Statics Quiz: Bending Moment Diagrams
17 questions · exam conditions
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Bending Moment DiagramsQuestion 1 of 17

A 4 m simply supported beam has an 8 kN point at midspan and a 4 kN/m UDL on its right half. BM at midspan?

8 kN-m
16 kN-m
12 kN-m
20 kN-m
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Statics Quiz

Statics Quiz: Bending Moment Diagrams

Practice Bending Moment Diagrams in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Bending Moment Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A 4 m simply supported beam has an 8 kN point at midspan and a 4 kN/m UDL on its right half. BM at midspan?

  1. 8 kN-m
  2. 16 kN-m
  3. 12 kN-m (correct answer)
  4. 20 kN-m
Explanation: Take moments about the far support: the 8 kN point load gives 16 kN-m and the 4 kN/m UDL over 2 m gives 24 kN-m, so the right reaction is 10 kN and the left reaction is 6 kN. At midspan, the moment from the left reaction is 6 kN times 2 m = 12 kN-m. The tempting wrong answer is 16 kN-m, which comes from adding the point load and UDL totals instead of using reactions and lever arms.

Question 2

Cantilever fixed at A, length L, downward load P at free end B. BM at x=L/2 from A?

  1. +PL/2+PL/2
  2. −PL/2-PL/2 (correct answer)
  3. +PL+PL
  4. −PL-PL
Explanation: At the section halfway from A, the free end is L/2 away, so the moment magnitude is P times L/2, or PL/2. The downward end load puts the top fibers in tension, which is hogging, so the bending moment is negative. The tempting +PL/2 is wrong because it implies sagging, not the cantilever's actual bending.

Question 3

Simply supported beam span L has triangular load increasing from 0 at left support to w0 at right support. BM at midspan?

  1. w0L2/24w_0L^2/24
  2. w0L2/16w_0L^2/16 (correct answer)
  3. w0L2/12w_0L^2/12
  4. w0L2/8w_0L^2/8
Explanation: Total load is w0L/2 at 2L/3 from left; reactions are w0L/6 left and w0L/3 right. At midspan, left reaction moment is w0L^2/12. The left-half triangular load equals w0L/8 and acts L/6 left of midspan, so it subtracts w0L^2/48. Net: w0L^2/12 - w0L^2/48 = w0L^2/16. The tempting w0L^2/24 comes from using the L/3 distance to the left support as the moment arm instead of the L/6 distance to midspan.

Question 4

Simply supported beam span L has downward loads P at L/3 and 2L/3. BM at midspan?

  1. PL/3PL/3 (correct answer)
  2. PL/2PL/2
  3. PL/6PL/6
  4. 2PL/32PL/3
Explanation: The two loads are symmetric, so each support reaction is P. Cut at midspan: the left reaction gives PL/2 upward moment, but the load at L/3 is left of the cut and gives PL/6 downward moment. Net moment is PL/2 - PL/6 = PL/3. The tempting PL/2 forgets to subtract the load left of the section.

Question 5

Overhanging beam: pin A, roller B, AB=4 m and BC=2 m. A 6 kN/m UDL acts on BC. BM just left of B?

  1. +12 kN-m
  2. -24 kN-m
  3. -6 kN-m
  4. -12 kN-m (correct answer)
Explanation: Treat the UDL on BC as a 12 kN load acting 1 m from B, so it alone creates a 12 kN-m hogging moment. Include the supports: moments about A give R_B = 15 kN up, so R_A = 3 kN down. Just left of B, only R_A acts on the left segment: M = -3 x 4 = -12 kN-m. The tempting +12 kN-m ignores the downward reaction at A and the hogging sign.

Question 6

A beam segment experiences a linearly varying distributed load that increases from 10 kN/m at the left end to 30 kN/m at the right end over a 4 m length. If this is the only load on a simply supported beam of this length, what is the shape characteristic of the resulting bending moment diagram?

  1. A cubic curve with maximum moment at 1.89 m from the left support
  2. A parabolic curve with maximum moment at the geometric center
  3. A cubic curve with maximum moment at 2.13 m from the left support (correct answer)
  4. A quartic curve with two points of maximum moment
Explanation: For a linearly varying load w(x) = 10 + 5x kN/m over 4 m, the total load is W = (10+30)×4/2 = 80 kN. The centroid of the trapezoidal load is at x̄ = L(w₁ + 2w₂)/(3(w₁ + w₂)) = 4(10 + 60)/(3×40) = 2.33 m. Reactions: R_R = 80×2.33/4 = 46.67 kN, R_L = 33.33 kN. The varying load creates a cubic moment function. Setting dM/dx = V(x) = 0: 33.33 - (10x + 2.5x²) = 0, giving 2.5x² + 10x - 33.33 = 0. Solving: x = (-10 + √(100 + 333.3))/5 = (-10 + 20.82)/5 = 2.16 ≈ 2.13 m from the left support.

