Let's solve the problem step by step: --- ## **Given Data** - \( b = 3.750 \) in - \( c = 5.80 \) in - \( d = 1.800 \) in - \( h = 0.3125 \) in - \( \tau_{allow} = 25 \) ksi = 25,000 psi --- ## **Step 1: Find the Total Weld Length** The weld is around three sides (left, top, and right): \[ \text{Total Weld Length (L)} = b + 2c \] \[ L = 3.75 + 2 \times 5.80 = 15.35\ \text{in} \] --- ## **Step 2: Find Weld Throat Area** The throat of a fillet weld is: \[ \text{Throat} = 0.707h \] \[ \text{Area} = \text{Total Weld Length} \times \text{Throat} \] \[ \text{Area} = 15.35 \times 0.707 \times 0.3125 = 3.397\ \text{in}^2 \] --- ## **Step 3: Shear due to the Applied Force \( F \) (Direct Shear)** \[ \tau_{direct} = \frac{F}{\text{Weld Area}} \] --- ## **Step 4: Shear due to Moment (Eccentric Load)** The force \( F \) is applied at a distance \( d \) from the weld centroid, producing a moment: \[ M = F \times d \] Each infinitesimal segment of weld resists a part of this moment as secondary shear. The maximum secondary shear stress occurs at the farthest point from the centroid. For a rectangular weld group (approximate method): \[ r_{max} = \sqrt{\left(\frac{b}{2}\right)^2 + c^2} \] \[ J = \sum (l_i r_i^2) \] Since weld group is symmetric: \[ J = \frac{b^3}{12} + b \left(\frac{c}{2}\right)^2 + 2\left[\frac{c^3}{12} + c \left(\frac{b}{2}\right)^2\right] \] But for simplicity, since the weld is three sides (U-shape), let's use: \[ J = \sum l_i r_i^2 \] For each side: - Bottom: \( b \) at \( r = c \) - Sides: \( c \) at \( r = \sqrt{(b/2)^2 + c^2} \) But let's use a simplified version for classroom problems: \[ J = b c^2 + 2c \left[ \left(\frac{b}{2}\right)^2 \right] = b c^2 + c b^2/2 \] --- ## **Step 5: Maximum Shear Stress** The resultant shear stress at the most critical point: \[ \tau_{max} = \sqrt{ \left(\tau_{direct}\right)^2 + \left(\tau_{secondary}\right)^2 + 2 \tau_{direct} \tau_{secondary} \cos \theta } \] But if the directions are perpendicular, usually: \[ \tau_{max} = \tau_{direct} + \tau_{secondary} \] --- ## **Step 6: Set \( \tau_{max} = \tau_{allow} \) and Solve for \( F \)** Let's perform the calculations step-by-step. --- ### **Detailed Calculations** #### **Weld Length** \[ L = b + 2c = 3.75 + 2 \times 5.80 = 15.35\ \text{in} \] #### **Throat Area** \[ A = L \times 0.707 \times h = 15.35 \times 0.707 \times 0.3125 = 3.397\ \text{in}^2 \] #### **Shear due to Force \( F \)** \[ \tau_{direct} = \frac{F}{3.397} \] #### **Location of Centroid** For a U-shape weld, the centroid is offset from the left edge. But for this calculation, let's take the distance from the centroid to where \( F \) acts: \[ r_{max} = \sqrt{(b/2)^2 + c^2} = \sqrt{(3.75/2)^2 + (5.80)^2} = \sqrt{(1.875)^2 + (5.80)^2} = \sqrt{3.516 + 33.64} = \sqrt{37.156} = 6.10\ \text{in} \] #### **Polar Moment of Weld Group (\( J \)):** For a U-shaped weld: \[ J \approx b c^2 + \frac{b^3}{6} = 3.75 \times (5.80)^2 + \frac{(3.75)^3}{6} = 3.75 \times 33.64 + \frac{52.73}{6} = 126.15 + 8.79 = 134.94\ \text{in}^3 \] #### **Secondary Shear due to Moment** \[ \tau_{secondary} = \frac{M r_{max}}{J \cdot \text{throat}} = \frac{F d r_{max}}{J A / L} \] But, using the more common formula: \[ \tau_{secondary} = \frac{M r_{max}}{J \cdot \text{throat}} = \frac{F d r_{max}}{J \cdot 0.707 h} \] But since \( A = L \times 0.707 h \), and \( J \) is as above: \[ \tau_{secondary} = \frac{F d r_{max}}{J \cdot 0.707 h} \] Plug in values: - \( d = 1.8 \) in - \( r_{max} = 6.10 \) in - \( J = 134.94 \) in\(^3\) - \( h = 0.3125 \) in \[ \tau_{secondary} = \frac{F \times 1.8 \times 6.10}{134.94 \times 0.707 \times 0.3125} \] \[ = \frac{F \times 10.98}{29.89} = 0.3675 F \] #### **Total Shear** \[ \tau_{max} = \tau_{direct} + \tau_{secondary} = \frac{F}{3.397} + 0.3675 F \] Set \( \tau_{max} = \tau_{allow} = 25,000 \) psi: \[ \frac{F}{3.397} + 0.3675 F = 25,000 \] \[ F \left( \frac{1}{3.397} + 0.3675 \right) = 25,000 \] \[ \frac{1}{3.397} = 0.2943 \] \[ 0.2943 + 0.3675 = 0.6618 \] \[ 0.6618 F = 25,000 \] \[ F = \frac{25,000}{0.6618} = 37,781 \ \text{lbf} \] --- ### **Step 7: Convert to kips** \[ 1 \ \text{kip} = 1,000 \ \text{lbf} \] \[ F = \frac{37,781}{1,000} = 37.8 \ \text{kip} \] --- ## **Final Answer** **The safe force is:** \[ \boxed{37.8 \ \text{kip}} \] --- ### **Summary Table** | Step | Value | |------------------------------|-------------| | Total weld length \( L \) | 15.35 in | | Weld throat area \( A \) | 3.397 in² | | Polar moment \( J \) | 134.94 in³ | | Max distance \( r_{max} \) | 6.10 in | | Allowable shear \( \tau_{allow} \) | 25 ksi = 25,000 psi | | **Safe force \( F \)** | **37.8 kip**|