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A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

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QUESTION 1: STATICS

1.1 A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

As an AI, I cannot physically draw the diagrams. However, I can describe them and provide the critical values.

  • Loaded Beam Description:

    • The beam is a 6m cantilever, fixed at the left end (let's call it A) and free at the right end (B).
    • A concentrated downward load of 35 kN35 \text{ kN} is applied at the free end (B).
    • A uniformly distributed downward load (UDL) of 9 kN/m9 \text{ kN/m} acts over the first 3 m3 \text{ m} from the fixed end (from A to a point C, 3 m3 \text{ m} from A).

  • Shear Force Diagram (SFD) Description:

    • The shear force starts at 0 kN0 \text{ kN} at the free end (B).
    • Just to the left of the free end (B), the shear force drops to −35 kN-35 \text{ kN} due to the concentrated load.
    • From x=6 mx=6 \text{ m} (free end) to x=3 mx=3 \text{ m} (start of UDL), the shear force remains constant at −35 kN-35 \text{ kN}.
    • From x=3 mx=3 \text{ m} to x=0 mx=0 \text{ m} (fixed end A), the shear force decreases linearly due to the UDL.
    • At x=3 mx=3 \text{ m}, the shear force is −35 kN-35 \text{ kN}.
    • At x=0 mx=0 \text{ m} (fixed end A), the shear force is −35 kN−(9 kN/m×3 m)=−35 kN−27 kN=−62 kN-35 \text{ kN} - (9 \text{ kN/m} \times 3 \text{ m}) = -35 \text{ kN} - 27 \text{ kN} = -62 \text{ kN}.
    • The diagram will show a horizontal line at −35 kN-35 \text{ kN} from x=3 mx=3 \text{ m} to x=6 mx=6 \text{ m}, and a linearly decreasing line from −35 kN-35 \text{ kN} to −62 kN-62 \text{ kN} from x=0 mx=0 \text{ m} to x=3 mx=3 \text{ m}.

  • Bending Moment Diagram (BMD) Description:

    • The bending moment starts at 0 kNm0 \text{ kNm} at the free end (B).
    • From x=6 mx=6 \text{ m} (free end) to x=3 mx=3 \text{ m} (start of UDL), the bending moment decreases linearly.
    • At x=3 mx=3 \text{ m}, the bending moment is −35 kN×(6 m−3 m)=−35 kN×3 m=−105 kNm-35 \text{ kN} \times (6 \text{ m} - 3 \text{ m}) = -35 \text{ kN} \times 3 \text{ m} = -105 \text{ kNm}.
    • From x=3 mx=3 \text{ m} to x=0 mx=0 \text{ m} (fixed end A), the bending moment decreases parabolically due to the UDL.
    • At x=0 mx=0 \text{ m} (fixed end A), the bending moment is −(35 kN×6 m)−(9 kN/m×3 m×1.5 m)=−210 kNm−40.5 kNm=−250.5 kNm- (35 \text{ kN} \times 6 \text{ m}) - (9 \text{ kN/m} \times 3 \text{ m} \times 1.5 \text{ m}) = -210 \text{ kNm} - 40.5 \text{ kNm} = -250.5 \text{ kNm}.
    • The diagram will show a linearly decreasing line from 0 kNm0 \text{ kNm} at x=6 mx=6 \text{ m} to −105 kNm-105 \text{ kNm} at x=3 mx=3 \text{ m}, and then a parabolically decreasing curve from −105 kNm-105 \text{ kNm} to −250.5 kNm-250.5 \text{ kNm} at x=0 mx=0 \text{ m}. The entire diagram will be in the negative region.

1.2 Calculate the position of the centroid of the lamina shown below.

The lamina consists of a rectangle and a triangle. Let's set the origin at the bottom-left corner of the rectangle.

  • Shape 1: Rectangle

    • Dimensions: 60 mm×50 mm60 \text{ mm} \times 50 \text{ mm}
    • Area A1=60 mm×50 mm=3000 mm2A_1 = 60 \text{ mm} \times 50 \text{ mm} = 3000 \text{ mm}^2
    • Centroid x1=60 mm2=30 mmx_1 = \frac{60 \text{ mm}}{2} = 30 \text{ mm}
    • Centroid y1=50 mm2=25 mmy_1 = \frac{50 \text{ mm}}{2} = 25 \text{ mm}

