Structural Analysis

A. Diagrams (shear and moment)

Shear & Moment Diagrams

Build shear and moment diagrams from the load-shear-moment relations, locate maximum moment and inflection points, and read the AISC Table 3-23 standard cases.

Almost every structural problem on the PE eventually asks one question: where is the moment largest, and how big is it? Shear and moment diagrams answer it directly, and they are the gateway to flexural design, deflection, and indeterminate analysis. The PE Civil Reference Handbook treats this in §4.1.7 'Moment, Shear, and Deflection Diagrams' (its AISC Table 3-23 reproduction) and gives the governing calculus in §1.6.7 'Beams'. Points are most often lost not on the concept but on bookkeeping: a sign flip, a kip-vs-lb slip, or reading

wL^2/8

as the moment when the span carries a point load instead of a UDL. This concept makes the construction mechanical and the standard cases instant.

The load-shear-moment relationships

Three differential relations tie the loading

w(x)

, the shear

V(x)

, and the moment

M(x)

together. Following the handbook §1.6.7 convention, the slope of the shear diagram at any point equals the distributed-load term

w(x)

there; the slope of the moment diagram equals the shear. Integrating, the change in shear between two sections equals the area of the load diagram, and the change in moment equals the area under the shear diagram. These four facts let you draw both diagrams without writing a single equation of a section cut. (A downward applied load enters as a negative

w

in this convention, so the shear still drops across a downward UDL — just keep the signs consistent with whichever load-direction sign you adopt.)

\frac{dV}{dx} = w(x), \qquad \frac{dM}{dx} = V(x)

Building the shear diagram

Start at the left end and walk rightward. A point load (or a reaction) causes a vertical jump in the shear equal to that force; an upward reaction jumps the shear up, a downward applied load jumps it down. Between concentrated forces the change in shear equals the area under the load diagram,

V_2 - V_1 = \int w\,dx

— exactly the positive form in handbook §1.6.7 and in this concept's key equations. Keeping the signed convention (a downward load is a negative

w

), a downward UDL therefore makes the shear slope downward, because

w

is negative; with no distributed load the shear is constant (a flat step). The diagram must return to zero at the far end — that closure is your free check on the reactions.

V_2 - V_1 = \int_{x_1}^{x_2} w(x)\,dx = (\text{area under load diagram})

Building the moment diagram and finding the maximum

The moment at any section equals the running sum of the area under the shear diagram from the left end. Because

dM/dx = V

, the moment is a maximum (or minimum) exactly where the shear passes through zero — this is the single most useful fact for hand calculations. Under a UDL, the shear varies linearly, so the moment is parabolic; under no load the moment is linear. A concentrated couple (applied moment) produces a vertical jump in the moment diagram but leaves the shear unchanged.

M_2 - M_1 = \int_{x_1}^{x_2} V(x)\,dx = (\text{area under shear diagram})

Inflection points and sign convention

An inflection point is where the bending moment passes through zero and changes sign — the curvature reverses from concave-up (sagging, positive) to concave-down (hogging, negative). These points matter because they mark where flexural reinforcement can be cut off and where a beam behaves as if pinned. Use the handbook §1.6.7 convention: positive moment puts the bottom fiber in tension (beam sags); positive shear means the portion to the right of the cut tends to slide down relative to the left.

The AISC Table 3-23 standard cases

Handbook §4.1.7 reproduces AISC Steel Construction Manual (15th ed.) Table 3-23: closed-form shear, moment, and deflection for dozens of common spans and loadings. Memorize a handful. A simply supported span with a full UDL gives

V_{max}=wL/2

at the supports and

M_{max}=wL^2/8

at midspan. A simply supported span with a central point load gives

V_{max}=P/2

and

M_{max}=PL/4

. A cantilever with a tip load

P

gives

M_{max}=PL

at the fixed end; with a full UDL it gives

M_{max}=wL^2/2

. On the exam these are read straight off the table — but you must match the support and load case exactly, since

wL^2/8

,

wL^2/12

(fixed-fixed), and

wL^2/2

(cantilever) are three different beams.

\begin{aligned} \text{SS, UDL:} &\quad M_{max}=\tfrac{wL^2}{8},\ V_{max}=\tfrac{wL}{2} \\ \text{SS, central } P:&\quad M_{max}=\tfrac{PL}{4},\ V_{max}=\tfrac{P}{2} \\ \text{Cantilever, tip } P:&\quad M_{max}=PL,\ V_{max}=P \end{aligned}

Overhangs and continuous beams

An overhanging end carries the cantilever moment

wL_o^2/2

back into the support as a negative (hogging) moment, which can govern over the positive span moment. The maximum positive moment still sits where the shear crosses zero inside the back-span. For continuous beams the support moments come from indeterminate analysis (moment distribution, covered separately), but once those end moments are known, each span is drawn exactly like a simple beam carrying its load plus the two known end couples.

