text{max}} = F_{\text{max}} \approx 35630 \, \text{N} \] ## **Final Answer** The maximum allowable load \(P\) is: \[ \boxed{P_{\text{max}} = 35.6 \, \text{kN}} \] ## **Summary** - **Pin Diameter:** \(18 \, \text{mm}\) - **Max Shear Stress:** \(70 \, \text{MPa}\) - **Area of Pin:** \(2.545 \times 10^{-4} \, \text{m}^2\) - **Max Force per Pin:** \(35,630 \, \text{N}\) - **Max Load \(P\):** \(35.6 \, \text{kN}\) Feel free to ask for further clarification or detailed steps on any part!