# Solution: Maximum Shear Stress at Point A Let's break down the problem step by step to determine the **absolute maximum shear stress at point A**. --- ## 1. **Identify Forces and Moments at Point A** ### **Given:** - Point A is on the rectangular section (40 mm × 25 mm). - Various forces and a distributed load act on the structure. - Internal stresses at A will result from: - Shear force (\(V\)) - Bending moment (\(M\)) - Torsion (\(T\)) --- ## 2. **Internal Forces at A** Let's identify the types of stresses at point A: ### **A. Shear Force (\(V\))** - Direct shear force due to external loads. ### **B. Bending Moment (\(M\))** - Generates normal (bending) stresses, but for maximum shear, we focus on the shear due to transverse loading. ### **C. Torsion (\(T\))** - Torsional moment creates shear across the section. --- ## 3. **Shear Stress Calculation** ### **A. Shear Stress Due to Transverse Shear Force** For a **rectangular section**: \[ \tau_{\text{shear}} = \frac{V Q}{I b} \] Where: - \(V\) = Shear force at section - \(Q\) = First moment of area about the neutral axis - \(I\) = Moment of inertia - \(b\) = Width at point A ### **B. Shear Stress Due to Torsion** For a **rectangular section** (approximate): \[ \tau_{\text{torsion, max}} = \frac{T}{K} \] Where \(K\) is the torsional modulus for a rectangle But, **for maximum shear stress**, we check both from direct shear and from torsion and combine appropriately. --- ## 4. **Formulas and Section Properties** ### **A. Section Properties for Rectangle** - \(b = 25\, \text{mm} = .025\, \text{m}\) - \(h = 40\, \text{mm} = .04\, \text{m}\) - \(I = \frac{bh^3}{12} = \frac{.025 \times (.04)^3}{12} = 1.33 \times 10^{-6} \,\text{m}^4\) - \(Q = \frac{b h^2}{8} = \frac{.025 \times (.04)^2}{8} = 5 \times 10^{-5} \,\text{m}^3\) ### **B. Shear Force at A** - From the diagram, the 8 kN/m distributed load and the point loads will generate a reaction at A. Assume the section cut at A includes all left-side loads. - **Sum up vertical forces at A.** #### **Distributed Load Contribution:** \[ w = 8\, \text{kN/m} \times .5\, \text{m} = 4\, \text{kN} \] #### **Add Point Loads:** - 2 kN, 1 kN (as shown), and the 4 kN from distributed load. \[ V_A = 4\, \text{kN} + 2\, \text{kN} + 1\, \text{kN} = 7\, \text{kN} \] (Assuming all act vertically and in the same sense) --- ## 5. **Calculate Shear Stress at A** ### **A. Shear Stress from Shear Force** \[ \tau_{\text{shear}} = \frac{V Q}{I b} \] \[ Q = \frac{b h^2}{8} = \frac{.025 \times (.04)^2}{8} = .000005\, \text{m}^3 \] \[ I = \frac{bh^3}{12} = \frac{.025 \times (.04)^3}{12} \approx .00000133\, \text{m}^4 \] \[ b = .025\, \text{m} \] \[ V = 700\, \text{N} \] \[ \tau_{\text{shear}} = \frac{700 \times .000005}{.00000133 \times .025} \] \[ \tau_{\text{shear}} = \frac{.035}{.00003325} \] \[ \tau_{\text{shear}} \approx 1053.38\, \text{kPa} = 1.05\, \text{MPa} \] --- ### **B. Shear Stress from Torsion** For a **rectangular shaft**, the max torsional shear stress is: \[ \tau_{\text{torsion, max}} = \frac{T}{\frac{1}{3} b h^2} \] But from the diagram, A is on the rectangular section, and torsion appears to act mainly on the cylindrical bar, not the rectangular section. Thus, **torsional shear at A can be neglected**. --- ## 6. **Absolute Maximum Shear Stress** Since torsion at A is negligible (as per diagram setup), and the max shear occurs at the **neutral axis** for a rectangle: \[ \boxed{\tau_{\text{max}} \approx 1.05\, \text{MPa}} \] --- ## 7. **Summary Table** | Contribution | Shear Stress (MPa) | |----------------------|-------------------| | Shear force (V) | 1.05 | | Torsion (T) | (negligible) | --- ## **Final Answer** \[ \boxed{ \text{The absolute maximum shear stress at point A is approximately } 1.05\, \text{MPa}. } \] --- ### **If you need the calculation for a different section (cylindrical bar), let me know!**

# Maximum Shear Stress at Point A To calculate the absolute maximum shear stress at point A, we need to consider the internal forces acting on the section, specifically the shear force and the bending moment. --- ## 1. **Understanding the Problem** ### **Given Information:** - Dimensions of the section: - Width \( b = 25 \, \text{mm} = 0.025 \, \text{m} \) - Height \( h = 40 \, \text{mm} = 0.040 \, \text{m} \) - External loads: - Shear force \( V = 7 \, \text{kN} = 7000 \, \text{N} \) - Geometry is rectangular. --- ## 2. **Section Properties** ### **A. Moment of Inertia (\( I \))** For a rectangle: \[ I = \frac{b h^3}{12} = \frac{0.025 \times (0.040)^3}{12} \approx 1.333 \times 10^{-7} \, \text{m}^4 \] ### **B. First Moment of Area (\( Q \))** For the area above the neutral axis: \[ Q = \frac{b h^2}{8} = \frac{0.025 \times (0.040)^2}{8} = 5.0 \times 10^{-6} \, \text{m}^3 \] --- ## 3. **Calculating Shear Stress** ### **A. Shear Stress from Transverse Shear Force** Using the shear stress formula: \[ \tau_{\text{shear}} = \frac{V Q}{I b} \] Substituting the known values: \[ \tau_{\text{shear}} = \frac{7000 \times 5.0 \times 10^{-6}}{1.333 \times 10^{-7} \times 0.025} \] #### **Calculating Values:** - Numerator: \[ 7000 \times 5.0 \times 10^{-6} = 0.035 \, \text{N·m} \] - Denominator: \[ 1.333 \times 10^{-7} \times 0.025 = 3.333 \times 10^{-9} \, \text{m}^4 \] ### **B. Resulting Shear Stress** \[ \tau_{\text{shear}} = \frac{0.035}{3.333 \times 10^{-9}} \approx 10.5 \times 10^6 \, \text{Pa} = 10.5 \, \text{MPa} \] --- ## 4. **Torsional Shear Stress** If torsion is acting on point A, it should be considered as well. However, if it is negligible due to the cross-section being primarily loaded by shear and bending moments, we conclude that: ### **Final Result** \[ \boxed{\tau_{\max} \approx 10.5 \, \text{MPa}} \] ### **Conclusion** The maximum shear stress at point A is approximately **10.5 MPa** based on transverse shear alone. If torsion must be considered, further calculations would be necessary to combine the shear stresses appropriately.