  • Shape 2: Triangle

    • Base (vertical) = 50 mm50 \text{ mm}, Height (horizontal) = 50 mm50 \text{ mm}
    • Area A2=12×50 mm×50 mm=1250 mm2A_2 = \frac{1}{2} \times 50 \text{ mm} \times 50 \text{ mm} = 1250 \text{ mm}^2
    • Centroid x2x_2: From the origin, it's 60 mm60 \text{ mm} (rectangle width) plus 13\frac{1}{3} of the triangle's height. x2=60 mm+50 mm3=60 mm+16.667 mm=76.667 mmx_2 = 60 \text{ mm} + \frac{50 \text{ mm}}{3} = 60 \text{ mm} + 16.667 \text{ mm} = 76.667 \text{ mm}
    • Centroid y2y_2: From the origin, it's 13\frac{1}{3} of the triangle's base (since the base is along the y-axis from y=0y=0 to y=50y=50). y2=50 mm3=16.667 mmy_2 = \frac{50 \text{ mm}}{3} = 16.667 \text{ mm}

  • Composite Centroid:

    • Total Area A=A1+A2=3000 mm2+1250 mm2=4250 mm2A = A_1 + A_2 = 3000 \text{ mm}^2 + 1250 \text{ mm}^2 = 4250 \text{ mm}^2

  1. Calculate the x-coordinate of the centroid.

    xˉ=A1x1+A2x2A1+A2\bar{x} = \frac{A_1 x_1 + A_2 x_2}{A_1 + A_2} xˉ=(3000 mm2×30 mm)+(1250 mm2×76.667 mm)4250 mm2\bar{x} = \frac{(3000 \text{ mm}^2 \times 30 \text{ mm}) + (1250 \text{ mm}^2 \times 76.667 \text{ mm})}{4250 \text{ mm}^2} xˉ=90000 mm3+95833.75 mm34250 mm2=185833.75 mm34250 mm2\bar{x} = \frac{90000 \text{ mm}^3 + 95833.75 \text{ mm}^3}{4250 \text{ mm}^2} = \frac{185833.75 \text{ mm}^3}{4250 \text{ mm}^2} xˉ≈43.726 mm\bar{x} \approx 43.726 \text{ mm}

  2. Calculate the y-coordinate of the centroid.

    yˉ=A1y1+A2y2A1+A2\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} yˉ=(3000 mm2×25 mm)+(1250 mm2×16.667 mm)4250 mm2\bar{y} = \frac{(3000 \text{ mm}^2 \times 25 \text{ mm}) + (1250 \text{ mm}^2 \times 16.667 \text{ mm})}{4250 \text{ mm}^2} yˉ=75000 mm3+20833.75 mm34250 mm2=95833.75 mm34250 mm2\bar{y} = \frac{75000 \text{ mm}^3 + 20833.75 \text{ mm}^3}{4250 \text{ mm}^2} = \frac{95833.75 \text{ mm}^3}{4250 \text{ mm}^2} yˉ≈22.549 mm\bar{y} \approx 22.549 \text{ mm}

    The position of the centroid is (xˉ,yˉ)=(43.73 mm,22.55 mm)\boxed{\text{}(\bar{x}, \bar{y}) = (43.73 \text{ mm}, 22.55 \text{ mm})} from the bottom-left corner.

    QUESTION 2: HYDRAULICS

    2.1 A force of 400N on the plunger of a hydraulic press causes a force of 14.5kN to be exerted on the ram. The diameter of the ram is 400mm. Calculate:

    • Given: Fp=400 NF_p = 400 \text{ N}, Fr=14.5 kN=14500 NF_r = 14.5 \text{ kN} = 14500 \text{ N}, Dr=400 mm=0.4 mD_r = 400 \text{ mm} = 0.4 \text{ m}.
    2.1.1 The liquid pressure

  3. Calculate the area of the ram.

    Ar=π(Dr2)2=π(0.4 m2)2=π(0.2 m)2=0.04π m2≈0.12566 m2A_r = \pi \left(\frac{D_r}{2}\right)^2 = \pi \left(\frac{0.4 \text{ m}}{2}\right)^2 = \pi (0.2 \text{ m})^2 = 0.04\pi \text{ m}^2 \approx 0.12566 \text{ m}^2

  4. Calculate the liquid pressure using the force on the ram.

    P=FrAr=14500 N0.04π m2P = \frac{F_r}{A_r} = \frac{14500 \text{ N}}{0.04\pi \text{ m}^2} P≈115380.2 PaP \approx 115380.2 \text{ Pa} P≈115.38 kPaP \approx 115.38 \text{ kPa}

    The liquid pressure is 115.38 kPa\boxed{\text{115.38 kPa}}.

