Statics and Dynamics Quiz: Load Shear Moment Relationships

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Load Shear Moment RelationshipsQuestion 1 of 14

A cantilever beam of length $L$ is fixed at $x = 0$ (left end) and free at $x = L$ (right end). It carries a uniformly distributed load $w_0$ (downward, positive by the convention $dV/dx = -w$ ) over its entire length. The free-end boundary conditions are $V(L) = 0$ and $M(L) = 0$ .

Working from the free end toward the fixed end using $dV/dx = -w_0$ and $dM/dx = V$ , what is the bending moment at the fixed support $x = 0$ ?

M(0) = -w_0 L^2 / 2

, obtained by integrating

V(x) = -w_0(L-x)

using

dM/dx = V

with the boundary condition

M(L) = 0

, yielding a negative (hogging) parabolic moment that reaches its peak magnitude at the wall.

M(0) = +w_0 L^2 / 2

, because the fixed end must supply an upward moment reaction to the beam; integrating the shear from the free end inward produces a positive internal moment at the wall under the sign convention used.

M(0) = -w_0 L^2 / 6

, because the triangular shear diagram has its resultant located at

L/3

from the free end, giving a net moment arm of

L/3

and a fixed-end moment of

-w_0 L \cdot L/3 = -w_0 L^2/3

, which is then halved to account for parabolic weighting.

M(0) = +w_0 L^2 / 8

, because the standard maximum-moment formula for a uniformly loaded beam is

w_0 L^2/8

; the cantilever boundary conditions are accounted for by applying this formula directly to the full span.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Load Shear Moment Relationships

Practice Load Shear Moment Relationships 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 Load Shear Moment Relationships, 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.

All questions

Question 1

Working from the free end toward the fixed end using $dV/dx = -w_0$ and $dM/dx = V$ , what is the bending moment at the fixed support $x = 0$ ?

$M(0) = -w_0 L^2 / 2$ , obtained by integrating $V(x) = -w_0(L-x)$ using $dM/dx = V$ with the boundary condition $M(L) = 0$ , yielding a negative (hogging) parabolic moment that reaches its peak magnitude at the wall. (correct answer)
$M(0) = +w_0 L^2 / 2$ , because the fixed end must supply an upward moment reaction to the beam; integrating the shear from the free end inward produces a positive internal moment at the wall under the sign convention used.
$M(0) = -w_0 L^2 / 6$ , because the triangular shear diagram has its resultant located at $L/3$ from the free end, giving a net moment arm of $L/3$ and a fixed-end moment of $-w_0 L \cdot L/3 = -w_0 L^2/3$ , which is then halved to account for parabolic weighting.
$M(0) = +w_0 L^2 / 8$ , because the standard maximum-moment formula for a uniformly loaded beam is $w_0 L^2/8$ ; the cantilever boundary conditions are accounted for by applying this formula directly to the full span.

Explanation: When analyzing a cantilever beam with a distributed load, the key is working systematically from the known boundary conditions (free end) toward the unknown reactions (fixed end), integrating the load-shear-moment relationships step by step. Starting at the free end where

V(L) = 0

and

M(L) = 0

, integrate

dV/dx = -w_0

to get

V(x) = -w_0(x - L) = -w_0(L - x)

. Notice this satisfies

V(L) = 0

correctly. Then integrate

dM/dx = V(x) = -w_0(L-x)

with

M(L) = 0

M(x) = \int -w_0(L-x)\,dx = -w_0\left(Lx - \frac{x^2}{2}\right) + C

Applying

M(L) = 0

gives

C = w_0 L^2/2

, so

M(x) = -w_0 Lx + \frac{w_0 x^2}{2} + \frac{w_0 L^2}{2}

. At

x = 0

M(0) = \frac{w_0 L^2}{2} \cdot(-1) \cdot (-1)

... simplifying directly:

M(0) = -w_0 L^2/2

. This negative (hogging) result is physically correct — the wall pulls the top fiber in tension, which is the hallmark of cantilever bending. Answer A is correct. Answer B incorrectly assigns a positive sign, confusing the reaction moment the wall exerts on the structure with the internal bending moment at the section, which is negative under this convention. Answer C misapplies the centroid argument — the resultant of a uniform (not triangular) load acts at

L/2

, giving

w_0 L \cdot L/2

, not

L/3

. The extra halving is fabricated. Answer D applies the simply-supported midspan formula

w_0 L^2/8

to a cantilever, which is a completely different boundary condition producing a fundamentally different moment diagram. Your study tip: always track your sign convention explicitly. Integrate from the known end, confirm boundary conditions at each step, and remember that negative internal moment in a cantilever means hogging — it's physically expected, not a calculation error.

Question 2

A simply supported beam of length $L$ carries a linearly varying distributed load that increases from $w = 0$ at the left support to $w = w_0$ at the right support. The left reaction is $R_A = w_0 L/6$ and the right reaction is $R_B = w_0 L/3$ .

Using the relationship $dV/dx = -w(x)$ , where $w(x) = w_0 x / L$ , which of the following correctly describes the shear diagram for this beam?

The shear diagram is a straight line with negative slope, since $w$ is constant and $dV/dx$ is therefore constant and negative throughout the span.
The shear diagram is a downward-opening parabola, since $dV/dx = -w_0 x/L$ is linear in $x$ , causing $V$ to vary as a quadratic function that starts at $+w_0 L/6$ , curves downward, and ends at $-w_0 L/3$ . (correct answer)
The shear diagram is an upward-opening parabola, since the load increases from left to right and the beam therefore stores increasing strain energy, causing the shear to rise parabolically toward midspan before falling.
The shear diagram is a cubic curve, since integrating the linearly varying load once yields a quadratic shear and integrating again yields a cubic moment; the shear is therefore cubic rather than quadratic across the span.

Explanation: Whenever you see a question linking a distributed load to a shear diagram, your instinct should be to apply the differential relationship

dV/dx = -w(x)

directly. The shape of the shear diagram is determined by integrating the negative load function — so identifying the type of function

w(x)

is the first step. Here, the load varies linearly:

w(x) = w_0 x / L

. Substituting into the relationship gives

dV/dx = -w_0 x / L

, which is linear in

x

. Integrating once yields

V(x) = R_A - \frac{w_0 x^2}{2L} = \frac{w_0 L}{6} - \frac{w_0 x^2}{2L}

, a quadratic (parabolic) function. At

x = 0

V = +w_0 L/6

; at

x = L

V = w_0 L/6 - w_0 L/2 = -w_0 L/3

, matching

-R_B

. Because the coefficient of

x^2

is negative, the parabola opens downward. This confirms B. A is wrong because it assumes

w

is constant, which would only be true for a uniformly distributed load. Here

w

is linear, so

dV/dx

is not constant and the shear is not a straight line. C is wrong in both its conclusion and its reasoning. "Strain energy storage" does not govern the shape of a shear diagram — that is not a valid statics principle, making this a classic distractor built on plausible-sounding but irrelevant physics. D confuses shear with moment. Integrating the linear load once gives a quadratic shear; integrating again gives a cubic moment. The shear is quadratic, not cubic. Your study tip: always count integrations. One integration of

w(x)

gives

V(x)

; a second gives

M(x)

. The degree increases by one at each step.

Question 3

A simply supported beam of span $L$ carries two equal and opposite concentrated moments (couples) $M_0$ applied at $x = L/4$ and $-M_0$ applied at $x = 3L/4$ , with no other loads. The reactions at both supports are zero.

Using $dM/dx = V$ , what is the shear force in the region $L/4 < x < 3L/4$ ?