Exam strategy

Three steps, every time: (1) compute reactions and verify the shear diagram closes to zero at the far end; (2) find where

V=0

— that is your

M_{max}

location; (3) sum the shear area up to that point. Keep units rigid:

M=wL^2/8

needs

w

in

\text{k/ft}

and

L

in

\text{ft}

to give

\text{kip-ft}

— if loads come in plf, convert or expect a 1000× error. When a span matches a Table 3-23 case exactly, read the answer off the table; when it does not (mixed loads, overhangs), build the diagrams by area. Always sketch, even roughly — the picture catches sign errors a formula will not.

Key equations

Load-shear relation

\dfrac{dV}{dx} = w(x)

Slope of the shear diagram equals the distributed-load term $w$ (force/length), per handbook §1.6.7. A downward applied load enters as a negative $w$ , so the shear drops across it. $w$ in $\text{k/ft}$ , $V$ in $\text{kip}$ .

Shear-moment relation

\dfrac{dM}{dx} = V(x)

Slope of the moment diagram equals the shear. Maximum moment occurs where $V=0$ . $M$ in $\text{kip-ft}$ .

Shear from load area

V_2 - V_1 = \int_{x_1}^{x_2}\! w\,dx

Change in shear between two sections equals the area of the load diagram (handbook §1.6.7). With a downward load taken as negative $w$ , the integral is negative, so the shear drops by the magnitude of that area.

Moment from shear area

M_2 - M_1 = \int_{x_1}^{x_2}\! V\,dx

Change in moment equals the area under the shear diagram between two sections.

Simply supported, full UDL

V_{max}=\dfrac{wL}{2},\quad M_{max}=\dfrac{wL^2}{8}

$M_{max}$ at midspan. $w$ in $\text{k/ft}$ , $L$ in $\text{ft}$ . AISC Table 3-23.

Simply supported, central point load

V_{max}=\dfrac{P}{2},\quad M_{max}=\dfrac{PL}{4}

$M_{max}$ at midspan under the load. $P$ in $\text{kip}$ .

Simply supported, off-center point load

M_{max}=\dfrac{P\,a\,b}{L}

Load $P$ at distance $a$ from left, $b$ from right, $a+b=L$ . Max moment under the load.

Cantilever, tip point load

V_{max}=P,\quad M_{max}=PL

Maximum (hogging) moment at the fixed end. $L$ = cantilever length.

Cantilever, full UDL

V_{max}=wL,\quad M_{max}=\dfrac{wL^2}{2}

Hogging at the fixed end. Note the factor is $1/2$ , not $1/8$ .

Overhang back-moment

M_{support}= -\dfrac{w\,L_o^{2}}{2}

Hogging moment delivered to the support by an overhang of length $L_o$ carrying UDL $w$ .

Worked examples

Simply supported beam under a uniform load

Problem. A simply supported beam spans $L = 24\,\text{ft}$ and carries a uniform load $w = 3\,\text{k/ft}$ over the full span (self-weight plus service load, already combined). Find the support reactions, the maximum shear, and the maximum moment, and state where each occurs.

Solution. By symmetry each reaction carries half the total load:

R_A = R_B = \dfrac{wL}{2} = \dfrac{3(24)}{2} = 36.0\,\text{kip}

. Maximum shear is at the supports and equals the reaction:

V_{max} = 36.0\,\text{kip}

. The shear varies linearly from

+36

at

A

to

-36

at

B

, crossing zero at midspan (

x = 12\,\text{ft}

). Maximum moment is where

V = 0

, i.e. midspan:

M_{max} = \dfrac{wL^2}{8} = \dfrac{3(24)^2}{8} = \dfrac{3(576)}{8} = 216\,\text{kip-ft}

. Sanity check: the moment equals the area of the shear triangle from

A

to midspan,

\tfrac12 (36\,\text{kip})(12\,\text{ft}) = 216\,\text{kip-ft}

. Units:

\text{k/ft}\times\text{ft}^2 = \text{kip-ft}

. Both methods agree:

\boxed{216\,\text{kip-ft}}

.

M_{max} = \frac{wL^2}{8} = \frac{3(24)^2}{8} = 216\ \text{kip-ft}

Overhanging beam — positive span moment and negative support moment

Problem. A beam carries a uniform load $w = 2\,\text{k/ft}$ over its entire $24\,\text{ft}$ length. It is supported by a pin at $A$ ( $x=0$ ) and a roller at $B$ ( $x=16\,\text{ft}$ ), leaving an $8\,\text{ft}$ overhang past $B$ . Find the reactions, the maximum positive moment, and the negative moment over support $B$ .

