1. **Problem Statement:** Determine the minimum width $b$ of a beam subjected to a distributed load of 2000 N/m over 4 m total length, a point load of 5000 N at 1 m from the left support, such that the maximum flexural stress does not exceed 10 MPa. The beam has a rectangular cross-section with height $h=200$ mm and width $b$ (unknown). 2. **Given Data:** - Distributed load $w = 2000$ N/m over 4 m - Point load $P = 5000$ N at 1 m from left support - Beam spans: 1 m + 2 m + 1 m = 4 m total - Cross-section height $h = 200$ mm = 0.2 m - Maximum allowable flexural stress $\sigma_{max} = 10$ MPa = $10 \times 10^{6}$ Pa 3. **Step 1: Calculate reactions $R_1$ and $R_2$ at supports** - Total distributed load $W = w \times L = 2000 \times 4 = 8000$ N - Let left support be at $x=0$, right support at $x=4$ m - Sum of vertical forces: $R_1 + R_2 = W + P = 8000 + 5000 = 13000$ N - Taking moments about left support: $$ R_2 \times 4 = (w \times L) \times \frac{L}{2} + P \times 1 = 8000 \times 2 + 5000 \times 1 = 16000 + 5000 = 21000 $$ $$ R_2 = \frac{21000}{4} = 5250 \text{ N} $$ - Then, $$ R_1 = 13000 - 5250 = 7750 \text{ N} $$ 4. **Step 2: Find the maximum bending moment $M_{max}$** - The bending moment will be maximum where shear force changes or under point load. - Calculate bending moment at key points: At $x=1$ m (under point load): $$ M(1) = R_1 \times 1 - w \times 1 \times \frac{1}{2} = 7750 \times 1 - 2000 \times 1 \times 0.5 = 7750 - 1000 = 6750 \text{ Nm} $$ At $x=3$ m (just before right support): $$ M(3) = R_1 \times 3 - w \times 3 \times \frac{3}{2} - P \times (3 - 1) = 7750 \times 3 - 2000 \times 3 \times 1.5 - 5000 \times 2 = 23250 - 9000 - 10000 = 4250 \text{ Nm} $$ Maximum bending moment is $M_{max} = 6750$ Nm. 5. **Step 3: Use flexural stress formula to find $b$** - Flexural stress formula: $$ \sigma = \frac{M c}{I} $$ where - $M$ = bending moment - $c = \frac{h}{2} = 0.1$ m (distance from neutral axis to outer fiber) - $I = \frac{b h^3}{12}$ (moment of inertia for rectangular section) Rearranged to solve for $b$: $$ \sigma = \frac{M c}{I} = \frac{M c}{\frac{b h^3}{12}} = \frac{12 M c}{b h^3} \implies b = \frac{12 M c}{\sigma h^3} $$ 6. **Step 4: Substitute values** $$ b = \frac{12 \times 6750 \times 0.1}{10 \times 10^{6} \times (0.2)^3} = \frac{8100}{10 \times 10^{6} \times 0.008} = \frac{8100}{80000} = 0.10125 \text{ m} = 101.25 \text{ mm} $$ 7. **Final answer:** The minimum width $b$ of the beam is approximately **101.25 mm** to ensure the flexural stress does not exceed 10 MPa.