1. **Problem Statement:** Determine the maximum shearing stress in a simply supported W310x74 steel beam of length 9 m carrying a dead load of 3 kN/m. The beam is made of A36 steel with yield strength $F_y = 248$ MPa and has full lateral support. 2. **Given Data:** - Span length, $L = 9$ m = 9000 mm - Dead load, $w = 3$ kN/m = 3 N/mm - Beam properties: - Depth, $d = 305$ mm - Flange width, $b_f = 139$ mm - Flange thickness, $t_f = 16.7$ mm - Web thickness, $t_w = 17.4$ mm - Section modulus, $S_x = 833 \times 10^3$ mm$^3$ - Moment of inertia, $I_x = 127 \times 10^6$ mm$^4$ 3. **Step 1: Calculate the maximum bending moment $M_{max}$ for a simply supported beam with uniform load:** $$ M_{max} = \frac{wL^2}{8} $$ Convert load to N/mm: $$ w = 3 \text{ kN/m} = 3 \times 10^3 \text{ N/m} = \frac{3 \times 10^3}{1000} = 3 \text{ N/mm} $$ Calculate $M_{max}$: $$ M_{max} = \frac{3 \times (9000)^2}{8} = \frac{3 \times 81 \times 10^6}{8} = \frac{243 \times 10^6}{8} = 30.375 \times 10^6 \text{ Nmm} $$ 4. **Step 2: Calculate the maximum shear force $V_{max}$ at the supports:** $$ V_{max} = \frac{wL}{2} = \frac{3 \times 9000}{2} = 13500 \text{ N} $$ 5. **Step 3: Calculate the maximum bending stress $\sigma_{max}$ using section modulus $S_x$:** $$ \sigma_{max} = \frac{M_{max}}{S_x} = \frac{30.375 \times 10^6}{833 \times 10^3} = 36.47 \text{ MPa} $$ 6. **Step 4: Calculate the maximum shear stress $\tau_{max}$ in the web:** The maximum shear stress in an I-beam occurs in the web and is given by: $$ \tau_{max} = \frac{VQ}{Ib} $$ where: - $V$ = shear force = $V_{max} = 13500$ N - $Q$ = first moment of area about neutral axis for the area above (or below) the point where shear is calculated - $I$ = moment of inertia = $127 \times 10^6$ mm$^4$ - $b$ = web thickness = $t_w = 17.4$ mm 7. **Step 5: Calculate $Q$ for the web at neutral axis:** The area above the neutral axis is the flange plus half the web height: $$ Q = A' \times \bar{y} $$ where $A'$ is the area above neutral axis and $\bar{y}$ is the distance from neutral axis to centroid of $A'$. Calculate $A'$: - Flange area: $A_f = b_f \times t_f = 139 \times 16.7 = 2321.3$ mm$^2$ - Web area above neutral axis: $A_w = t_w \times \frac{d - 2t_f}{2} = 17.4 \times \frac{305 - 2 \times 16.7}{2} = 17.4 \times \frac{305 - 33.4}{2} = 17.4 \times 135.8 = 2362.92$ mm$^2$ Total $A' = A_f + A_w = 2321.3 + 2362.92 = 4684.22$ mm$^2$ Calculate $\bar{y}$: - Distance from neutral axis to centroid of flange area: $y_f = \frac{d}{2} - \frac{t_f}{2} = \frac{305}{2} - \frac{16.7}{2} = 152.5 - 8.35 = 144.15$ mm - Distance from neutral axis to centroid of web area above neutral axis: $y_w = \frac{d}{4} = \frac{305}{4} = 76.25$ mm Calculate combined centroid: $$ \bar{y} = \frac{A_f y_f + A_w y_w}{A_f + A_w} = \frac{2321.3 \times 144.15 + 2362.92 \times 76.25}{4684.22} = \frac{334,693.7 + 180,147.5}{4684.22} = \frac{514,841.2}{4684.22} = 109.9 \text{ mm} $$ Calculate $Q$: $$ Q = A' \times \bar{y} = 4684.22 \times 109.9 = 514,841.2 \text{ mm}^3 $$ 8. **Step 6: Calculate $\tau_{max}$:** $$ \tau_{max} = \frac{VQ}{Ib} = \frac{13500 \times 514,841.2}{127 \times 10^6 \times 17.4} = \frac{6.95 \times 10^9}{2.2098 \times 10^9} = 3.14 \text{ MPa} $$ **Final Answer:** The maximum shearing stress in the beam is approximately **3.14 MPa**.