Statics and Dynamics Quiz: Distributed Loads

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Distributed LoadsQuestion 1 of 10

A beam of length $10 \text{ m}$ is fixed at the left end ( $x = 0$ ) and free at the right end ( $x = 10 \text{ m}$ ). It carries two distributed loads: (1) a uniform load of $5 \text{ kN/m}$ acting over the entire span, and (2) a triangular load decreasing linearly from $10 \text{ kN/m}$ at $x = 0$ to $0$ at $x = 10 \text{ m}$ .

What is the magnitude of the fixed-end moment $M_A$ (the reaction moment at the wall) required for equilibrium? Take clockwise moments as positive.

M_A = 375 \text{ kN·m}

, found by correctly computing the combined trapezoidal resultant

R = \frac{1}{2}(15 + 5)(10) = 100 \text{ kN}

but locating it using the centroid formula with

w_1

and

w_2

swapped, placing it at

\frac{10}{3}\cdot\frac{5 + 2(15)}{5 + 15} = 3.75 \text{ m}

from the fixed end, giving

M_A = 100(3.75) = 375 \text{ kN·m}

M_A = 583.3 \text{ kN·m}

, found by computing

F_1 = 50 \text{ kN}

5 \text{ m}

from the wall and

F_2 = 50 \text{ kN}

placed at

\bar{x}_2 = \frac{2}{3}(10) = 6.67 \text{ m}

from the fixed end (incorrectly applying two-thirds from the heavy end rather than one-third), giving

M_A = 50(5) + 50(6.67) = 583.3 \text{ kN·m}

M_A = 500 \text{ kN·m}

, found by noting that the combined loading is trapezoidal (

15 \text{ kN/m}

at the fixed end,

5 \text{ kN/m}

at the free end) with average intensity

10 \text{ kN/m}

, giving

R = 10(10) = 100 \text{ kN}

placed at mid-span

x = 5 \text{ m}

(incorrectly assuming the average intensity always acts at mid-span), so

M_A = 100(5) = 500 \text{ kN·m}

M_A = 416.7 \text{ kN·m}

, found by computing the uniform-load resultant

F_1 = 5(10) = 50 \text{ kN}

acting at

\bar{x}_1 = 5 \text{ m}

from the wall, and the triangular-load resultant

F_2 = \frac{1}{2}(10)(10) = 50 \text{ kN}

acting at

\bar{x}_2 = \frac{1}{3}(10) = 3.33 \text{ m}

from the wall (centroid at one-third from the heavy end), giving

M_A = 50(5) + 50(3.33) = 416.7 \text{ kN·m}

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Distributed Loads

Practice Distributed Loads 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 Distributed Loads, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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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.

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Question 1

What is the magnitude of the fixed-end moment $M_A$ (the reaction moment at the wall) required for equilibrium? Take clockwise moments as positive.

$M_A = 375 \text{ kN·m}$ , found by correctly computing the combined trapezoidal resultant $R = \frac{1}{2}(15 + 5)(10) = 100 \text{ kN}$ but locating it using the centroid formula with $w_1$ and $w_2$ swapped, placing it at $\frac{10}{3}\cdot\frac{5 + 2(15)}{5 + 15} = 3.75 \text{ m}$ from the fixed end, giving $M_A = 100(3.75) = 375 \text{ kN·m}$ .
$M_A = 583.3 \text{ kN·m}$ , found by computing $F_1 = 50 \text{ kN}$ at $5 \text{ m}$ from the wall and $F_2 = 50 \text{ kN}$ placed at $\bar{x}_2 = \frac{2}{3}(10) = 6.67 \text{ m}$ from the fixed end (incorrectly applying two-thirds from the heavy end rather than one-third), giving $M_A = 50(5) + 50(6.67) = 583.3 \text{ kN·m}$ .
$M_A = 500 \text{ kN·m}$ , found by noting that the combined loading is trapezoidal ( $15 \text{ kN/m}$ at the fixed end, $5 \text{ kN/m}$ at the free end) with average intensity $10 \text{ kN/m}$ , giving $R = 10(10) = 100 \text{ kN}$ placed at mid-span $x = 5 \text{ m}$ (incorrectly assuming the average intensity always acts at mid-span), so $M_A = 100(5) = 500 \text{ kN·m}$ .
$M_A = 416.7 \text{ kN·m}$ , found by computing the uniform-load resultant $F_1 = 5(10) = 50 \text{ kN}$ acting at $\bar{x}_1 = 5 \text{ m}$ from the wall, and the triangular-load resultant $F_2 = \frac{1}{2}(10)(10) = 50 \text{ kN}$ acting at $\bar{x}_2 = \frac{1}{3}(10) = 3.33 \text{ m}$ from the wall (centroid at one-third from the heavy end), giving $M_A = 50(5) + 50(3.33) = 416.7 \text{ kN·m}$ . (correct answer)

