Statics and Dynamics Quiz: Support Reactions

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Support ReactionsQuestion 1 of 3

A rigid frame consists of two members: a horizontal beam AC of length $6 \text{ m}$ and a vertical column CB of length $4 \text{ m}$ , connected rigidly at C. The base B is a pin support, and the end A has a roller support that permits horizontal displacement (the roller reaction is therefore vertical). A horizontal wind load of $F = 20 \text{ kN}$ acts to the right at joint C.

Determine the horizontal reaction at pin B, $B_x$ .

B_x = 20 \text{ kN}

to the left, because the roller at A provides only a vertical reaction, so the pin at B must supply the entire horizontal equilibrium force to balance the wind load at C.

B_x = 0 \text{ kN}

, because the moment from the wind load is balanced by the vertical reactions at A and B, and horizontal equilibrium is satisfied by the wind load alone without any horizontal reaction at B.

B_x = 13.3 \text{ kN}

to the left, found by distributing the horizontal wind force between A and B in proportion to the distances from C to each support along the frame.

B_x = 20 \text{ kN}

to the right, because the pin at B must react in the same direction as the applied load to maintain rotational equilibrium of the column member CB about joint C.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Support Reactions

Practice Support Reactions in Statics and Dynamics 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 and Dynamics.

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

Determine the horizontal reaction at pin B, $B_x$ .

$B_x = 20 \text{ kN}$ to the left, because the roller at A provides only a vertical reaction, so the pin at B must supply the entire horizontal equilibrium force to balance the wind load at C. (correct answer)
$B_x = 0 \text{ kN}$ , because the moment from the wind load is balanced by the vertical reactions at A and B, and horizontal equilibrium is satisfied by the wind load alone without any horizontal reaction at B.
$B_x = 13.3 \text{ kN}$ to the left, found by distributing the horizontal wind force between A and B in proportion to the distances from C to each support along the frame.
$B_x = 20 \text{ kN}$ to the right, because the pin at B must react in the same direction as the applied load to maintain rotational equilibrium of the column member CB about joint C.

Explanation: When analyzing a structure with mixed support types, your first move should always be to catalog what reactions each support can provide. A roller permits displacement in one direction, meaning it can only push perpendicular to that direction — here, the roller at A allows horizontal movement, so it delivers a vertical reaction only. The pin at B, by contrast, is fully fixed and can provide reactions in both horizontal and vertical directions. With that established, apply the equations of equilibrium. Summing forces in the horizontal direction: the only external horizontal force is the

20 \text{ kN}

wind load at C (pointing right). The roller at A contributes nothing horizontally. Therefore, for

\sum F_x = 0

, the pin at B must supply a

20 \text{ kN}

horizontal reaction pointing left. That confirms answer A is correct. Answer B is tempting but wrong — it misreads the equilibrium equation. Saying "horizontal equilibrium is satisfied by the wind load alone" is nonsensical; an unbalanced external force would cause the structure to accelerate, violating static equilibrium entirely.

B_x

cannot be zero. Answer C fabricates a proportional distribution rule that doesn't exist in statics. You don't split forces based on geometric distances unless you're working with distributed loads or influence lines — neither applies here. Answer D gets the magnitude right but the direction wrong, and for a flawed reason. The pin at B must oppose the wind load (reaction to the left), not match its direction. Reacting in the same direction would add to the unbalance, not correct it. Study tip: Always identify support types before writing equilibrium equations — the support conditions define which reaction components exist, and that shapes every equation that follows.

Question 2

A propped cantilever beam of length $L$ is fixed at the left end A and supported by a roller at the right end B. A single concentrated load $P$ is applied downward at the midpoint. A student claims that the three equilibrium equations ( $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ) are sufficient to solve for all support reactions uniquely.

Which statement best describes why the student's claim is incorrect, and what consequence this has for analysis?

The claim is incorrect because the fixed support at A provides three reaction components (horizontal force, vertical force, and moment), and together with the vertical roller reaction at B, the structure has four unknown reactions but only three equilibrium equations, making it statically indeterminate to the first degree and requiring a compatibility condition for solution. (correct answer)
The claim is incorrect because a roller support provides two unknown reaction components (vertical and horizontal), raising the total unknown count to five, which exceeds the three equilibrium equations by two degrees, making the structure statically indeterminate to the second degree.
The claim is correct; a propped cantilever with a fixed end and a roller has exactly three unknown reactions—one moment at A, one vertical force at A, and one vertical force at B—which are uniquely determined by the three equilibrium equations without additional conditions.
The claim is incorrect because the fixed end at A provides only a moment reaction and no force reactions, giving three unknowns total, but the moment equation is redundant with the force equations, leaving the system with only two independent equations and thus statically indeterminate to the first degree.

