Let's break down and solve the given problem step by step. --- ## **Given Data** - **Yield strength of aluminum strip (\(\sigma_{yield}\))**: 73 ksi - **Thickness (t)**: .25 in - **Length (L)**: 15 in - **Bending moment (M)**: Not directly given, but strip is bent as shown - **Assume:** Maximum stress occurs at the outer fiber. - **You need to find:** 1. The maximum stress and the angle of twist (in degrees) if the strip is assumed to be a thin rectangular section (shear stress) 2. The same if the strip is assumed solid rectangular section under torsional shear. --- ## **Step 1: Understanding the Geometry & Loading** - The strip is bent as shown in the image, so *torsion* is involved. - For maximum shear stress in a **thin rectangular section** (like a thin-walled tube), the formula changes compared to a **solid rectangular section**. Let’s **assume the force is applied at the end to produce torsion** about the longitudinal axis (since the problem mentions torsion and angle of twist). --- ## **Step 2: Thin Rectangular Strip (Shear Stress and Angle of Twist)** #### **A. Shear Stress in Thin-Walled Rectangular Section** For a thin-walled rectangular section: \[ \tau_{max} = \frac{T}{2 t A_m} \] Where: - \(T\) = applied torque (unknown, but let's relate to yield strength) - \(t\) = wall thickness = .25" - \(A_m\) = mean area enclosed by the median line But more commonly (for a thin strip of width \(b\) and thickness \(t\)): \[ \tau_{max} = \frac{T}{b t^2} \] But we need the **maximum torque** that causes the strip to reach yield, i.e., \(\tau_{max} = \sigma_{yield}\): \[ \tau_{max} = \sigma_{yield} = 73 \text{ ksi} = 73,000 \text{ psi} \] So, \[ T_{max} = \tau_{max} \cdot b t^2 \] But we don’t have the width (\(b\)), but if this is a strip, let’s say the width is much greater than thickness (thin-walled approximation). #### **B. Angle of Twist (in degrees)** The angle of twist for a thin-walled section: \[ \theta = \frac{TL}{G J} \] Where: - \(T\): Applied torque - \(L\): Length (\(15 \text{ in.}\)) - \(G\): Shear modulus for aluminum (\(\approx 3.8 \times 10^6\) psi) - \(J\): Torsional constant for thin rectangle (\(J \approx \frac{1}{3} b t^3\)) But again, we need width (\(b\)). If width is not specified, we may leave the answer in terms of \(b\). --- ## **Step 3: Solid Rectangular Section (Shear Stress and Angle of Twist)** #### **A. Shear Stress in Solid Rectangular Section** Maximum torsional shear stress for a solid rectangle: \[ \tau_{max} = \frac{T}{c W t^2} \] But more accurately, maximum shear stress for a solid rectangle: \[ \tau_{max} = \frac{T}{\alpha b t^2} \] Where \(\alpha\) is a coefficient (for a rectangle, \(\alpha\) depends on the aspect ratio). Or, using the approximation: \[ \tau_{max} = \frac{T}{J} \cdot c \] Where \(c\) is the distance from the center to the outside edge, \(J\) is the torsion constant: \[ J = \frac{1}{3} b t^3 \text{ (thin wall)} \] \[ J \approx .141 b t^3 \text{ (solid rectangle, width \(b\) > thickness \(t\))} \] #### **B. Angle of Twist** \[ \theta = \frac{TL}{GJ} \] --- ## **Step 4: Final Answers (in terms of width \(b\)), with Explanations** ### **(a) Thin Rectangular Section** **Maximum Shear Stress:** \[ \tau_{max} = \frac{T}{b t^2} \] Set \(\tau_{max} = \sigma_{yield}\): \[ T_{max} = \sigma_{yield} \cdot b t^2 = 73,000 \cdot b \cdot (.25)^2 = 73,000 \cdot b \cdot .0625 = 4,562.5b \text{ in-lb} \] **Angle of Twist:** \[ \theta = \frac{T_{max} L}{G J} \] For thin rectangle: \(J = \frac{1}{3} b t^3\) \[ J = \frac{1}{3} \cdot b \cdot (.25)^3 = \frac{1}{3} \cdot b \cdot .015625 = .005208b \text{ in}^4 \] \[ \theta = \frac{4,562.5b \cdot 15}{3.8 \times 10^6 \cdot .005208b} \] \(b\) cancels out: \[ \theta = \frac{4,562.5 \times 15}{3.8 \times 10^6 \times .005208} \] \[ = \frac{68,437.5}{19,790} \] \[ \approx 3.46 \text{ radians} \] Convert to degrees: \[ \theta_{deg} = 3.46 \times \frac{180}{\pi} \approx 198.3^\circ \] --- ### **(b) Solid Rectangular Section** **Maximum Shear Stress:** \[ \tau_{max} = \frac{T_{max} c}{J} \] Where \(c = t/2 = .125\) in For solid rectangle: \(J = .141 b t^3\) \[ T_{max} = \tau_{max} \cdot \frac{J}{c} = \sigma_{yield} \cdot \frac{.141 b t^3}{.125} \] \[ T_{max} = 73,000 \cdot \frac{.141b(.25)^3}{.125} \] \[ = 73,000 \cdot \frac{.141b \cdot .015625}{.125} \] \[ = 73,000 \cdot \frac{.0022b}{.125} \] \[ = 73,000 \cdot .0176b \] \[ = 1,284.8b \text{ in-lb} \] **Angle of Twist:** \[ \theta = \frac{T_{max} L}{G J} \] \[ J = .141 b t^3 = .141 b (.25)^3 = .0022b \text{ in}^4 \] \[ \theta = \frac{1,284.8b \cdot 15}{3.8 \times 10^6 \cdot .0022b} \] \(b\) cancels out: \[ \theta = \frac{19,272}{8,360} \] \[ = 2.31 \text{ radians} \] Convert to degrees: \[ \theta_{deg} = 2.31 \times \frac{180}{\pi} \approx 132.4^\circ \] --- ## **Final Answers (Summarized)** ### **(a) Thin Rectangular Section** - **Maximum Torque:** \(T_{max} = 4,562.5b\) in-lb - **Angle of Twist:** \(\theta \approx 3.46\) radians, or \(\boxed{198.3^\circ}\) ### **(b) Solid Rectangular Section** - **Maximum Torque:** \(T_{max} = 1,284.8b\) in-lb - **Angle of Twist:** \(\theta \approx 2.31\) radians, or \(\boxed{132.4^\circ}\) --- **Assumptions:** - Width (\(b\)) was not given, so answers are in terms of \(b\). - Shear modulus of aluminum (\(G\)) approximated as \(3.8 \times 10^6\) psi. If you have the width, you can substitute \(b\) to get numeric values! If you need further clarification or numeric answers for a specific width, let me know!