Statics Quiz: Support Reactions
8 questions · exam conditions
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Support ReactionsQuestion 1 of 8

A compound beam consists of beam AC (8 m long) pinned at A and connected by an internal hinge at B (6 m from A) to beam BC (4 m long) which has a roller support at C. A downward load of 16 kN acts at B. What is the moment transmitted through the hinge at B?

0 kN⋅m
16 kN⋅m
24 kN⋅m
32 kN⋅m
48 kN⋅m
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Statics Quiz

Statics Quiz: Support Reactions

Practice Support Reactions 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 Support Reactions, 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 compound beam consists of beam AC (8 m long) pinned at A and connected by an internal hinge at B (6 m from A) to beam BC (4 m long) which has a roller support at C. A downward load of 16 kN acts at B. What is the moment transmitted through the hinge at B?

  1. 0 kN⋅m (correct answer)
  2. 16 kN⋅m
  3. 24 kN⋅m
  4. 32 kN⋅m
  5. 48 kN⋅m
Explanation: When analyzing compound beams with internal hinges, remember that hinges are special connection points that cannot transmit moments - they can only transmit forces. This is the fundamental property that makes compound beam analysis possible. To solve this problem, you need to recognize that an internal hinge at point B means the moment at that location must be zero. Think of a hinge like a door hinge - it allows rotation but doesn't resist it, so no moment can be transmitted through it. You can verify this by analyzing beam BC separately. Since BC is simply supported (hinge at B, roller at C) with no external loads applied to it, the moment at every point along BC must be zero. Taking moments about point C: the reaction at B creates no moment about C, confirming that the internal forces create no bending moment anywhere in segment BC, including at point B. Looking at the wrong answers: B) 16 kN⋅m might seem logical if you incorrectly tried to calculate a moment using the 16 kN load and some arbitrary distance. C) 24 kN⋅m could result from multiplying the 16 kN load by the 1.5 m distance from B to the midpoint of BC. D) 32 kN⋅m might come from using the 16 kN load with the 2 m distance from B to C, but these calculations miss the fundamental point about hinges. Study tip: Whenever you see an internal hinge in a compound beam problem, immediately write "moment = 0" at that location. This constraint is often the key to solving the entire structure and is one of the most commonly tested concepts in statics.

Question 2

A cantilever beam of length 5 m is fixed at point A and supports two concentrated loads: 15 kN downward at 2 m from A, and 25 kN downward at 4 m from A. The beam also has a uniformly distributed load of 3 kN/m over its entire length. What is the horizontal reaction at the fixed support A?

  1. 0 kN (correct answer)
  2. 5 kN to the right
  3. 10 kN to the left
  4. 15 kN to the right
  5. 20 kN to the left
Explanation: When analyzing forces on a cantilever beam, you must consider all applied loads and determine what reactions are needed at the fixed support to maintain equilibrium. The key insight here is identifying the direction of all applied forces. Looking at this problem, you have three downward forces acting on the beam: a 15 kN load at 2m, a 25 kN load at 4m, and a distributed load of 3 kN/m over the entire 5m length (totaling 15 kN downward). Notice that every single applied force acts vertically downward - there are no horizontal forces whatsoever. Since equilibrium requires that the sum of all forces in any direction equals zero (∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0), and there are no horizontal forces applied to the beam, there can be no horizontal reaction force at the fixed support. The horizontal reaction must be 0 kN to satisfy horizontal equilibrium. Choice A (0 kN) is correct because no horizontal forces are applied to the beam. Choice B (5 kN to the right) incorrectly assumes some horizontal force is needed, perhaps confusing force magnitudes with directions. Choice C (10 kN to the left) makes a similar error, possibly misinterpreting one of the vertical loads as horizontal. Choice D (15 kN to the right) might stem from incorrectly thinking the distributed load creates a horizontal component. Remember this fundamental principle: reaction forces only develop in directions where external forces are applied. Always identify the direction of applied loads first - if all loads are vertical, horizontal reactions will always be zero.

Question 3

A cantilever beam of length 4 m is fixed at point A and free at point B. A triangular distributed load varies linearly from zero at A to 8 kN/m at B. What is the magnitude of the reaction moment at the fixed support A?