Explanation: When a beam carries multiple distributed loads, the most reliable approach is to resolve each load into its resultant force and locate that force at the load's centroid, then sum moments about the fixed end for equilibrium. For this cantilever, you have two loads. The uniform load gives

F_1 = 5 \times 10 = 50 \text{ kN}

, acting at the midpoint

\bar{x}_1 = 5 \text{ m}

from the wall — straightforward. The triangular load decreases from

10 \text{ kN/m}

at the wall to zero at the free end, giving

F_2 = \frac{1}{2}(10)(10) = 50 \text{ kN}

. The critical step: the centroid of a triangle lies one-third of the length from the heavy end, so

\bar{x}_2 = \frac{1}{3}(10) = 3.33 \text{ m}

from the wall. Summing moments:

M_A = 50(5) + 50(3.33) = 250 + 166.7 = 416.7 \text{ kN·m}

, confirming answer D. Answer A correctly computes the total resultant but treats the loading as a single trapezoid and misapplies the centroid formula by swapping

w_1

and

w_2

, pushing the centroid too far from the wall. Answer B correctly finds both resultants but places the triangular load's centroid at

\frac{2}{3}

from the wall — confusing "two-thirds from the light end" with "two-thirds from the heavy end." Answer C correctly finds the average intensity but incorrectly assumes a non-uniform load's resultant always acts at mid-span; that only holds for truly uniform loads. Study tip: For any triangular load, always ask yourself "one-third from which end?" — the answer is always one-third from the heavy end, or equivalently two-thirds from the zero end.

Question 2

A simply supported beam of length $L = 6 \text{ m}$ carries a trapezoidal distributed load that varies linearly from $w_1 = 4 \text{ kN/m}$ at the left support ( $x = 0$ ) to $w_2 = 10 \text{ kN/m}$ at the right support ( $x = L$ ).

To replace the trapezoidal load with a single equivalent resultant force, a student decomposes it into a uniform load of $4 \text{ kN/m}$ plus a triangular load that increases from zero to $6 \text{ kN/m}$ . Where does the resultant of the combined (trapezoidal) loading act, measured from the left support?

$x = 3.00 \text{ m}$ (midpoint), because the average intensity of the trapezoidal load acts at the centroid of the rectangular portion, which dominates the moment calculation when the two sub-resultants are combined.
$x = 3.43 \text{ m}$ , found by taking moments of the two sub-resultants: the uniform block resultant of $24 \text{ kN}$ acts at $3 \text{ m}$ and the triangular resultant of $18 \text{ kN}$ acts at $4 \text{ m}$ , giving $\bar{x} = (24 \times 3 + 18 \times 4)/(24 + 18)$ . (correct answer)
$x = 3.67 \text{ m}$ , found by noting that the heavier loading is at the right end, so the resultant must be located at two-thirds of the span from the left support, regardless of the magnitude of the uniform component.
$x = 2.57 \text{ m}$ , found by taking moments of the two sub-resultants but incorrectly placing the triangular resultant at $L/3 = 2 \text{ m}$ from the left (the lighter end) and the uniform resultant at $L/2 = 3 \text{ m}$ , giving $\bar{x} = (24 \times 3 + 18 \times 2)/(24 + 18)$ .

Explanation: When a distributed load has a non-uniform shape, a powerful strategy is to decompose it into simpler shapes — typically a rectangle plus a triangle — find each sub-resultant, then combine them using the moment-weighted average to locate the overall resultant. Here, the trapezoidal load breaks into a uniform block of

4 \text{ kN/m}

and a triangular load rising from

0

6 \text{ kN/m}

. The uniform resultant is

F_1 = 4 \times 6 = 24 \text{ kN}

, acting at the rectangle's centroid:

x_1 = L/2 = 3 \text{ m}

. The triangular resultant is

F_2 = \frac{1}{2}(6)(6) = 18 \text{ kN}

. Critically, because the triangle's peak is at the right end, its centroid lies at two-thirds of the span from the left (the lighter end), so

x_2 = \frac{2}{3}(6) = 4 \text{ m}

. Taking the moment-weighted average:

