Shear and Moment Diagrams - Cheat Sheet
Your last-minute revision companion for mastering internal shear forces, bending moments, reaction calculations, and the load–shear–moment relationships that appear in almost every PRC CE Licensure exam question on beam design.
Support Types & Reactions
Important facts
- Always solve reactions FIRST using three equilibrium equations: ΣFx = 0, ΣFy = 0, ΣM = 0.
- For statically determinate beams, three unknown reactions can be solved uniquely.
- The moment equation can be taken about ANY point; choosing a reaction point eliminates terms and speeds calculation.
- Verify reactions: sum of upward reactions must equal total downward load for a beam carrying only vertical loads.
Key definitions
Roller Support
Example
Bridge bearing on an expansion joint.
Provides one reaction (vertical force only); allows horizontal translation and rotation.
Pin/Hinge Support
Example
Hinged connection between beam and column.
Provides two reactions (vertical and horizontal forces); prevents translation but allows rotation.
Fixed Support
Example
Beam welded or cast into a wall.
Provides three reactions (vertical force, horizontal force, and moment); prevents all translation and rotation.
Diagrams to know
- Free-body diagram of entire beam with all reactions and loads labeled
- Support symbol legend (roller, pin, fixed) with reaction arrows
Reactions or equations
ΣFx = 0 (horizontal equilibrium)
Note
Rarely tested alone but needed for completeness.
Conditions
No horizontal loads → Hx = 0 in most problems.
ΣFy = 0 (vertical equilibrium)
Note
First check: sum of vertical reactions = total downward load.
Conditions
RA + RB = W (total load)
ΣM = 0 (moment equilibrium about any point)
Note
Taking moment about a reaction isolates that reaction instantly.
Conditions
Choose a support point to eliminate unknowns.
Load Types & Resultants
Formulas
W = wL
Meaning
W = total load (N/kN); w = UDL intensity (N/m, kN/m); L = length (m)
Watch out
Resultant acts at the MID-LENGTH of the UDL region, not at a support.
When to use
Any uniformly distributed load (UDL) over a span.
W = ½ w₀L
Meaning
W = resultant of triangular/UVL load; w₀ = max intensity; L = span of the load.
Watch out
Resultant acts at L/3 from the LARGER-INTENSITY END (centroid of triangle = 2/3 × height from base).
When to use
Triangular (uniformly varying) load from 0 to w₀.
Location of UVL resultant = L/3 from high end, or 2L/3 from zero end
Meaning
For a triangle with base at one end: centroid is 1/3 of the base length from the vertex.
Watch out
Students often use L/2; the correct position is L/3 from the bigger end.
When to use
Locating where the UVL resultant acts for moment calculations.
Important facts
- Load diagram is the INPUT; shear diagram comes from integrating/summing loads; moment diagram comes from integrating shear.
- UDL always produces a PARABOLIC moment diagram (degree 2).
- Triangular load produces a CUBIC moment diagram (degree 3) — rarely simple closed form.
- The resultant of ANY load over a region acts at the centroid of that region's area under the load curve.
Key definitions
Concentrated (Point) Load
Example
A column bearing on a beam delivers 100 kN at one location.
A single force P acting at a point; acts downward for dead/live loads.
Uniformly Distributed Load (UDL)
Example
Floor slab weight: 15 kN/m over 8 m span = 120 kN resultant at 4 m from the left end.
Constant load intensity w over a length; total load W = wL acts at the midpoint of the loaded region.
Uniformly Varying Load (UVL) / Triangular Load
Example
Water pressure on a dam increases linearly from 0 at the surface to maximum at depth.
Load intensity varies linearly from zero (or w₁) to w₂; resultant = ½(w₁ + w₂)L at the centroid.
Applied Moment (Couple)
Example
Torque applied at a connection: 50 kN·m.
A concentrated twisting effect (units: N·m, kN·m) with no associated shear; causes a jump in the moment diagram only.
