**Sub-Subject:** Mechanics of Materials **Topic:** Bending of Beams – Self-Weight (Deflection and Stress) --- # Detailed Step-by-Step Solution ## **Restatement of the Problem** A *simply supported* steel rod of **diameter** \( d = 25 \) mm is supported over a span of \( L \) metres. The only load is its own weight. The **allowable bending stress** is \( 120 \) MPa. The **Young’s modulus** is \( E = 210 \) GPa, and the **density** of steel is \( 7800 \) kg/m³. The tasks are: 1. **Calculate the maximum span \( L \)** such that the bending stress does not exceed the allowable limit. 2. **Determine the deflection at the centre** of the span. 3. **Discuss the implications** of these results. --- ## **Step 1: Calculate the Weight per Unit Length of the Rod** The **cross-sectional area** \( A \) of the rod is: \[ A = \frac{\pi d^2}{4} \] Given \( d = 25 \) mm \( = 0.025 \) m, \[ A = \frac{\pi}{4} (0.025)^2 = \frac{\pi}{4} \times 0.000625 = 0.0004909 \text{ m}^2 \] The **weight per unit length** (\( w \), in N/m) is: \[ w = \text{mass per unit length} \times g = (\rho A)g \] where \( \rho = 7800 \) kg/m³ and \( g = 9.81 \) m/s², \[ w = 7800 \times 0.0004909 \times 9.81 = 3.829 \times 9.81 = 37.58 \text{ N/m} \] --- ## **Step 2: Maximum Bending Moment for a Simply Supported Beam with Uniform Load** For a simply supported beam of length \( L \) carrying a **uniformly distributed load** \( w \) (N/m), the **maximum bending moment** occurs at the centre: \[ M_{\text{max}} = \frac{wL^2}{8} \] --- ## **Step 3: Maximum Bending Stress** The **maximum bending stress** is given by: \[ \sigma_{\text{max}} = \frac{M_{\text{max}}y_{\text{max}}}{I} \] where - \( y_{\text{max}} = \) distance from the neutral axis to the extreme fibre \( = \frac{d}{2} \) - \( I = \) second moment of area (for a solid circle): \[ I = \frac{\pi d^4}{64} \] Substitute \( M_{\text{max}} \) and \( I \): \[ \sigma_{\text{max}} = \frac{\frac{wL^2}{8} \cdot \frac{d}{2}}{\frac{\pi d^4}{64}} = \frac{wL^2 d}{16} \cdot \frac{64}{\pi d^4} = \frac{4wL^2}{\pi d^3} \] Set \( \sigma_{\text{max}} = 120 \) MPa \( = 120 \times 10^6 \) N/m²: \[ 120 \times 10^6 = \frac{4wL^2}{\pi d^3} \] --- ## **Step 4: Solve for the Maximum Span \( L \)** Rearrange for \( L \): \[ L^2 = \frac{120 \times 10^6 \times \pi \times d^3}{4w} \] \[ L = \sqrt{ \frac{120 \times 10^6 \times \pi \times d^3}{4w} } \] Substitute values: - \( d = 0.025 \) m - \( w = 37.58 \) N/m Calculate \( d^3 \): \[ d^3 = (0.025)^3 = 0.000015625 \text{ m}^3 \] Calculate numerator: \[ 120 \times 10^6 \times \pi \times 0.000015625 = 120 \times 10^6 \times 3.1416 \times 0.000015625 \] \[ = 120 \times 10^6 \times 0.0000490874 = 5,890.49 \] Now divide by \( 4w \): \[ 4w = 4 \times 37.58 = 150.32 \] \[ L^2 = \frac{5,890.49}{150.32} = 39.19 \] \[ L = \sqrt{39.19} = 6.26 \text{ m} \] **Final answer for part 1:** \[ \boxed{L = 6.26 \text{ m}} \] --- ## **Step 5: Calculate the Maximum Deflection at the Centre** For a simply supported beam with a **uniform load**: \[ \delta_{\text{max}} = \frac{5wL^4}{384EI} \] Where: - \( E = 210 \) GPa \( = 210 \times 10^9 \) N/m² - \( I = \frac{\pi d^4}{64} \) Calculate \( d^4 \): \[ d^4 = (0.025)^4 = 0.000000390625 \text{ m}^4 \] \[ I = \frac{\pi}{64} \times 0.000000390625 = 0.04909 \times 0.000000390625 = 0.000000019202 \text{ m}^4 \] \[ I = 1.227 \times 10^{-8} \text{ m}^4 \] Now substitute all values: \[ \delta_{\text{max}} = \frac{5 \times 37.58 \times (6.26)^4}{384 \times 210 \times 10^9 \times 1.227 \times 10^{-8}} \] First, calculate \( (6.26)^4 \): \[ 6.26^4 = (6.26^2)^2 = (39.1876)^2 = 1,535.66 \] Calculate the numerator: \[ 5 \times 37.58 \times 1,535.66 = 187.9 \times 1,535.66 = 288,669 \] Calculate the denominator: \[ 384 \times 210 \times 10^9 \times 1.227 \times 10^{-8} \] First, \( 384 \times 210 = 80,640 \) Now, \( 10^9 \times 10^{-8} = 10^1 = 10 \) So, \[ 80,640 \times 1.227 \times 10 = 80,640 \times 12.27 = 988,332 \] Now, \[ \delta_{\text{max}} = \frac{288,669}{988,332} = 0.292 \text{ m} \] --- ## **Step 6: Discussion and Deductions** The **maximum span** is \( 6.26 \) m before exceeding the allowable stress, but the **deflection** at this span is approximately \( 0.29 \) m (29 cm), which is extremely large for any practical structural application. Typically, acceptable deflections are much less (often span/250 to span/360 for beams), so the rod would be **too flexible** even though it is safe from a stress perspective. **Conclusion:** - The *stress* limit allows a relatively long span, - However, the *deflection* is unacceptably high, - **Deflection, not stress, will usually govern the allowable span for slender beams loaded by self-weight.** --- ## **Final Answers** 1. **Maximum allowable span:** \[ \boxed{L = 6.26 \text{ m}} \] 2. **Deflection at centre:** \[ \boxed{\delta_{\text{max}} = 0.29 \text{ m}} \] 3. **Deduction:** The stress limit permits a much longer span than is practical, due to the large deflection. For slender rods under their own weight, *deflection* is the controlling factor, not *failure by bending stress*. --- **Key Formulas Used:** - Area: \( A = \frac{\pi d^2}{4} \) - Second moment of area: \( I = \frac{\pi d^4}{64} \) - Max bending moment (UDL): \( M = \frac{wL^2}{8} \) - Max stress: \( \sigma = \frac{M y}{I} \) - Max deflection (UDL): \( \delta = \frac{5wL^4}{384EI} \)