Statics and Dynamics Quiz: Shear And Bending Moment Diagrams

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Shear And Bending Moment DiagramsQuestion 1 of 3

A simply supported beam (pin at $A$ , roller at $B$ ) has span $L$ . It is subjected to a uniformly distributed load $w$ over its entire span and a concentrated load $P$ at the quarter-span point ( $x = L/4$ from $A$ ). A student constructs the shear force diagram and identifies what they believe is the location of maximum bending moment.

The student claims: 'The maximum bending moment always occurs at the quarter-span point where the concentrated load is applied, because the shear diagram must cross zero at a concentrated load.' Which of the following statements best evaluates this claim?

The claim is correct: whenever a concentrated load is present, the shear diagram crosses zero at that load's position, making it the location of maximum bending moment regardless of the magnitudes of

P

and

w

The claim is incorrect: the shear diagram experiences a jump discontinuity at the concentrated load, but the zero crossing of the shear (and hence the maximum moment) occurs at a different location that depends on the relative magnitudes of

P

and

w

, and may or may not coincide with

x = L/4

The claim is incorrect: for a beam with both a uniform load and a concentrated load, the shear diagram is always continuous and has exactly one zero crossing at the midspan

x = L/2

, making the maximum moment always occur at midspan regardless of load positions.

The claim is partially correct: the shear diagram does jump at

x = L/4

, and if the jump is large enough to change the shear sign, the maximum moment is at the quarter point; however, if the shear does not change sign there, the maximum moment must be at

x = L/2

by symmetry of the distributed load.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Shear And Bending Moment Diagrams

Practice Shear And Bending Moment Diagrams 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 Shear And Bending Moment Diagrams, 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

The claim is correct: whenever a concentrated load is present, the shear diagram crosses zero at that load's position, making it the location of maximum bending moment regardless of the magnitudes of $P$ and $w$ .
The claim is incorrect: the shear diagram experiences a jump discontinuity at the concentrated load, but the zero crossing of the shear (and hence the maximum moment) occurs at a different location that depends on the relative magnitudes of $P$ and $w$ , and may or may not coincide with $x = L/4$ . (correct answer)
The claim is incorrect: for a beam with both a uniform load and a concentrated load, the shear diagram is always continuous and has exactly one zero crossing at the midspan $x = L/2$ , making the maximum moment always occur at midspan regardless of load positions.
The claim is partially correct: the shear diagram does jump at $x = L/4$ , and if the jump is large enough to change the shear sign, the maximum moment is at the quarter point; however, if the shear does not change sign there, the maximum moment must be at $x = L/2$ by symmetry of the distributed load.

Explanation: When analyzing beams with mixed loading, your goal is always to find where the shear force equals zero — because that's where the bending moment reaches its maximum (or minimum). The key distinction is between a jump in the shear diagram and a zero crossing. At a concentrated load, the shear diagram experiences a sudden jump discontinuity equal in magnitude to

P

. Before the jump, shear might be positive; after, it could be positive, negative, or exactly zero — depending entirely on the relative magnitudes of

P

and

w

, and the support reactions. The reactions themselves depend on both loads, so the actual zero crossing of shear (and thus the location of maximum moment) must be determined by calculation for each specific case. This is why B is correct: the zero crossing may or may not land at

x = L/4

, and its location is governed by the numbers, not the mere presence of a concentrated load. A commits the student's exact error — it treats the jump discontinuity as synonymous with a zero crossing. These are different things. A jump changes the shear value, but the shear only equals zero at one specific location determined by equilibrium. C is wrong on two counts: the shear diagram is not continuous when a concentrated load is present (it has a jump), and midspan is only the zero crossing when loading is symmetric — which it isn't here since

P

is at

L/4

. D sounds reasonable but is flawed because it assumes the fallback zero crossing must be at

x = L/2

. With asymmetric loading, there's no symmetry argument to anchor the zero crossing at midspan. Study tip: Always draw the full shear diagram using calculated reactions before identifying the maximum moment location — never assume it coincides with a load application point.

Question 2

For a beam in static equilibrium, which of the following statements correctly describes the necessary and sufficient conditions for a local maximum or minimum in the bending moment diagram at an interior point of the beam (not at a support or free end)?

The bending moment is maximum or minimum wherever the distributed load intensity $w(x)$ equals zero, because the second derivative of the moment equals the load, so a zero load implies a stationary point in the moment diagram.
The bending moment is maximum or minimum wherever the shear force $V = 0$ , provided the shear force changes sign at that point, because $dM/dx = V$ means a zero shear with sign reversal is both necessary and sufficient for a local extremum in the moment diagram. (correct answer)
The bending moment is maximum or minimum wherever the shear force $V = 0$ , because $dM/dx = V$ and a zero derivative is the necessary and sufficient condition for an extremum at any interior point, regardless of whether the shear changes sign.
The bending moment is maximum or minimum wherever the curvature of the elastic curve changes sign, because the moment is proportional to curvature via $M = EI\kappa$ , and an inflection point in deflection corresponds to an extremum in the moment diagram.

