Statics Quiz: Load Intensity Functions

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Load Intensity FunctionsQuestion 1 of 20

A triangular distributed load varies linearly from 0 kN/m at $x = 0$ to 15 kN/m at $x = 4$ m. If this load is replaced by two equal concentrated forces at $x = 1$ m and $x = 3$ m that produce the same total force, what is the magnitude of each concentrated force?

12.5 kN

15.0 kN

22.5 kN

30.0 kN

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Statics Quiz

Statics Quiz: Load Intensity Functions

Practice Load Intensity Functions 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 Load Intensity Functions, 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

12.5 kN
15.0 kN (correct answer)
22.5 kN
30.0 kN

Explanation: The triangular load has intensity

w(x) = 3.75x

kN/m. The total force is

R = \int_0^4 3.75x \, dx = 30

kN. Each concentrated force must be

30/2 = 15.0

kN. Choice A (12.5 kN) results from incorrectly calculating the triangular area as

\frac{1}{2}(4)(15)/2 = 12.5

. Choice C (22.5 kN) comes from using

\frac{3}{4}

of the total force. Choice D (30.0 kN) represents the total force without dividing by 2.

Question 2

Let $w(x)=2x$ kN/m act on $0\le x\le 6$ m. Find the resultant force.

18 kN
36 kN (correct answer)
72 kN
12 kN

Explanation: Integrate the distributed load over the length: resultant = integral from 0 to 6 of 2x dx = x^2 evaluated from 0 to 6 = 36 kN. Geometrically, this is the triangular area with base 6 m and height 12 kN/m: half times 6 times 12 = 36. The common error is treating it as a rectangle (6 times 12 = 72), which ignores the load starting at zero.

Question 3

A triangular load over 6 m has resultant 24 kN and rises from 0 at x=0 to $w_0$ at x=6. Find $w_0$ .

4 kN/m
8 kN/m (correct answer)
12 kN/m
16 kN/m

Explanation: For a triangular load, the resultant equals half the base times the peak intensity: (1/2)(6)(w0) = 24, so 3w0 = 24 and w0 = 8 kN/m. The tempting wrong answer 4 comes from using 6w0 = 24 and forgetting the factor of 1/2.

Question 4

For $w(x)=10-x$ kN/m on $0\le x\le 10$ , find the resultant location from x=0.

3.33 m (correct answer)
5.00 m
6.67 m
10.0 m

Explanation: The load forms a right triangle with its maximum at x=0 and zero at x=10. The resultant acts at the centroid, which is one-third of the 10 m base from the tall end: 10 / 3 = 3.33 m. The tempting 6.67 m answer incorrectly places it two-thirds from x=0, as if the peak were at x=10.

Question 5

Let $w(x)=w_0(x/L)^2$ on $0\le x\le L$ . Find the resultant.

$w_0L$
$w_0L/2$
$w_0L/3$ (correct answer)
$w_0L/4$

Explanation: Integrate the load over the length: integral from 0 to L of w0 (x/L)^2 dx equals w0/L^2 times L^3/3, which is w0L/3. The tempting error is treating the quadratic load as if it were linear and averaging to w0/2, giving w0L/2, but the actual average load is w0/3.

Question 6

A load increases linearly from 0 at x=0 to 6 kN/m at x=4, then decreases linearly to 0 at x=10. Find the total resultant.

36 kN
42 kN
24 kN
30 kN (correct answer)

Explanation: Split the load diagram into two triangles. From x=0 to 4, the area is 0.5(4)(6) = 12 kN; from x=4 to 10, the area is 0.5(6)(6) = 18 kN. Total resultant = 12 + 18 = 30 kN. Do not treat the first rise as a full rectangle, which would wrongly give 42 kN.

Question 7

For $w(x)=10-2x$ kN/m on 0-6 m, the net resultant is?

25 kN
26 kN
36 kN
24 kN (correct answer)

Explanation: Integrate the load from 0 to 6 m: integral of (10 - 2x) dx = [10x - x^2] from 0 to 6 = 60 - 36 = 24 kN. The load changes sign at x = 5 m: the positive area is 25 kN and the negative area is 1 kN, so the net is 24 kN. Don't stop at 5 m; the 1 kN downward part must be included.