\bar{x} = \frac{24(3) + 18(4)}{24 + 18} = \frac{72 + 72}{42} = \frac{144}{42} \approx 3.43 \text{ m}

This confirms B is correct. A is wrong because the resultant doesn't simply land at the midpoint — the triangular portion shifts it rightward toward the heavier end. C is wrong because

2L/3 = 4 \text{ m}

applies only to the triangular sub-resultant alone, not the combined trapezoidal load. D places the triangular resultant at

L/3 = 2 \text{ m}

from the left — but

L/3

from the heavy end equals

2L/3

from the light end; reversing this is the classic trap here. Study tip: For any triangular load, always identify which end is zero and which is the peak — the centroid is always at

\frac{1}{3}

from the peak (or

\frac{2}{3}

from the zero end). Getting this direction wrong is the most common error on distributed-load problems.

Question 3

A beam of length $L$ is pinned at the left end and roller-supported at the right end. A linearly varying distributed load acts on the beam: the intensity is $w_1$ at the left support and $w_2$ at the right support, with $w_1 \neq w_2$ . The standard centroid formula for a trapezoidal load places the equivalent resultant at $\bar{x} = \frac{L}{3}\cdot\frac{w_1 + 2w_2}{w_1 + w_2}$ from the left end.

A student uses this formula with $w_1 = 0$ (triangular load, zero at left, peak $w_2 = w_0$ at right) and obtains $\bar{x} = \frac{2L}{3}$ . A second student uses the formula for the same load but measures $x$ from the right support, obtaining $\bar{x}' = \frac{L}{3}$ from the right, i.e., $\bar{x} = \frac{2L}{3}$ from the left. The students compare their results. Which statement is correct?

Both students obtain the same physical location for the resultant ( $2L/3$ from the left/zero end), but only the second student's method is dimensionally self-consistent because measuring from the heavier end is the convention required by the formula.
Both students correctly locate the resultant at $2L/3$ from the left support (the zero end), confirming that the formula is self-consistent regardless of which end is used as the reference, as long as the intensities $w_1$ and $w_2$ are assigned to match the chosen reference end. (correct answer)
The first student makes an error: substituting $w_1 = 0$ into the formula assumes the zero end is the left support, but the formula is derived with $w_1$ at the right support, so the correct location from the left is $\bar{x} = L/3$ , not $2L/3$ .
The two students get contradictory results because the trapezoidal centroid formula changes form depending on the reference end, and without a universal sign convention the formula yields an ambiguous location that can be either $L/3$ or $2L/3$ from the left.

Explanation: Whenever you encounter problems involving equivalent resultants of distributed loads, the key question is: does the physical answer change if I shift my reference point? It shouldn't — and verifying this is exactly what this problem tests. The trapezoidal centroid formula

\bar{x} = \frac{L}{3}\cdot\frac{w_1 + 2w_2}{w_1 + w_2}

measures the resultant's distance from the end where intensity equals

w_1

. For a triangular load (zero at left, peak

w_0

at right), the first student assigns

w_1 = 0

and

w_2 = w_0

, measuring from the left (zero) end:

\bar{x} = \frac{L}{3}\cdot\frac{0 + 2w_0}{0 + w_0} = \frac{2L}{3}

from the left. The second student flips the reference — now

w_1 = w_0

(at the right end) and

w_2 = 0

— and gets

\bar{x}' = \frac{L}{3}\cdot\frac{w_0 + 0}{w_0 + 0} = \frac{L}{3}

from the right, which is identically

\frac{2L}{3}

from the left. Both students locate the same physical point. B is correct: the formula is self-consistent as long as

w_1

and

w_2

are assigned to match whichever end you measure from. A is wrong because it falsely claims only one direction is dimensionally valid — the formula works from either end with no special convention required. C is wrong because it invents a fictitious constraint about which end

w_1

belongs to; the labeling is entirely up to you. D is wrong because there's no ambiguity — both approaches converge on the same physical location. As a study tip: whenever a formula involves a reference direction, verify consistency by solving from both ends. If the physics is the same, your formula is robust — and that's a reliable check on exam problems involving distributed loads.

Question 4

A cantilever beam is fixed at $x = 0$ and free at $x = 4 \text{ m}$ . It supports a triangular distributed load that is zero at the free end and increases to $w_0 = 12 \text{ kN/m}$ at the fixed end. A second, uniform distributed load of $3 \text{ kN/m}$ acts over only the right half of the beam ( $2 \text{ m} \leq x \leq 4 \text{ m}$ ).