Explanation: Whenever you see a question about extrema in bending moment diagrams, anchor your thinking to the fundamental differential relationship between shear and moment:

dM/dx = V

. This single equation is the key to every part of this problem. Because

dM/dx = V

, a local maximum or minimum in

M

requires

V = 0

— that's basic calculus. However, a zero derivative alone doesn't guarantee an extremum; it could also be an inflection point. You need the derivative to change sign, which here means the shear force must change sign at that zero crossing. When

V

goes from positive to negative (or vice versa),

M

transitions from increasing to decreasing (or vice versa), confirming a true local extremum. This makes B correct:

V = 0

with a sign change is both necessary and sufficient for a local extremum in the moment diagram. A is wrong because

d^2M/dx^2 = -w(x)

, so

w = 0

affects the curvature of the moment diagram, not its slope. Zero load doesn't make

V = 0

; it just means shear isn't changing at that point. C contains a subtle but critical flaw:

V = 0

alone is necessary but not sufficient. If shear touches zero without changing sign (like a tangent point),

M

has an inflection point, not an extremum. The sign change requirement is essential. D confuses inflection points with extrema. Where curvature changes sign,

M = 0

(or changes sign), not where

M

is maximum or minimum — exactly the opposite relationship. Study tip: Remember the chain

w \to V \to M

: each is the derivative of the next. For extrema in

M

, always check that

V = 0

and changes sign — both conditions together, never just one.

Question 3

A simply supported beam (span $L = 12\text{ m}$ , pin at $A$ , roller at $B$ ) carries a triangular distributed load that has zero intensity at $A$ and maximum intensity $w_0 = 9\text{ kN/m}$ at $B$ . The load intensity at position $x$ (measured from $A$ ) is $w(x) = \frac{w_0 x}{L} = \frac{3x}{4}\text{ kN/m}$ .

Find the location $x_{max}$ (measured from $A$ ) where the bending moment is maximum, and determine the maximum bending moment $M_{max}$ .

$x_{max} = 6.0\text{ m}$ (midspan) and $M_{max} = 81\text{ kN·m}$ , because the triangular load's centroid at $2L/3$ from $A$ creates a symmetric-like condition that places the maximum moment at midspan.
$x_{max} = 4\sqrt{3} \approx 6.93\text{ m}$ from $A$ and $M_{max} \approx 83.1\text{ kN·m}$ , found by setting the shear expression $V(x) = R_A - \frac{w_0 x^2}{2L} = 0$ and computing the moment at that location. (correct answer)
$x_{max} = 8.0\text{ m}$ from $A$ and $M_{max} = 72\text{ kN·m}$ , found by locating the centroid of the triangular load at $2L/3 = 8\text{ m}$ from $A$ and treating that point as the zero-shear location.
$x_{max} = 4\sqrt{3} \approx 6.93\text{ m}$ from $A$ and $M_{max} \approx 72.0\text{ kN·m}$ , found by setting the shear force equal to zero and integrating the moment, but using the total load resultant's moment arm rather than integrating the parabolic shear diagram.

Explanation: Whenever you see a beam with a non-uniform distributed load, the key is remembering that maximum bending moment occurs where shear force equals zero — not necessarily at midspan or at the load's centroid. Start by finding the reactions. The total load resultant is

\frac{1}{2}w_0 L = \frac{1}{2}(9)(12) = 54\text{ kN}

, acting at

\frac{2L}{3} = 8\text{ m}

from

A

. Taking moments about

B

R_A = \frac{54 \times (12-8)}{12} = 18\text{ kN}

. The shear at position

x

is the reaction minus the load accumulated from

A

x

V(x) = R_A - \int_0^x \frac{w_0 t}{L}\,dt = 18 - \frac{9x^2}{2(12)} = 18 - \frac{3x^2}{8}

. Setting

V(x)=0

x^2 = 48

, so

x_{max} = 4\sqrt{3} \approx 6.93\text{ m}

. The moment there is

M = 18(6.93) - \frac{3(6.93)^3}{24} \approx 124.7 - 41.6 \approx 83.1\text{ kN·m}

. This confirms B is correct. Choice A incorrectly assumes symmetry places the maximum moment at midspan — triangular loads are inherently asymmetric, and midspan is not the zero-shear point. Choice C confuses the centroid location (

2L/3 = 8\text{ m}

) with the zero-shear location; the centroid is used only to find the resultant force, not to locate maximum moment. Choice D finds the correct

x_{max}

but then improperly computes the moment using the resultant's moment arm instead of integrating the actual moment expression, producing the wrong

M_{max}

. Your study tip: always write

V(x)

explicitly by integrating the load, set it to zero to find

x_{max}

, then integrate

V(x)

to get

M(x_{max})

. Never shortcut by using the centroid for both steps.