  1. 21.3 kN⋅m clockwise
  2. 32.0 kN⋅m clockwise
  3. 42.7 kN⋅m clockwise (correct answer)
  4. 48.0 kN⋅m clockwise
  5. 64.0 kN⋅m clockwise
Explanation: When analyzing cantilever beams with distributed loads, you need to find the resultant force and its location, then apply equilibrium equations. The triangular load creates both a vertical reaction force and a reaction moment at the fixed support. For a triangular distributed load varying from 0 to 8 kN/m over 4 m, the resultant force equals the area of the triangle: R=12×8×4=16 kNR = \frac{1}{2} \times 8 \times 4 = 16 \text{ kN}. This resultant acts at the centroid of the triangle, located at 23\frac{2}{3} of the length from the zero end, or 23×4=2.67 m\frac{2}{3} \times 4 = 2.67 \text{ m} from point A. To find the reaction moment at A, take moments about point A. The moment caused by the resultant force is: MA=16×2.67=42.7 kN⋅mM_A = 16 \times 2.67 = 42.7 \text{ kN⋅m} clockwise. Answer A (21.3 kN⋅m) represents a common error where students incorrectly place the resultant at the midpoint (2 m) instead of the centroid: 16×2=3216 \times 2 = 32, then make an additional calculation error. Answer B (32.0 kN⋅m) occurs when using the midpoint location correctly but stops there. Answer D (48.0 kN⋅m) results from incorrectly assuming the resultant acts at the free end: 16×3=4816 \times 3 = 48. The correct answer is C (42.7 kN⋅m clockwise). Remember: for triangular loads, the resultant always acts at the centroid, which is 23\frac{2}{3} of the distance from the zero-load end toward the maximum-load end.

Question 4

A cantilever beam of length 3 m is fixed at the left end and free at the right end. It supports a uniformly distributed load of 6 kN/m over the right half of the beam only. What is the vertical reaction at the fixed support?

  1. 4.5 kN upward
  2. 9.0 kN upward (correct answer)
  3. 13.5 kN upward
  4. 18.0 kN upward
  5. 27.0 kN upward
Explanation: When analyzing cantilever beams with distributed loads, you need to apply equilibrium principles systematically. The key insight is that the fixed support must provide whatever vertical force is needed to balance all downward loads on the beam. First, calculate the total load on the beam. The uniformly distributed load of 6 kN/m acts only over the right half of the 3 m beam, so it covers a 1.5 m length. The total downward force is: 6 kN/m×1.5 m=9 kN6 \text{ kN/m} \times 1.5 \text{ m} = 9 \text{ kN} For vertical equilibrium, the sum of all vertical forces must equal zero. Since there's a 9 kN downward load and no other vertical forces except the reaction at the fixed support, the reaction must be 9 kN upward to maintain equilibrium. Looking at the wrong answers: Choice A (4.5 kN) represents half the correct value—you might get this if you incorrectly calculated the distributed load as acting over only 0.75 m instead of 1.5 m. Choice C (13.5 kN) could result from mistakenly applying the 6 kN/m load over 2.25 m instead of 1.5 m. Choice D (18 kN) would occur if you incorrectly assumed the distributed load acted over the entire 3 m length of the beam. Study tip: For cantilever beam problems, always carefully identify the length over which distributed loads act, then apply vertical force equilibrium. The reaction at a fixed support equals the total downward load when considering only vertical forces.

Question 5

A beam of length 10 m is simply supported at both ends. It carries a uniformly distributed load of 2 kN/m over its entire length and a concentrated moment of 15 kN⋅m applied clockwise at 7 m from the left support. What is the reaction at the left support?