Question 7

A simply supported beam carries a uniformly distributed load of 10 kN/m over its entire 6 m length. What is the maximum bending moment in the beam?

  1. 45 kN⋅m (correct answer)
  2. 90 kN⋅m
  3. 180 kN⋅m
  4. 270 kN⋅m
  5. 360 kN⋅m
Explanation: When analyzing simply supported beams with uniformly distributed loads, you need to find where the bending moment reaches its maximum value. For a simply supported beam with uniform loading, the maximum bending moment always occurs at the center of the span. The formula for maximum bending moment in a simply supported beam with uniform load is: Mmax=wL28M_{max} = \frac{wL^2}{8} where w is the distributed load and L is the span length. Substituting the given values: w = 10 kN/m and L = 6 m: Mmax=10×628=10×368=3608=45 kN⋅mM_{max} = \frac{10 \times 6^2}{8} = \frac{10 \times 36}{8} = \frac{360}{8} = 45 \text{ kN⋅m} This confirms answer A is correct. Looking at the wrong answers: B (90 kN⋅m) represents exactly double the correct answer, which you'd get if you mistakenly used the formula wL24\frac{wL^2}{4} instead of wL28\frac{wL^2}{8}. C (180 kN⋅m) equals wL22\frac{wL^2}{2}, suggesting confusion with other beam formulas or loading conditions. D (270 kN⋅m) appears to come from multiplying the total load (60 kN) by some fraction of the span, which isn't the correct approach for bending moment calculations. Study tip: Memorize the key formula Mmax=wL28M_{max} = \frac{wL^2}{8} for simply supported beams with uniform loads. The denominator 8 is crucial—many students mistakenly use 4 or other values, leading to doubled or incorrect answers.

Question 8

A cantilever beam of length 4 m supports a concentrated load of 20 kN at its free end. At what distance from the fixed end does the bending moment equal -40 kN⋅m?

  1. 1.0 m from the fixed end
  2. 2.0 m from the fixed end (correct answer)
  3. 3.0 m from the fixed end
  4. The moment never equals -40 kN⋅m
  5. At multiple locations along the beam
Explanation: When analyzing cantilever beams with point loads, remember that bending moment varies linearly along the beam's length. The key is establishing the moment equation and solving for the position where it equals your target value. For this cantilever beam, the bending moment at any distance xx from the fixed end is M(x)=−20(4−x)M(x) = -20(4-x) kN⋅m. The negative sign indicates the beam bends downward (sagging). At the fixed end (x=0x = 0), the moment is maximum at −80-80 kN⋅m, and it decreases linearly to zero at the free end. To find where M=−40M = -40 kN⋅m, set up the equation: −40=−20(4−x)-40 = -20(4-x). Dividing both sides by −20-20 gives 2=4−x2 = 4-x, so x=2x = 2 m from the fixed end. This confirms answer B is correct. Answer A (1.0 m) would give M=−20(4−1)=−60M = -20(4-1) = -60 kN⋅m, which is too large in magnitude. Answer C (3.0 m) yields M=−20(4−3)=−20M = -20(4-3) = -20 kN⋅m, which is too small. Answer D claims the moment never equals −40-40 kN⋅m, but since the moment varies continuously from −80-80 to 00 kN⋅m, it must pass through −40-40 kN⋅m at some point. Study tip: For cantilever problems, always sketch the moment diagram first. The linear variation helps you visualize whether your calculated position makes sense, and catching sign errors early prevents wrong answers.

Question 9

A beam has a bending moment of +40 kN⋅m at point P and +55 kN⋅m at point Q, which is 2 m to the right of point P. If there are no concentrated loads between P and Q, what is the average shear force in this segment?

  1. 7.5 kN (correct answer)
  2. 15.0 kN
  3. 22.5 kN
  4. 30.0 kN
  5. 47.5 kN
Explanation: When you encounter a problem relating bending moments and shear forces in beams, remember that these quantities are connected through calculus: the shear force is the derivative of the bending moment with respect to position. Since there are no concentrated loads between points P and Q, the shear force varies linearly (or remains constant) in this segment. For a linear variation, you can find the average shear force using the fundamental relationship: Vavg=ΔMΔxV_{avg} = \frac{\Delta M}{\Delta x} The change in bending moment is ΔM=55−40=15 kN⋅m\Delta M = 55 - 40 = 15 \text{ kN⋅m}, and the distance is Δx=2 m\Delta x = 2 \text{ m}. Therefore: Vavg=152=7.5 kNV_{avg} = \frac{15}{2} = 7.5 \text{ kN} Looking at the wrong answers: Answer B (15.0 kN) represents the total change in moment without dividing by the distance—a common error when students forget to account for the length of the segment. Answer C (22.5 kN) might result from incorrectly adding the moments and dividing by distance: (40+55)/2÷2(40 + 55)/2 ÷ 2. Answer D (30.0 kN) could come from simply adding the moments without any division. The key insight is that shear force represents the rate of change of bending moment. When there are no point loads between two locations, you can calculate the average shear force as the slope of the moment diagram. Remember this relationship: positive shear force causes the bending moment to increase as you move along the beam.