$V = 0$ throughout the region $L/4 < x < 3L/4$ , since the reactions are zero and no transverse forces are applied anywhere on the beam, making the shear identically zero over the entire span.
$V = +2M_0/L$ throughout the region $L/4 < x < 3L/4$ , because the moment diagram jumps by $M_0$ at $x = L/4$ and by $-M_0$ at $x = 3L/4$ , creating a constant positive slope $dM/dx = V$ between the two couples.
$V$ varies linearly from $+M_0/L$ at $x = L/4$ to $-M_0/L$ at $x = 3L/4$ , because applied couples create a linearly distributed equivalent shear between their points of application, consistent with $dM/dx = V$ .
$V = -2M_0/L$ throughout the region $L/4 < x < 3L/4$ , because the moment decreases from $M_0$ to $0$ over the half-span $L/2$ between the two couples, giving $dM/dx = -M_0/(L/2) = -2M_0/L = V$ . (correct answer)

Explanation: When analyzing beams loaded only by concentrated couples (no transverse forces), the key relationship to lean on is

dM/dx = V

. Since no transverse loads act anywhere, the shear force must be constant in any region between load discontinuities — but "no transverse forces" does not mean shear is zero everywhere. Concentrated moments cause jumps in the bending moment diagram, and those jumps define the shear between them. Here's the reasoning for the correct answer, D: In the outer regions (

0 < x < L/4

and

3L/4 < x < L

), reactions are zero and no loads act, so

M = 0

throughout both outer zones. At

x = L/4

, the applied moment

M_0

causes the bending moment to jump up to

+M_0

. At

x = 3L/4

, the moment

-M_0

pulls it back to

0

. So between the two couples,

M

drops from

+M_0

0

over a distance of

L/2

. Applying

V = dM/dx

V = \frac{0 - M_0}{L/2} = -\frac{2M_0}{L}

. This constant negative shear is consistent with no transverse loads in that region (constant

M

slope = constant

V

). A is tempting but wrong — zero reactions don't imply zero shear. Shear can exist internally even without support reactions. B gets the magnitude right but the sign wrong; the moment decreases from

M_0

0

, giving a negative slope, not positive. C is a fabricated concept — couples don't produce linearly varying shear; shear is constant between point loads/couples. Your strategy: always sketch the moment diagram first, then differentiate it to find shear. Sign discipline on

dM/dx

is where most errors occur.

Question 4

The shear diagram crosses zero from positive to negative. The bending-moment diagram there is:

Local minimum
Local maximum (correct answer)
Inflection point
Jump discontinuity

Explanation: The slope of the bending-moment diagram equals the shear. When the shear crosses from positive to negative, the moment's slope changes from rising to falling, so the moment has a horizontal tangent and is at a local maximum. The tempting wrong choice is inflection point, but that requires a peak or valley in the shear diagram, not simply crossing zero.

Question 5

A beam segment carries a downward distributed load that increases linearly to the right. The bending-moment diagram is:

Quadratic curve
Straight line
Cubic curve (correct answer)
Horizontal line

Explanation: Integrate the linearly increasing load once to get shear, which is quadratic; integrate again to get bending moment, so it is cubic. The tempting wrong answer is quadratic because that is the shear diagram, not the moment diagram. Since the slope of the moment diagram equals shear, a quadratic shear produces a cubic moment.

Question 6

A downward concentrated force P acts on a beam. What is true of the bending-moment diagram at that section?

Has a kink, no jump (correct answer)
Jumps downward by P
Jumps upward by P
Is a parabola locally

Explanation: At a concentrated load, shear force jumps by P, and since the slope of the bending moment equals shear, the moment diagram's slope changes suddenly at that section. That creates a kink while the moment itself stays continuous, so there is no jump. The tempting jump-by-P answer describes the shear diagram, not the bending moment.

Question 7

At a concentrated couple M0, the shear diagram:

Jumps by M0
Drops to zero
Changes slope
Does not change (correct answer)

Explanation: A concentrated couple M0 changes the bending moment, not the shear force. Since shear is the slope of the moment diagram, a sudden jump in moment has no corresponding jump in shear; the shear diagram remains continuous across the point. The tempting wrong answer is 'jumps by M0,' which confuses the moment jump with shear.

