Statics and Dynamics Quiz: Trusses Method Of Joints

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Trusses Method Of JointsQuestion 1 of 7

A simple planar truss has the following geometry described by joints and members. Joint A is at the origin (0, 0) and is pin-supported. Joint B is at (4 m, 0) and is roller-supported (vertical reaction only). Joint C is at (2 m, 2 m). A single vertical downward load of 10 kN is applied at joint C. The three members are AC, BC, and AB.

Using the method of joints, what is the force in member AC, and is it in tension or compression?

F_{AC} = 5\sqrt{2} \text{ kN, tension}

; reactions are

A_y = B_y = 5 \text{ kN}

, and at joint A, vertical equilibrium gives

F_{AC}\sin 45° = 5 \text{ kN}

, so the member pulls joint A toward C (tension).

F_{AC} = 5\sqrt{2} \text{ kN, compression}

; reactions are

A_y = B_y = 5 \text{ kN}

, and at joint A, vertical equilibrium gives

F_{AC}\sin 45° = 5 \text{ kN}

, but the member pushes into the joint (compression).

F_{AC} = 10 \text{ kN, compression}

; isolating joint C and summing vertical forces, both diagonal members carry equal vertical components summing to 10 kN, making each member force equal to 10 kN.

F_{AC} = 5 \text{ kN, tension}

; summing moments about B gives

A_y = 5 \text{ kN}

, and member AC is treated as carrying only the vertical component of the reaction, omitting the geometric angle factor.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Trusses Method Of Joints

Practice Trusses Method Of Joints in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Trusses Method Of Joints, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Question 1

Using the method of joints, what is the force in member AC, and is it in tension or compression?

$F_{AC} = 5\sqrt{2} \text{ kN, tension}$ ; reactions are $A_y = B_y = 5 \text{ kN}$ , and at joint A, vertical equilibrium gives $F_{AC}\sin 45° = 5 \text{ kN}$ , so the member pulls joint A toward C (tension). (correct answer)
$F_{AC} = 5\sqrt{2} \text{ kN, compression}$ ; reactions are $A_y = B_y = 5 \text{ kN}$ , and at joint A, vertical equilibrium gives $F_{AC}\sin 45° = 5 \text{ kN}$ , but the member pushes into the joint (compression).
$F_{AC} = 10 \text{ kN, compression}$ ; isolating joint C and summing vertical forces, both diagonal members carry equal vertical components summing to 10 kN, making each member force equal to 10 kN.
$F_{AC} = 5 \text{ kN, tension}$ ; summing moments about B gives $A_y = 5 \text{ kN}$ , and member AC is treated as carrying only the vertical component of the reaction, omitting the geometric angle factor.

Explanation: When tackling truss problems, your first move is always global equilibrium to find support reactions, then you isolate individual joints to find member forces — that's the heart of the method of joints. Start with reactions. Summing moments about A gives

B_y \cdot 4 = 10 \cdot 2

, so

B_y = 5 \text{ kN}

. By vertical equilibrium,

A_y = 5 \text{ kN}

. With no horizontal load and a roller at B,

A_x = 0

. Now isolate joint A. Member AC runs from (0,0) to (2,2), making a 45° angle. Member AB runs horizontally. Assuming members are in tension (forces pull away from the joint), vertical equilibrium gives

F_{AC}\sin 45° = 5 \text{ kN}

, so

F_{AC} = \frac{5}{\sin 45°} = 5\sqrt{2} \text{ kN}

. Since the assumed tension direction was correct (the member pulls joint A upward toward C), AC is in tension. Choice A is correct. Choice B makes the right calculation but wrongly labels AC as compression. If

F_{AC}

came out positive under a tension assumption, the member is in tension — a positive result never flips the sign convention. Choice C skips the angle correction entirely: isolating joint C, each diagonal has a vertical component of 5 kN, but the actual member force is

5/\sin 45° = 5\sqrt{2} \text{ kN}

, not 10 kN. Choice D correctly finds

A_y = 5 \text{ kN}

but then assigns that value directly to

F_{AC}

, forgetting that a diagonal member carries a force larger than its vertical component by a factor of

1/\sin\theta

. Always assume tension in every member before solving — a negative result simply means compression. This keeps your sign convention consistent and eliminates guesswork.

Question 2

In a statically determinate truss, an engineer applies the method of joints and finds that at a certain joint, only two members meet and no external load is applied. The engineer concludes both members are zero-force members. Under what condition is this conclusion incorrect?

