Let's break down the problem step by step: ### **1. Find the Force in Member AB** First, we need to determine the force in member **AB**. This can be done using the method of joints. The vertical load of 1000 lbs is applied at joint **C**. Due to symmetry, the reactions at **A** and **D** will be equal and the truss will be symmetric. #### **Step 1: Calculate the reactions at A and D** Sum of moments about \( D \): \[ \sum M_D = 0 \implies R_A \times 10 - 1000 \times 6 = 0 \] \[ R_A = \frac{6000}{10} = 600 \text{ lb} \] Similarly, \( R_D = 1000 - R_A = 400 \) lb. #### **Step 2: Analyze Joint A** At joint **A**, considering only vertical and horizontal components: Let \( F_{AB} \) be the force in member AB, and \( F_{AC} \) in AC. For vertical equilibrium at joint A: \[ \sum F_y = 0 \implies R_A + F_{AB} \sin \theta = 0 \] \[ 600 + F_{AB} \sin \theta = 0 \] \[ F_{AB} = -\frac{600}{\sin \theta} \] Find \( \theta \) (angle between AB and the horizontal): \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{\sqrt{4^2+6^2}} = \frac{6}{\sqrt{16+36}} = \frac{6}{\sqrt{52}} = \frac{6}{7.211} = 0.832 \] \[ F_{AB} = -\frac{600}{0.832} = -721 \text{ lb} \] (The negative sign indicates compression.) --- ### **2. Euler Buckling Formula** The critical buckling load for a pin-ended column is: \[ P_{cr} = \frac{\pi^2 E I}{(K L)^2} \] For pin-ended, \( K = 1 \). - \( E = 10 \times 10^6 \) psi - \( L \) (length AB): \( \sqrt{4^2 + 6^2} = 7.211 \) ft = \( 86.53 \) in. --- ### **3. Factor of Safety** Required \( P_{cr} \) with a FOS of 2: \[ P_{cr} \geq 2 \times |F_{AB}| = 2 \times 721 = 1442 \text{ lb} \] --- ### **4. Moment of Inertia (I) in Terms of Radius of Gyration (r)** For a solid circular cross-section: \[ I = A r^2 \] But for buckling: \[ r = \sqrt{\frac{I}{A}} \] \[ P_{cr} = \frac{\pi^2 E I}{L^2} \] \[ I = \frac{P_{cr} L^2}{\pi^2 E} \] --- ### **5. Solve for Minimum Radius of Gyration \( r \)** \[ r = \sqrt{\frac{I}{A}} \] But since \( I = A r^2 \), and we don't know A, let's use the alternate form: \[ P_{cr} = \frac{\pi^2 E A r^2}{L^2} \] \[ r^2 = \frac{P_{cr} L^2}{\pi^2 E A} \] But for buckling, the minimum radius of gyration required is: \[ r_{min} = \frac{K L}{\pi} \sqrt{\frac{FOS \cdot P}{E A}} \] Instead, since \( P_{cr} = \frac{\pi^2 E I}{L^2} \), and \( I = A r^2 \), rearrange: \[ r = \frac{L}{\pi} \sqrt{\frac{P_{cr}}{E A}} \] But since A cancels for solid circular sections, textbooks often use: \[ \frac{L}{r} \leq \sqrt{\frac{\pi^2 E}{F_{max}}} \] Where \( F_{max} = \frac{P_{cr}}{A} \). But let's use: \[ I = \frac{P_{cr} L^2}{\pi^2 E} \] For a solid circular section: \[ I = \frac{\pi}{64} d^4 \] \[ A = \frac{\pi}{4} d^2 \] \[ r = \sqrt{\frac{I}{A}} = \sqrt{\frac{\frac{\pi}{64} d^4}{\frac{\pi}{4} d^2}} = \sqrt{\frac{1}{16} d^2} = \frac{d}{4} \] So, \[ I = A r^2 \] So finally, \[ r = \sqrt{\frac{I}{A}} \] But let's use: \[ r = \sqrt{\frac{I}{A}} = \frac{L}{\pi} \sqrt{\frac{P_{cr}}{E A}} \] But since the minimum radius of gyration is independent of A for a minimum requirement: \[ r_{min} = \sqrt{\frac{P_{cr} L^2}{\pi^2 E A}} \] But you can also use: \[ P_{cr} = \frac{\pi^2 E A r^2}{L^2} \] So, \[ r^2 = \frac{P_{cr} L^2}{\pi^2 E A} \] But for minimum, let’s just use: \[ r_{min} = \frac{L}{\pi} \sqrt{\frac{P_{cr}}{E}} \] --- ### **6. Plug in Values** - \( L = 7.211 \) ft = \( 86.53 \) in - \( E = 10 \times 10^6 \) psi - \( P_{cr} = 1442 \) lb \[ r_{min} = \frac{86.53}{\pi} \sqrt{\frac{1442}{10 \times 10^6}} \] Calculate: \[ \frac{86.53}{\pi} = 27.54 \] \[ \frac{1442}{10,000,000} = 0.0001442 \] \[ \sqrt{0.0001442} = 0.01201 \] \[ r_{min} = 27.54 \times 0.01201 = 0.331 \text{ in} \] --- ## **Final Answer** **The minimum radius of gyration of member AB to prevent Euler buckling with a factor of safety of 2 is:** \[ \boxed{0.33 \text{ in}} \] If you need the minimum diameter for a solid circular section, recall \( r = \frac{d}{4} \), so: \[ d_{min} = 4 \times r_{min} = 4 \times 0.33 = 1.32 \text{ in} \]