Statics and Dynamics Quiz: Internal Forces Via Section Cuts

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Internal Forces Via Section CutsQuestion 1 of 4

A beam of length $L$ is simply supported at both ends. It carries a single downward point load $P$ at distance $a$ from the left support ( $a < L/2$ , so the load is in the left half). A student makes a section cut at distance $x$ from the left support, where $a < x < L$ .

Which expression correctly gives the bending moment $M(x)$ in the region $a < x < L$ using the left free-body diagram, and what is the physical significance of the result being linear in $x$ ?

M(x) = \dfrac{P(L-a)}{L}\,x - P(x-a)

; this is linear in

x

because there are no distributed loads in this region, so the shear force — which equals

dM/dx

— is constant between the point load and the right support.

M(x) = \dfrac{Pa}{L}(L-x)

; this is linear in

x

because the right reaction

R_B

alone governs this region when the right FBD is used, and

R_B

is a fixed value that creates a moment proportional to the remaining distance

(L-x)

to the right support.

M(x) = \dfrac{P(L-a)}{L}\,x

; this is linear in

x

because no external loads act between A and the cut when

x < a

, meaning the moment grows proportionally with distance from A — a formula valid throughout the entire left half of the beam.

M(x) = \dfrac{Pa}{L}\,x - P(x-a)

; this is linear in

x

because the only contribution to the moment comes from the reaction at A, and using

R_A = Pa/L

(the smaller reaction) reflects the fact that the load is closer to A than to B.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Internal Forces Via Section Cuts

Practice Internal Forces Via Section Cuts 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 Internal Forces Via Section Cuts, 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

Which expression correctly gives the bending moment $M(x)$ in the region $a < x < L$ using the left free-body diagram, and what is the physical significance of the result being linear in $x$ ?

$M(x) = \dfrac{P(L-a)}{L}\,x - P(x-a)$ ; this is linear in $x$ because there are no distributed loads in this region, so the shear force — which equals $dM/dx$ — is constant between the point load and the right support. (correct answer)
$M(x) = \dfrac{Pa}{L}(L-x)$ ; this is linear in $x$ because the right reaction $R_B$ alone governs this region when the right FBD is used, and $R_B$ is a fixed value that creates a moment proportional to the remaining distance $(L-x)$ to the right support.
$M(x) = \dfrac{P(L-a)}{L}\,x$ ; this is linear in $x$ because no external loads act between A and the cut when $x < a$ , meaning the moment grows proportionally with distance from A — a formula valid throughout the entire left half of the beam.
$M(x) = \dfrac{Pa}{L}\,x - P(x-a)$ ; this is linear in $x$ because the only contribution to the moment comes from the reaction at A, and using $R_A = Pa/L$ (the smaller reaction) reflects the fact that the load is closer to A than to B.

Explanation: When analyzing internal bending moments using the method of sections, your first step is always to identify which external forces act on your chosen free-body diagram (FBD), then sum moments about the cut face. For the region

a < x < L

, the left FBD includes two forces: the left reaction

R_A = \frac{P(L-a)}{L}

and the point load

P

at position

a

. Summing moments about the cut (taking counterclockwise as positive):

M(x) = R_A \cdot x - P(x - a) = \frac{P(L-a)}{L}\,x - P(x-a)

This is answer A. Expanding and simplifying confirms the expression is linear in

x

— specifically because no distributed loads act in this region. Since shear force

V = dM/dx

is constant wherever no distributed loads exist, the moment must change at a constant rate, i.e., linearly. Answer B arrives at a numerically equivalent expression (you can verify algebraically), but its reasoning is misleading — it frames the result as coming from the right FBD and

R_B

alone, which is a valid method but doesn't explain linearity the way the question asks. More importantly, the explanation misattributes why linearity occurs. Answer C uses the formula for the region

x < a

(before the point load) and incorrectly claims it applies throughout the left half — a region-boundary error that ignores the discontinuity introduced by

P

x = a

. Answer D uses

R_A = \frac{Pa}{L}

, which is actually

R_B

, the right reaction. This is a classic trap: remembering which reaction is "larger" without checking the moment equilibrium carefully. Study tip: Always re-derive reactions from scratch using

\sum M = 0

. Memorizing which reaction is "larger" leads to sign and labeling errors — the algebra will catch your mistakes if you set it up correctly.

Question 2

A beam is loaded such that the bending moment diagram (BMD) is known: it is zero at both simply supported ends, rises linearly to $+M_1$ at $x = a$ , then drops linearly to $-M_2$ at $x = L - b$ , then returns to zero at $x = L$ . Here $M_1 > 0$ (sagging) and $M_2 > 0$ (the beam hogs in the right region). The shear force diagram (SFD) has already been sketched.

