Internal Forces via Section Cuts - Statics and Dynamics

1. 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.)
   Answer: $M = +4\text{ kN}\cdot\text{m}$, found from the left FBD as $M = R_A(3) - M_A - w(3)(1.5) = 30 - 8 - 18 = +4\text{ kN}\cdot\text{m}$, confirming that the clockwise fixed-end moment acts on the left segment and reduces the net sagging moment. 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 $M_A = 8\text{ kN}\cdot\text{m}$ clockwise, and the distributed load $w = 4\text{ kN/m}$ downward. Taking moments about the cut (positive = sagging, meaning counterclockwise from the left): $M = R_A(3) - M_A - w(3)(1.5) = 30 - 8 - 18 = +4\text{ kN}\cdot\text{m}$ 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 $+4\text{ kN}\cdot\text{m}$ 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.
2. 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?
   Answer: 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. 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.
3. 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 $x$?
   Answer: $M(x) = \dfrac{P(L-a)}{L}\,x - P(x-a)$; 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. 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 $P$ at $x = a$. 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.
4. 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?
   Answer: Both classmates are partially correct: shear is zero at midspan by symmetry, and the bending moment is $M = +60\text{ kN}\cdot\text{m}$, computed from the left FBD as $M = R_A(4) - P(2) = 120 - 60 = +60\text{ kN}\cdot\text{m}$. 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 $x = 0$ to $x = 4\text{ m}$), 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: $M = R_A(4) - P(2) = 30(4) - 30(2) = 120 - 60 = +60\text{ kN}\cdot\text{m}$. 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 $+120\text{ kN}\cdot\text{m}$. Choice D arrives at $+80\text{ kN}\cdot\text{m}$ through a miscalculation using the right FBD; correctly applied, the right FBD gives $M = 30(4) - 30(2) = +60\text{ kN}\cdot\text{m}$ 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.

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