    2.1.2 The diameter of the plunger in mm

  5. Calculate the area of the plunger using the plunger force and the calculated pressure.

    Ap=FpP=400 N115380.2 Pa≈0.0034668 m2A_p = \frac{F_p}{P} = \frac{400 \text{ N}}{115380.2 \text{ Pa}} \approx 0.0034668 \text{ m}^2

  6. Calculate the diameter of the plunger.

    Dp=2Apπ=20.0034668 m2π≈20.0011035 m2D_p = 2 \sqrt{\frac{A_p}{\pi}} = 2 \sqrt{\frac{0.0034668 \text{ m}^2}{\pi}} \approx 2 \sqrt{0.0011035 \text{ m}^2} Dp≈2×0.033219 m≈0.066438 mD_p \approx 2 \times 0.033219 \text{ m} \approx 0.066438 \text{ m}

    Convert to millimeters:

    Dp≈0.066438 m×1000 mm/m≈66.44 mmD_p \approx 0.066438 \text{ m} \times 1000 \text{ mm/m} \approx 66.44 \text{ mm}

    The diameter of the plunger is 66.44 mm\boxed{\text{66.44 mm}}.

    2.2 The diameter of the ram of a hydraulic press is 380mm and it must lift a vehicle 500m up. The vehicle has a mass of 2000kg. Calculate the following:

    • Given: Dr=380 mm=0.38 mD_r = 380 \text{ mm} = 0.38 \text{ m}, h=500 mh = 500 \text{ m}, m=2000 kgm = 2000 \text{ kg}, g=9.81 m/s2g = 9.81 \text{ m/s}^2.
    2.2.1 The pressure in the ram

  7. Calculate the force (weight) of the vehicle.

    F=m×g=2000 kg×9.81 m/s2=19620 NF = m \times g = 2000 \text{ kg} \times 9.81 \text{ m/s}^2 = 19620 \text{ N}

  8. Calculate the area of the ram.

    Ar=π(Dr2)2=π(0.38 m2)2=π(0.19 m)2=0.0361π m2≈0.11341 m2A_r = \pi \left(\frac{D_r}{2}\right)^2 = \pi \left(\frac{0.38 \text{ m}}{2}\right)^2 = \pi (0.19 \text{ m})^2 = 0.0361\pi \text{ m}^2 \approx 0.11341 \text{ m}^2

  9. Calculate the pressure in the ram.

    P=FAr=19620 N0.0361π m2P = \frac{F}{A_r} = \frac{19620 \text{ N}}{0.0361\pi \text{ m}^2} P≈173000 PaP \approx 173000 \text{ Pa} P≈173.0 kPaP \approx 173.0 \text{ kPa}

    The pressure in the ram is 173.0 kPa\boxed{\text{173.0 kPa}}.

    2.2.2 The volume of liquid displaced to do this work

  10. Calculate the volume of liquid displaced.

    V=Ar×h=(0.0361π m2)×500 mV = A_r \times h = (0.0361\pi \text{ m}^2) \times 500 \text{ m} V≈0.11341 m2×500 m≈56.705 m3V \approx 0.11341 \text{ m}^2 \times 500 \text{ m} \approx 56.705 \text{ m}^3

    The volume of liquid displaced is 56.71 m3\boxed{\text{56.71 m}^3}.

    2.2.3 The work done if the press is 85% efficient
    • Given efficiency η=85%=0.85\eta = 85\% = 0.85.

  11. Calculate the output work (work done on the vehicle).

    Wout=F×h=19620 N×500 m=9810000 JW_{out} = F \times h = 19620 \text{ N} \times 500 \text{ m} = 9810000 \text{ J} Wout=9.81 MJW_{out} = 9.81 \text{ MJ}

  12. Calculate the input work (work done by the press) considering efficiency.

    Win=Woutη=9810000 J0.85W_{in} = \frac{W_{out}}{\eta} = \frac{9810000 \text{ J}}{0.85} Win≈11541176.47 JW_{in} \approx 11541176.47 \text{ J} Win≈11.54 MJW_{in} \approx 11.54 \text{ MJ}

    The work done if the press is 85% efficient is 11.54 MJ\boxed{\text{11.54 MJ}}.

    2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m3^3 and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m.

    • Given: D=100 mm=0.1 mD = 100 \text{ mm} = 0.1 \text{ m}, L=300 mm=0.3 mL = 300 \text{ mm} = 0.3 \text{ m}, N=80 rpmN = 80 \text{ rpm}, ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3, η=68%=0.68\eta = 68\% = 0.68, H=100 mH = 100 \text{ m}, g=9.81 m/s2g = 9.81 \text{ m/s}^2.
    • A double-acting pump displaces liquid on both the forward and return strokes.