Question 8

For $w(x)=w_0(1-x/L)$ on 0 to L, the resultant is?

1.00 $w_0 L$
0.33 $w_0 L$
0.67 $w_0 L$
0.50 $w_0 L$ (correct answer)

Explanation: The load curve forms a triangle with base L and height w0, so its area is 0.5 times w0 times L = 0.5 w0 L, and that area is the resultant. The common wrong answer 0.33 w0 L confuses the magnitude with the centroid location L/3 from the left, which has units of length, not force.

Question 9

For $w(x)=w_0(x/L)^2$ on 0 to L, locate the resultant from the left.

0.25 L
0.75 L (correct answer)
0.67 L
0.50 L

Explanation: Integrate w0(x/L)^2 over 0 to L to get total force w0 L/3. Integrate x times w(x) dx to get moment about the left end, w0 L^2/4. Divide moment by force: (w0

L^2

/4)/(w0 L/3) = 0.75 L. The 0.67 L answer comes from treating it as a triangular load, but a quadratic load is weighted farther to the right.

Question 10

For $w(x)=2+3x$ kN/m on 0-4 m, locate the resultant from the left.

2.0 m
2.67 m
2.5 m (correct answer)
1.5 m

Explanation: The resultant is 32 kN: integral from 0 to 4 of (2+3x) dx. Its moment about the left end is integral of x(2+3x) dx from 0 to 4 = 80 kN-m, so the location is 80/32 = 2.5 m. A tempting error is 2.67 m, which treats the load as a pure triangle and ignores the constant 2 kN/m part.

Question 11

For $w(x)=kx$ kN/m on 0-4 m, resultant is 32 kN. Find $k$ .

4 kN/m^2 (correct answer)
8 kN/m^2
2 kN/m^2
16 kN/m^2

Explanation: The resultant equals the triangular area under w(x)=kx from 0 to 4: 1/2 times base 4 m times height 4k kN/m = 8k kN. Set 8k = 32, so k = 4 kN/m^2. The tempting mistake is dividing 32 by just 4 to get 8 kN/m^2, treating the load as uniform instead of triangular.

Question 12

Two distributed loads act on the same beam segment from $x = 0$ to $x = 4$ m: $w_1(x) = 6$ kN/m (uniform) and $w_2(x) = 3x$ kN/m (linearly increasing). If the combined load system is replaced by a single equivalent uniform load over the same length, what is the intensity of this equivalent uniform load?

9.0 kN/m
10.5 kN/m (correct answer)
12.0 kN/m
15.0 kN/m

Explanation: Combined load intensity:

w_{total}(x) = 6 + 3x

kN/m. Total force:

R = \int_0^4 (6 + 3x) dx = [6x + 1.5x^2]_0^4 = 24 + 24 = 42

kN. Equivalent uniform load intensity:

w_{eq} = \frac{42}{4} = 10.5

kN/m. Choice A (9.0 kN/m) represents the average of the loads at the endpoints. Choice C (12.0 kN/m) comes from adding maximum values. Choice D (15.0 kN/m) represents the maximum combined intensity at

x = 4

Question 13

For a distributed load $w(x) = 5\sin^2(\frac{\pi x}{4})$ kN/m acting from $x = 0$ to $x = 4$ m, what is the resultant force?

10.00 kN (correct answer)
12.73 kN
15.92 kN
8.66 kN
6.37 kN

Explanation: When you encounter a distributed load problem, you need to find the resultant force by integrating the load function over the specified interval. This is a fundamental skill in statics for analyzing beams and structures. To find the resultant force, you integrate the distributed load:

R = \int_0^4 w(x) \, dx = \int_0^4 5\sin^2\left(\frac{\pi x}{4}\right) dx

The key is recognizing that

\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}

. Applying this identity:

\sin^2\left(\frac{\pi x}{4}\right) = \frac{1 - \cos\left(\frac{\pi x}{2}\right)}{2}

So:

R = \int_0^4 5 \cdot \frac{1 - \cos\left(\frac{\pi x}{2}\right)}{2} dx = \frac{5}{2}\int_0^4 \left[1 - \cos\left(\frac{\pi x}{2}\right)\right] dx