The conclusion is incorrect if the two members are not collinear, because non-collinear members at an unloaded joint must each independently satisfy equilibrium in perpendicular directions, forcing both to zero — so the conclusion is always correct for non-collinear cases and the rule never fails.
The conclusion is incorrect if the two members are collinear, because collinear members can transmit equal forces through the joint without violating equilibrium, meaning both could be nonzero — the zero-force rule guarantees both are zero only when the two members are non-collinear.
The conclusion is incorrect if the joint is a support joint where a reaction force acts, because the reaction effectively constitutes an external load at that joint, violating the 'no external load' premise and allowing the members to carry nonzero forces. (correct answer)
The conclusion is incorrect if one member is oriented vertically, because vertical members must resist the self-weight of the truss even in a nominally unloaded analysis, so they can never be true zero-force members regardless of joint equilibrium conditions.

Explanation: When applying the zero-force member rules, you must carefully check whether the "no external load" condition is truly satisfied — and reaction forces count as external loads. The two-member, no-load rule works like this: if exactly two members meet at an unloaded joint, resolving forces in the direction perpendicular to one member forces the other to zero, and then the first must also be zero. This logic is airtight — but only when no external load exists at that joint. A support joint carries a reaction force from the ground (or wall), which is absolutely an external load acting at that joint. With that reaction present, the equilibrium equations are no longer homogeneous; both members can carry nonzero forces to balance the reaction. The "no external load" premise has been violated, so the zero-force conclusion fails. This is exactly what answer C identifies, making it correct. Answer A is wrong in its conclusion. Non-collinear members at a truly unloaded joint do both become zero-force members — that part is correct — but the answer ignores the support-joint exception entirely, claiming the rule "never fails," which is false. Answer B describes a real and important rule (collinear members can pass force through a joint without violating equilibrium), but it misapplies it here. The question asks when the two-member rule fails, not when collinear members can be nonzero. Collinearity is a separate zero-force scenario, not a counterexample to the one being tested. Answer D invents a false exception. Self-weight is typically neglected in introductory truss analysis, and vertical orientation alone carries no special immunity from zero-force classification. Study tip: Always scan for support joints before labeling zero-force members — reactions are invisible loads that invalidate the rule.

Question 3

An analyst is solving a truss using the method of joints and encounters a joint where four members meet. Two of the members (call them 1 and 2) are collinear along the x-axis, and the other two (members 3 and 4) are collinear along the y-axis. No external load acts at this joint.

Which of the following statements correctly describes what the method of joints equilibrium equations reveal about the member forces at this joint?

$F_1 = F_2$ and $F_3 = F_4$ , but neither pair is necessarily zero; the x-equilibrium constrains the two x-axis members to carry equal forces, and y-equilibrium constrains the two y-axis members to carry equal forces, with magnitudes determined by conditions at other joints. (correct answer)
All four member forces must equal zero; when four members meet at an unloaded joint with two pairs of collinear members, the only solution to both equilibrium equations is the trivial solution $F_1 = F_2 = F_3 = F_4 = 0$ .
$F_1 = -F_2$ and $F_3 = -F_4$ , meaning the two collinear members in each direction carry forces of equal magnitude but opposite type (one tension, one compression), which is required to satisfy equilibrium at the joint without an external load.
The system is indeterminate at this joint because four unknowns exist but only two equilibrium equations are available; the method of joints cannot determine any of the four member forces without additional information from adjacent joints or a compatibility condition.

Explanation: When you apply the method of joints to an unloaded joint where four members meet in two collinear pairs, your two equilibrium equations (

\sum F_x = 0

and

\sum F_y = 0

) each involve exactly two unknowns. Writing them out explicitly:

F_1 - F_2 = 0

and

F_3 - F_4 = 0

, which gives

F_1 = F_2

and

F_3 = F_4

. This is the essence of answer A — each collinear pair must carry the same force magnitude and type (both tension or both compression), but that shared value is set by conditions elsewhere in the truss, not forced to zero at this joint alone. Answer B is the most tempting trap. It confuses this situation with the zero-force member rule. That rule applies when only two non-collinear members meet at an unloaded joint — there, both forces must be zero. With two collinear pairs, equilibrium is fully satisfied by equal (nonzero) forces, so the trivial solution is not the only one. Answer C has the sign convention backwards. If both members on the x-axis pull away from the joint (tension), they both contribute in opposite directions, yielding

F_1 - F_2 = 0

, not

F_1 = -F_2

. Equal magnitude with opposite type would violate equilibrium, not satisfy it. Answer D misidentifies this as statically indeterminate at the joint level. You do have four unknowns and two equations, but the collinearity reduces each equation to one unknown relationship — the system is perfectly determinate at this joint. Study tip: Always ask whether collinear members reduce your effective unknowns before declaring a joint indeterminate — this is a classic exam trap.

