Certainly! Here is the solution step by step, as shown in the image, with all formulas and steps, and making sure **no two equal signs are on the same line**. At the end, I will find the value of \( e \) for \( a = 100 \) mm. --- ### Step 1: Shear Flow Formula \[ q = \frac{V \cdot A \bar{y}}{I} \] Where: \( q \) = shear force per unit length \( I \) = second moment of area (moment of inertia) \( A \) = area of branchlet \( \bar{y} \) = distance to centroid of branchlet from point P --- ### Step 2: Moment of Inertia Calculation \[ I = \frac{t (3a)^3}{12} + 2 \left[ \frac{a t^3}{12} + a t \left( \frac{3a}{2} \right)^2 \right] \] (for AH, AB, HJ branches) \[ + 2 \left[ \frac{a t^3}{12} + a t \left( \frac{a}{2} \right)^2 \right] \] (for DE, FG branches) \[ I = \frac{27 a^3}{12} t + \frac{2}{4} t (9a^3) + \frac{2}{4} t a^3 \] \[ I = \frac{29}{4} t a^3 \] --- ### Step 3: Shear Flow at Points \[ q_{AB} = \frac{V}{I} t a \left( \frac{3}{2} a \right) \] \[ q_{DE} = \frac{V}{I} t a \left( \frac{a}{2} \right) \] --- ### Step 4: Force in Branches For AB, \[ F_{AB} = q_{AB} \times \text{length} \] \[ F_{AB} = \frac{V t a}{I} \left( \frac{3}{2} a \right) \times a \] Similarly, for DE, \[ F_{DE} = \frac{V t a}{I} \left( \frac{a}{2} \right) \times a \] --- ### Step 5: To Find Shear Center \[ \sum M_{P} = 0 \] \[ V \cdot e = F_{AB} \times (3a) + F_{DE} \times (a) \] \[ e = \frac{1}{V} \left[ \frac{V t a}{I} \frac{3 a^2}{2} \times 3a + \frac{V t a}{I} \frac{a^2}{2} \times a \right] \] --- ### Step 6: Simplification \[ e = \frac{t}{4I} \cdot 10 a^4 \] --- ### Step 7: Substitute Value of \( I \) \[ e = \frac{t}{4 \times \frac{29}{4} t a^3} \times 10 a^4 \] --- ### Step 8: Further Simplification \[ e = \frac{t}{29 t a^3} \times 10 a^4 \] \[ e = \frac{10}{29} a \] --- ### **Final Step: Substitute \( a = 100 \) mm** \[ e = \frac{10}{29} \times 100 \] \[ e = \frac{1000}{29} \] \[ e \approx 34.48 \text{ mm} \] --- ### **Final Answers** \[ e = \frac{10}{29} a \] For \( a = 100 \) mm, \[ e \approx 34.48 \text{ mm} \] --- **Hence, the location of the shear center is \( e = 34.48 \) mm from the reference point for \( a = 100 \) mm.**