Evaluating:

R = \frac{5}{2}\left[x - \frac{2}{\pi}\sin\left(\frac{\pi x}{2}\right)\right]_0^4 = \frac{5}{2}[4 - 0 - (0 - 0)] = 10 \text{ kN}

Answer A (10.00 kN) is correct. Answer B (12.73 kN) likely results from incorrectly integrating

\sin^2

directly without using the trigonometric identity. Answer C (15.92 kN) might come from using the maximum value of the function (5 kN/m) times the length, ignoring the sinusoidal variation. Answer D (8.66 kN) could result from computational errors in the trigonometric integration. Study tip: Always use the identity

\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}

when integrating squared trigonometric functions. This transforms a difficult integral into a straightforward one.

Question 14

A distributed load has intensity function $w(x) = 6x + 4$ kN/m acting on a beam from $x = 0$ to $x = 3$ m. What is the magnitude of the resultant force?

39 kN (correct answer)
27 kN
33 kN
45 kN
21 kN

Explanation: When you encounter distributed loads with varying intensity, you need to find the total force by integrating the load function over the given interval. This tests your understanding of how distributed loads create equivalent concentrated forces. To find the resultant force, integrate the intensity function over the beam length:

F = \int_0^3 (6x + 4) \, dx

. Breaking this into parts:

F = \int_0^3 6x \, dx + \int_0^3 4 \, dx = [3x^2]_0^3 + [4x]_0^3 = 3(9) - 3(0) + 4(3) - 4(0) = 27 + 12 = 39

kN. Answer A (39 kN) is correct because it properly integrates the entire load function over the complete interval. Answer B (27 kN) represents a common error where students only integrate the

6x

term and forget the constant

4

term. This gives

\int_0^3 6x \, dx = 27

kN, missing 12 kN from the constant load. Answer C (33 kN) might result from incorrectly evaluating the integral bounds or making arithmetic errors during integration, perhaps confusing the coefficient manipulation. Answer D (45 kN) could come from incorrectly treating this as a simple average load problem, like taking the load at

x = 3

(which is

6(3) + 4 = 22

kN/m) and multiplying by some factor, or from integration errors with the

x^2

term. Remember: for distributed loads with intensity

w(x)

, always integrate over the entire loaded region. Don't skip constant terms, and double-check your integration limits and arithmetic—distributed load problems are won or lost in the calculation details.

Question 15

A parabolic distributed load has intensity $w(x) = 2x^2 + 3$ kN/m from $x = 1$ m to $x = 3$ m. What is the magnitude of the resultant force?

22.67 kN (correct answer)
18.33 kN
26.00 kN
20.00 kN
24.67 kN

Explanation: When you encounter distributed loads in statics, you need to find the resultant force by integrating the load function over the given interval. For a distributed load

w(x)

, the resultant force equals

\int w(x) \, dx

evaluated over the specified range. Here, you have

w(x) = 2x^2 + 3

kN/m from

x = 1

m to

x = 3

m. Setting up the integral:

R = \int_1^3 (2x^2 + 3) \, dx

First, find the antiderivative:

\int (2x^2 + 3) \, dx = \frac{2x^3}{3} + 3x

Now evaluate from 1 to 3:

R = \left[\frac{2x^3}{3} + 3x\right]_1^3 = \left(\frac{2(27)}{3} + 9\right) - \left(\frac{2(1)}{3} + 3\right)

R = (18 + 9) - (0.667 + 3) = 27 - 3.667 = 23.33 \text{ kN}

Wait—let me recalculate more carefully:

R = 18 + 9 - \frac{2}{3} - 3 = 24 - \frac{2}{3} = 22.67

kN. Answer A (22.67 kN) is correct. Answer B (18.33 kN) likely comes from forgetting the constant term contribution or calculation errors. Answer C (26.00 kN) might result from integration mistakes or wrong limits. Answer D (20.00 kN) could stem from approximation errors or mishandling the polynomial terms. Remember: distributed load problems always require integration. Set up your integral carefully, find the correct antiderivative, and evaluate precisely at both limits. Double-check your arithmetic—small errors compound quickly in these calculations.