Question 8

For a beam under uniform downward load, the slope of the shear diagram is:

Constant positive
Decreasing linearly
Zero everywhere
Constant negative (correct answer)

Explanation: For a beam, the slope of the shear diagram equals the negative of the distributed load. A uniform downward load is a constant negative value, so the slope of the shear diagram is constant negative. The tempting error is 'decreasing linearly', but that describes a linearly varying load, not a uniform one.

Question 9

A beam's bending moment diagram shows a region where the moment is increasing (positive slope) and the beam is known to be loaded only by a distributed load (no point forces in this region). Which of the following is the most complete and accurate conclusion that can be drawn solely from the relationships $dV/dx = -w$ and $dM/dx = V$ ?

The shear force $V$ must be positive in this region, and the distributed load $w$ must be negative (upward), since an increasing moment requires positive shear by $dM/dx = V > 0$ , and positive shear can only be sustained against a distributed load if the load acts upward.
The shear force $V$ must be positive in this region (since $dM/dx = V > 0$ ), but no conclusion about the sign of $w$ can be drawn from this alone, because $V$ may be positive yet increasing, decreasing, or constant depending on the magnitude and direction of $w$ . (correct answer)
The distributed load $w$ must be upward (negative) throughout the region, and the shear $V$ must be both positive and increasing, because an increasing moment requires $V > 0$ and a decreasing load intensity is necessary for the moment to continue increasing.
The shear force $V$ must be positive in this region, and the distributed load $w$ must be downward (positive), since a positive and increasing shear is required to sustain an increasing moment, and downward loads drive shear in the positive direction by $dV/dx = -w$ .

Explanation: Whenever you see a bending moment diagram problem tied to the beam equilibrium relationships, your job is to trace the logic carefully through each equation — and stop at exactly what the math tells you, nothing more. Here, the moment is increasing, so

dM/dx > 0

. Since

dM/dx = V

, you can immediately conclude that

V > 0

in this region. That's a solid, direct conclusion. Now ask: what does that tell you about

w

? The relationship

dV/dx = -w

connects the rate of change of shear to the load — not the sign of shear to the load. A positive

V

is entirely compatible with any sign of

w

: if

w > 0

(downward), shear decreases; if

w < 0

(upward), shear increases; if

w = 0

, shear is constant. In all three cases,

V

can remain positive. Therefore, B is correct — you can conclude

V > 0

, but the sign of

w

is indeterminate from this information alone. Choice A incorrectly claims that positive shear requires an upward load. This confuses the sign of

V

with the sign of

dV/dx

. Choice C makes an even broader error, asserting that

V

must also be increasing — but

dM/dx > 0

only requires

V > 0

, not

dV/dx > 0

. Choice D similarly conflates a positive shear with a need for downward loading, again misapplying

dV/dx = -w

. The key study tip: distinguish between the value of a quantity and its rate of change. These relationships chain together, but each equation only constrains exactly what it says — nothing else travels for free.

Question 10

A beam's shear diagram shows a region where $V(x)$ is a positive constant $V_0$ over an interval $[a, b]$ . Immediately to the right of $x = b$ , the shear diagram jumps discontinuously downward by an amount $P$ . Which of the following correctly identifies the loading conditions in and at the boundaries of this interval?

The interval $[a,b]$ carries no distributed load (since $dV/dx = -w = 0$ implies $w = 0$ ), and a concentrated downward point force of magnitude $P$ is applied at $x = b$ , producing the discontinuous drop in shear. (correct answer)
The interval $[a,b]$ carries a uniform upward distributed load that maintains constant shear, and a concentrated upward point force $P$ at $x = b$ causes the downward jump as the beam transitions to a new equilibrium state.
The interval $[a,b]$ carries no distributed load, and a concentrated upward point force of magnitude $P$ is applied at $x = b$ , because a jump upward in the load diagram always corresponds to a jump downward in the shear diagram.
The interval $[a,b]$ must carry a linearly increasing distributed load to sustain constant shear, and the jump at $b$ represents a concentrated moment rather than a point force, since moments cause discontinuities in shear diagrams.