What is the magnitude of the single equivalent resultant force for the entire distributed loading on this beam?

$R = 30 \text{ kN}$ , computed as the area of the full triangular load $\frac{1}{2}(12)(4) = 24 \text{ kN}$ plus the area of the partial uniform load $3(2) = 6 \text{ kN}$ , for a total of $30 \text{ kN}$ . (correct answer)
$R = 27 \text{ kN}$ , computed by treating the triangular load as a uniform load of average intensity $6 \text{ kN/m}$ over the full $4 \text{ m}$ giving $24 \text{ kN}$ , then adding $3 \text{ kN/m}$ over $2 \text{ m}$ giving $6 \text{ kN}$ , but subtracting $3 \text{ kN}$ for the overlap region where both loads act simultaneously.
$R = 36 \text{ kN}$ , computed as $12 \times 4 = 48 \text{ kN}$ for the triangular load (using the peak intensity times the full length instead of the area formula) minus $3(4) = 12 \text{ kN}$ for the uniform load extended over the full beam.
$R = 24 \text{ kN}$ , found by correctly computing the triangular load resultant as $\frac{1}{2}(12)(4) = 24 \text{ kN}$ but neglecting the partial uniform load because it acts over only half the span and is considered secondary.

Explanation: When a beam carries multiple distributed loads, your job is to find the resultant of each load separately, then sum them — no mixing of intensities or subtracting overlaps. For a triangular load, the resultant equals the area of the triangle:

R_1 = \frac{1}{2} \times w_0 \times L = \frac{1}{2}(12)(4) = 24 \text{ kN}

. For a uniform load acting over a partial length, the resultant is simply intensity times that length:

R_2 = 3 \times 2 = 6 \text{ kN}

. These two loads are independent — they act on different (or overlapping) regions of the beam, but their resultants simply add together. The total equivalent force is

R = 24 + 6 = 30 \text{ kN}

, confirming answer A. Answer B is wrong because it invents a "correction" for overlap. The two loads are not competing forces — they both push down on the beam. There is no reason to subtract anything. This reflects a fundamental misunderstanding of superposition. Answer C uses the wrong area formula for the triangular load, multiplying peak intensity by full length (

12 \times 4 = 48

) instead of using the

\frac{1}{2}

factor. It then incorrectly extends the partial uniform load to the full beam and subtracts it, compounding two separate errors. Answer D correctly computes the triangular resultant but ignores the uniform load entirely. No distributed load is ever "secondary" — every load contributes to the resultant, regardless of how short its span is. Study tip: Whenever you see multiple distributed loads, treat each one as its own resultant (area under the load diagram), then superimpose. Never subtract loads just because they share the same region of the beam.

Question 5

A beam of length $L$ supports a distributed load whose intensity is $w(x) = w_0\left(\frac{x}{L}\right)^2$ , varying parabolically from zero at $x = 0$ to $w_0$ at $x = L$ . Which of the following gives the correct location $\bar{x}$ of the equivalent resultant force measured from $x = 0$ ?

$\bar{x} = \dfrac{L}{2}$ , obtained by noting that the squared intensity function is symmetric about the midpoint when viewed as an area, so the centroid of the parabolic arch must fall at $L/2$ by symmetry of the squared function.
$\bar{x} = \dfrac{2L}{3}$ , obtained by applying the standard triangular-load centroid formula (two-thirds of the span from the zero end) to the parabolic load, since both loads are zero at the left and maximum at the right, making the centroid location identical regardless of the curvature of the profile.
$\bar{x} = \dfrac{3L}{4}$ , obtained by computing $\bar{x} = \dfrac{\int_0^L x \cdot w_0(x/L)^2\,dx}{\int_0^L w_0(x/L)^2\,dx} = \dfrac{w_0 L^{-2}\int_0^L x^3\,dx}{w_0 L^{-2}\int_0^L x^2\,dx} = \dfrac{L^4/4}{L^3/3} = \dfrac{3L}{4}$ . (correct answer)
$\bar{x} = \dfrac{4L}{5}$ , obtained by mistakenly treating the load as $w(x) = w_0(x/L)^3$ (a cubic profile) instead of the given quadratic, which shifts the centroid calculation to $\dfrac{\int_0^L x(x/L)^3\,dx}{\int_0^L (x/L)^3\,dx} = \dfrac{L^5/5}{L^4/4} \cdot \dfrac{1}{L} = \dfrac{4L}{5}$ .