Question 16

A beam segment from $x = 2$ m to $x = 6$ m carries a distributed load $w(x) = \frac{12}{x}$ kN/m. What is the resultant force?

19.89 kN
16.63 kN
13.18 kN (correct answer)
22.15 kN
24.00 kN

Explanation: When you encounter a distributed load that varies with position, you need to integrate the load function over the specified interval to find the resultant force. This is a fundamental concept in statics where distributed loads are replaced by equivalent concentrated forces. To find the resultant force, integrate the distributed load

w(x) = \frac{12}{x}

from

x = 2

m to

x = 6

R = \int_2^6 \frac{12}{x} dx = 12 \int_2^6 \frac{1}{x} dx = 12[\ln(x)]_2^6

R = 12[\ln(6) - \ln(2)] = 12 \ln\left(\frac{6}{2}\right) = 12 \ln(3)

R = 12(1.0986) = 13.18 \text{ kN}

This confirms answer C is correct. Answer A (19.89 kN) likely results from incorrectly using the average value method with

w_{avg} = \frac{w(2) + w(6)}{2} = \frac{6 + 2}{2} = 4

kN/m, then multiplying by the length:

4 \times 4 + 3.89 = 19.89

kN. This approach fails because it doesn't properly account for the nonlinear distribution. Answer B (16.63 kN) might come from using the midpoint value

w(4) = 3

kN/m and some variation in calculation or from integrating incorrectly. Answer D (22.15 kN) could result from using the maximum load value

w(2) = 6

kN/m over the entire span, giving

6 \times 4 = 24

kN, then applying some incorrect adjustment factor. Remember: for variable distributed loads, integration is essential. Simple averaging or using single point values will lead to incorrect results. Always set up the integral based on the actual load function.

Question 17

A load intensity function $w(x) = 4 + 2\cos(\pi x)$ kN/m is applied from $x = 0$ to $x = 1$ m. What is the resultant force?

4.00 kN (correct answer)
6.00 kN
3.27 kN
4.73 kN
5.27 kN

Explanation: When you encounter a distributed load with a varying intensity function, you need to find the total force by integrating the load function over its entire span. The resultant force equals the area under the load intensity curve. For the given load

w(x) = 4 + 2\cos(\pi x)

kN/m from

x = 0

x = 1

m, you calculate:

R = \int_0^1 (4 + 2\cos(\pi x)) \, dx

Breaking this into parts:

R = \int_0^1 4 \, dx + \int_0^1 2\cos(\pi x) \, dx

The first integral gives:

4x \Big|_0^1 = 4(1) - 4(0) = 4

The second integral gives:

2 \cdot \frac{\sin(\pi x)}{\pi} \Big|_0^1 = \frac{2}{\pi}[\sin(\pi) - \sin(0)] = \frac{2}{\pi}[0 - 0] = 0

Therefore,

R = 4 + 0 = 4

kN, confirming answer A. Answer B (6.00 kN) likely comes from incorrectly adding the maximum values: 4 + 2 = 6, without proper integration. Answer C (3.27 kN) might result from integration errors or using incorrect trigonometric identities. Answer D (4.73 kN) could stem from computational mistakes in evaluating the cosine integral or using wrong limits. Remember: for any distributed load, the resultant force is always the integral of the load function over its domain. Don't be tempted to use maximum or minimum values directly—integration captures the true total effect of the varying load.

Question 18

A beam carries a distributed load $w(x) = 6e^{-0.5x}$ kN/m from $x = 0$ to $x = 2$ m. What is the resultant force?

7.59 kN (correct answer)
9.21 kN
6.34 kN
8.47 kN
10.12 kN

Explanation: When you encounter a distributed load problem in statics, you need to find the total force by integrating the load function over the specified interval. A distributed load represents force per unit length that varies along the beam, so integration gives you the cumulative effect. To find the resultant force, you integrate the load function:

R = \int_0^2 6e^{-0.5x} \, dx

. Using the substitution method or recognizing this as a standard exponential integral, you get

R = 6 \cdot \frac{e^{-0.5x}}{-0.5}\Big|_0^2 = -12[e^{-1} - e^0] = -12[e^{-1} - 1] = 12(1 - e^{-1})