  1. 8.5 kN upward
  2. 10.0 kN upward
  3. 11.5 kN upward (correct answer)
  4. 13.0 kN upward
  5. 15.0 kN upward
Explanation: When analyzing simply supported beams with combined loading, you need to apply equilibrium conditions systematically. The key insight is that applied moments affect the distribution of reactions even though the total upward reaction remains equal to the total downward load. Start by finding the total distributed load: 2 kN/m×10 m=20 kN2 \text{ kN/m} \times 10 \text{ m} = 20 \text{ kN} acting downward at the beam's center (5 m from left support). Now apply moment equilibrium about the right support to find the left reaction RLR_L: ∑Mright=0\sum M_{\text{right}} = 0 RL×10−20×5−15=0R_L \times 10 - 20 \times 5 - 15 = 0 RL=100+1510=11.5 kNR_L = \frac{100 + 15}{10} = 11.5 \text{ kN} The applied moment increases the left reaction because it creates additional moment that must be balanced by the support reactions. Choice A (8.5 kN) represents subtracting the moment effect instead of adding it—a common sign error. Choice B (10.0 kN) is simply half the total load, ignoring the applied moment entirely. This would be correct only if there were no applied moment. Choice D (13.0 kN) appears to incorrectly account for the moment's position or magnitude. You can verify: the right reaction is 20−11.5=8.5 kN20 - 11.5 = 8.5 \text{ kN}, and checking moment equilibrium about the left support confirms this solution. Study tip: Applied moments always affect reaction distribution in statically determinate beams. Set up your moment equation carefully, ensuring consistent sign conventions, and remember that clockwise moments typically require counterclockwise moment reactions to maintain equilibrium.

Question 6

A beam is supported by three supports: a pin at A, a roller at B, and a roller at C. The beam is loaded with a triangular distributed load that varies from 0 at point A to 400 N/m at point C. The distances are AB = 3 m and BC = 2 m. Due to settlement, support B drops by 5 mm while supports A and C remain at their original positions. What is the new reaction at support B?

  1. The reaction becomes zero since settlement removes contact with the beam (correct answer)
  2. The reaction decreases proportionally to the settlement ratio of 5 mm
  3. The reaction remains 400 N upward based on original static analysis
  4. The reaction increases to 533 N upward due to load redistribution
Explanation: This is a statically indeterminate system that becomes determinate when support B settles and loses contact. The beam spans from A to C as a simply supported beam with the triangular load. Without contact at B, the reaction there becomes zero. The loads redistribute to supports A and C. Choice B misapplies proportional reasoning. Choice C ignores the settlement effect. Choice D incorrectly assumes the support remains effective.

Question 7

A bracket is attached to a wall with two bolts arranged vertically. The bracket supports a load of 1000 N applied horizontally at a distance of 300 mm from the wall. The bolts are spaced 200 mm apart vertically. If the upper bolt can only resist tension and the lower bolt can resist both tension and compression, what is the force in the upper bolt?

  1. 1500 N tension based on moment equilibrium about lower bolt (correct answer)
  2. 750 N tension from equal load distribution assumption
  3. 2000 N tension considering maximum moment resistance
  4. 1000 N tension equal to applied horizontal load magnitude
Explanation: Taking moments about the lower bolt: Fupper×0.2=1000×0.3F_{upper} \times 0.2 = 1000 \times 0.3, which gives Fupper=1500F_{upper} = 1500 N tension. The constraint that the upper bolt can only resist tension is satisfied. The lower bolt provides the horizontal reaction of 1000 N and a vertical compression. Choice B assumes equal distribution ignoring moment equilibrium. Choice C overestimates the required resistance. Choice D confuses horizontal and bolt forces.

Question 8

A simply supported beam of length 6 m carries a uniformly distributed load of 200 N/m over its entire length and a concentrated load P at the center. If the reaction at the left support is twice the reaction at the right support, what is the magnitude of the concentrated load P?

  1. 600 N downward creating the specified reaction ratio
  2. 400 N upward balancing the distributed loading
  3. 600 N upward achieving the required support conditions (correct answer)
  4. 300 N downward based on equilibrium requirements
Explanation: Let R_L and R_R be the left and right reactions. The distributed load creates a total downward force of 1200 N. Given R_L = 2R_R and vertical equilibrium R_L + R_R = 1200 - P (taking upward P as positive), we get 3R_R = 1200 - P, so R_R = 400 - P/3. Taking moments about the left support: R_R × 6 = 1200 × 3 - P × 3. Substituting: (400 - P/3) × 6 = 3600 - 3P. Solving: 2400 - 2P = 3600 - 3P, which gives P = 600 N upward.