Explanation: When a distributed load varies continuously along a beam, you cannot simply "eyeball" its centroid — you must use the weighted-average integral formula:

\bar{x} = \dfrac{\int_0^L x\, w(x)\,dx}{\int_0^L w(x)\,dx}

. This is the same principle as finding the centroid of an area, where the load intensity plays the role of the area density. For

w(x) = w_0(x/L)^2

, the denominator is

\int_0^L w_0\dfrac{x^2}{L^2}\,dx = \dfrac{w_0}{L^2}\cdot\dfrac{L^3}{3} = \dfrac{w_0 L}{3}

, and the numerator is

\int_0^L x\cdot w_0\dfrac{x^2}{L^2}\,dx = \dfrac{w_0}{L^2}\cdot\dfrac{L^4}{4} = \dfrac{w_0 L^2}{4}

. Dividing gives

\bar{x} = \dfrac{w_0 L^2/4}{w_0 L/3} = \dfrac{3L}{4}

, confirming C is correct. Choice A is wrong because

w(x) = w_0(x/L)^2

is not symmetric about

L/2

— it rises steeply near

x = L

, pulling the centroid to the right of center. Choice B commits a common trap: assuming any load that is zero on one end and maximum on the other behaves like a triangle. The centroid location depends critically on how fast the load grows, not just its endpoints. The

2L/3

result applies only to a linear (triangular) profile. Choice D uses a cubic profile

w_0(x/L)^3

instead of the given quadratic, illustrating how even a small misread of the load function shifts the answer significantly. Your study tip: always integrate — never assume the centroid of a non-uniform load matches a simpler shape just because the endpoints look similar.

Question 6

A beam segment from $x = 2 \text{ m}$ to $x = 5 \text{ m}$ (length $3 \text{ m}$ ) carries a uniform distributed load of $8 \text{ kN/m}$ . The rest of the beam (total length $8 \text{ m}$ , pinned at $x = 0$ , roller at $x = 8 \text{ m}$ ) is unloaded.

When replacing the distributed load with its equivalent resultant for the purpose of computing support reactions, which of the following correctly states both the resultant force magnitude and its point of application?

$R = 24 \text{ kN}$ acting at $x = 2.5 \text{ m}$ from the pin, because the point of application is the half-length of the loaded segment ( $3/2 = 1.5 \text{ m}$ ) measured from the start of the load at $x = 2 \text{ m}$ , but is then re-referenced to the midpoint of the unloaded left portion at $x = 1 \text{ m}$ , yielding $x = 3.5 - 1 = 2.5 \text{ m}$ .
$R = 24 \text{ kN}$ acting at $x = 4 \text{ m}$ from the pin, because even though the load is offset from center, the resultant of a uniformly distributed load on any segment of a simply supported beam always acts at the beam's mid-span for the purposes of reaction calculation.
$R = 64 \text{ kN}$ acting at $x = 3.5 \text{ m}$ , because the resultant must be computed over the full beam length using the load intensity ( $8 \text{ kN/m} \times 8 \text{ m} = 64 \text{ kN}$ ) even when the load acts over only part of the span, with the action point still at the midpoint of the loaded segment.
$R = 24 \text{ kN}$ acting at $x = 3.5 \text{ m}$ from the pin, because the resultant of a uniform load over a segment equals the intensity times the segment length ( $8 \times 3 = 24 \text{ kN}$ ), and it acts at the midpoint of that segment (midpoint of $[2, 5]$ is $x = 3.5 \text{ m}$ ). (correct answer)

Explanation: When a distributed load acts over only part of a beam, your job is to replace it with a statically equivalent single force — one that produces the same net force and the same moment about any point. Two rules govern this: the resultant magnitude equals the load intensity times the loaded length (not the full beam length), and it acts at the centroid of the loaded region (the midpoint for a uniform load). Here, the load covers

x = 2\text{ m}

x = 5\text{ m}

, a segment of length

3\text{ m}

. So the resultant is

R = 8\text{ kN/m} \times 3\text{ m} = 24\text{ kN}

, acting at the midpoint of that segment:

x = \frac{2+5}{2} = 3.5\text{ m}

from the pin. That's exactly what D states, making it correct. A gets the magnitude right but then invents a bizarre re-referencing procedure — subtracting the "midpoint of the unloaded portion" — which has no basis in statics. The centroid of the loaded segment is simply

x = 3.5\text{ m}

, full stop. B correctly computes

R = 24\text{ kN}

but falsely claims the resultant always acts at mid-span of the full beam. That rule doesn't exist; the action point depends on where the load actually is, not the beam's geometry. C makes the critical error of integrating over the full

8\text{ m}

beam length, yielding an inflated

64\text{ kN}

. You only integrate over the region where the load is nonzero. Study tip: Always sketch the loaded segment separately, compute

R = w \times L_{\text{loaded}}

, and place it at that segment's midpoint — never the full beam's midpoint.