Explanation: Whenever you encounter a structural analysis question, your first move should be to count unknowns vs. equations. The three equilibrium equations (

\sum F_x = 0

\sum F_y = 0

\sum M = 0

) can only solve for three unknowns. If a structure has more, it's statically indeterminate, and you need additional compatibility (deformation) conditions. For this propped cantilever, catalog the reactions carefully. A fixed support at A resists horizontal translation, vertical translation, and rotation — giving three reaction components:

A_x

A_y

, and

M_A

. A roller at B resists only perpendicular displacement, giving one reaction:

B_y

. That's four unknowns total against only three equilibrium equations, making the structure statically indeterminate to the first degree (

4 - 3 = 1

). One compatibility condition — typically derived from a deflection constraint — is required. Answer A captures this correctly. Answer B is wrong because a roller provides only one reaction component (perpendicular to its rolling surface), not two. Confusing a roller with a pin is a classic mistake — pins resist motion in two directions; rollers resist only one. Answer C is wrong because it undercounts the fixed support's contributions. A fixed end provides three reactions (

A_x

A_y

M_A

), not two forces and a moment summing to three — the total with the roller is four, not three. Answer D is wrong because it mischaracterizes the fixed support as providing only a moment. Fixed supports always provide both force and moment reactions. Study tip: Memorize reaction counts — pin = 2, roller = 1, fixed = 3 — and always subtract from total unknowns to find the degree of indeterminacy before attempting any analysis.

Question 3

A beam of length $12 \text{ m}$ is supported by a pin at A (left end) and a roller at C, where C is located $9 \text{ m}$ from A. The beam extends $3 \text{ m}$ beyond C to a free end D. A uniformly distributed load of $w = 5 \text{ kN/m}$ acts downward over the entire length of the beam (from A to D). A student computes the roller reaction at C by summing moments about A for the entire beam.

Which of the following correctly evaluates the student's approach and identifies the correct value of $R_C$ ?

There is no error in the approach; summing moments about A for the entire free body is valid. The total load resultant is $5(12) = 60 \text{ kN}$ acting at $x = 6 \text{ m}$ from A, giving $R_C(9) = 60(6) \Rightarrow R_C = 40 \text{ kN upward}$ . (correct answer)
The approach is valid, but the student likely placed the full load resultant at $x = 4.5 \text{ m}$ (midpoint of AC) instead of $x = 6 \text{ m}$ (midpoint of the full beam), yielding the incorrect result $R_C = \frac{60(4.5)}{9} = 30 \text{ kN upward}$ .
The approach is valid, but the overhanging portion CD beyond the roller at C must be analyzed with a separate free-body diagram; including the overhang load in the global moment equation double-counts its effect and artificially inflates $R_C$ .
The approach is invalid because the overhanging beam with a pin and roller has four unknown reaction components (two at the pin, one at the roller, plus an unknown at the free end), making the structure statically indeterminate and preventing direct use of the moment equation.

Explanation: When analyzing a statically determinate beam — pin at one support, roller at another, with a free overhang — your most powerful tool is the global free-body diagram combined with moment equilibrium. The key insight is that you can always sum moments about any point on the entire free body, as long as you account for every external force correctly. For this beam, the UDL covers the full 12 m, so the total resultant is

w \cdot L = 5(12) = 60 \text{ kN}

, acting at the centroid of that load, which is the midpoint of the full beam:

x = 6 \text{ m}

from A. Summing moments about A eliminates the pin reactions entirely (they act at A), leaving only

R_C

\sum M_A = 0: \quad R_C(9) - 60(6) = 0 \implies R_C = \frac{360}{9} = 40 \text{ kN upward}

This confirms A is correct. B is wrong because it misplaces the resultant at

x = 4.5 \text{ m}

, which is only the midpoint of segment AC — a classic error of forgetting that the UDL extends over the full beam, not just between the supports. C is wrong because it invents a fictional "double-counting" problem. There is no rule requiring you to isolate the overhang; including the entire load in a single global moment equation is perfectly valid and standard procedure. D is wrong because it misidentifies the structure as indeterminate. A pin provides two reaction components and a roller provides one — that's three unknowns total, exactly matching the three equilibrium equations available for a 2D system. Study tip: Always locate the UDL resultant at the midpoint of its full extent, not just between the supports — misplacing it is one of the most common errors on beam equilibrium problems.