1. **Problem Statement:** Analyze the simply supported steel beam of span $L=6.0$ m under a uniformly distributed load (UDL) $w=3.0$ kN/m and a concentrated load $P=10.0$ kN at mid-span. 2. **Given Data:** - Span, $L=6.0$ m - UDL, $w=3.0$ kN/m - Concentrated load, $P=10.0$ kN at mid-span - Beam cross-section: rectangular, width $b=120$ mm $=0.12$ m, depth $h=240$ mm $=0.24$ m - Modulus of elasticity, $E=200$ GPa $=200\times10^9$ Pa - Allowable bending stress, $\sigma_{allow}=150$ MPa $=150\times10^6$ Pa - Allowable deflection limit, $\delta_{allow}=\frac{L}{360}=\frac{6.0}{360}=0.0167$ m 3. **Step 1: Calculate section properties** - Moment of inertia for rectangular section: $$I=\frac{b h^3}{12}=\frac{0.12 \times (0.24)^3}{12}=1.3824 \times 10^{-4} \text{ m}^4$$ - Distance from neutral axis to outer fiber: $$c=\frac{h}{2}=0.12 \text{ m}$$ 4. **Step 2: Calculate maximum bending moment** - Maximum bending moment due to UDL at mid-span: $$M_w=\frac{w L^2}{8}=\frac{3.0 \times 6.0^2}{8}=13.5 \text{ kNm}=13,500 \text{ Nm}$$ - Maximum bending moment due to concentrated load at mid-span: $$M_P=\frac{P L}{4}=\frac{10.0 \times 6.0}{4}=15.0 \text{ kNm}=15,000 \text{ Nm}$$ - Total maximum bending moment: $$M_{max}=M_w + M_P=13,500 + 15,000=28,500 \text{ Nm}$$ 5. **Step 3: Calculate maximum bending stress** Using bending stress formula: $$\sigma=\frac{M c}{I}=\frac{28,500 \times 0.12}{1.3824 \times 10^{-4}}=24,720,000 \text{ Pa}=24.72 \text{ MPa}$$ 6. **Step 4: Calculate maximum shear force and shear stress** - Maximum shear force due to UDL and point load at support: $$V_{max}=\frac{w L}{2} + \frac{P}{2}=\frac{3.0 \times 6.0}{2} + \frac{10.0}{2}=9 + 5=14 \text{ kN}=14,000 \text{ N}$$ - Maximum shear stress for rectangular section: $$\tau_{max}=\frac{3 V}{2 A}$$ where cross-sectional area: $$A=b h=0.12 \times 0.24=0.0288 \text{ m}^2$$ Calculate shear stress: $$\tau_{max}=\frac{3 \times 14,000}{2 \times 0.0288}=728,000 \text{ Pa}=0.728 \text{ MPa}$$ 7. **Step 5: Calculate deflection at mid-span** - Deflection due to UDL: $$\delta_w=\frac{5 w L^4}{384 E I}=\frac{5 \times 3,000 \times 6.0^4}{384 \times 200 \times 10^9 \times 1.3824 \times 10^{-4}}=0.0073 \text{ m}$$ - Deflection due to concentrated load: $$\delta_P=\frac{P L^3}{48 E I}=\frac{10,000 \times 6.0^3}{48 \times 200 \times 10^9 \times 1.3824 \times 10^{-4}}=0.0032 \text{ m}$$ - Total deflection: $$\delta_{max}=\delta_w + \delta_P=0.0073 + 0.0032=0.0105 \text{ m}$$ 8. **Step 6: Compare with allowable limits** - Bending stress $24.72$ MPa $< 150$ MPa (safe) - Deflection $0.0105$ m $< 0.0167$ m (within limit) **Final answers:** - Maximum bending stress: $24.72$ MPa - Maximum shear stress: $0.728$ MPa - Maximum deflection at mid-span: $0.0105$ m These values indicate the beam is safe under the given loads with the rectangular cross-section. **Slug:** "bridge beam analysis" **Subject:** "structural engineering" **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 1