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Sample Quiz: Strength of Materials Chapter 6

Beam Deflections — Strength of Materials (Andrew Pytel)

10 questions · 23 points · answer key included

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Multiple Choice

  1. ApplyIn the area-moment method for finding midspan deflections, what does the deviation tC/B represent when C is at midspan and B is at the support? (2 PTS)
    1. The actual deflection at midspan
    2. Twice the actual midspan deflection
    3. The bending moment at midspan
    4. The slope at the support
  2. ApplyWhen applying the area-moment theorem to find midspan deflection using the transformation to symmetry method, which portion of the beam's moment diagram is needed? (2 PTS)
    1. Only the right half of the beam's moment diagram
    2. The entire beam's moment diagram
    3. Either the left or right half, as they are identical
    4. Only the left half of the beam's moment diagram
  3. ApplyFor a uniformly distributed load of intensity w0 over the entire span of a simply supported beam of length L, what is the midspan deflection in terms of EIδ? (2 PTS)
    1. EIδ = (w0 L4)/192
    2. EIδ = (w0 L4)/384
    3. EIδ = (w0 L4)/256
    4. EIδ = (5w0 L4)/768
  4. ApplyWhen using the method of superposition to find deflections, what is the primary restriction on this method? (2 PTS)
    1. Each separate load must not cause an excessive change in the original shape or length of the beam
    2. It requires the beam to be symmetrically loaded
    3. It cannot be used with distributed loads
    4. It can only be applied to cantilever beams
  5. In the double integration method, the differential equation of the elastic curve EId2y/dx2=M is derived by equating two expressions for the curvature 1/ρ. Which of the following correctly identifies these two expressions? (2 PTS)
    1. d3y/dx3 from the shear equation and M/EI from the moment equation
    2. d2y/dx2 from geometry and M/EI from the flexure formula
    3. dy/dx from the slope equation and M/EI from bending stress
    4. 1/ρ=dθ/ds and 1/ρ=M/EI where θ is the slope angle

True or False

  1. UnderstandThe area-moment method requires knowledge of the complete slope and deflection equations before finding deflection at a particular point. (1 PTS)
  2. The area-moment method requires that the complete slope and deflection equations be derived before finding the deflection at a particular point on the beam. (1 PTS)
  3. In a cantilever beam with a fixed support, the tangential deviation at any point measured from a tangent drawn at the fixed end equals the actual deflection at that point. (1 PTS)

Problem Solving

  1. ApplyA simply supported beam of length 9 ft carries two concentrated loads: 80 lb at 3 ft from the left support and 100 lb at 2 ft from the right support. Using the method of superposition and the formula EIδ = Pb/48(3L2 - 4b2), determine the midspan deflection EIδ. (5 PTS)
  2. ApplyA simply supported beam of length L carries a uniformly distributed load that varies linearly from zero at the left end to w0 at the right end. Using the transformation to symmetry method, show that the midspan deflection is EIδ = (5w0 L4)/768. (5 PTS)

Answer key

Multiple Choice

  1. 1.B — The worked example states that the deviation of C from the midspan tangent drawn at B equals 2δ, or twice the actual midspan deflection.
  2. 2.D — The worked example explicitly states that 'we need the M diagram for only half the beam' and shows the moment diagram drawn for the left half of the beam.
  3. 3.D — The worked example for a uniformly varying load transformed to uniform distribution shows that EIδ = 5/768w0 L4.
  4. 4.A — The concept of superposition explicitly states that the only restriction is that each separate load must not cause an excessive change in the original shape or length of the beam, ensuring that effects are independent.
  5. 5.D — The differential equation is derived by equating the geometric expression for curvature 1/ρ=dθ/dsapproxd2y/dx2 with the flexure formula expression 1/ρ=M/EI, as stated in the principle on double integration.

True or False

  1. 1.False — The area-moment method is generally more direct than the double-integration method, especially when deflection at a particular position is desired, and does not require complete equations.
  2. 2.False — The area-moment method is described as 'the most direct of any, especially when the deflection at a specific location is desired,' allowing calculation of deflection at particular points without deriving complete equations, unlike the double integration method.
  3. 3.True — The concept section on cantilever deflections states that because the wall is perfectly fixed with a horizontal tangent, the tangential deviation tA/B measured from the tangent at the fixed support equals the deflection δA at point A.

Problem Solving

  1. 1.For the 80 lb load at 3 ft from left support: b = 3 ft (smaller segment), so EIδ1 = 80(3)/48[3(9)2 - 4(3)2] = 240/48[243 - 36] = 5(207) = 1035 lb·ft3. For the 100 lb load at 2 ft from right support (7 ft from left): b = 2 ft, so EIδ2 = 100(2)/48[3(9)2 - 4(2)2] = 200/48[243 - 16] = 4.167(227) = 945 lb·ft3. Total: EIδ = 1035 + 945 = 1980 lb·ft3. — This solution correctly applies the superposition method by calculating the midspan deflection for each load separately using the given formula, then summing the results. The calculation properly identifies the smaller segment for each load.
  2. 2.Transform to symmetry by adding a load varying from w0 at left to zero at right, creating uniform load w0 over entire span. The deviation at midspan C from the support tangent at B equals 2δ. Using the area-moment theorem on the M diagram of half the beam: 2EIδ = (w0L2/8 · L/2)(1/2 · L/2) - (1/3 · w0L2/8 · L/2)(1/4 · L/2) = 5w0L4/384. Therefore, EIδ = 5w0L4/768. — This solution correctly applies the transformation to symmetry method, uses the area-moment theorem on the appropriate half-beam moment diagram, and arrives at the stated formula through proper algebraic manipulation.