Question 4

A Pratt truss has top chord joints at A (0,0), C (2 m, 2 m), E (4 m, 2 m), and bottom chord joints at B (0,0)... [Revised simpler geometry below.] A symmetric planar truss spans 6 m with joints at: L (0,0) pin, M (3,0), R (6,0) roller, and apex P (3, 3). Members: LP, MP, RP, LM, MR. A 24 kN downward load acts at P.

After determining support reactions, an analyst isolates joint M to find the force in member MP. Members LM and MR are both horizontal, and MP is vertical. No external load acts at M. The analyst uses the zero-force member rule to declare $F_{MP} = 0$ . However, joint P equilibrium gives a different result. What is the actual force in MP, and why does the analyst's application of the zero-force rule fail?

$F_{MP} = 12 \text{ kN, compression}$ ; the zero-force rule fails because it only applies when exactly two members meet at a joint or when two are collinear and no external load is present — at joint M the rule is valid and MP = 0 is correct, but joint P analysis overrides it, revealing a contradiction that means the truss is improperly constrained.
$F_{MP} = 12 \text{ kN, compression}$ ; the zero-force rule is correctly stated but the analyst misidentifies which joint to apply it to — at joint M with no external load and LM/MR collinear, MP is indeed zero-force, but the load at P propagates downward into MP making it carry half the applied load.
$F_{MP} = 0$ ; the zero-force rule is correctly applied at joint M, and the 24 kN load at P is carried entirely by LP and RP through the diagonal members, consistent with joint P equilibrium showing $F_{LP}\sin 45° + F_{RP}\sin 45° = 24 \text{ kN}$ with no contribution from MP. (correct answer)
$F_{MP} = 24 \text{ kN, compression}$ ; the zero-force rule fails because the analyst ignores that MP is the only vertical member at joint M, so by vertical equilibrium at P, MP must single-handedly carry the entire applied load rather than sharing it with the diagonal members.

Explanation: When analyzing trusses, the zero-force member rule is one of the most powerful shortcuts available — but only when applied correctly. The rule states that if two non-collinear members meet at an unloaded joint, both are zero-force; alternatively, if two members at an unloaded joint are collinear, the third member is zero-force. The key word throughout is unloaded joint. At joint M, no external load acts, LM and MR are collinear (both horizontal), and MP is the only non-collinear member — so the rule correctly identifies MP as zero-force. This is not a misapplication. Now check joint P: the 24 kN load acts downward, and members LP and RP each rise at 45° (since the apex is at (3,3) and supports are at (0,0) and (6,0)). Vertical equilibrium at P gives

F_{LP}\sin 45° + F_{RP}\sin 45° = 24 \text{ kN}

, which yields

F_{LP} = F_{RP} = 12\sqrt{2} \text{ kN}

. MP contributes nothing to this equation —

F_{MP} = 0

is fully consistent. Answer C is correct. Choice A is wrong because it invents a contradiction and claims the truss is improperly constrained — there is no such contradiction here. Choice B incorrectly claims the load "propagates into MP," confusing this geometry with a case where MP would actually be needed for equilibrium. Choice D overstates MP's role; LP and RP have vertical components that together balance the full 24 kN, leaving MP with nothing to carry. Your study tip: always verify zero-force conclusions at a second joint. If both joints are consistent — as they are here — trust the rule. A contradiction signals a geometry or loading error, not a flaw in the rule itself.

Question 5

A student is analyzing a truss using the method of joints and reaches a joint where the equilibrium equations yield a unique solution. The student then checks the solution by attempting to apply the method of joints at a different joint — one that now has all member forces known from previous joints. The student finds the equations are satisfied identically (0 = 0) at this check joint.

What does this result — equilibrium equations satisfied identically at the check joint — most rigorously indicate about the truss and the analysis?