Question 7

A horizontal beam of length $8 \text{ m}$ is pinned at $A$ (left end) and roller-supported at $B$ (right end). It carries a uniformly distributed load of $6 \text{ kN/m}$ over the left half ( $0 \leq x \leq 4 \text{ m}$ ) and a separate uniformly distributed load of $6 \text{ kN/m}$ over the right half ( $4 \text{ m} \leq x \leq 8 \text{ m}$ ).

A student argues that because both partial loads have the same intensity and together span the full beam, they can be immediately replaced by a single equivalent resultant of $48 \text{ kN}$ acting at mid-span ( $x = 4 \text{ m}$ ) before computing reactions. A second student keeps the two loads separate, computes two resultants, then sums moments. Which statement best evaluates these two approaches in the context of finding support reactions?

Both approaches yield identical reactions, because combining two equal-intensity loads that together cover the full span into a single resultant at the midpoint is mathematically equivalent to summing the moments of the two separate resultants; the combined centroid at mid-span follows directly from the symmetry and equal magnitudes of the two sub-resultants. (correct answer)
The first student's approach is incorrect because the equivalent resultant of two separate distributed loads must always be found by integration of the combined load function; a simple arithmetic combination is only valid when both loads share the same start and end points on the beam.
The first student's approach gives the correct answer here only by coincidence; if the two load intensities were unequal, combining them into a single resultant at mid-span would always place the resultant at the wrong location, so the method is not generally valid for partial distributed loads.
The first student's approach undercounts the total resultant force because combining partial distributed loads requires integrating over each sub-region and then applying a correction factor for the gap at $x = 4 \text{ m}$ where the two loads meet.

Explanation: Whenever you see distributed loads on a beam, the key concept is equivalent resultants: any distributed load can be replaced by a single force equal to the area under the load diagram, acting at that area's centroid. The critical insight is that this substitution is valid before or after combining loads, as long as you correctly locate each resultant. Here, both loads have intensity

6 \text{ kN/m}

over

4 \text{ m}

, giving two resultants of

6 \times 4 = 24 \text{ kN}

. The left resultant acts at

x = 2 \text{ m}

, the right at

x = 6 \text{ m}

. The combined resultant is

48 \text{ kN}

, and its location is the weighted centroid:

\bar{x} = \frac{24(2) + 24(6)}{48} = \frac{48 + 144}{48} = 4 \text{ m}

. Because the two sub-resultants are equal in magnitude, the combined centroid falls exactly at mid-span. The first student's shortcut works perfectly here — not by coincidence, but by correct application of centroid symmetry. A is correct. B is wrong because there is no rule requiring integration over identical start/end points. Superposition of resultants is always valid; you simply track each resultant's location and magnitude separately before combining. C is the most tempting distractor. It claims the method only works by luck, but in reality the first student correctly identified the combined centroid through symmetry. If intensities were unequal, a careful student would recompute

\bar{x}

— but the method itself remains valid. D is wrong because there is no "correction factor" for loads meeting at a point. Distributed loads are additive across subregions with no penalty at shared boundaries. Study tip: Always ask two questions about any resultant: How big is it? (area under diagram) and Where does it act? (centroid). If you answer both correctly, combining partial loads before summing moments is always legitimate.

Question 8

Two engineers are designing a retaining wall that experiences a hydrostatic (triangular) pressure distribution on one face. The pressure varies linearly from $p = 0$ at the top ( $y = H$ ) to $p = \gamma H$ at the base ( $y = 0$ ), where $\gamma$ is the fluid specific weight and $H$ is the wall height. Engineer A replaces the pressure distribution with an equivalent resultant force for structural analysis. Engineer B argues that a distributed load can only be replaced by an equivalent resultant for global equilibrium checks (i.e., finding support reactions) and that doing so changes the internal stress distribution in the wall, making the replacement invalid for internal force analysis.

Which of the following statements most accurately evaluates Engineer B's claim?