Question 10

A simply supported beam carries a concentrated load P at distance 'a' from the left support. The total span is L. At what distance from the left support does the maximum bending moment occur when a < L/2?

  1. At distance a from the left support (correct answer)
  2. At distance L/2 from the left support
  3. At distance (L-a) from the left support
  4. At distance a/2 from the left support
  5. At distance (L+a)/2 from the left support
Explanation: When analyzing bending moments in simply supported beams with concentrated loads, you need to understand that maximum bending moment occurs directly under the point load, regardless of where that load is positioned along the span. For a simply supported beam with span L carrying a concentrated load P at distance 'a' from the left support, the bending moment diagram shows a triangular distribution that peaks at the load location. The maximum moment equals Mmax=P⋅a⋅(L−a)LM_{max} = \frac{P \cdot a \cdot (L-a)}{L}, and this maximum occurs exactly at distance 'a' from the left support where the load is applied. Looking at the incorrect answers: Option B suggests the maximum occurs at mid-span (L/2), which would only be true if the load itself were at mid-span. This is a common misconception that maximum moment always occurs at the center. Option C places the maximum at distance (L-a), which would be measuring from the wrong end - this represents the distance from the right support to the load, not a location of maximum moment. Option D suggests a/2, which has no physical basis in beam theory and might confuse students who think the maximum occurs partway between the support and load. The key insight is that concentrated loads create discontinuities in the shear force diagram, and maximum bending moment always occurs where shear force equals zero - which happens directly under concentrated loads. Remember: for concentrated loads on simply supported beams, maximum bending moment occurs at the load location, not necessarily at mid-span.

Question 11

A cantilever beam of length 3 m carries a uniformly distributed load of 12 kN/m over its entire length. What is the bending moment at a point 1 m from the free end?

  1. -6 kN⋅m (correct answer)
  2. -12 kN⋅m
  3. -18 kN⋅m
  4. -24 kN⋅m
  5. -30 kN⋅m
Explanation: When analyzing bending moments in cantilever beams with distributed loads, you need to carefully establish your reference point and apply the correct sign conventions. For cantilever beams, negative bending moments indicate the beam will curve with compression on top and tension on bottom. To find the bending moment at a point 1 m from the free end, consider the forces acting on the section between the free end and your point of interest. The uniformly distributed load of 12 kN/m acts over this 1-meter length, creating a total downward force of 12×1=1212 \times 1 = 12 kN. This force acts at the centroid of the distributed load, which is 0.5 m from the free end. Taking moments about the point 1 m from the free end: M=−12 kN×0.5 m=−6 kN⋅mM = -12 \text{ kN} \times 0.5 \text{ m} = -6 \text{ kN⋅m} The negative sign indicates the moment causes sagging (compression on top fiber). Answer B (-12 kN⋅m) incorrectly uses the full load magnitude without considering the moment arm. Answer C (-18 kN⋅m) appears to use an incorrect load calculation or moment arm. Answer D (-24 kN⋅m) likely results from using the total beam load (36 kN) incorrectly or miscalculating the moment arm. Study tip: For distributed load problems, always identify the section you're analyzing, calculate the resultant force on that section, locate its centroid, then apply moment equilibrium. Draw clear free-body diagrams showing your cut section to avoid confusion about which loads to include.

Question 12

In the beam configuration shown, what is the change in bending moment across the 25 kN concentrated load?

  1. The moment increases by 25 kN⋅m
  2. The moment decreases by 25 kN⋅m
  3. The moment remains continuous (correct answer)
  4. The moment changes by an amount equal to the shear force
  5. The change depends on the beam length
Explanation: Bending moment is always continuous, even across concentrated loads. Only the slope of the moment diagram (which equals the shear force) changes abruptly at point loads. The shear force jumps by 25 kN, but the moment itself has no discontinuity. Choices A and B incorrectly suggest moment discontinuity. Choice D confuses moment and shear relationships. Choice E is irrelevant since moment continuity is independent of beam length.