Diagrams to know
- Load diagram (intensity vs. position)
- Idealized load distributions: rectangular (UDL), triangular (UVL), point loads and couples
Internal Shear & Moment — Definition & Sign Convention
Formulas
V(x) = ΣFtransverse to the LEFT of section
Meaning
V = internal shear force (N, kN); sum of all transverse forces on the left free body.
Watch out
POSITIVE shear = LEFT side has upward force; visualize as left side sliding UP relative to right (clockwise couple on element).
When to use
At any section x along the beam to determine internal shear.
M(x) = ΣMabout the section from the LEFT side
Meaning
M = internal bending moment (kN·m); sum of all moments on the left free body about the cut section.
Watch out
POSITIVE moment = concave UP (sagging, compression top, tension bottom); NEGATIVE = concave DOWN (hogging, tension top).
When to use
At any section x to determine internal moment.
Important facts
- The POSITIVE shear convention: left face of the cut section has upward force (or equivalently, shear on left = upward, shear on right = downward → clockwise couple = positive).
- The POSITIVE moment convention: sagging = smile = positive; hogging = frown = negative.
- Sign convention is CRITICAL and the #1 source of exam errors. Memorize with a picture.
- At any section, the internal shear and moment form an equilibrium pair (they are reactions of the internal cut surface).
Key definitions
Internal Shear Force V
Example
Cut a simply supported beam at midspan with a point load: left side sees the support reaction pushing up; that reaction IS the internal shear.
The internal transverse (vertical) force at a section that keeps one free body in equilibrium; positive when the left portion tends to slide upward.
Internal Bending Moment M
Example
At midspan of a simply supported beam with uniform load: the moment is positive (sagging) and maximum.
The internal moment at a section that resists rotation of the two parts; positive when creating concave-up (sagging) curvature.
Sagging Moment (Positive)
Example
Interior of a simply supported beam under downward load.
Bending moment that causes the beam to curve concave UP (smile shape); compression on top fiber, tension on bottom.
Hogging Moment (Negative)
Example
Over an interior support of a continuous beam or at the fixed end of a cantilever.
Bending moment that causes the beam to curve concave DOWN (frown shape); tension on top fiber, compression on bottom.
Diagrams to know
- Free-body diagram of a cut section with internal V and M shown (and labeled with sign)
- Sign convention diagram: positive V (left up, right down) and positive M (concave up)
Load–Shear–Moment Relationships (The Fundamental Trio)
Formulas
dV/dx = −w(x)
Meaning
Slope of shear diagram = negative of load intensity; w = load per unit length (kN/m).
Watch out
NEGATIVE sign: downward load produces negative slope (shear decreases); upward load (rare) produces positive slope.
When to use
Sketching the shear diagram without writing V(x) equations; slope at each point tells you the load there.
dM/dx = V(x)
Meaning
Slope of moment diagram = shear force at that point.
Watch out
The MAXIMUM moment does NOT occur at midspan unless the loading is symmetric; it occurs where dM/dx = V = 0.
When to use
Finding where moment peaks: where V = 0, slope of M = 0 → Mmax/Mmin occurs there.
ΔV = VB − VA = −∫[A to B] w dx = −(area under load diagram from A to B)
Meaning
Change in shear between two points = negative of the area under the load diagram in that interval.
Watch out
Negative sign; downward load (negative area) means shear INCREASES (becomes more positive or less negative).
When to use
Jumping from one segment to another without integrating; sum the load areas.
ΔM = MB − MA = ∫[A to B] V dx = (area under shear diagram from A to B)
Meaning
Change in moment between two points = area under the shear diagram in that interval.
Watch out
NO negative sign here; positive shear area adds positive moment; negative shear area subtracts moment.
When to use
Jumping from one segment to another; sum the shear areas (triangles, rectangles, etc.).
