Solve for me Question C and D in the screenshot

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2 parts
c

22.284 MPa22.284 \text{ MPa}

2.846 MPa2.846 \text{ MPa}

45.25 mm45.25 \text{ mm}

d

480 MPa480 \text{ MPa}

52.083 m52.083 \text{ m}

230.4 MPa230.4 \text{ MPa}

Step-by-step solution

10 steps

let's solve questions C and D.

Part c· 5 steps

Column Analysis

Given:
Load, P=75 kN=75×103 NP = 75 \text{ kN} = 75 \times 10^3 \text{ N}
External diameter, D=200 mm=0.200 mD = 200 \text{ mm} = 0.200 \text{ m}
Internal diameter, d=180 mm=0.180 md = 180 \text{ mm} = 0.180 \text{ m}
Eccentricity, e=35 mm=0.035 me = 35 \text{ mm} = 0.035 \text{ m}

  1. Calculate the cross-sectional area (AA) and moment of inertia (II) of the hollow circular column.

    A=π4(D2−d2)=π4((0.200 m)2−(0.180 m)2)A = \frac{\pi}{4} (D^2 - d^2) = \frac{\pi}{4} ((0.200 \text{ m})^2 - (0.180 \text{ m})^2) A=π4(0.0400−0.0324) m2=π4(0.0076) m2=0.005969 m2A = \frac{\pi}{4} (0.0400 - 0.0324) \text{ m}^2 = \frac{\pi}{4} (0.0076) \text{ m}^2 = 0.005969 \text{ m}^2 I=π64(D4−d4)=π64((0.200 m)4−(0.180 m)4)I = \frac{\pi}{64} (D^4 - d^4) = \frac{\pi}{64} ((0.200 \text{ m})^4 - (0.180 \text{ m})^4) I=π64(0.0016−0.00104976) m4=π64(0.00055024) m4=2.701×10−5 m4I = \frac{\pi}{64} (0.0016 - 0.00104976) \text{ m}^4 = \frac{\pi}{64} (0.00055024) \text{ m}^4 = 2.701 \times 10^{-5} \text{ m}^4

  2. Calculate the section modulus (ZZ).

    The maximum distance from the neutral axis to the extreme fiber is ymax=D2=0.200 m2=0.100 my_{max} = \frac{D}{2} = \frac{0.200 \text{ m}}{2} = 0.100 \text{ m}.

    Z=Iymax=2.701×10−5 m40.100 m=2.701×10−4 m3Z = \frac{I}{y_{max}} = \frac{2.701 \times 10^{-5} \text{ m}^4}{0.100 \text{ m}} = 2.701 \times 10^{-4} \text{ m}^3

    C (i) The maximum and minimum stress intensities.

  3. Calculate the direct stress (σd\sigma_d) and bending stress (σb\sigma_b).

    σd=PA=75×103 N0.005969 m2=12.565×106 N/m2=12.565 MPa\sigma_d = \frac{P}{A} = \frac{75 \times 10^3 \text{ N}}{0.005969 \text{ m}^2} = 12.565 \times 10^6 \text{ N/m}^2 = 12.565 \text{ MPa}

    The bending moment is M=P×e=75×103 N×0.035 m=2625 N⋅mM = P \times e = 75 \times 10^3 \text{ N} \times 0.035 \text{ m} = 2625 \text{ N} \cdot \text{m}.

    σb=MZ=2625 N⋅m2.701×10−4 m3=9.719×106 N/m2=9.719 MPa\sigma_b = \frac{M}{Z} = \frac{2625 \text{ N} \cdot \text{m}}{2.701 \times 10^{-4} \text{ m}^3} = 9.719 \times 10^6 \text{ N/m}^2 = 9.719 \text{ MPa}

  4. Calculate the maximum and minimum stress intensities.

    σmax=σd+σb=12.565 MPa+9.719 MPa=22.284 MPa\sigma_{max} = \sigma_d + \sigma_b = 12.565 \text{ MPa} + 9.719 \text{ MPa} = \boxed{22.284 \text{ MPa}} σmin=σd−σb=12.565 MPa−9.719 MPa=2.846 MPa\sigma_{min} = \sigma_d - \sigma_b = 12.565 \text{ MPa} - 9.719 \text{ MPa} = \boxed{2.846 \text{ MPa}}

    C (ii) Up to what eccentricity there is no tensile stress in the column?

  5. For no tensile stress, the minimum stress must be zero or positive (σmin≥0\sigma_{min} \ge 0).

    This occurs when σd=σb\sigma_d = \sigma_b.

