Shear & Bending Moment Diagrams - Statics and Dynamics

1. Find the location $x_{max}$ (measured from $A$) where the bending moment is maximum, and determine the maximum bending moment $M_{max}$.
   Answer: $x_{max} = 4\sqrt{3} \approx 6.93\text{ m}$ from $A$ and $M_{max} \approx 83.1\text{ kN}\cdot\text{m}$, found by setting the shear expression $V(x) = R_A - \frac{w_0 x^2}{2L} = 0$ and computing the moment at that location. 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 $A$ to $x$: $V(x) = R_A - \int_0^x \frac{w_0 t}{L}\,dt = 18 - \frac{9x^2}{2(12)} = 18 - \frac{3x^2}{8}$. Setting $V(x)=0$: $x^2 = 48$, so $x_{max} = 4\sqrt{3} \approx 6.93\text{ m}$. The moment there is $M = 18(6.93) - \frac{3(6.93)^3}{24} \approx 124.7 - 41.6 \approx 83.1\text{ kN}\cdot\text{m}$. 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.
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)?
   Answer: 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. 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.
3. 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?
   Answer: 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 $x = L/4$. 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.

Start this quiz