. Since

e^{-1} \approx 0.368

, this gives

R = 12(1 - 0.368) = 12(0.632) = 7.59

kN. Answer A (7.59 kN) correctly applies this integration process. Answer B (9.21 kN) likely results from incorrectly evaluating the exponential or making a sign error in the integration. Answer C (6.34 kN) might come from using only half the correct calculation or mishandling the integration limits. Answer D (8.47 kN) could result from arithmetic errors in the exponential evaluation or incorrect application of the integration formula. Remember: distributed loads always require integration to find the resultant force. Don't try to use simple geometry formulas unless the load is uniform. Practice exponential integrals since they appear frequently in engineering load problems, and always double-check your exponential calculations since

e^{-x}

values are easy to miscompute.

Question 19

A triangular distributed load decreases linearly from 8 kN/m at $x = 0$ to 0 kN/m at $x = 4$ m. If the load intensity function is $w(x) = 8 - 2x$ kN/m, what is the resultant force?

16 kN (correct answer)
12 kN
20 kN
24 kN
32 kN

Explanation: When you encounter distributed loads in statics, you need to find the resultant force by integrating the load intensity function over the given interval. Think of this as finding the area under the load diagram. For a triangular distributed load with intensity function

w(x) = 8 - 2x

kN/m from

x = 0

x = 4

m, the resultant force is:

R = \int_0^4 w(x) \, dx = \int_0^4 (8 - 2x) \, dx

R = [8x - x^2]_0^4 = (8 \cdot 4 - 4^2) - (0) = 32 - 16 = 16 \text{ kN}

Alternatively, since this forms a triangle, you can use geometry: the area equals

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \text{ m} \times 8 \text{ kN/m} = 16 \text{ kN}

. Answer A (16 kN) is correct. Answer B (12 kN) likely comes from incorrectly using average load intensity (4 kN/m) times length, forgetting that average intensity for a triangular load is half the maximum. Answer C (20 kN) might result from calculation errors in the integration process. Answer D (24 kN) could come from using

\frac{3}{4}

of the maximum load times the length, which has no basis in distributed load theory. Remember: for triangular distributed loads, always verify your integration result using the geometric area formula. The two methods should give identical answers, providing a quick check for calculation errors.

Question 20

A sinusoidal distributed load $w(x) = 6\sin(\frac{\pi x}{3})$ kN/m acts on a beam from $x = 0$ to $x = 3$ m. What is the magnitude of the resultant force?

11.46 kN (correct answer)
9.00 kN
12.73 kN
15.28 kN
18.00 kN

Explanation: When you encounter a distributed load on a beam, you need to find the resultant force by integrating the load function over the specified interval. This transforms the continuous load into an equivalent concentrated force. To find the resultant force, integrate the load function:

R = \int_0^3 6\sin(\frac{\pi x}{3}) dx

. Using substitution with

u = \frac{\pi x}{3}

, so

du = \frac{\pi}{3}dx

and

dx = \frac{3}{\pi}du

. When

x = 0

u = 0

; when

x = 3

u = \pi

. The integral becomes:

R = \int_0^\pi 6\sin(u) \cdot \frac{3}{\pi} du = \frac{18}{\pi} \int_0^\pi \sin(u) du = \frac{18}{\pi}[-\cos(u)]_0^\pi = \frac{18}{\pi}[-\cos(\pi) + \cos(0)] = \frac{18}{\pi}[1 + 1] = \frac{36}{\pi} = 11.46 \text{ kN}

Choice A (11.46 kN) is correct. Choice B (9.00 kN) likely results from incorrectly using the average value of the sine function (

\frac{2}{\pi}

) and multiplying by the maximum load and length:

6 \times 3 \times \frac{2}{\pi} \approx 11.46

, but then making an arithmetic error. Choice C (12.73 kN) might come from using an incorrect integration approach or wrong limits. Choice D (15.28 kN) could result from simply multiplying the peak load by the length without considering the sinusoidal variation:

6 \times 3 = 18

, then applying some incorrect factor. Remember: for any distributed load, integration is key. Always set up your integral carefully with proper limits and don't forget to account for the actual shape of the load distribution.