At the location $x = a$ where the bending moment reaches its local maximum $M_1$ , a student claims: 'The shear force must be zero at $x = a$ because this is a maximum of the BMD.' Under what condition is this claim incorrect, and what is the correct statement?

The claim is incorrect when a concentrated moment (couple) is applied at $x = a$ . In that case, the BMD has a jump discontinuity at $x = a$ rather than a smooth peak, so the condition $dM/dx = V = 0$ does not apply at a jump; the shear can be nonzero at $x = a$ even though the moment appears to peak.
The claim is incorrect when a point load is applied at $x = a$ . At a concentrated force, the shear force is discontinuous (jumps), so $dM/dx$ is undefined at that point. The moment may still reach a local maximum at $x = a$ even though the shear is nonzero on both sides of the cut; the peak occurs because the shear changes sign at the load point. (correct answer)
The claim is always incorrect: a maximum of the BMD implies that $dM/dx$ changes from positive to negative, which means $V$ changes sign from positive to negative, but this does not require $V = 0$ at $x = a$ — the shear can jump through zero at a point load without equaling zero at that exact location.
The claim is incorrect only when $M_1 = M_2$ , because in that symmetric case the shear diagram is antisymmetric about midspan and the zero-shear point does not coincide with the moment peak; instead, the moment is maximum where the shear changes sign between two equal loads at symmetric positions.

Explanation: Whenever you encounter a question linking shear force and bending moment, anchor your thinking in the fundamental relationship

V = dM/dx

. This tells you that shear is the slope of the moment diagram — but this relationship assumes a smooth, differentiable moment curve. That assumption breaks down at specific loading conditions. At a point (concentrated) load, the shear force jumps instantaneously. Because

V

is discontinuous there,

dM/dx

is undefined at that exact point. The moment diagram still has a kink — a sharp peak — at

x = a

, meaning the moment reaches a local maximum. However, the shear is nonzero on both sides of that location (positive just to the left, negative just to the right, or vice versa). The peak occurs not because

V = 0

, but because

V

changes sign across the concentrated load. This is precisely why B is correct: the student's claim fails at a point load, where the shear jumps through zero without ever equaling zero at the exact cut location. A is tempting but describes the wrong loading. A concentrated moment causes a jump in the bending moment diagram, not a smooth peak — so the scenario described in the passage (a linear rise to

M_1

) doesn't fit that case. C overstates things by calling the claim "always incorrect." In a distributed-load beam with no point loads at

x = a

V = dM/dx = 0

genuinely holds at a smooth moment maximum, so the claim can be correct. D introduces a symmetry argument that is irrelevant to the core question about the mathematical condition at

x = a

. Study tip: Always ask whether a moment peak is smooth (distributed load region →

V = 0

there) or a kink (point load →

V

jumps, undefined at that point). The shape of the BMD tells you which case applies.

Question 3

A propped cantilever beam (fixed at A, roller at B) of length $4\text{ m}$ is statically indeterminate. Using the compatibility method, the roller reaction is determined to be $R_B = \tfrac{3wL}{8} = 6\text{ kN}$ (upward) for a uniform downward load $w = 4\text{ kN/m}$ over the full span. The fixed-end reactions are then: $R_A = wL - R_B = 10\text{ kN}$ upward and $M_A = 8\text{ kN}\cdot\text{m}$ clockwise (i.e., the wall applies a clockwise moment on the beam at A).

Using a section cut at $x = 3\text{ m}$ from the fixed support A and the left free-body diagram, what is the internal bending moment $M$ at the cut? (Positive moment = sagging.)

$M = +12\text{ kN·m}$ , found from the left FBD as $M = R_A(3) - w(3)(1.5) = 30 - 18 = +12\text{ kN·m}$ , because the fixed-end moment $M_A$ acts on the wall, not on the beam, and therefore does not appear in the free-body diagram of the beam segment.
$M = +4\text{ kN·m}$ , found from the left FBD as $M = R_A(3) - M_A - w(3)(1.5) = 30 - 8 - 18 = +4\text{ kN·m}$ , confirming that the clockwise fixed-end moment acts on the left segment and reduces the net sagging moment. (correct answer)
$M = +22\text{ kN·m}$ , found from the left FBD as $M = R_A(3) + M_A - w(3)(1.5) = 30 + 8 - 18 = +20\text{ kN·m}$ — wait, recalculating: using $w(3)^2/2 = 18$ and adding $M_A$ as a sagging contribution gives $30 + 8 - 18 = +20\text{ kN·m}$ , because the clockwise fixed-end moment is on the same side as sagging.
$M = -4\text{ kN·m}$ , found from the left FBD as $M = R_A(3) - M_A - w(3)(1.5) = 30 - 8 - 18 = +4\text{ kN·m}$ ; however, since the beam is a propped cantilever it primarily hogs near the fixed end, so the sign must be reversed to give a hogging (negative) result.

