This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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let's solve questions C and D. Question C: Column Analysis Given: Load, P = 75 kN = 75 × 10^3 N External diameter, D = 200 mm = 0.200 m Internal diameter, d = 180 mm = 0.180 m Eccentricity, e = 35 mm = 0.035 m Step 1: Calculate the cross-sectional area (A) and moment of inertia (I) of the hollow circular column. A = ()/(4) (D^2 - d^2) = ()/(4) ((0.200 m)^2 - (0.180 m)^2) A = ()/(4) (0.0400 - 0.0324) m^2 = ()/(4) (0.0076) m^2 = 0.005969 m^2 I = ()/(64) (D^4 - d^4) = ()/(64) ((0.200 m)^4 - (0.180 m)^4) I = ()/(64) (0.0016 - 0.00104976) m^4 = ()/(64) (0.00055024) m^4 = 2.701 × 10^-5 m^4 Step 2: Calculate the section modulus (Z). The maximum distance from the neutral axis to the extreme fiber is y_max = (D)/(2) = 0.200 m2 = 0.100 m. Z = (I)/(y_max) = 2.701 × 10^-5 m^40.100 m = 2.701 × 10^-4 m^3 C (i) The maximum and minimum stress intensities. Step 3: Calculate the direct stress (_d) and bending stress (_b). _d = (P)/(A) = 75 × 10^3 N0.005969 m^2 = 12.565 × 10^6 N/m^2 = 12.565 MPa The bending moment is M = P × e = 75 × 10^3 N × 0.035 m = 2625 N · m. _b = (M)/(Z) = 2625 N · m2.701 × 10^-4 m^3 = 9.719 × 10^6 N/m^2 = 9.719 MPa Step 4: Calculate the maximum and minimum stress intensities. _max = _d + _b = 12.565 MPa + 9.719 MPa = 22.284 MPa _min = _d - _b = 12.565 MPa - 9.719 MPa = 2.846 MPa C (ii) Up to what eccentricity there is no tensile stress in the column? Step 5: For no tensile stress, the minimum stress must be zero or positive (_min 0). This occurs when _d = _b. (P)/(A) = P × e_maxZ e_max = (Z)/(A) e_max = 2.701 × 10^-4 m^30.005969 m^2 = 0.04525 m e_max = 45.25 mm Question D: Rectangular Beam Analysis Given: Depth, d = 250 mm = 0.250 m Width, b = 150 mm = 0.150 m Maximum bending moment, M = 750 kNm = 750 × 10^3 N · m Modulus of Elasticity, E = 200 GN/m^2 = 200 × 10^9 N/m^2 Step 1: Calculate the moment of inertia (I) for the rectangular beam. I = (bd^3)/(12) = 0.150 m × (0.250 m)^312 I = (0.150 × 0.015625)/(12) m^4 = 1.953125 × 10^-4 m^4 D (i) The maximum stress in the beam. Step 2: Calculate the section modulus (Z) and maximum bending stress (_max). The maximum distance from the neutral axis to the extreme fiber is y_max = (d)/(2) = 0.250 m2 = 0.125 m. Z = (I)/(y_max) = 1.953125 × 10^-4 m^40.125 m = 1.5625 × 10^-3 m^3 _max = (M)/(Z) = 750 × 10^3 N · m1.5625 × 10^-3 m^3 = 480 × 10^6 N/m^2 _max = 480 MPa D (ii) If the value of E for the beam material is 200 GN/m², calculate the radius of curvature for that portion of the beam where the bending is maximum. Step 3: Use the bending equation (M)/(I) = (E)/(R) to find the radius of curvature (R). R = (EI)/(M) R = (200 × 10^9 N/m^2) × (1.953125 × 10^-4 m^4)750 × 10^3 N · m R = (39062500)/(750000) m = 52.083 m D (iii) The value of the longitudinal stress at a distance of 65 mm from the top surface of the beam. Step 4: Determine the distance (y) from the neutral axis. The neutral axis is at the mid-depth, which is 250 mm2 = 125 mm from the top surface. The distance from the top surface is 65 mm. So, y = 125 mm - 65 mm = 60 mm = 0.060 m. Step 5: Calculate the longitudinal stress () at this distance. = (M y)/(I) = (750 × 10^3 N · m) × (0.060 m)1.953125 × 10^-4 m^4 = (45000)/(1.953125 × 10^-4) N/m^2 = 230.4 × 10^6 N/m^2 = 230.4 MPa What's next?