**Sub-Subject:** Mechanics of Materials **Topic:** Bending of Beams (Self-weight Analysis, Maximum Stress, Deflection) --- ## Restatement of the Problem A **steel rod** with a **diameter of 25 mm** is **simply supported** over a span of \( L \) meters. The **only loading** is due to its **own mass**. - The **allowable stress** in the rod is **120 MPa**. - **Young’s modulus, \( E = 210 \) GPa** - **Density of steel, \( \rho = 7800 \) kg/m³** **Tasks:** 1. Calculate the **span \( L \)** such that the **maximum stress** does **not exceed 120 MPa**. 2. Find the **deflection at the center** of the span. 3. State any **deductions** that can be made from these answers. --- ## Step-by-Step Solution ### **Step 1: Calculation of Self-Weight per Unit Length** The rod is **circular** with a **diameter, \( d = 25 \) mm = 0.025 m**. - **Cross-sectional area:** \[ A = \frac{\pi d^2}{4} = \frac{\pi}{4} \times (0.025)^2 = 4.9087 \times 10^{-4} \text{ m}^2 \] - **Weight per unit length, \( w \):** \[ w = \text{mass per unit length} \times g = (\rho A)g \] Where: - \( \rho = 7800\, \text{kg/m}^3 \) - \( g = 9.81\, \text{m/s}^2 \) \[ w = 7800 \times 4.9087 \times 10^{-4} \times 9.81 = 37.53\, \text{N/m} \] --- ### **Step 2: Maximum Bending Moment Due to Self-Weight** For a simply supported beam, **uniformly distributed load** (\( w \)), **maximum bending moment** at center: \[ M_{max} = \frac{w L^2}{8} \] --- ### **Step 3: Section Properties** - **Section modulus, \( Z \) (for a solid circular section):** \[ Z = \frac{\pi d^3}{32} \] \[ Z = \frac{\pi}{32} \times (0.025)^3 = 1.534 \times 10^{-5} \text{ m}^3 \] --- ### **Step 4: Maximum Bending Stress** The **maximum bending stress** is: \[ \sigma_{max} = \frac{M_{max}}{Z} \] Set \( \sigma_{max} = 120 \) MPa \( = 120 \times 10^6 \) Pa, and solve for \( L \): \[ \sigma_{max} = \frac{w L^2}{8Z} \] \[ L^2 = \frac{8Z \sigma_{max}}{w} \] \[ L = \sqrt{\frac{8Z \sigma_{max}}{w}} \] Plug in values: - \( Z = 1.534 \times 10^{-5} \) m³ - \( \sigma_{max} = 120 \times 10^6 \) Pa - \( w = 37.53 \) N/m \[ L = \sqrt{ \frac{8 \times 1.534 \times 10^{-5} \times 120 \times 10^6}{37.53} } \] \[ = \sqrt{ \frac{8 \times 1.534 \times 120 \times 10^1 }{37.53} } \] Calculate numerator: \[ 8 \times 1.534 \times 10^{-5} = 1.2272 \times 10^{-4} \] \[ 1.2272 \times 10^{-4} \times 120 \times 10^6 = 1.2272 \times 120 \times 10^{2} = 147.264 \times 10^{2} = 14726.4 \] But, including the powers: \[ 1.534 \times 10^{-5} \times 120 \times 10^6 = 1.534 \times 120 \times 10^1 = 184.08 \] \[ 8 \times 184.08 = 1472.64 \] So numerator is \( 1472.64 \) Now divide by \( w \): \[ L^2 = \frac{1472.64}{37.53} = 39.25 \] \[ L = \sqrt{39.25} = 6.27\, \text{m} \] --- ### **Step 5: Maximum Deflection at Center** For a simply supported beam with uniformly distributed load: \[ \delta_{max} = \frac{5 w L^4}{384 E I} \] Where: - \( I = \) Second moment of area for circle (\( I = \frac{\pi d^4}{64} \)) - \( E = 210\, \text{GPa} = 210 \times 10^9\, \text{Pa} \) Calculate \( I \): \[ I = \frac{\pi}{64} \times (0.025)^4 = \frac{\pi}{64} \times 3.90625 \times 10^{-7} \] \[ = 1.918 \times 10^{-8} \text{ m}^4 \] Plug in all values: - \( w = 37.53 \) N/m - \( L = 6.27 \) m - \( E = 210 \times 10^9 \) Pa - \( I = 1.918 \times 10^{-8} \) m⁴ Calculate \( L^4 \): \[ L^4 = (6.27)^4 = (6.27 \times 6.27) \times (6.27 \times 6.27) = 39.3 \times 39.3 = 1544 \] Now substitute in the deflection formula: \[ \delta_{max} = \frac{5 \times 37.53 \times 1544}{384 \times 210 \times 10^9 \times 1.918 \times 10^{-8}} \] Calculate numerator: \[ 5 \times 37.53 \times 1544 = 5 \times 57941 = 289705 \] Calculate denominator: \[ 384 \times 210 \times 10^9 \times 1.918 \times 10^{-8} \] \[ 384 \times 210 = 80640 \] \[ 80640 \times 1.918 \times 10^{-8} = 154716 \times 10^{-8} = 1.54716 \times 10^{-3} \] \[ 1.54716 \times 10^{-3} \times 10^9 = 1.54716 \times 10^{6} \] Therefore, \[ \delta_{max} = \frac{289705}{1.54716 \times 10^{6}} = 0.1873\, \text{m} = 187.3\, \text{mm} \] --- ### **Step 6: Deductions** - The **maximum span** based on *allowable stress* is \( 6.27\, \text{m} \). - The **central deflection** is extremely large (\( 187.3\, \text{mm} \)), which is *unacceptable* for most engineering applications. - **Conclusion:** **Deflection**, not *stress*, governs the design for beams subjected to self-weight, especially for slender members. In practice, serviceability (deflection limits) will require a much shorter span or a stiffer section. --- ## **Final Answers** 1. **Maximum span \( L \) for allowable stress of 120 MPa:** \[ \boxed{6.27\, \text{m}} \] 2. **Deflection at the center of the span:** \[ \boxed{187.3\, \text{mm}} \] 3. **Deduction:** While the span is limited by *allowable stress*, the resulting deflection is excessively large, indicating that *deflection* (serviceability) is the true limiting factor for such long, slender beams under self-weight. **Designs should be checked for both strength and serviceability.**