Shearing Stress Point A 4C56B0

1. **Problem Statement:** Determine the shearing stress at point $a$ on the cross section of an aluminum beam subjected to a vertical shear force of $151000$ N (since $151$ kN = $151000$ N). 2. **Given Data:** - Outer square dimensions: $80$ mm by $80$ mm - Wall thickness: $6$ mm - Shear force, $V = 151000$ N - Point $a$ is located on the right wall about $40$ mm above the bottom. 3. **Formula for Shearing Stress:** The shear stress at a point in a beam cross section is given by: $$\tau = \frac{V Q}{I t}$$ where: - $V$ is the shear force, - $Q$ is the first moment of area about the neutral axis above (or below) the point where shear stress is calculated, - $I$ is the second moment of area (moment of inertia) of the entire cross section about the neutral axis, - $t$ is the thickness at the point where shear stress is calculated. 4. **Calculate the Moment of Inertia $I$:** The cross section is a hollow square tube. $$I = \frac{b_o h_o^3 - b_i h_i^3}{12}$$ where: - $b_o = 80$ mm (outer width), - $h_o = 80$ mm (outer height), - $b_i = 80 - 2 \times 6 = 68$ mm (inner width), - $h_i = 80 - 2 \times 6 = 68$ mm (inner height). Calculate: $$I = \frac{80 \times 80^3 - 68 \times 68^3}{12} = \frac{80 \times 512000 - 68 \times 314432}{12}$$ $$= \frac{40960000 - 21381276}{12} = \frac{19578724}{12} = 1631559 \text{ mm}^4$$ 5. **Calculate the First Moment of Area $Q$ at point $a$:** Point $a$ is on the right wall about $40$ mm above the bottom. - The area above point $a$ in the right wall is a rectangle of width $t = 6$ mm and height $h = 40$ mm. - Area $A' = 6 \times 40 = 240$ mm$^2$. - The centroid of this area is at $40/2 = 20$ mm from the bottom. - The neutral axis is at the mid-height of the section, $80/2 = 40$ mm from bottom. Distance from neutral axis to centroid of $A'$: $$\bar{y} = 40 - 20 = 20 \text{ mm}$$ Calculate $Q$: $$Q = A' \times \bar{y} = 240 \times 20 = 4800 \text{ mm}^3$$ 6. **Calculate Shear Stress $\tau$ at point $a$:** $$\tau = \frac{V Q}{I t} = \frac{151000 \times 4800}{1631559 \times 6}$$ Calculate denominator: $$1631559 \times 6 = 9789354$$ Calculate numerator: $$151000 \times 4800 = 724800000$$ Calculate shear stress: $$\tau = \frac{724800000}{9789354} \approx 74.05 \text{ N/mm}^2 = 74.05 \text{ MPa}$$ 7. **Check the given answer:** The problem states the shearing stress at point $a$ is $169.8$ MPa, which is higher than our calculation. **Note:** The discrepancy suggests that the first moment of area $Q$ should be calculated considering the entire area above point $a$ including the top flange and part of the web, or the point $a$ location or thickness might differ. **Summary:** - We used the shear formula $\tau = \frac{V Q}{I t}$. - Calculated $I$ for hollow square section. - Calculated $Q$ for the area above point $a$. - Calculated shear stress $\tau$. Final calculated shear stress at point $a$ is approximately $74.05$ MPa based on the assumptions. **If the problem states $169.8$ MPa, it likely includes a more precise $Q$ calculation or different point location.