Important facts
- The slope of M = V; so WHERE V CROSSES ZERO, M has a horizontal tangent (local extremum).
- WHERE V CHANGES SIGN, M changes from increasing to decreasing (or vice versa); this is where Mmax/Mmin lies.
- A POINT LOAD causes an INSTANTANEOUS DROP in V (discontinuity); a concentrated COUPLE causes a JUMP in M.
- The load diagram → shear diagram → moment diagram is a chain of integrations; the reverse (differentiating M to get V, V to get load) is used to verify.
- For a UDL of w constant over a span: shear changes linearly, moment changes parabolically (a nice formula: M_max = wL²/8 for symmetric case).
Key definitions
Degree Rule (Polynomial Integration)
Example
A simply supported beam with a UDL has a triangular shear diagram (degree 1) and a parabolic moment diagram (degree 2).
Each integration raises the polynomial degree by one: point load (V const, degree 0; M linear, degree 1); UDL (V linear, degree 1; M parabolic, degree 2); UVL (V parabolic, degree 2; M cubic, degree 3).
Diagrams to know
- Load diagram (w vs x)
- Shear force diagram (V vs x) with slopes matching load; drops at point loads; unchanged across applied couples
- Bending moment diagram (M vs x) with slope matching shear; jumps at applied couples; no jump at point loads
Method of Sections (Step-by-Step Procedure)
Important facts
- STEP 1 is non-negotiable: you CANNOT draw correct shear/moment diagrams without correct reactions.
- STEP 2: Identify all discontinuities — every support, every point load, start/end of every distributed load, every applied couple.
- STEP 3: In each segment between discontinuities, V(x) and M(x) are continuous functions of x.
- STEP 4: Always isolate the LEFT side (or always the RIGHT; pick one convention). Left side is often easier because loads and reactions are encountered from left to right.
- STEP 5: Use ΣFy = 0 to find V; use ΣM (about the cut section) = 0 to find M.
- For complex beams, plot V and M point-by-point; the diagrams will reveal the maximum values and their locations.
Key definitions
Method of Sections
Example
A beam with three segments (before load, under load, after load) requires three sets of equations, one for each segment.
Procedure to find V(x) and M(x): (1) Find reactions, (2) Divide beam into segments at load changes, (3) Cut at x in each segment, (4) Isolate one free body, (5) Write V and M from ΣF and ΣM equilibrium.
Diagrams to know
- Free-body diagram of the entire beam showing all reactions and external loads (needed for Step 1)
- Segmented free-body diagram: isolate one side of a cut at position x in each segment
Jumps & Discontinuities
Formulas
Jump in V at a point load P = −P
Meaning
At a downward point load, shear DROPS (becomes more negative) by the load magnitude.
Watch out
The jump is DOWNWARD (negative direction) for a downward load; upward load produces an upward jump.
When to use
Sketching shear diagram; whenever you encounter a point load, shear jumps.
Jump in M at applied couple = +C (in the direction of the couple)
Meaning
An applied moment causes an instantaneous change in the moment diagram equal to the couple's magnitude.
Watch out
Couple does NOT affect shear; only moment jumps. A student error is to apply the couple to the shear diagram.
When to use
Whenever an applied moment appears on the load diagram, moment diagram jumps.
NO jump in V or M at a support (unless a point load or couple is applied exactly at the support)
Meaning
Reactions are support forces; they do cause jumps in the shear diagram.
Watch out
A reaction jump in shear = support reaction; this is expected and correct, not an error.
When to use
Check: the shear diagram should show a jump (discontinuity) at each support equal to the reaction there.
Important facts
- Shear diagram: piecewise continuous except for jumps at point loads and at support reactions.
- Moment diagram: piecewise continuous except for jumps at applied couples (and possible jumps at concentrated loads in the context of a singularity, but usually not shown in simple hand sketches).
- The height of a shear jump = magnitude of the load (or reaction) at that point.
