Let's break down and solve the problem **step-by-step**. --- ## **Problem Summary** Two wooden planks (80 × 170 mm cross-section) are glued together at an angle (45°), and a force \( P \) is applied. - Allowable **shear stress in the glue**: \( 850 \) kPa - Allowable **normal (tensile/compressive) stress in the wood**: \( 12 \) MPa - Find the **largest centric load \( P \)** that can be applied. --- ## **Step 1: Area of Cross Section** Cross-sectional area \( A \): \[ A = 80\,\text{mm} \times 170\,\text{mm} = 13600\,\text{mm}^2 = 0.0136\,\text{m}^2 \] --- ## **Step 2: Stresses on the Glue Joint (Inclined at 45°)** When the load \( P \) is applied **axially**, the stresses on the inclined plane (joint) are: - **Normal stress on the joint**: \[ \sigma = \frac{P}{A} \cos^2\theta \] - **Shear stress on the joint**: \[ \tau = \frac{P}{A} \sin\theta \cos\theta \] But both are derived from the transformation equations. For an angle \( \theta = 45^\circ \): \[ \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \] So: \[ \cos^2 45^\circ = \sin^2 45^\circ = \frac{1}{2} \] \[ \sin 45^\circ \cos 45^\circ = \frac{1}{2} \] So for the inclined plane: - **Normal stress**: \( \sigma = \frac{P}{A} \cdot \frac{1}{2} \) - **Shear stress**: \( \tau = \frac{P}{A} \cdot \frac{1}{2} \) --- ## **Step 3: Allowable Stresses** - **Allowable shear**: \( \tau_{allow} = 850\,\text{kPa} = 0.85\,\text{MPa} \) - **Allowable normal**: \( \sigma_{allow} = 12\,\text{MPa} \) --- ## **Step 4: Maximum Load Based on Each Limit** ### **(A) Based on Shear in Glue** \[ \tau = \frac{P}{A} \cdot \frac{1}{2} \leq 0.85\,\text{MPa} \] \[ \frac{P}{A} \leq 2 \times 0.85 = 1.7\,\text{MPa} \] \[ P \leq 1.7\,\text{MPa} \times 0.0136\,\text{m}^2 \times 10^6 \,(\text{to convert to N}) \] \[ P \leq 1.7 \times 13,600 = 23,120\,\text{N} \] --- ### **(B) Based on Normal Stress in Wood** \[ \sigma = \frac{P}{A} \leq 12\,\text{MPa} \] \[ P \leq 12 \times 13,600 = 163,200\,\text{N} \] --- ### **(C) Based on Normal Stress on the Joint** But on the joint (inclined plane): \[ \sigma_{joint} = \frac{P}{A} \cdot \frac{1}{2} \leq 12\,\text{MPa} \] \[ \frac{P}{A} \leq 24\,\text{MPa} \] \[ P \leq 24 \times 13,600 = 326,400\,\text{N} \] But this is *higher* than the axial limit, so not critical. --- ## **Step 5: Pick the Smallest Limiting Value** - **Shear in glue joint**: \( 23,120\,\text{N} \) - **Axial normal stress**: \( 163,200\,\text{N} \) **Shear in the glue joint is the critical limit.** --- ## **Final Answer** \[ \boxed{23.1\,\text{kN}} \] This matches the answer shown in the image. --- ### **Summary Table** | Limiting Factor | Max Load \(P\) (N) | |-----------------------|------------------| | Shear in glue | 23,120 | | Normal in wood (axial)| 163,200 | **The largest centric load \(P\) that can be applied is \( \boxed{23.1\,\text{kN}} \) (rounded to 3 significant digits).** --- ### **Why?** The glue joint fails first in shear, so that's the limiting factor. Always check all possible modes of failure! --- Let me know if you want an explanation of the stress transformation formula derivations!