Explanation: When analyzing internal forces using a free-body diagram of a cut segment, every external force and moment acting on that segment must appear in your equilibrium equation — no more, no less. For the left segment (0 to 3 m), three things act on it: the upward reaction

R_A = 10\text{ kN}

, the fixed-end moment

M_A = 8\text{ kN·m}

clockwise, and the distributed load

w = 4\text{ kN/m}

downward. Taking moments about the cut (positive = sagging, meaning counterclockwise from the left):

M = R_A(3) - M_A - w(3)(1.5) = 30 - 8 - 18 = +4\text{ kN·m}

The clockwise

M_A

on the left segment opposes sagging, so it subtracts. This confirms B is correct. A is wrong because it omits

M_A

entirely, claiming the fixed-end moment acts on the wall rather than the beam. This is backwards — the wall exerts

M_A

onto the beam at A, so it absolutely appears in any left-segment FBD. Dropping it overstates the moment by 8 kN·m. C contains a sign error, adding

M_A

as if it promotes sagging. A clockwise moment on the left face of a cut segment is hogging in tendency, not sagging — it must be subtracted, not added. D correctly computes

+4\text{ kN·m}

but then arbitrarily reverses the sign based on a general expectation about propped cantilevers. Never override a correctly derived equilibrium result with intuition — let the math speak. Study tip: Always draw the FBD first and mark every reaction (forces and moments) acting on the cut segment. Sign errors almost always come from either omitting a reaction or misidentifying its rotational direction relative to your positive-moment convention.

Question 4

A simply supported beam of span $L = 8\text{ m}$ carries two equal point loads $P = 30\text{ kN}$ each, placed symmetrically at $x = 2\text{ m}$ and $x = 6\text{ m}$ from the left support. A student is asked to find the internal forces at $x = 4\text{ m}$ (midspan).

A classmate argues: 'By symmetry, the shear force at midspan must be zero, so I only need to compute the bending moment.' A second classmate counters: 'The shear force is zero only if the section cut is not at the location of a point load; since $x=4\text{ m}$ is between the two loads, the shear is indeed zero there, but the bending moment must be computed carefully.' Which of the following statements about the internal forces at $x = 4\text{ m}$ is correct?

Both classmates are partially correct: shear is zero at midspan by symmetry, and the bending moment is $M = +60\text{ kN·m}$ , computed from the left FBD as $M = R_A(4) - P(2) = 120 - 60 = +60\text{ kN·m}$ . (correct answer)
The first classmate is wrong: symmetry implies zero shear only at the exact midpoint of a symmetric loading, but a section cut at $x=4\text{ m}$ is between two loads so the shear is $V = +30\text{ kN}$ (the unbalanced load from the left). The bending moment is $M = +90\text{ kN·m}$ .
Both classmates are correct: shear is zero at midspan, and the bending moment equals $M = R_A(4) = 30(4) = +120\text{ kN·m}$ , because the two point loads are outside the left free-body diagram when the cut is made exactly at midspan.
The second classmate's caution is misplaced: since the cut is not at a load point, the shear force is zero and the bending moment is $M = +80\text{ kN·m}$ , found by using the right FBD and summing moments about the cut from $R_B$ and the load at $x=6\text{ m}$ .

Explanation: When analyzing internal forces on a symmetric beam, always start by finding reactions, then draw a free-body diagram (FBD) to the left or right of your cut. Here, symmetry gives equal reactions:

R_A = R_B = \frac{2(30)}{2} = 30\text{ kN}

. The two loads sit at

x = 2\text{ m}

and

x = 6\text{ m}

, so the cut at

x = 4\text{ m}

falls between them — no load acts exactly at the cut. For the left FBD (from

x = 0

x = 4\text{ m}

), the forces present are

R_A = 30\text{ kN}

upward and

P = 30\text{ kN}

downward at

x = 2\text{ m}

. Summing vertical forces:

V = 30 - 30 = 0\text{ kN}

. This confirms the shear is zero — consistent with symmetry, and also consistent with the second classmate's point that the cut isn't at a load location. Summing moments about the cut:

M = R_A(4) - P(2) = 30(4) - 30(2) = 120 - 60 = +60\text{ kN·m}

. That makes A correct. Choice B is wrong because it claims

V = +30\text{ kN}

, forgetting to subtract the load at

x = 2\text{ m}

that's already inside the left FBD. Choice C incorrectly excludes the load at

x = 2\text{ m}

from the left FBD, which is inside the cut region and must be included — giving an inflated moment of

+120\text{ kN·m}

. Choice D arrives at

+80\text{ kN·m}

through a miscalculation using the right FBD; correctly applied, the right FBD gives

M = 30(4) - 30(2) = +60\text{ kN·m}

as well. Always sketch your FBD explicitly and list every force between the support and the cut — it's the single most common source of errors on internal-force problems.