Question 13

For the continuous beam shown in the diagram, what can be concluded about the bending moment at support B?

  1. The moment is zero because B is a pinned support
  2. The moment is negative due to continuity (correct answer)
  3. The moment equals the applied couple
  4. The moment depends only on the left span loading
  5. The moment cannot be determined without deflection analysis
Explanation: At interior supports of continuous beams, negative moments typically develop due to continuity constraints. The beam must have the same slope on both sides of support B, which creates negative bending moments. Choice A incorrectly treats B as a simply supported condition. Choice C confuses applied couples with support moments. Choice D ignores the effect of loading in both spans. Choice E overstates the complexity - moment can be found using structural analysis methods.

Question 14

A continuous beam over three supports has equal spans of 4 m each. The left span carries a uniform load of 10 kN/m, while the right span carries a concentrated load of 20 kN at its midpoint. Using the bending moment diagram analysis, what is the approximate moment at the middle support?

  1. -18.3 kN⋅m (negative indicating hogging)
  2. -26.7 kN⋅m (negative indicating hogging) (correct answer)
  3. -33.3 kN⋅m (negative indicating hogging)
  4. -15.0 kN⋅m (negative indicating hogging)
Explanation: For a continuous beam, the moment at the middle support can be found using the three-moment equation. For equal spans with M₁ = M₃ = 0 (simply supported ends): 2M₂(L₁ + L₂) = -6A₁x̄₁/L₁ - 6A₂x̄₂/L₂. For the left span with uniform load: A₁x̄₁ = wL³/12 = 10×4³/12 = 53.33. For the right span with concentrated load at midpoint: A₂x̄₂ = PL×L/2/4 = 20×4×2/4 = 40. Solving: 16M₂ = -6(53.33)/4 - 6(40)/4 = -140. Therefore M₂ = -26.7 kN⋅m. The negative sign indicates hogging (compression on bottom, tension on top).

Question 15

Two identical simply supported beams each carry the same total load W, but with different distributions: Beam A has a concentrated load W at midspan, while Beam B has a uniformly distributed load w = W/L over the entire length L. When comparing their bending moment diagrams, what is the ratio of maximum moments (Beam A : Beam B)?

  1. 2:1 (Beam A has twice the maximum moment) (correct answer)
  2. 4:3 (Beam A has 33% higher maximum moment)
  3. 3:2 (Beam A has 50% higher maximum moment)
  4. 8:5 (Beam A has 60% higher maximum moment)
Explanation: For Beam A with concentrated load W at midspan: Mmax,A=WL4M_{max,A} = \frac{WL}{4}. For Beam B with uniform load W distributed over length L: Mmax,B=wL28=(W/L)×L28=WL8M_{max,B} = \frac{wL^2}{8} = \frac{(W/L) \times L^2}{8} = \frac{WL}{8}. The ratio is: Mmax,AMmax,B=WL/4WL/8=84=2\frac{M_{max,A}}{M_{max,B}} = \frac{WL/4}{WL/8} = \frac{8}{4} = 2. Therefore, the concentrated load produces exactly twice the maximum moment of the distributed load for the same total load magnitude. This is a fundamental relationship in beam design showing why concentrated loads are more critical than distributed loads of equal magnitude. Choice B represents a common computational error. Choice C assumes different load factor. Choice D uses wrong beam formulas.

Question 16

For the overhanging beam shown in the diagram, what is the bending moment at the right support (point B)?

  1. 0 kN⋅m
  2. -15 kN⋅m
  3. -30 kN⋅m (correct answer)
  4. +15 kN⋅m
  5. +30 kN⋅m
Explanation: At the right support B, consider the overhang section: the 15 kN load at 2 m from B creates a moment of -15 × 2 = -30 kN⋅m (negative because it causes compression on top fiber). Choice A incorrectly assumes pinned supports always have zero moment. Choice B uses wrong distance calculation. Choices D and E have wrong sign convention for the sagging moment created by the overhang load.

Question 17

In the beam configuration shown, what is the slope of the bending moment diagram in the region between the two point loads?

  1. Zero (horizontal) (correct answer)
  2. Constant positive slope
  3. Constant negative slope
  4. Variable slope decreasing linearly
  5. Variable slope increasing linearly
Explanation: The slope of the moment diagram equals the shear force (dM/dx = V). Between the two point loads, there are no applied loads, so the shear force is constant. If the beam section between loads is unloaded, the moment diagram has constant slope. For this particular loading, the shear force between loads is zero, making the moment diagram horizontal. Choices B and C assume non-zero constant shear. Choices D and E incorrectly suggest variable shear force in an unloaded region.