It indicates the truss is statically indeterminate; an identically satisfied equation provides no new information, which is the defining characteristic of a redundant constraint, and the student must use compatibility equations to complete the solution.
It indicates the truss is statically determinate and the solution is consistent; in a properly solved determinate truss, the final joint used as a check will have its equilibrium identically satisfied by the already-found forces, confirming correctness — this is the expected outcome, not an error. (correct answer)
It indicates the truss has a mechanism (kinematic instability); when the equilibrium check yields 0 = 0, the stiffness matrix is singular, meaning the truss has insufficient members to prevent rigid-body motion under the applied loads.
It indicates the student made compensating errors in earlier joints; a truly correct solution would produce nonzero residuals at the check joint that can be compared against known loads, and an identity result means the check joint carries no load, which is only possible if the student's earlier forces are mutually inconsistent.

Explanation: When analyzing a truss using the method of joints, you systematically apply equilibrium (

\sum F_x = 0

\sum F_y = 0

) at each joint until all member forces are determined. For a statically determinate truss satisfying

m + r = 2j

(members plus reactions equals twice the joints), you have exactly enough equations to solve for all unknowns — no more, no less. This means by the time you reach the final "check" joint, every force acting on it has already been found from previous joints. Plugging those known values into the check joint's equilibrium equations must yield

0 = 0

identically — this is the expected, correct outcome, confirming your solution is self-consistent. Answer B captures this precisely. Answer A is a tempting trap because "an equation that provides no new information" sounds like redundancy, which you might associate with indeterminacy. But indeterminacy means you cannot solve the system using equilibrium alone — it doesn't describe what happens after a valid solution is already complete. The identity at the check joint arises from sufficiency, not redundancy. Answer C confuses the check-joint identity with a singular stiffness matrix. A mechanism means the truss can move without deforming — a structural deficiency detectable before solving, not by getting

0 = 0

after a successful solution. Answer D inverts the logic entirely. Compensating errors would likely produce nonzero residuals at the check joint, not an identity. Getting

0 = 0

is confirmation of correctness, not a red flag. Study tip: Always count

m + r = 2j

before starting. If it holds, expect the final joint to be your verification — a clean

0 = 0

means you solved it correctly.

Question 6

A planar truss has five joints: A (0,0) pin, B (3,0), C (6,0) roller, D (3,4), and E (0,4). Members: AB, BC, AE, ED, BD, BE, CD. A 20 kN downward load acts at D and a 10 kN downward load acts at B. No other external loads exist.

After computing reactions, a student begins the method of joints at joint A, then proceeds to joint E. At joint E, three members meet: AE (horizontal), ED (horizontal), and BE (diagonal from B(3,0) to E(0,4)). The student writes: 'Since AE and ED are collinear, BE is a zero-force member.' Is the student correct, and what is $F_{BE}$ ?

The student is correct; AE and ED are collinear horizontal members and BE is the non-collinear member at a joint with no external load, so $F_{BE} = 0$ by the zero-force member rule, regardless of the load magnitudes applied elsewhere in the truss.
The student is incorrect; the zero-force rule requires verifying that no external load acts at joint E, which is satisfied here, but AE and ED point in opposite horizontal directions from E, meaning they are collinear only if you consider E as an interior point — in this configuration the rule does apply, so $F_{BE} = 0$ is correct, but the student's reasoning contains a subtle flaw in not checking the external load condition explicitly.
The student is incorrect; even though no external load acts at E, the zero-force rule for a three-member joint with two collinear members requires that the support reactions have already been verified as zero at that joint before it can be applied, and since A is a pin support adjacent to E, this condition is not met, making $F_{BE}$ nonzero.
The student is incorrect; AE runs from A(0,0) to E(0,4), making it vertical, not horizontal, so AE and ED are not collinear. With three non-collinear members at E and no external load, equilibrium requires all three forces to be zero only if the geometry is symmetric, which it is not, so $F_{BE}$ must be solved from equilibrium equations at E. (correct answer)

Explanation: Whenever you apply the zero-force member rule at a joint, your very first step must be to correctly identify the directions of every member meeting there — a geometry error at this stage invalidates everything that follows. At joint E(0,4), let's trace each member carefully. Member ED runs from E(0,4) to D(3,4) — that's a horizontal member. Member BE runs from B(3,0) to E(0,4) — that's a diagonal. Now examine member AE: it runs from A(0,0) to E(0,4). Both points share the same x-coordinate, so AE is vertical, not horizontal. AE and ED therefore point in perpendicular directions — they are absolutely not collinear. This makes D the correct answer. With three non-collinear members and no external load at E, you must write and solve the full equilibrium equations (