Explanation: When analyzing shear and moment diagrams, your foundational tool is the differential relationship

dV/dx = -w(x)

, where

w(x)

is the distributed load intensity (positive upward). This equation tells you everything about how the shear diagram's slope connects to the loading. Over the interval

[a, b]

, the shear is constant at

V_0

, meaning

dV/dx = 0

. Substituting into the governing equation gives

-w = 0

, so

w = 0

— no distributed load exists on that interval. At

x = b

, the shear drops discontinuously by

P

. A sudden jump in the shear diagram is the signature of a concentrated point force: specifically, a downward force of magnitude

P

causes a downward jump of

P

in the shear diagram. This makes A correct. B is wrong on two counts: a uniform upward distributed load would produce a rising slope in the shear diagram (not constant shear), and an upward point force would cause an upward jump, not a downward one. C correctly identifies that no distributed load exists, but then reverses the sign convention. A concentrated upward force of magnitude

P

would produce an upward jump in shear — not the downward drop described. The direction matters. D confuses what causes what. A linearly increasing load would produce a linearly changing shear (not constant), and concentrated moments cause discontinuities in the moment diagram, not the shear diagram. Study tip: Memorize these two rules cold —

dV/dx = -w

governs slope, and a point force

P

downward produces a downward jump of magnitude

P

in shear. Sign errors and slope-vs-jump confusion are the most common traps on these questions.

Question 11

A beam is loaded such that its bending moment diagram is a perfect sine curve: $M(x) = M_0 \sin(\pi x / L)$ for $0 \leq x \leq L$ , with $M(0) = M(L) = 0$ .

Using $dM/dx = V$ and $dV/dx = -w$ , what distributed load intensity $w(x)$ is required to produce this moment distribution?

$w(x) = \dfrac{M_0 \pi^2}{L^2} \cos\!\left(\dfrac{\pi x}{L}\right)$ , obtained by differentiating $M$ twice under the incorrect belief that the second derivative of $\sin(\pi x/L)$ is proportional to $\cos(\pi x/L)$ .
$w(x) = \dfrac{M_0 \pi}{L} \cos\!\left(\dfrac{\pi x}{L}\right)$ , obtained by differentiating $M$ once to get $V$ and equating $w = V$ , since the distributed load is assumed to equal the shear force directly.
$w(x) = -\dfrac{M_0 \pi^2}{L^2} \sin\!\left(\dfrac{\pi x}{L}\right)$ , obtained by computing $d^2M/dx^2$ without negating, on the assumption that $w = +d^2M/dx^2$ for downward-positive loading.
$w(x) = \dfrac{M_0 \pi^2}{L^2} \sin\!\left(\dfrac{\pi x}{L}\right)$ , obtained by differentiating $M$ twice and applying $w = -d^2M/dx^2$ . (correct answer)

Explanation: Whenever a beam problem gives you a moment distribution and asks for the loading, your roadmap comes from the differential relationships:

dM/dx = V

and

dV/dx = -w

. Combining these gives

w = -d^2M/dx^2

. That negative sign is critical — it reflects the sign convention where downward load is positive. Starting with

M(x) = M_0 \sin(\pi x / L)

, differentiate once:

V = dM/dx = \frac{M_0 \pi}{L}\cos(\pi x/L)

. Differentiate again:

dV/dx = -\frac{M_0 \pi^2}{L^2}\sin(\pi x/L)

. Since

w = -dV/dx

, you get

w(x) = \frac{M_0 \pi^2}{L^2}\sin(\pi x/L)

, which is answer D — positive, sinusoidal, and physically sensible (downward load over the span). Each distractor reflects a specific, common mistake. A applies the second derivative incorrectly, claiming

d^2[\sin(\pi x/L)]/dx^2

is proportional to cosine — it isn't; the second derivative of sine is negative sine. B stops after one differentiation and sets

w = V

, confusing the shear force with the distributed load; these are related by a derivative, not equality. C computes

d^2M/dx^2

correctly as

-\frac{M_0\pi^2}{L^2}\sin(\pi x/L)

but then sets

w = +d^2M/dx^2

, forgetting the required negation — getting the right magnitude but the wrong sign. Your study tip: memorize the chain

w \to V \to M

as integration going left, and differentiation going right, with the sign flip

w = -d^2M/dx^2

. When working backwards from

M

, always differentiate twice and flip the sign.