    PA=P×emaxZ\frac{P}{A} = \frac{P \times e_{max}}{Z} emax=ZAe_{max} = \frac{Z}{A} emax=2.701×10−4 m30.005969 m2=0.04525 me_{max} = \frac{2.701 \times 10^{-4} \text{ m}^3}{0.005969 \text{ m}^2} = 0.04525 \text{ m} emax=45.25 mme_{max} = \boxed{45.25 \text{ mm}}

Part d

480 MPa480 \text{ MPa}

Rectangular Beam Analysis

Given:
Depth, d=250 mm=0.250 md = 250 \text{ mm} = 0.250 \text{ m}
Width, b=150 mm=0.150 mb = 150 \text{ mm} = 0.150 \text{ m}
Maximum bending moment, M=750 kNm=750×103 N⋅mM = 750 \text{ kNm} = 750 \times 10^3 \text{ N} \cdot \text{m}
Modulus of Elasticity, E=200 GN/m2=200×109 N/m2E = 200 \text{ GN/m}^2 = 200 \times 10^9 \text{ N/m}^2

  1. Calculate the moment of inertia (II) for the rectangular beam.

    I=bd312=0.150 m×(0.250 m)312I = \frac{bd^3}{12} = \frac{0.150 \text{ m} \times (0.250 \text{ m})^3}{12} I=0.150×0.01562512 m4=1.953125×10−4 m4I = \frac{0.150 \times 0.015625}{12} \text{ m}^4 = 1.953125 \times 10^{-4} \text{ m}^4

    D (i) The maximum stress in the beam.

  2. Calculate the section modulus (ZZ) and maximum bending stress (σmax\sigma_{max}).

    The maximum distance from the neutral axis to the extreme fiber is ymax=d2=0.250 m2=0.125 my_{max} = \frac{d}{2} = \frac{0.250 \text{ m}}{2} = 0.125 \text{ m}.

    Z=Iymax=1.953125×10−4 m40.125 m=1.5625×10−3 m3Z = \frac{I}{y_{max}} = \frac{1.953125 \times 10^{-4} \text{ m}^4}{0.125 \text{ m}} = 1.5625 \times 10^{-3} \text{ m}^3 σmax=MZ=750×103 N⋅m1.5625×10−3 m3=480×106 N/m2\sigma_{max} = \frac{M}{Z} = \frac{750 \times 10^3 \text{ N} \cdot \text{m}}{1.5625 \times 10^{-3} \text{ m}^3} = 480 \times 10^6 \text{ N/m}^2 σmax=480 MPa\sigma_{max} = \boxed{480 \text{ 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.

  3. Use the bending equation MI=ER\frac{M}{I} = \frac{E}{R} to find the radius of curvature (RR).

    R=EIMR = \frac{EI}{M} R=(200×109 N/m2)×(1.953125×10−4 m4)750×103 N⋅mR = \frac{(200 \times 10^9 \text{ N/m}^2) \times (1.953125 \times 10^{-4} \text{ m}^4)}{750 \times 10^3 \text{ N} \cdot \text{m}} R=39062500750000 m=52.083 mR = \frac{39062500}{750000} \text{ m} = \boxed{52.083 \text{ m}}

    D (iii) The value of the longitudinal stress at a distance of 65 mm from the top surface of the beam.

  4. Determine the distance (yy) from the neutral axis.

    The neutral axis is at the mid-depth, which is 250 mm2=125 mm\frac{250 \text{ mm}}{2} = 125 \text{ mm} from the top surface.
    The distance from the top surface is 65 mm65 \text{ mm}.
    So, y=125 mm−65 mm=60 mm=0.060 my = 125 \text{ mm} - 65 \text{ mm} = 60 \text{ mm} = 0.060 \text{ m}.

  5. Calculate the longitudinal stress (σ\sigma) at this distance.

    σ=MyI=(750×103 N⋅m)×(0.060 m)1.953125×10−4 m4\sigma = \frac{M y}{I} = \frac{(750 \times 10^3 \text{ N} \cdot \text{m}) \times (0.060 \text{ m})}{1.953125 \times 10^{-4} \text{ m}^4} σ=450001.953125×10−4 N/m2=230.4×106 N/m2\sigma = \frac{45000}{1.953125 \times 10^{-4}} \text{ N/m}^2 = 230.4 \times 10^6 \text{ N/m}^2 σ=230.4 MPa\sigma = \boxed{230.4 \text{ MPa}}
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