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Sample Quiz: Strength of Materials Chapter 4

Shear and Moment in Beams — Strength of Materials (Andrew Pytel)

10 questions · 23 points · answer key included

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Multiple Choice

  1. ApplyWhen determining the maximum bending moment under a particular load in a system of moving wheel loads crossing a simply supported beam, the correct positioning rule is that: (2 PTS)
    1. The resultant load should be directly over one of the supports
    2. The centerline of the span should be midway between that load and the resultant of all loads on the span
    3. The load should be positioned at the center of the span
    4. All loads should be equally spaced across the span
  2. ApplyFor a cantilever beam with the wall at the right end, when writing shear and moment equations, the shear at any section is computed as V = (∑ Fy)L. For a uniformly varying triangular load with maximum intensity w0 at the free end, what is the form of the shear equation in the loaded region? (2 PTS)
    1. V = -(w0 x2)/2L
    2. V = w0 x2
    3. V = -w0 x
    4. V = -w0 L
  3. ApplyFor a beam segment where the shear is constant (no distributed load), the bending moment diagram in that segment will have what shape? (2 PTS)
    1. A horizontal line
    2. A parabolic curve
    3. A cubic curve
    4. A straight line with slope equal to the shear value
  4. ApplyAccording to the sign convention for bending moment, a bending moment is positive when it: (2 PTS)
    1. Produces clockwise rotation on the left segment of the beam
    2. Acts in the counterclockwise direction on the right segment
    3. Is caused by upward-acting external forces
    4. Bends the beam concave downward
  5. ApplyA simply supported beam carries a concentrated load P at distance a from the left support. When an exploratory section is taken at distance x from the left support (where x > a), which expression correctly represents the bending moment using the left-side convention? (2 PTS)
    1. M = P(x - a) - R1 x
    2. M = R1 x + P(x - a)
    3. M = R1 x - P(x - a)
    4. M = R2(L - x)

True or False

  1. UnderstandA positive shearing force tends to move the left segment of a beam upward with respect to the right segment. (1 PTS)
  2. UnderstandThe shearing force V at any section of a beam can be computed by summing the vertical forces acting either to the left or to the right of that section, and both methods will yield the same numerical result. (1 PTS)
  3. UnderstandThe bending moment at any section of a beam can be computed by taking moments about that section using either the forces to the left or the forces to the right of the section, and both methods will yield the same result. (1 PTS)

Problem Solving

  1. ApplyA truck with axle loads of 40 kN and 60 kN separated by 5 m crosses a simply supported span of 10 m. Using the principle that maximum moment under a load occurs when the span centerline is midway between that load and the resultant, determine the position of the loads and calculate the maximum bending moment under the 60 kN load. (5 PTS)
  2. ApplyFor a simply supported beam of length 14 m carrying a uniformly distributed load of 20 kN/m over the entire span, write the shear and moment equations and determine the location and magnitude of the maximum bending moment. (5 PTS)

Answer key

Multiple Choice

  1. 1.B — Equation (4-7) and the principle in the Moving Loads section establish that maximum moment under a load occurs when the span centerline is midway between that load and the resultant of all loads on the span.
  2. 2.A — For a triangular load varying from w0 at the free end to zero at distance x, the resultant force equals the area of the triangle, which is 1/2w0 x · x/L = (w0 x2)/2L, giving negative shear in the cantilever convention.
  3. 3.D — The relationship V = dM/dx shows that when shear is constant, the moment changes linearly; the slope of the moment diagram equals the constant shear value.
  4. 4.C — The sign convention states that upward-acting external forces cause positive bending moments regardless of whether they act to the left or right of the exploratory section, as established in the Negative Bending section.
  5. 5.C — The bending moment at any section equals the sum of moments of all forces to the left; the upward reaction R1 produces positive moment, and the downward load P at distance (x - a) from the section produces negative moment.

True or False

  1. 1.True — The definition of positive shear in the Shear and Moment section states that a positive shearing force tends to move the left segment upward with respect to the right.
  2. 2.False — The definition V = (∑ Fy)L restricts the summation to forces on the left side only; summing forces on the right side would give the negative of the shear, not the same result.
  3. 3.True — The definition M = (∑ M)L = (∑ M)R explicitly states that bending moment may be computed using forces from either side of the section, and both give equal results.

Problem Solving

  1. 1.The resultant load is R = 40 + 60 = 100 kN, located 2.5 m from the 60 kN load (toward the 40 kN load). For maximum moment under the 60 kN load, position the loads so the span centerline (5 m from left support) is midway between the 60 kN load and R. This means the 60 kN load is at distance x from the left support where 5 = (x + (x - 2.5))/2, giving x = 6.25 m. The 40 kN load is at 6.25 - 5 = 1.25 m from the left support. Taking moments about the right support: 10 R1 = 100(10 - 6.25) = 375 kN·m, so R1 = 37.5 kN. The bending moment under the 60 kN load (at 6.25 m) is M = 37.5(6.25) - 40(6.25 - 1.25) = 234.375 - 200 = 34.375 kN·m. However, checking the configuration with only the 60 kN load at midspan gives M = 60(5) = 150 kN·m, which is larger. The maximum moment is 150 kN·m. — This problem applies the moving loads principle and the positioning rule to find maximum moment, requiring calculation of the resultant, positioning using the midspan rule, reaction calculation, and moment computation.
  2. 2.Taking the left support as origin with x measured from the left: The reaction at each support is R1 = R2 = (20 × 14)/2 = 140 kN. For the entire span (0 ≤ x ≤ 14): V = 140 - 20x kN and M = 140x - 10x2 kN·m. The shear becomes zero when 140 - 20x = 0, giving x = 7 m (at midspan). The maximum moment occurs at this location: Mmax = 140(7) - 10(7)2 = 980 - 490 = 490 kN·m. Alternatively, using the semigraphical method: the area under the load diagram from 0 to 7 m is 20 × 7 = 140 kN, which equals the change in shear from 140 to 0. The area under the shear diagram from 0 to 7 m is a triangle with base 7 and height 140, giving area = 1/2(7)(140) = 490 kN·m, which is the moment at midspan. — This problem requires writing shear and moment equations for a uniformly distributed load, finding zero shear, calculating maximum moment, and demonstrates the relationship between load, shear, and moment.