Question 12

A beam's shear diagram has a local maximum (positive peak) at interior point $x = c$ . A student claims this means the bending moment must also have a local maximum at $x = c$ . A second student disagrees and says the moment has an inflection point at $x = c$ instead. Which student, if either, is correct, and why?

Neither student is correct. A local maximum of $V$ at $x = c$ means $dV/dx = 0$ there, implying $w(c) = 0$ by $dV/dx = -w$ . This tells us only about the load at that point and is insufficient to classify the moment's behavior without knowing $M$ 's value.
The first student is correct. Because $dM/dx = V$ and $V$ is at a positive local maximum at $x = c$ , the moment is increasing at its fastest rate there; a maximum rate of increase defines a local maximum of the function itself.
The second student is correct. A local maximum of $V$ at $x = c$ requires $dV/dx = 0$ , so $d^2M/dx^2 = dV/dx = 0$ at $x = c$ . A zero second derivative of $M$ , provided it changes sign there, is the condition for an inflection point of $M$ . (correct answer)
The first student is correct. Since $V(c) > 0$ , we have $dM/dx = V(c) > 0$ , confirming $M$ is increasing at $x = c$ ; because the shear is at its peak value, the moment's rate of increase is greatest there, which constitutes a local maximum of $M$ .

Explanation: Whenever you see a question linking shear and moment diagrams, your anchor should be the two fundamental differential relationships:

dM/dx = V

and

dV/dx = -w

. These chain together beautifully — the load governs shear, and shear governs moment. Here's the key insight for this problem. If

V

has a local maximum at

x = c

, then by definition

dV/dx = 0

at that point. Now chain the relationships: since

d^2M/dx^2 = dV/dx

, you immediately get

d^2M/dx^2 = 0

x = c

. A zero second derivative is precisely the signature of an inflection point of

M

— provided the second derivative changes sign on either side of

c

(which it does when

V

genuinely peaks and then decreases). The moment is neither maximized nor minimized there; it's changing concavity. This confirms C is correct. A is wrong because it correctly identifies that

w(c) = 0

, but then retreats unnecessarily — the chain of derivatives actually gives us enough information to classify

M

's behavior. The reasoning is incomplete, not the math. B and D share the same critical flaw: they confuse the rate of change of

M

with the value of

M

. Yes,

dM/dx = V(c) > 0

, meaning

M

is increasing at

x = c

— but a maximum rate of increase is an inflection point of

M

, not a local maximum. A local maximum of

M

requires

dM/dx = V = 0

, not

V > 0

. Study tip: Remember that local extrema of

M

occur where

V = 0

, and inflection points of

M

occur where

V

is itself at an extremum. These are a favorite exam trap precisely because students conflate the function's peak with its derivative's peak.

Question 13

A beam segment carries no distributed load ( $w = 0$ ) and no concentrated forces or couples within the segment. The shear at the left end of the segment is $V_L = -8$ kN. Which of the following statements about this segment is correct?

The shear varies linearly from $-8$ kN at the left end to some value at the right end depending on the span length, because $dV/dx = -w = 0$ means $V$ is constant only when the load is upward.
The shear is $-8$ kN throughout the segment, and the moment is constant throughout, since a negative shear indicates a region of decreasing moment that eventually reaches a constant value.
The shear is $-8$ kN throughout the segment, the moment varies linearly with negative slope ( $dM/dx = -8$ kN), and the segment cannot contain a point of zero moment unless the moment at the left end is positive. (correct answer)
The shear is $-8$ kN throughout the segment, the moment varies linearly with positive slope ( $dM/dx = +8$ kN), since the sign convention requires $dM/dx = -V$ for downward-positive shear definitions.

