Let's break down the problem step by step to find the correct answer for the shear stress at point \( A \). --- ### Given Data - **Load \( P \) = 40 kN** - **Cross-section dimensions:** - Top flange: 200 mm wide, 20 mm thick - Web: 20 mm wide, 300 mm high - Bottom flange: 200 mm wide, 20 mm thick - **Point A:** Located at the junction of the web and top flange. --- ### Step 1: Calculate Shear Force at Section n-n Since the load \( P \) is at midspan and the beam is symmetrically supported, the reaction at each support is: \[ R = \frac{P}{2} = \frac{40\,\text{kN}}{2} = 20\,\text{kN} \] At section n-n (just to the left or right of the load), the shear force \( V \) is: \[ V = 20\,\text{kN} = 20,000\,\text{N} \] --- ### Step 2: Shear Stress Formula The general formula for shear stress at a point in a cross-section is: \[ \tau = \frac{VQ}{Ib} \] - \( V \) = Shear force at the section (\( 20,000 \) N) - \( Q \) = First moment of area about the neutral axis above point A - \( I \) = Second moment of area (moment of inertia) of the whole section about the neutral axis - \( b \) = Width of the section at point A (which is the width of the web, 20 mm) --- ### Step 3: Calculate \( I \) (Moment of Inertia) Let's approximate \( I \) for the given I-beam: #### For the web: - Height = 300 mm - Width = 20 mm #### For the flanges: - Each flange: 200 mm wide, 20 mm thick #### Place neutral axis at the center (since it's symmetric): - Total height = 20 + 300 + 20 = **340 mm** Calculate the centroid position \( y_{NA} \): - Area of flange = \( 200 \times 20 = 4000 \) mm² (2 flanges) - Area of web = \( 300 \times 20 = 6000 \) mm² - Distance from base to bottom of web: 20 mm - Neutral axis from bottom: \[ y_{NA} = \frac{(4000 \times 10) + (6000 \times (20+150)) + (4000 \times (320+10))}{4000 + 6000 + 4000} \] \[ = \frac{(40000) + (6000 \times 170) + (4000 \times 330)}{14000} \] \[ = \frac{40000 + 1020000 + 1320000}{14000} = \frac{2380000}{14000} \approx 170\,\text{mm} \] So the NA is 170 mm above the bottom. #### Moment of Inertia \( I \) (approximate): Sum for each part (using parallel axis theorem): - **Web:** \( I_{web} = \frac{1}{12} b h^3 \) - \( = \frac{1}{12} \times 20 \times (300)^3 = \frac{1}{12} \times 20 \times 27,000,000 = 45,000,000 \) mm⁴ - **Top flange:** \( I_{flange} = \frac{1}{12} b h^3 + Ad^2 \) - \( = \frac{1}{12} \times 200 \times (20)^3 + 4000 \times (170-10)^2 \) - \( = \frac{1}{12} \times 200 \times 8000 + 4000 \times 160^2 \) - \( = 133,333.33 + 4000 \times 25,600 = 133,333.33 + 102,400,000 = 102,533,333.33 \) mm⁴ - **Bottom flange:** \( I_{flange} = \frac{1}{12} \times 200 \times 8000 + 4000 \times (320-170)^2 \) - \( = 133,333.33 + 4000 \times 150^2 = 133,333.33 + 4000 \times 22,500 = 133,333.33 + 90,000,000 = 90,133,333.33 \) mm⁴ - **Total \( I \):** - \( = 45,000,000 + 102,533,333.33 + 90,133,333.33 = 237,666,666.66 \) mm⁴ --- ### Step 4: Calculate \( Q \) at Point A First moment of area \( Q \) at point A (just below top flange): - Area above A = area of top flange - Distance from NA to centroid of this area = (170 - 10) = 160 mm \[ Q = 4000 \times 160 = 640,000\,\text{mm}^3 \] --- ### Step 5: Substitute Values Shear force \( V = 20,000\,\text{N} \) Width at A (\( b \)) = 20 mm = 0.02 m = 20 mm \[ \tau = \frac{V Q}{I b} = \frac{20,000 \times 640,000}{237,666,666.66 \times 20} \] Calculate denominator: \[ 237,666,666.66 \times 20 = 4,753,333,333.2 \] Calculate numerator: \[ 20,000 \times 640,000 = 12,800,000,000 \] \[ \tau = \frac{12,800,000,000}{4,753,333,333.2} = 2.693\,\text{MPa} \] --- ### Step 6: Closest Answer Given the options, **2.56 MPa** is the closest value. --- ## **Final Answer:** **Shear stress at point A is 2.56 MPa.** --- If you have any questions or want to see a more detailed calculation for any step, let me know!