\sum F_x = 0

and

\sum F_y = 0

) to find

F_{BE}

; it will generally be nonzero. Choice A is wrong because it accepts the student's false premise that AE is horizontal. The zero-force rule is being misapplied to a geometry that doesn't satisfy it. Choice B is similarly flawed — it acknowledges a "subtle" reasoning error but still arrives at

F_{BE} = 0

, which is unjustified since the collinearity condition is simply not met. Choice C invents a nonexistent requirement about support reactions at adjacent joints; the zero-force member rule only demands no external load at the joint in question and two collinear members — there is no condition about neighboring supports. Study tip: Before invoking any zero-force rule, sketch the member directions and confirm collinearity with coordinates — never assume a member is horizontal or vertical without checking both endpoints.

Question 7

A planar truss is loaded with a single 30 kN downward force at midspan joint M. The truss is symmetric and simply supported at joints L (left pin) and R (right roller). By symmetry, each support carries 15 kN upward. A student applies the method of joints starting at joint L. At L, two members meet: LM (inclined at 30° above horizontal toward M) and LB (horizontal toward B, which is the bottom chord joint directly below M).

At joint L, the student assumes both member forces are tensile and sets up equilibrium. The vertical equation gives $F_{LM}\sin 30° = 15$ kN, yielding $F_{LM} = 30$ kN. The horizontal equation gives $F_{LB} = -F_{LM}\cos 30° = -30\cos 30°$ . The student reports $F_{LB} = -25.98$ kN and concludes LB is in tension because the assumed direction was tension. What is the correct conclusion about $F_{LB}$ ?

$F_{LB} = 25.98 \text{ kN, tension}$ ; the negative sign confirms the assumed tensile direction is correct because the x-axis is oriented leftward, so the negative result in the rightward-positive convention actually indicates the member pulls the joint to the right, which is tension.
$F_{LB} = 25.98 \text{ kN, compression}$ ; when equilibrium yields a negative value for an assumed-tension force, the actual force is compressive — the member pushes the joint rather than pulls it, so the magnitude is correct but the sign convention indicates compression. (correct answer)
$F_{LB} = 25.98 \text{ kN, tension}$ ; the sign convention in method of joints defines positive as tension by assumption, so a negative result means the student made an error in the equilibrium equation — the correct equation should give a positive value confirming tension, and the student should re-examine the geometry.
$F_{LB} = 25.98 \text{ kN, tension}$ ; the horizontal equilibrium equation should be written as $F_{LB} + F_{LM}\cos 30° = 0$ only if there is a horizontal reaction at L, but since A_x = 0, the equation $F_{LB} = F_{LM}\cos 30°$ gives a positive value directly, meaning the student's sign error arose from incorrectly including a zero reaction term.

Explanation: Whenever you apply the method of joints, you assume unknown member forces are tensile — meaning the member pulls the joint outward. If equilibrium then returns a negative value, that assumption was wrong: the member actually pushes the joint, meaning it's in compression. The sign is everything. Here, at joint L, assuming both members tensile and summing forces horizontally (rightward positive): the horizontal component of

F_{LM}

points rightward (toward M), so

F_{LB} + F_{LM}\cos 30° = 0

, giving

F_{LB} = -30\cos 30° \approx -25.98 \text{ kN}

. The negative sign means the actual force is opposite to the assumed tensile direction — the member compresses the joint rather than pulling it. The magnitude is correct at 25.98 kN, but the correct classification is compression. This confirms answer B. A is wrong because the x-axis orientation doesn't flip the physical meaning of tension versus compression. A negative result in a rightward-positive convention means the force acts leftward on the joint — that's the member pushing in, which is compression, not tension. C is wrong because a negative result is not a signal of an equation error. It's a perfectly valid and expected outcome that simply tells you to reverse the assumed direction. No re-examination of geometry is needed. D is wrong because the pin at L carries no horizontal reaction (roller is at R, pin at L with no horizontal load means

A_x = 0

), so there's no missing reaction term. The student's equilibrium equation is set up correctly; the only issue is misinterpreting the negative sign. Study tip: Memorize this rule — negative assumed-tension force = compression, positive = tension. Never second-guess the equation; trust the sign.