Let's solve this **step-by-step**. We need to determine the **allowable load \(P\)** of the joint by checking **all failure modes**: - Plate failure in tension (net section) - Bolt shear failure - Plate bearing failure **Given:** - Plate thickness, \( t = 12\,\text{mm} \) - Plate width, \( b = 400\,\text{mm} \) - Bolt diameter, \( d = 20\,\text{mm} \) - Allowable stresses: - Plate tension: \( \sigma_t = 125\,\text{MPa} \) - Shear: \( \tau = 100\,\text{MPa} \) - Bearing: \( \sigma_b = 320\,\text{MPa} \) - Plates are in **single shear**. - Assume number of bolts \( n \) (count the number in the figure—let’s estimate **8 bolts** based on the figure, but confirm for your actual case). --- ## 1. **Plate Net Section Tension Failure** The net area is the total area minus the area removed by bolt holes. \[ A_{\text{net}} = (b - n d) \times t \] \[ P_{\text{tension}} = \sigma_t \times A_{\text{net}} \] Assuming \( n = 8 \): \[ A_{\text{net}} = [400 - (8 \times 20)] \times 12 = (400 - 160) \times 12 = 240 \times 12 = 288\,\text{mm}^2 \] \[ P_{\text{tension}} = 125 \times 288 = 360,000\,\text{N} = \boxed{360\,\text{kN}} \] --- ## 2. **Bolt Shear Failure** Each bolt is in single shear: \[ A_{\text{bolt}} = \frac{\pi}{4} d^2 = \frac{\pi}{4} \times 20^2 = 314.16\,\text{mm}^2 \] \[ P_{\text{shear, per\,bolt}} = \tau \times A_{\text{bolt}} = 100 \times 314.16 = 31,416\,\text{N} \] For \( n = 8 \) bolts: \[ P_{\text{shear, total}} = 8 \times 31,416 = 251,328\,\text{N} = \boxed{251.3\,\text{kN}} \] --- ## 3. **Plate Bearing Failure** \[ P_{\text{bearing, per\,bolt}} = \sigma_b \times d \times t = 320 \times 20 \times 12 = 76,800\,\text{N} \] For \( n = 8 \) bolts: \[ P_{\text{bearing, total}} = 8 \times 76,800 = 614,400\,\text{N} = \boxed{614.4\,\text{kN}} \] --- ## 4. **Determine Allowable Load \(P\)** **Allowable load \(P\) is governed by the smallest failure load:** \[ P_{allowable} = \min\left( P_{\text{tension}}, P_{\text{shear}}, P_{\text{bearing}} \right) \] \[ P_{allowable} = \min\left( 360, 251.3, 614.4 \right)\ \text{kN} = \boxed{251.3\,\text{kN}} \] --- ## **Summary Table** | Failure Mode | Allowable Load (kN) | |---------------------|--------------------| | Net Section Tension | 360. | | Bolt Shear | 251.3 | | Plate Bearing | 614.4 | --- ## **Final Answer** \[ \boxed{P_{allowable} = 251.3\,\text{kN}} \] The allowable load \(P\) of the joint is **251.3 kN**, governed by **bolt shear**. Double-check the number of bolts for your case and recalculate if necessary!