Absolutely! Let's break this down step by step. We’ll focus on the mechanics and **how to solve for the normal stress** in link DE for both cases. --- ## **Given Data** - **Thickness of DE, \( t \) = .125 in** - **Width of DE, \( w \) = .9 in** - **Force at F = 60 lb** (acts at angle θ) - Find: **Normal stress in DE** for: - (a) \( \theta = ^\circ \) - (b) \( \theta = 90^\circ \) --- ## **Step 1: Cross-sectional Area of DE** \[ A = \text{width} \times \text{thickness} = .9\, \text{in} \times .125\, \text{in} = .1125\, \text{in}^2 \] --- ## **Step 2: Free Body Diagram (FBD) of CEF** To analyze the force in link DE, we **isolate member CEF**. The 60 lb force at F can be broken into components based on θ: - **Horizontal component:** \( 60 \cos\theta \) - **Vertical component:** \( 60 \sin\theta \) --- ## **Step 3: Find the Force in DE (F_DE)** ### **Geometry** - Link DE is oriented **vertically** (from D to E, parallel to y-axis). - The force in DE, \( F_{DE} \), will be either in **tension or compression** depending on θ. --- ### **Case (a): \( \theta = ^\circ \)** - Force at F acts **horizontally** to the right. #### **Sum of Forces in x-direction:** Since DE is vertical, it **does not carry horizontal force**; the horizontal force is carried by other members. Therefore, the force in DE is due only to vertical equilibrium. #### **Sum of Forces in y-direction:** - Only vertical components contribute to DE. - For \( \theta = ^\circ \), \( \sin ^\circ = \). - So, **no vertical force at F**. ##### **Conclusion:** \[ F_{DE} = \] \[ \sigma_{DE} = \frac{F_{DE}}{A} = \, \text{psi} \] --- ### **Case (b): \( \theta = 90^\circ \)** - Force at F acts **vertically upward**. #### **Sum of Forces in y-direction:** Now, the full 60 lb acts vertically and must be balanced by DE (since DE is the only vertical member in CEF). \[ F_{DE} = 60\, \text{lb} \quad \text{(in compression)} \] \[ \sigma_{DE} = \frac{F_{DE}}{A} = \frac{60}{.1125} = 533.3\, \text{psi (compression)} \] --- ## **Step 4: Final Answers** ### (a) **θ = °** \[ \sigma_{DE} = \, \text{psi} \] ### (b) **θ = 90°** \[ \sigma_{DE} = 533\, \text{psi (compression)} \] --- ## **Summary Table** | θ | \( F_{DE} \) (lb) | \( \sigma_{DE} \) (psi) | |------------|-------------------|-------------------------| | \( ^\circ \) | | | | \( 90^\circ \) | 60 | 533 | --- **Let me know if you want to see the equilibrium equations or a free-body diagram!**

# Explanation of Normal Stress in Link DE Let's analyze the normal stress in link DE using the given data and breaking it down into steps. ## Given Data - **Thickness of DE, \( t \) = 0.125 in** - **Width of DE, \( w \) = 0.9 in** - **Force at F = 60 lb** (acts at angle \( \theta \)) ### Objective We will find the normal stress in DE for two cases: 1. \( \theta = 0^\circ \) 2. \( \theta = 90^\circ \) ## Step 1: Cross-sectional Area of DE The cross-sectional area \( A \) of DE is calculated as follows: \[ A = w \times t = 0.9\, \text{in} \times 0.125\, \text{in} = 0.1125\, \text{in}^2 \] ## Step 2: Free Body Diagram (FBD) of CEF To isolate the forces in link DE, we consider the force \( F \) acting at an angle \( \theta \): - **Horizontal component:** \( F_x = 60 \cos\theta \) - **Vertical component:** \( F_y = 60 \sin\theta \) ## Step 3: Force in DE (\( F_{DE} \)) ### Geometry Link DE is oriented vertically. The force in DE, \( F_{DE} \), is determined based on the angle \( \theta \). ### Case (a): \( \theta = 0^\circ \) Here, the force acts horizontally to the right. As DE is vertical, it does not carry horizontal forces. Thus, the vertical force from \( F \) is zero: \[ F_{DE} = 0 \] Calculating the normal stress: \[ \sigma_{DE} = \frac{F_{DE}}{A} = \frac{0}{0.1125} = 0\, \text{psi} \] ### Case (b): \( \theta = 90^\circ \) In this case, the force acts vertically upward, fully contributing to the load in DE: \[ F_{DE} = 60\, \text{lb} \quad \text{(in compression)} \] Calculating the normal stress: \[ \sigma_{DE} = \frac{F_{DE}}{A} = \frac{60}{0.1125} = 533.3\, \text{psi (compression)} \] ## Final Answers - For \( \theta = 0^\circ \): - \( \sigma_{DE} = 0\, \text{psi} \) - For \( \theta = 90^\circ \): - \( \sigma_{DE} = 533.3\, \text{psi (compression)} \) Feel free to ask if you need more details or clarifications!