Engineer B is correct: replacing a distributed load with an equivalent resultant is only valid for external reaction calculations. The internal shear and moment diagrams of a beam or wall depend on the actual load distribution, and using the resultant instead of the distributed load will yield incorrect internal forces at all cross-sections except possibly the endpoints.
Engineer B is partially correct: the equivalent resultant gives accurate internal forces only at the cross-section where the resultant is applied, but produces errors proportional to the eccentricity of the resultant from the centroid of each cross-section at all other locations along the wall.
Engineer B is incorrect for external reactions but correct for internal forces: the equivalent resultant preserves the net force and moment about any external point, so global equilibrium is maintained; however, the shear force and bending moment at any interior cross-section depend on the actual distributed load and will differ from those computed using the lumped resultant. (correct answer)
Engineer B is entirely incorrect: replacing a distributed load with its statically equivalent resultant is valid for both external reaction calculations and internal force analysis at every cross-section, because static equivalence preserves all force and moment resultants globally and locally throughout the member.

Explanation: Whenever you see a question about equivalent resultant forces, ask yourself: at what scale is the equivalence valid? Static equivalence means the resultant preserves the same net force and net moment as the original distribution — but only when evaluated over the entire system. Once you cut the member at an interior cross-section, you're no longer analyzing the whole system, and the equivalence breaks down. Here's why C is correct: For external reactions (support forces), you integrate the distributed load or use the resultant — the answers are identical, because both produce the same net force and moment about any external point. However, when you compute internal shear

V(y)

or bending moment

M(y)

at a cross-section, you must consider only the loads acting on one side of the cut. A triangular pressure distribution contributes continuously varying loads between the cut and the boundary. Replacing it with a single lumped force changes what you "see" on that sub-body, yielding incorrect shear and moment diagrams at interior locations. A overclaims by saying the replacement "will yield incorrect internal forces at all cross-sections except possibly the endpoints." In fact, it's correct at the endpoints and potentially consistent with global checks, but the specific error description misrepresents the mechanism. B introduces a false concept — "eccentricity from the centroid of each cross-section" — that is not how internal force errors from load-replacement actually arise. This answer conflates stress distribution with force/moment resultants. D is the classic trap: confusing global static equivalence with local equivalence. Static equivalence is not preserved at every interior cross-section — only across the structure as a whole. Study tip: Always ask whether you're analyzing the full structure (resultant is fine) or a cut sub-body (you need the actual load distribution). That distinction separates reaction problems from internal force problems.

Question 9

A beam of length $L$ is subjected to a distributed load whose intensity varies as $w(x) = w_0 \sin\!\left(\frac{\pi x}{L}\right)$ , where $x$ is measured from the left support and $w_0$ is the peak intensity at mid-span.

Which of the following correctly identifies both the magnitude of the equivalent resultant force and the location of its line of action measured from the left support?

$R = \frac{2w_0 L}{\pi}$ acting at $x = \frac{L}{2}$ , because the integral of $\sin$ over a full half-period equals $2/\pi$ times the amplitude times the base, and the load is symmetric about mid-span so the centroid falls at the midpoint. (correct answer)
$R = \frac{w_0 L}{2}$ acting at $x = \frac{L}{2}$ , because the average value of $\sin$ over $[0, L]$ is $\frac{1}{2}$ , giving a resultant equal to the average intensity times the length, and symmetry places the action point at mid-span.
$R = \frac{2w_0 L}{\pi}$ acting at $x = \frac{L}{\pi}$ , because the centroid of a sine arch is located at $L/\pi$ from the left end, analogous to how the centroid of a triangle is at one-third of the base from the heavy end.
$R = w_0 L$ acting at $x = \frac{L}{2}$ , because integrating $w_0 \sin(\pi x / L)$ over $[0, L]$ without applying the antiderivative scaling factor yields $w_0 L$ , and symmetry places the resultant at mid-span.

Explanation: When a distributed load varies continuously over a beam, you find its equivalent resultant by integrating the load function, and you locate the resultant by finding the centroid of the load diagram — exactly like finding the center of mass of a shape. For

w(x) = w_0 \sin\!\left(\frac{\pi x}{L}\right)

, the resultant magnitude is:

R = \int_0^L w_0 \sin\!\left(\frac{\pi x}{L}\right)dx = w_0\left[-\frac{L}{\pi}\cos\!\left(\frac{\pi x}{L}\right)\right]_0^L = w_0 \cdot \frac{L}{\pi}\left[-\cos(\pi)+\cos(0)\right] = \frac{2w_0 L}{\pi}

For the location, since

\sin(\pi x/L)

is perfectly symmetric about

x = L/2

, the centroid of the load diagram falls exactly at mid-span:

\bar{x} = L/2

. This makes answer A correct. Answer B is tempting but wrong — the average value of

\sin

over a half-period is

2/\pi \approx 0.637

, not

1/2

. Using

1/2

underestimates the resultant. Answer C gets the magnitude right but misplaces the centroid at

L/\pi \approx 0.318L

. The centroid of a symmetric arch is at its midpoint, not

L/\pi

. That value comes from confusing the antiderivative scaling factor with a geometric centroid formula. Answer D omits the

L/\pi

scaling factor that emerges from the chain rule when integrating

\sin(\pi x/L)

, inflating the resultant to

w_0 L

. A reliable strategy: always carry the antiderivative scaling factor through your integration, and immediately check whether the load shape has symmetry — if it does, the centroid is at the midpoint, saving you a moment-integral calculation.

Question 10

A beam extends from $x = 0$ to $x = 9 \text{ m}$ . A distributed load acts on the beam with intensity $w(x) = 2x \text{ kN/m}$ , where $x$ is in meters. The beam is pinned at $x = 0$ and has a roller at $x = 9 \text{ m}$ .

The equivalent resultant of this distributed load has magnitude $R$ and acts at location $\bar{x}$ from the pin. Using these values, what is the vertical reaction at the roller (at $x = 9 \text{ m}$ )?

$B_y = 54 \text{ kN}$ , found by computing $R = \int_0^9 2x\,dx = 81 \text{ kN}$ acting at $\bar{x} = \frac{\int_0^9 2x^2\,dx}{81} = 6 \text{ m}$ , then taking moments about the pin: $B_y = 81(6)/9 = 54 \text{ kN}$ . (correct answer)
$B_y = 40.5 \text{ kN}$ , found by treating the linearly varying load as equivalent to a uniform load of average intensity $w_{avg} = 2(9)/2 = 9 \text{ kN/m}$ , giving $R = 81 \text{ kN}$ acting at $\bar{x} = 4.5 \text{ m}$ (the midpoint), then $B_y = 81(4.5)/9 = 40.5 \text{ kN}$ .
$B_y = 60.75 \text{ kN}$ , found by computing the resultant as $R = \frac{1}{2}(w_{max})(L) = \frac{1}{2}(18)(9) = 81 \text{ kN}$ and placing it at $\bar{x} = \frac{2}{3}(9) = 6.75 \text{ m}$ from the pin (two-thirds from the zero end at $x=0$ ), giving $B_y = 81(6.75)/9 = 60.75 \text{ kN}$ .
$B_y = 27 \text{ kN}$ , found by computing the resultant as $R = \frac{1}{2}(18)(9) = 81 \text{ kN}$ acting at $\bar{x} = \frac{1}{3}(9) = 3 \text{ m}$ from the pin (one-third from the zero end, confusing the centroid rule for a triangle), giving $B_y = 81(3)/9 = 27 \text{ kN}$ .

Explanation: When a distributed load varies with position, you must find both the magnitude of the resultant and its location using integration — not geometric shortcuts or averages. The resultant magnitude is

R = \int_0^9 2x\,dx = [x^2]_0^9 = 81 \text{ kN}

, and its location is

\bar{x} = \frac{\int_0^9 x \cdot 2x\,dx}{R} = \frac{\int_0^9 2x^2\,dx}{81} = \frac{[2x^3/3]_0^9}{81} = \frac{486}{81} = 6 \text{ m}

. Taking moments about the pin gives

B_y(9) = 81(6)

, so

B_y = 54 \text{ kN}

. That's answer A, and the logic is exact. Answer B fails because it places the resultant at the midpoint

\bar{x} = 4.5 \text{ m}

. The midpoint is only correct for a uniform load. For a linearly increasing load, the centroid shifts toward the heavier end — you cannot use a simple average here. Answer C applies the two-thirds rule incorrectly. For a triangular load that starts at zero and increases, the centroid is located at

\frac{2}{3}L

from the zero-intensity end, which is indeed

x = 6 \text{ m}

— but answer C mistakenly uses

\frac{2}{3}(9) = 6.75 \text{ m}

, confusing where the "two-thirds" fraction is measured from. Answer D uses the one-third rule, placing the resultant at

3 \text{ m}

. This would apply if the load were largest at

x = 0

and tapered to zero — the opposite of what's given. Your study tip: always sketch the load profile and ask yourself which end is heavier. For

w(x) = 2x

, intensity grows with

x

, so the centroid must be past the midpoint — closer to

x = 9

. That sanity check alone eliminates B, C, and D.