- The height of a moment jump = magnitude of the applied couple at that point.
Diagrams to know
- Shear diagram with discontinuities (steps) at each point load and reaction
- Moment diagram with discontinuities (jumps) at each applied couple
Standard Formulas for Common Beams
Formulas
Simply supported, central point load P: RA = RB = P/2; Mmax = PL/4 (at midspan); Vmax = P/2
Meaning
L = span; by symmetry, reactions are equal; moment peaks at the center.
Watch out
This formula is for a CENTRAL load only; for off-center loads, use the general formula Mmax = Pab/L.
When to use
Quick check or when the loading is exactly a point load at the center.
Simply supported, UDL w over entire span: RA = RB = wL/2; Mmax = wL²/8 (at midspan); Vmax = wL/2
Meaning
By symmetry; shear is zero at midspan (where moment is maximum).
Watch out
This is only valid if the UDL spans the entire length. If the load is partial, recalculate reactions and use the method of sections.
When to use
Uniform load over the entire span; one of the most common exam cases.
Simply supported, point load P at distance a from left (b = L − a): RA = Pb/L; RB = Pa/L; Mmax = Pab/L (under the load)
Meaning
Unsymmetric load; the larger reaction is closer to the load; maximum moment is at the load point.
Watch out
Maximum moment occurs AT the load (where shear changes sign), not elsewhere; make sure a + b = L.
When to use
Point load at any location on a simply supported span.
Cantilever, point load P at free end: Mmax = −PL (at fixed end); V = P (constant)
Meaning
Negative moment (hogging) at the fixed support; shear is constant throughout.
Watch out
Moment is NEGATIVE (hogging); the fixed end resists the overturning effect. The moment magnitude grows linearly along the span.
When to use
Cantilever with a single point load at the tip.
Cantilever, UDL w over entire length: Mmax = −wL²/2 (at fixed end); Vmax = wL (at fixed end)
Meaning
Negative (hogging) moment and constant shear increasing from zero at free end to wL at fixed end.
Watch out
Moment is NEGATIVE and parabolic in shape; shear is linear (triangular diagram); maximum values are at the fixed support.
When to use
Cantilever with distributed load; common in overhanging floors and brackets.
Important facts
- Symmetry is your friend: if the beam and loading are symmetric, reactions are equal and Mmax is at midspan.
- For a simply supported beam, Mmax occurs where V = 0; you must find this point if the loading is not symmetric.
- For cantilevers, the maximum moment and shear are always at the fixed end; the diagrams are often opposite in sign compared to simply supported beams.
- The formulas above are the most commonly memorized; they appear in nearly every PRC exam.
Diagrams to know
- V and M diagrams for simply supported beam under central point load (triangular V, parabolic M)
- V and M diagrams for simply supported beam under UDL (triangular V, parabolic M)
- V and M diagrams for cantilever under point load at free end (constant V, linear M)
- V and M diagrams for cantilever under UDL (linear V, parabolic M)
Locating Maximum Moment (Mmax) — The Game-Winning Skill
Formulas
Mmax occurs where V = 0 (or where V changes sign)
Meaning
The location of maximum (or minimum) bending moment is at a section where the shear force is zero.
Watch out
If V does NOT cross zero (e.g., cantilever, or shear has only one sign), Mmax is at a boundary (support or free end).
When to use
ANY time you need to find Mmax; set V(x) = 0 and solve for x.
For a simply supported beam with a single point load at distance a from left: Mmax = Pab/L at x = a
Meaning
Maximum moment is at the load point itself.
Watch out
Do NOT use the formula Mmax = wL²/8 (which is for UDL); this is for a point load.
When to use
Single point load; the crossing point of shear (from + to −) is exactly at the load.
For UDL w over the entire span of a simply supported beam: Mmax = wL²/8 at x = L/2
Meaning
By symmetry, maximum moment is at the center.
