10.3 Shear and Moment Diagrams

10.3 Shear and Moment Diagrams

The NCEES Strength of Materials specification lists "Shear and moment diagrams" as a distinct sub-topic, and for good reason: the bending stress formula $\sigma = Mc/I$ and the transverse shear formula $\tau = VQ/(It)$ both require $M$ and $V$ at the critical section, and finding that section is what the diagrams are for. On a CBT exam you rarely draw the full diagram — you extract one value — so learning the shortcuts matters more than learning to sketch.

Sign Conventions

Positive moment producing tension on the bottom is why reinforcing steel goes in the bottom of a simply supported concrete beam and in the top over a continuous support.

The Differential and Integral Relationships

These four relationships are the entire toolkit:

$$\frac{dV}{dx} = -w(x) \qquad\qquad \frac{dM}{dx} = V(x)$$ $$\Delta V = -\int w\,dx = -(\text{area under the load diagram})$$ $$\Delta M = \int V\,dx = (\text{area under the shear diagram})$$

Read them as a degree ladder. Each integration raises the polynomial degree by one:

This table alone answers many exam items. "What is the shape of the moment diagram for a uniformly loaded simple beam?" — parabolic, without any calculation.

The Zero-Shear Rule

Since $dM/dx = V$, the moment is stationary where $V = 0$:

The maximum bending moment occurs where the shear diagram crosses zero (or at a support, or at a point of applied couple).

This is the single most useful fact in the topic. Rather than writing $M(x)$ and differentiating, you find where $V = 0$ and take moments there.

Discontinuities

Trap: an applied couple leaves the shear diagram completely unaffected while stepping the moment diagram vertically. Candidates who put a jump in the shear diagram at an applied moment corrupt everything downstream.

Standard Results Worth Memorizing

Note the factor-of-four difference between a simply supported beam under a central point load ($PL/4$) and a cantilever with the same load at its tip ($PL$). Support conditions matter more than load magnitude.

Worked Example 1: Simple Span with Uniform and Point Loads

A simply supported beam spans $L = 8.0$ m with a uniform load $w = 12$ kN/m over the full span plus a 40 kN point load at 3.0 m from the left support $A$. Find $V_{\max}$, the location of zero shear, and $M_{\max}$.

Reactions. Total uniform load $= 12(8.0) = 96$ kN at midspan (4.0 m).

$$\sum M_A = 0: \quad R_B(8.0) = 96(4.0) + 40(3.0) = 384 + 120 = 504$$ $$R_B = 63.0\ \text{kN}, \qquad R_A = 96 + 40 - 63.0 = 73.0\ \text{kN}$$

Shear at key points (from the left):

$$V(0^+) = +73.0\ \text{kN}$$ $$V(3.0^-) = 73.0 - 12(3.0) = 73.0 - 36.0 = +37.0\ \text{kN}$$ $$V(3.0^+) = 37.0 - 40 = -3.0\ \text{kN} \quad\text{(jump of 40 kN at the point load)}$$

Shear crosses zero within $0 < x < 3.0$ m, since $V$ goes from $+73.0$ to $+37.0$ and only then jumps negative. It does not cross zero before 3.0 m — it jumps through zero at the point load. So the zero crossing is at $x = 3.0$ m, at the point load itself.

$$M_{\max} = M(3.0) = 73.0(3.0) - 12(3.0)\left(\frac{3.0}{2}\right) = 219.0 - 54.0 = \boxed{165.0\ \text{kN}\cdot\text{m}}$$

Verification from the right end, which must give the same value:

$$M(3.0) = 63.0(5.0) - 12(5.0)(2.5) = 315.0 - 150.0 = 165.0\ \text{kN}\cdot\text{m}\ ✓$$

$V_{\max} = 73.0$ kN at support $A$.

Worked Example 2: Locating Zero Shear Inside a Span

A simply supported beam spans 6.0 m with a load increasing linearly from zero at $A$ to 18 kN/m at $B$. Find $M_{\max}$ and its location.

Total load $= \tfrac{1}{2}(18)(6.0) = 54$ kN, acting at $\tfrac{2}{3}(6.0) = 4.0$ m from $A$.

$$\sum M_A = 0: \quad R_B(6.0) = 54(4.0) = 216 \;\Rightarrow\; R_B = 36.0\ \text{kN}, \quad R_A = 18.0\ \text{kN}$$

The load intensity at distance $x$ is $w(x) = 3x$ kN/m, so the load carried up to $x$ is $\tfrac{1}{2}(3x)(x) = 1.5x^2$:

$$V(x) = 18.0 - 1.5x^2$$

Set $V = 0$:

$$x = \sqrt{\frac{18.0}{1.5}} = \sqrt{12} = 3.464\ \text{m}$$ $$M_{\max} = 18.0(3.464) - 1.5(3.464)^2\left(\frac{3.464}{3}\right) = 62.35 - 1.5(12.0)(1.1547) = 62.35 - 20.78 = \boxed{41.6\ \text{kN}\cdot\text{m}}$$

Note where the maximum sits: at 3.46 m, past midspan and toward the heavier end — exactly where the shear diagram's parabola crosses zero. Assuming the maximum moment is always at midspan gives $M(3.0) = 54.0 - 13.5 = 40.5$ kN·m, which is 2.6% low and offered as a distractor. For non-symmetric loading, always locate zero shear rather than assuming midspan.