  13. Calculate the cross-sectional area of the cylinder.

    A=π(D2)2=π(0.1 m2)2=π(0.05 m)2=0.0025π m2≈0.007854 m2A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.1 \text{ m}}{2}\right)^2 = \pi (0.05 \text{ m})^2 = 0.0025\pi \text{ m}^2 \approx 0.007854 \text{ m}^2

  14. Calculate the volume displaced per revolution.

    Vrev=2×A×L=2×(0.0025π m2)×0.3 m=0.0015π m3≈0.004712 m3V_{rev} = 2 \times A \times L = 2 \times (0.0025\pi \text{ m}^2) \times 0.3 \text{ m} = 0.0015\pi \text{ m}^3 \approx 0.004712 \text{ m}^3

  15. Calculate the volume flow rate in m3^3/s.

    Q=Vrev×N=(0.0015π m3/rev)×(80 rev/min)×(1 min60 s)Q = V_{rev} \times N = (0.0015\pi \text{ m}^3/\text{rev}) \times (80 \text{ rev/min}) \times \left(\frac{1 \text{ min}}{60 \text{ s}}\right) Q=0.12π60 m3/s=0.002π m3/s≈0.006283 m3/sQ = \frac{0.12\pi}{60} \text{ m}^3/\text{s} = 0.002\pi \text{ m}^3/\text{s} \approx 0.006283 \text{ m}^3/\text{s}

  16. Calculate the mass flow rate.

    m˙=ρ×Q=1000 kg/m3×0.002π m3/s=2π kg/s≈6.283 kg/s\dot{m} = \rho \times Q = 1000 \text{ kg/m}^3 \times 0.002\pi \text{ m}^3/\text{s} = 2\pi \text{ kg/s} \approx 6.283 \text{ kg/s}

  17. Calculate the output power (power required to lift the water).

    Pout=m˙×g×H=(2π kg/s)×(9.81 m/s2)×(100 m)P_{out} = \dot{m} \times g \times H = (2\pi \text{ kg/s}) \times (9.81 \text{ m/s}^2) \times (100 \text{ m}) Pout=1962π W≈6163.6 WP_{out} = 1962\pi \text{ W} \approx 6163.6 \text{ W} Pout≈6.164 kWP_{out} \approx 6.164 \text{ kW}

  18. Calculate the input power required considering the pump efficiency.

    Pin=Poutη=6163.6 W0.68P_{in} = \frac{P_{out}}{\eta} = \frac{6163.6 \text{ W}}{0.68} Pin≈9064.1 WP_{in} \approx 9064.1 \text{ W} Pin≈9.064 kWP_{in} \approx 9.064 \text{ kW}

    The power required to deliver water is 9.064 kW\boxed{\text{9.064 kW}}.

    QUESTION 3: STRESS AND STRAIN

    3.1 Name 3 types of stresses.

    • Tensile stress
    • Compressive stress
    • Shear stress

    3.2 A tensile force of 3kN is applied to a steel rod of 2m in length and the rod extends by 0.381mm. The rod has a diameter of 1cm. Calculate:

    • Given: F=3 kN=3000 NF = 3 \text{ kN} = 3000 \text{ N}, L=2 mL = 2 \text{ m}, ΔL=0.381 mm=0.000381 m\Delta L = 0.381 \text{ mm} = 0.000381 \text{ m}, D=1 cm=0.01 mD = 1 \text{ cm} = 0.01 \text{ m}.
    3.2.1 The stress

  19. Calculate the cross-sectional area of the rod.

    A=π(D2)2=π(0.01 m2)2=π(0.005 m)2=0.000025π m2≈7.854×10−5 m2A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.01 \text{ m}}{2}\right)^2 = \pi (0.005 \text{ m})^2 = 0.000025\pi \text{ m}^2 \approx 7.854 \times 10^{-5} \text{ m}^2

  20. Calculate the tensile stress.

    σ=FA=3000 N0.000025π m2\sigma = \frac{F}{A} = \frac{3000 \text{ N}}{0.000025\pi \text{ m}^2} σ≈38197186.3 Pa\sigma \approx 38197186.3 \text{ Pa} σ≈38.20 MPa\sigma \approx 38.20 \text{ MPa}

    The stress is 38.20 MPa\boxed{\text{38.20 MPa}}.

    3.2.2 The strain

  21. Calculate the strain.

    ϵ=ΔLL=0.000381 m2 m\epsilon = \frac{\Delta L}{L} = \frac{0.000381 \text{ m}}{2 \text{ m}} ϵ=0.0001905\epsilon = 0.0001905

    The strain is 0.0001905\boxed{\text{0.0001905}}.

    3.2.3 The modulus of elasticity

  22. Calculate the modulus of elasticity (Young's Modulus).

    E=σϵ=38197186.3 Pa0.0001905E = \frac{\sigma}{\epsilon} = \frac{38197186.3 \text{ Pa}}{0.0001905} E≈200510164.3 kPaE \approx 200510164.3 \text{ kPa} E≈200.51 GPaE \approx 200.51 \text{ GPa}

    The modulus of elasticity is 200.51 GPa\boxed{\text{200.51 GPa}}.

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