Explanation: When analyzing a beam segment with no distributed load, your starting point is always the differential equilibrium relationships:

dV/dx = -w

and

dM/dx = V

. When

w = 0

, the first equation tells you shear is constant across the entire segment — no exceptions, no conditions about load direction. Since

V = -8

kN everywhere in the segment, the moment equation becomes

dM/dx = V = -8

kN, meaning moment decreases linearly as you move rightward. Whether the segment actually contains a zero-moment point depends entirely on the moment value at the left end — specifically, the moment reaches zero only if

M_L > 0

, since the function

M(x) = M_L - 8x

crosses zero only when

M_L

is positive. This is exactly what C states, making it the correct answer. A is wrong because

dV/dx = 0

always means constant shear — the direction of the load is irrelevant. Shear doesn't "vary linearly" when

w = 0

; that happens when

w \neq 0

. B contains a subtle trap: while it correctly identifies constant shear, it wrongly concludes the moment is also constant. A nonzero shear always produces a changing moment (

dM/dx = -8 \neq 0

). The reasoning about "negative shear causing decreasing moment" is garbled — it confuses the sign of the slope with whether a slope exists. D flips the sign of the slope, claiming

dM/dx = +8

kN. This would only be true if

V = +8

kN. The standard convention gives

dM/dx = V

, not

-V

. Your study tip: memorize

dV/dx = -w

and

dM/dx = V

as a pair. On exam day, zero load → constant shear → linearly changing moment with slope equal to that shear value.

Question 14

A beam segment is in a region where the bending moment $M(x)$ is a positive constant (i.e., $dM/dx = 0$ ) and the moment does not equal zero. Which of the following must be true about this segment?

The shear force $V$ is zero throughout the segment, and the distributed load $w$ is also zero throughout the segment, since $dM/dx = V = 0$ and $dV/dx = -w = 0$ . (correct answer)
The shear force $V$ is zero throughout the segment, but the distributed load $w$ may be nonzero, since only the moment equation $dM/dx = V$ can be applied here and the load-shear relationship is independent.
The distributed load $w$ is zero throughout the segment, but the shear force $V$ could be any nonzero constant, since a constant moment is consistent with any constant shear as long as the beam is in equilibrium.
The shear force $V$ must equal the magnitude of the applied distributed load at each cross-section, since equilibrium of an infinitesimal element requires $V = w \cdot \Delta x$ when the moment is constant.

Explanation: Whenever you encounter a question about bending moments and shear forces, your first instinct should be to reach for the two fundamental differential relationships that govern beam behavior:

dM/dx = V

and

dV/dx = -w

. These two equations form a chain — moment connects to shear, and shear connects to distributed load — and that chain is the key to this problem. If the bending moment is a positive constant, then by definition

dM/dx = 0

. Substituting into the first relationship gives

V = 0

throughout the segment. Now apply the second relationship: since

V = 0

everywhere in the segment,

dV/dx = 0

as well, which means

-w = 0

, so

w = 0

. Both conclusions follow directly from the same chain of equations, making A correct. The fact that

M \neq 0

is irrelevant — the value of the moment doesn't affect these derivative relationships, only its rate of change does. B is wrong because it treats the two differential equations as independent. They are not — once you establish

V = 0

, the second equation immediately constrains

w

as well. C is wrong because it claims a constant moment is consistent with any constant shear. This directly contradicts

dM/dx = V

: if

dM/dx = 0

, then

V

must be zero, not "any nonzero constant." D is wrong because it invents a false equilibrium condition. The relationship

V = w \cdot \Delta x

is dimensionally inconsistent and has no basis in the standard beam equations. A useful memory anchor: think of the differential equations as a domino chain — a zero derivative in

M

topples the shear to zero, which then topples the load to zero.