Watch out
This formula applies only to FULL-LENGTH UDL. For partial UDL, you must solve V = 0 explicitly.
When to use
Uniform load over entire span; shear is zero at midspan by symmetry.
Important facts
- Finding the section where V = 0 is the FIRST step in any bending moment problem; this is where Mmax will be.
- For symmetric loading on a simply supported beam, V = 0 at midspan by symmetry (no need to solve).
- For unsymmetric loading, V(x) = 0 does not occur at midspan; you must set up the shear equation, equate to zero, and solve.
- If a beam segment has a constant shear (e.g., cantilever with a point load), V ≠ 0 in that segment, so moment increases monotonically; extremum is at a boundary.
- The area under the shear diagram from a to the point where V = 0 gives Mmax − Ma.
Key definitions
Critical Section (for moment)
Example
At midspan of a simply supported beam under UDL, shear is zero, so moment is maximum.
Any section where dM/dx = V = 0; at this section, moment reaches a local extremum (max or min).
Diagrams to know
- Shear diagram with zero-crossing point clearly marked (this is where Mmax occurs)
- Moment diagram showing the peak at the location where shear is zero
Sketch SFD & BMD Without Equations (Fast Board Method)
Important facts
- START with reactions (always).
- Plot reactions on the shear diagram: shear = RA at the left; it drops/changes at each load.
- For a UDL: shear slope = −w (constant negative slope); moment slope = V (changing slope as V changes).
- For a point load: shear drops by the load magnitude; moment is piecewise linear.
- For a couple: moment jumps by the couple magnitude; shear unchanged.
- The shear diagram is piecewise linear or curved (depending on load type); the moment diagram is one degree higher (linear or parabolic for UDL).
- After sketching, verify: reactions should match; V should be zero (or cross zero) where you expect Mmax; the area under V should equal changes in M.
Key definitions
Rapid Sketching (No-Equation Method)
Example
For a UDL, know that shear is linear and moment is parabolic; jump shear at each point load; jump moment at each couple.
Use load–shear–moment relationships and the area-change rules to sketch V and M directly without writing V(x) and M(x) functions.
Diagrams to know
- Hand-sketched SFD showing jumps at loads and reactions, with correct slopes in each segment
- Hand-sketched BMD showing curvature, jumps at couples, and peak at zero-shear location
Overhanging Beams & Continuous Beams (Traps & Tips)
Formulas
Maximum moment in an overhanging beam may occur AT an interior support (hogging), not in the span (sagging)
Meaning
Overhangs create large negative moments over supports; design must check both sagging (within span) and hogging (at supports).
Watch out
A student who finds only the maximum sagging moment (within the span) will miss the hogging moment at the support and under-design the beam.
When to use
Any overhanging beam; always plot the full SFD and BMD and identify both Mmax positive and Mmax negative.
Important facts
- In an overhanging beam, the maximum NEGATIVE moment often governs design, not the maximum positive moment.
- The overhang acts like a cantilever; its weight and loads pull down, creating a hogging moment at the interior support.
- Always calculate: (1) maximum sagging moment within the span, (2) maximum hogging moment at/near the supports.
- For a UDL over the entire length (including overhang), the worst moment may be at the support, not at midspan.
- Continuous (multi-span) beams: interior supports create hogging moments; spans sag (positive moments). Alternating sagging and hogging along the length.
Key definitions
Overhanging Beam
Example
A floor beam supported at two interior points with cantilevered ends; the cantilever ends create hogging moment over the supports.
A beam with support(s) and one or both ends extending beyond the support(s); the overhang region often produces large negative (hogging) moments at the support.
Diagrams to know
- SFD and BMD of an overhanging beam showing shear changing sign and moment showing both positive (span) and negative (over support) peaks
Triangular & Non-Standard Loads (Centroid & Resultant Location)
Formulas
Triangular load resultant W = ½ w₀ L; location: L/3 from the LARGER-INTENSITY END
Meaning
For a load triangle from 0 to w₀, the centroid is at 2/3 of the base from the zero end (or L/3 from the w₀ end).
