Let's solve each part step-by-step: --- ### **Problem 9** #### **Given Data** - **Bar width, \( b \) = 50 mm** - **Bar depth, \( d \) = 30 mm** - **Weld throat size, \( h \) = 5 mm** - **Allowable shear stress, \( \tau_{allow} \) = 140 MPa** - **Force \( F \) is horizontal (tension)** #### **Step 1: Find the Weld Throat Area** There are two welds, both of length \( b \) and \( d \). - **Total weld length** = \( b + d \) - **Throat area** = (total length) × (throat size) \[ A_{throat} = (b + d) \cdot h \] \[ A_{throat} = (50\, \text{mm} + 30\, \text{mm}) \cdot 5\, \text{mm} \] \[ A_{throat} = 80\, \text{mm} \cdot 5\, \text{mm} \] \[ A_{throat} = 400\, \text{mm}^2 \] #### **Step 2: Relate Shear Stress and Force** Allowable shear stress: \[ \tau_{allow} = \frac{F}{A_{throat}} \] Solve for \( F \): \[ F = \tau_{allow} \cdot A_{throat} \] \[ F = 140\, \text{MPa} \cdot 400\, \text{mm}^2 \] Remember: \( 1\, \text{MPa} = 1\, \text{N}/\text{mm}^2 \) \[ F = 140\, \text{N}/\text{mm}^2 \times 400\, \text{mm}^2 \] \[ F = 56,000\, \text{N} \] #### **Final Answer (Problem 9):** \[ \boxed{F = 56,\!000\, \text{N}} \] --- ### **Problem 10** Now, the force is vertical (bending force) and the total beam length is 200 mm. #### **Step 1: Find the Maximum Bending Moment** For a cantilever beam with a force \( F \) applied at the free end: \[ M = F \cdot L \] Where \( L = 200\, \text{mm} \) #### **Step 2: Calculate Maximum Shear Due to Bending** For welds subjected to bending moment, the maximum shear at the throat is: \[ \tau_{max} = \frac{M \cdot y_{max}}{J} \] Where: - \( y_{max} \): Maximum distance from the neutral axis (for rectangle, it's half the height) - \( J \): Polar moment of inertia of the weld group about the centroid ##### **Weld Layout** Assume both welds are on the sides (vertical), each of length \( d \): - **Total effective length**: \( 2d \) - Distance from centroid to each weld: \( b/2 \) ##### **Polar Moment of Inertia \( J \):** \[ J = 2 \cdot h \cdot d \cdot \left(\frac{b}{2}\right)^2 \] \[ J = 2 \cdot 5\, \text{mm} \cdot 30\, \text{mm} \cdot (25\, \text{mm})^2 \] \[ J = 2 \cdot 150 \cdot 625 \] \[ J = 2 \cdot 93,750 = 187,500\, \text{mm}^4 \] ##### **Maximum Shear Stress:** \[ \tau_{max} = \frac{M \cdot y_{max}}{J} \] Where \( y_{max} = b/2 = 25\, \text{mm} \), \( M = F \cdot 200\, \text{mm} \): \[ \tau_{allow} = \frac{F \cdot 200 \cdot 25}{187,500} \] Solve for \( F \): \[ F = \frac{\tau_{allow} \cdot 187,500}{200 \cdot 25} \] \[ F = \frac{140 \cdot 187,500}{5,000} \] \[ F = \frac{26,250,000}{5,000} \] \[ F = 5,250\, \text{N} \] #### **Final Answer (Problem 10):** \[ \boxed{F = 5,\!250\, \text{N}} \] --- ### **Summary** - **Problem 9 (Horizontal Force, Tension):** \(\boxed{56,\!000\, \text{N}}\) - **Problem 10 (Vertical Force, Bending):** \(\boxed{5,\!250\, \text{N}}\)