Solution. Total load

W = wL = 2(24) = 48\,\text{kip}

acting at the centroid

x = 12\,\text{ft}

. Sum moments about

A

:

R_B(16) = 48(12) \Rightarrow R_B = 36.0\,\text{kip}

. Then

R_A = 48 - 36 = 12.0\,\text{kip}

. Shear from

A

:

V(x) = R_A - wx = 12 - 2x

, which is zero at

x = 6\,\text{ft}

. The maximum positive moment is there:

M^+ = R_A x - \dfrac{wx^2}{2} = 12(6) - \dfrac{2(6)^2}{2} = 72 - 36 = 36.0\,\text{kip-ft}

. The overhang acts as a cantilever delivering a hogging moment at

B

:

M_B = -\dfrac{w L_o^2}{2} = -\dfrac{2(8)^2}{2} = -64.0\,\text{kip-ft}

. Sanity check:

|M_B| = 64 > M^+ = 36

, so the negative moment governs design here — a classic overhang trap. Final:

\boxed{M^+ = 36.0,\ M_B = -64.0\,\text{kip-ft}}

.

Beam with a uniform load plus an off-center point load

Problem. A simply supported beam spans $L = 20\,\text{ft}$ , carries a UDL $w = 2\,\text{k/ft}$ over the full span, and a point load $P = 10\,\text{kip}$ at $a = 5\,\text{ft}$ from the left support. Find the reactions and the maximum moment.

Solution. Sum moments about

A

:

R_B(20) = wL\!\left(\tfrac{L}{2}\right) + P\,a = 2(20)(10) + 10(5) = 400 + 50 = 450 \Rightarrow R_B = 22.5\,\text{kip}

. Then

R_A = wL + P - R_B = 40 + 10 - 22.5 = 27.5\,\text{kip}

. Shear just left of

P

:

27.5 - 2(5) = 17.5\,\text{kip}

; just right of

P

:

17.5 - 10 = 7.5\,\text{kip}

(still positive). So shear does not cross zero at the load. Beyond

P

:

V(x) = 27.5 - 2x - 10 = 0 \Rightarrow x = 8.75\,\text{ft}

. Moment there:

M = R_A x - \dfrac{wx^2}{2} - P(x-a) = 27.5(8.75) - \dfrac{2(8.75)^2}{2} - 10(3.75) = 240.6 - 76.56 - 37.5 = 126.6\,\text{kip-ft}

. Sanity check: with

P=0

the UDL alone gives

wL^2/8 = 2(20)^2/8 = 100\,\text{kip-ft}

; adding the off-center point load raises and shifts the peak, consistent with

126.6 > 100

. Final:

\boxed{M_{max} \approx 127\,\text{kip-ft at } x = 8.75\,\text{ft}}

.

Common pitfalls

•Using $wL^2/8$ when the load is a point load, not a UDL. A central point load gives $PL/4$ , a cantilever UDL gives $wL^2/2$ — match the AISC Table 3-23 case (support and load type) exactly.
•Mixing $\text{kip}$ and $\text{lb}$ or $\text{ft}$ and $\text{in}$ . In $M=wL^2/8$ , keep $w$ in $\text{k/ft}$ and $L$ in $\text{ft}$ ; loads given in $\text{plf}$ or psf must be converted first or you are off by $1000$ .
•Looking for the maximum moment under the point load. The moment peaks where shear crosses zero; if the shear stays positive through a load, the peak is farther along the span (Example 3).
•Forgetting that an overhang creates a hogging support moment $wL_o^2/2$ that often governs over the positive span moment — design both regions.
•Sign-convention slips: handbook §1.6.7 takes positive moment as sagging (bottom fiber in tension). A concentrated applied couple jumps the moment diagram but does not change the shear.
•Flipping the sign of $V_2-V_1=\int w\,dx$ . The integral form is positive; the shear drops under a downward UDL because the signed $w$ is negative, not because the integral carries a minus sign.
•Not closing the diagrams. The shear must return to zero at the far end of the beam; if it does not, a reaction or load is wrong — fix it before drawing the moment diagram.

References

NCEES PE Civil Reference Handbook — §4.1.7 Moment, Shear, and Deflection Diagrams
NCEES PE Civil Reference Handbook — §1.6.7 Beams (load-shear-moment relations and sign conventions)
AISC Steel Construction Manual, 15th ed. — Table 3-23, Shears, Moments and Deflections — The source the handbook adapts for standard beam cases.

Static Determinacy, Stability & Indeterminacy

Count unknowns against equilibrium equations to classify beams, trusses, and frames as stable/unstable and determinate/indeterminate, and find the degree of indeterminacy.