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7.10: Questions

  • Page ID
    23531
  • Quick questions

    You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

    1. How do the axial stresses within a vaulting pole vary with distance from the neutral axis?
      a They are zero at the neutral axis, rising to a tensile maximum at the outer surface, and a compressive maximum at the inner surface.
      b They are zero at the inner surface and rise linearly to a maximum at the outer surface.
      c They reach a maximum at the neutral axis, falling to zero at the outer and inner surfaces.
      d They are constant throughout the section.
    Answer

    A

    It is important to maximise the beam stiffness when attempting to minimise the deflection of a beam (of given mass). Which of the following shapes, all with dimensions such that they have the same cross-sectional area, will have the highest beam stiffness?

    a An I-beam
    b A square section beam
    c A hollow cylindrical beam
    d A circular section beam
    Answer

    A. In general, an I-beam gives a high value of I, assuming that the central section and the flange both have a fairly high aspect ratio.

    In which of the following situations is torsion occurring?

    a A tree bending in the wind.
    b A wire hanging under its own weight.
    c A screwdriver being used to tighten a screw.
    d The wings on an aircraft acting as cantilever beams during flight.
    Answer

    C.

    How can the stress distribution in an elastoplastic beam undergoing bending be predicted?

    a The axial strain will vary linearly from the neutral axis to the free surfaces, and so the stress distribution should increase linearly in the same fashion.
    b The axial strain will vary linearly from the neutral axis to the free surfaces, and the stress distribution can be found from this information and the stress-strain curve of the material.
    c The stress distribution cannot be calculated theoretically and must be found by experiment.
    d The beam will be fully plastic and so the stress will be of constant magnitude throughout the section. This stress can be predicted from the Young's modulus and the yield strain of the material.
    Answer

    B.

    Deeper questions

    The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP

    Which of the following sectional shapes will give the highest beam stiffness?
    a Hollow tube of outer diameter 15 cm and inner diameter 14 cm
    b I-beam with a central section 11 cm high by 2 cm wide and flanges 2 cm high by 10 cm wide (giving a total height of 15 cm)

    Answer

    The I-beam.

    The second moment of area of a shape is given by:

    \[I=\int_{0}^{y_{\max }} y^{2} \mathrm{d} A\]

    Therefore the second moment of area for the hollow tube is given by:

    \[I=\int_{0}^{D / 2} y^{2} \mathrm{d} A-\int_{0}^{d / 2} y^{2} \mathrm{d} A=\int_{0}^{d / 2} y^{2} \pi y \mathrm{d} y-\int_{0}^{D / 2} y^{2} \pi y \mathrm{d} y=\int_{d / 2}^{D / 2} \pi y^{3} \mathrm{d} y\]

    \[=\left[\frac{\pi y^{4}}{4}\right]_{a / 2}^{D / 2}=\frac{\pi\left((D / 2)^{4}-(d / 2)^{4}\right)}{4}=\frac{\pi}{64}\left(D^{4}-d^{4}\right)\]

    For the hollow tube of interest here

    \[I=\frac{\pi}{64}\left(15^{4}-14^{4}\right)=599 \mathrm{cm}^{4}\]

    The second moment of area of the I-beam is given by:

    \[I=\int_{0}^{H / 2} y^{2} \mathrm{d} A-\int_{0}^{(H / 2)-t} y^{2} \mathrm{d} A=2 \int_{0}^{H / 2} y^{2} W \mathrm{d} y-2 \int_{0}^{(H / 2)-t} y^{2}(W-t) \mathrm{d} y\]

    \[=2\left[\frac{y^{3} W}{3}\right]_{0}^{H / 2}-2\left[\frac{y^{3}(W-t)}{3}\right]_{0}^{(H / 2)-t}=\frac{1}{12}\left(W H^{3}-(W-t)(H-2 t)^{3}\right)\]

    Therefore the second moment of area of the I-beam shown is:

    \[I=\frac{1}{12}\left(10(15)^{3}-(10-2)(15-4)^{3}\right)=\frac{1}{12}(33750-10648)=1925 \mathrm{cm}^{4}\]

    A solid rectangular section beam of length, L = 100 cm, height, h = 5 cm and width, w = 1 cm, is loaded under symmetrical 4-point bending, with 1000 N downward forces applied at 40 cm in from both ends of the bar, which is supported at both ends. A deflection of 5 mm is measured at the centre of the beam. Using these data, calculate the Young's modulus of the beam. From your answer, suggest a likely material for the beam.

    hint: The equation for the deflection of a beam under symmetrical 4-point bending is:

    \[\delta=\frac{F L_{1}^{2}}{6 E I}\left(2 L_{1}+3 L_{2}\right)\]

    Answer

    E = 72 GPa . The beam is probably made of aluminium.

    The second moment of area of the beam is:

    \[I=\int_{0}^{h / 2} y^{2} \mathrm{d} \mathrm{A}=2 \int_{0}^{h / 2} y^{2} w \mathrm{d} y=2 w\left[\frac{y^{3}}{3}\right]_{0}^{h / 2}=\frac{w h^{3}}{12}=\frac{0.01(0.05)^{3}}{12}=1.04 \times 10^{-7} \mathrm{m}^{4}\]

    For symmetrical 4-point bending the equation for the deflection of the central region of the bar is:

    \[\delta=\frac{F L_{1}^{2}}{6 E l}\left(2 L_{1}+3 L_{2}\right)\]

    where 2L1 + L2 = L

    and in this case L1 =40 cm and L2 =20 cm
    Rearranging to obtain the Young's modulus:

    \[E=\frac{F L_{1}^{2}}{6 I \delta}\left(2 L_{1}+3 L_{2}\right)=\frac{1000(0.4)^{2}}{6\left(1.04 \times 10^{-7}\right)\left(5 \times 10^{-3}\right)}(2(0.4)+3(0.2))=72 \mathrm{GPa}\]

    The beam has a Young's modulus of 72 GPa, and so therefore is probably made from aluminium. (Glass also has a Young's modulus in the region of 70 GPa, but is unlikely to be used to produce a beam of this type).

    Calculate the shear modulus, G , of a material supplied in the form of a hollow tube (length 100 cm, outer diameter 5 cm, wall thickness 0.1 cm), given that, when it is subjected to an applied torque of 1000 N m, an angular twist of 0.10 radians is generated.

    hint: Start by calculating the polar moment of inertia for a hollow tube

    Answer

    G = 108 GPa.

    The polar moment of area of the beam is:

    \[I_{p}=\int_{A} \rho^{2} \mathrm{d} A=\int_{0}^{D / 2} \rho^{2} 2 \pi \rho \mathrm{d} \rho-\int_{0}^{d / 2} \rho^{2} 2 \pi \rho \mathrm{d} \rho=\left[\frac{2 \pi \rho^{4}}{4}\right]_{d / 2}^{D / 2}=\frac{\pi}{32}\left(D^{4}-d^{4}\right)\]

    Therefore for the beam dimensions given, the polar moment of inertia is:

    \[I_{\mu}=\frac{\pi}{32}\left(0.050^{4}-0.048^{4}\right)=9.24 \times 10^{-8} \mathrm{m}^{4}\]

    The equation for the angle of twist is:

    \[\theta=\frac{T L}{G I_{p}}\]

    Rearranging this to give the shear modulus, G

    \[G=\frac{T L}{\theta I_{p}}\]

    It follows that:

    \[G=\frac{1000(1)}{0.10\left(9.24 \times 10^{-8}\right)}=108 \mathrm{GPa}\]

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