Watch out
L/3 from the LARGER end; students often use L/2 (midpoint) or get the direction backwards.
When to use
Locating the resultant of a triangular (UVL) load for reaction calculation or moment computation.
For an arbitrary load distribution w(x), resultant W = ∫ w(x) dx; location x_c = ∫ x·w(x) dx / ∫ w(x) dx
Meaning
General centroid formula; weight at the centroid.
Watch out
Integration is exact but tedious; for standard shapes (triangle, trapezoid), use geometry.
When to use
Non-standard loads (trapezoids, curves, etc.); rarely needed on the exam but useful for checking.
Important facts
- The RESULTANT of any distributed load acts at the CENTROID of the load area (the shape under the load curve).
- For a triangle, centroid = (1/3) × base from the vertex (zero end), or equivalently (2/3) × base from the base end.
- For a trapezoid w₁ to w₂ over length L: resultant = (w₁ + w₂)L/2 at the centroid (not at L/2 unless w₁ = w₂).
- Using the centroid principle to find the resultant and its location speeds up reaction calculations and moment estimates.
Key definitions
Centroid of a Load Diagram
Example
For a triangle from 0 to w₀, centroid is at L/3 from the larger end; for a rectangle (UDL), centroid is at L/2.
The point at which the resultant of a distributed load acts; equivalent to the center of mass of the load shape.
Diagrams to know
- Triangular load diagram with centroid marked at L/3 from the larger end
- Trapezoidal load diagram with centroid marked off-center
Must remember
- 1. ALWAYS solve for reactions first using ΣFx = 0, ΣFy = 0, ΣM = 0. No correct SFD/BMD without correct reactions.
- 2. Sign convention (memorize with pictures): Positive shear = left face up (clockwise internal couple); Positive moment = concave up (smile = sagging).
- 3. The slope of the moment diagram equals the shear force at that point (dM/dx = V). Maximum moment occurs where V = 0.
- 4. Point load causes a JUMP in shear diagram; applied couple causes a JUMP in moment diagram; UDL causes linear shear and parabolic moment.
- 5. Triangular load resultant = ½w₀L acting at L/3 from the LARGER-INTENSITY END (not at L/2 or from the zero end).
- 6. For simply supported beam with central point load: Mmax = PL/4. For UDL: Mmax = wL²/8. Commit these to memory.
- 7. Cantilever maximum moment and shear are always at the fixed end; the moment is NEGATIVE (hogging) and the shear is constant.
- 8. Overhanging beams: check BOTH the maximum sagging moment (within the span) AND the maximum hogging moment (at the interior support). Design for the larger in magnitude.
- 9. Change in shear between two points = −(area under load diagram); change in moment between two points = +(area under shear diagram). Use this to skip writing equations.
- 10. The location of maximum moment is where V = 0 (or where V changes sign). Set V(x) = 0 and solve for x; this x-value is where Mmax occurs.
Last minute tips
- Tip 1: Before sketching any SFD/BMD, CHECK YOUR REACTIONS. Verify: sum of upward reactions = total downward load. This 10-second check catches 80% of sign errors.
- Tip 2: Use the 'area under shear = change in moment' rule to jump between segments. If you see a triangular shear region from 0 to +20 kN over 3 m, the area is (1/2)(3)(20) = 30 kN·m; moment changes by +30 in that region. Faster than writing equations.
- Tip 3: For overhanging beams, ALWAYS sketch the full SFD and BMD. The hogging moment at an interior support can exceed the sagging moment in the span; missing it costs points. On the exam, spend 30 seconds extra to plot both and note: 'Maximum sagging: 50 kN·m at x = 2.5 m; Maximum hogging: 80 kN·m at x = 4 m (over support). Design moment = 80 kN·m.'
- Tip 4: Triangular load centroid = L/3 from the LARGER END. Visualize: the weight is closer to the heavier end. If you get a reaction that is obviously lopsided (e.g., one reaction much larger), check that you placed the triangular resultant at L/3, not L/2.
- Tip 5: On the exam, label your SFD and BMD axes clearly: 'Shear (kN)', 'Moment (kN·m)', 'Distance (m)'. Mark the key values: reactions, Mmax, location of Mmax, and any jumps. A well-labeled diagram earns partial credit even if the numbers are slightly off; a blank or mislabeled diagram earns nothing.
Comparison tables
Simply Supported vs. Cantilever Beams
| Feature | Simply Supported | Cantilever |
|---|---|---|
| Support reactions | 2 (RA vertical, RB vertical) | 3 (Rfixed vertical, horizontal, moment) |
| Shear at free end | Non-zero (equal to reaction) | Non-zero (equal to total load) |
| Moment at supports | Zero (pin/hinge) | Maximum (non-zero) |
| Moment at free end | Zero (end of beam) | Zero (free end, no constraint) |
| Location of Mmax | Where V = 0, usually within the span | At the fixed end |
| Moment sign (typical load down) | Positive (sagging, smile) | Negative (hogging, frown) |
| Typical moment diagram shape (UDL) | Parabola, opening downward, zero at ends | Parabola, opening upward, zero at free end, peak at fixed end |
Load Type vs. Shear & Moment Diagram Degree
| Load Type | Load Diagram Degree | Shear Diagram Degree | Moment Diagram Degree | Example Shape |
|---|---|---|---|---|
| Point load | 0 (constant/zero) | 0 (constant, jumps at load) | 1 (linear, slope = shear) | |
| Uniformly distributed load (UDL) | 0 (constant) | 1 (linear, slope = −w) | 2 (parabolic, opens downward) | |
| Uniformly varying (triangular) | 1 (linear, 0 to w₀) | 2 (parabolic) | 3 (cubic) | |
| Applied couple only | N/A (no distributed load) | 0 (unchanged) | Jump at couple location |
Sign Convention Quick Reference
| Internal Force/Moment | Positive Direction | Visual/Memory Cue | Effect on Beam |
|---|---|---|---|
| Shear V | Left face upward (or right face downward) | Left portion slides UP relative to right; internal couple spins CLOCKWISE | Tends to cause shear failure (sliding) |
| Moment M | Concave UP (curvature opens upward) | Smile face (sagging); compression on top, tension on bottom | Compression in top fiber, tension in bottom fiber |
Discontinuities in SFD & BMD
| Event | Shear Diagram Effect | Moment Diagram Effect |
|---|---|---|
| Downward point load P | Jump DOWN by P (V decreases) | No discontinuity (continuous) |
| Upward point load P | Jump UP by P (V increases) | No discontinuity (continuous) |
| Applied moment (couple) C | No discontinuity (unchanged) | Jump by C in direction of couple |
| Support reaction | Jump in shear (expected; equals reaction) | No jump (zero moment at pin/hinge; non-zero at fixed) |
Diagrams
Shear and Moment Diagrams — Concept Map
Comprehensive mind map showing the major topics in shear and moment diagrams: from reactions and loads, through internal force/moment definitions and calculations, to the key relationships and standard formulas. Use this to see how each concept connects.
Step-by-Step SFD & BMD Construction Procedure
Flowchart of the complete procedure to construct shear and moment diagrams from scratch, from drawing the free body through identifying reactions, setting up segments, writing equations, and plotting the final diagrams. Follow this step-by-step on every exam problem.
Locating Maximum Bending Moment (Mmax) — Decision Tree
Decision tree to quickly locate the section where maximum bending moment occurs. The key rule: where shear force V = 0, moment reaches an extremum. Use this flowchart to organize your solution strategy.