CELE Strength of Materials — Shear and Moment DiagramsRevision Notes

Formulas

Exam Tips

• Draw a complete free-body diagram (FBD) of the entire beam showing all loads AND all reactions before writing any equilibrium equation.
• Always take moments about one support to find the other reaction — this eliminates two unknowns in one equation.
• For symmetric loading on a simply supported beam, each reaction = half the total load. Use this as a quick check.
• Label every load with its magnitude, direction, and point of application on your FBD — a missing label causes a wrong reaction.

Key Points

• A beam is a structural member that resists loads acting transverse (perpendicular) to its longitudinal axis primarily through bending and shear.
• Roller support: 1 unknown reaction (vertical only); Pin/Hinge: 2 unknowns (vertical + horizontal); Fixed support: 3 unknowns (vertical + horizontal + moment).
• Statically determinate beams (simply supported, cantilever, overhanging) are solved with the three equilibrium equations alone: ΣFx=0, ΣFy=0, ΣM=0.
• Statically indeterminate beams (propped cantilever, fixed-fixed, continuous) require compatibility methods (three-moment equation, moment distribution, etc.) covered in Structural Theory.
• A concentrated (point) load P has units kN. A uniformly distributed load (UDL) w has units kN/m with resultant W = wL at the midspan of the loaded length. A uniformly varying load (UVL/triangular) has intensity from 0 to w₀; resultant = ½w₀L at L/3 from the larger-intensity end.
• An applied couple M₀ (kN·m) is a concentrated moment; it does NOT create a shear jump — only a moment jump in the BMD.
• Always solve for reactions FIRST before drawing any SFD or BMD.

Definitions

Section Title

1. Beams, Supports, and Loads

Common Mistakes

• Placing the UVL (triangular load) resultant at midspan instead of L/3 from the larger-intensity (bigger) end.
• Forgetting to compute horizontal reaction at pin supports when horizontal loads are present — though for vertical-only loads, the horizontal reaction at a pin is zero.
• Treating a fixed support as a pin and missing the fixed-end moment reaction.
• Using the total span length when the UDL or UVL covers only a portion of the beam.

Formulas

Exam Tips

• Always work from the LEFT side whenever possible — reactions at the left support are known first, and positive shear = upward on the left, making the algebra more natural.
• Use the RIGHT side for overhanging or cantilever sections where the right side has fewer forces.
• Check boundary conditions: M = 0 at pins, rollers, and free ends; V = 0 at free ends. If your diagram violates these, find the error before moving on.
• Draw a small sketch of the cut element showing positive V and M directions BEFORE writing the equilibrium equations.

Key Points

• At any cross-section, expose the internal forces by cutting the beam and applying equilibrium to one isolated portion (free body). The cut face must carry V and M to maintain equilibrium.
• Shear force V at a section = algebraic sum of all transverse forces on one side of the cut (either side; use the simpler side).
• Bending moment M at a section = algebraic sum of moments of all forces on one side of the cut, taken about the centroid of the cut section.
• SIGN CONVENTION (standard beam convention — memorize): Positive shear V acts upward on the left face and downward on the right face of the cut element (the pair rotates CLOCKWISE). Positive moment M causes SAGGING (concave up, compression on top, tension at bottom — like a smile). Negative moment causes HOGGING (concave down — like a frown), typical over interior supports and in cantilevers.
• This is the most error-prone part: always sketch the assumed positive direction before writing the equation.
• At a free end (no load), V = 0 and M = 0. At a simply supported end (pin or roller), M = 0 but V is generally non-zero.

Definitions

Section Title

2. Internal Shear Force and Bending Moment – Definitions and Sign Convention

Common Mistakes

• Applying forces on the RIGHT side with a wrong sign because the student forgot that 'positive shear = upward force on the LEFT face' (the sign reverses when using the right side).
• Taking moments about a point other than the cut centroid — creates moment-arm errors.
• Confusing the direction of V and M drawn on the cut face; always draw them in the assumed POSITIVE direction and let the algebra give the sign.
• Forgetting that at a pin or roller support M = 0, and at a free end both V = 0 and M = 0 — these are boundary conditions for checking your diagrams.

Formulas

Exam Tips

• For the exam, identify the number of segments first (count loading changes), then write the shear and moment expressions only for the segment containing the point of interest.
• When asked to draw the FULL SFD and BMD, use the area-method (Section 4) — it is 3–5× faster than writing equations for every segment.
• Label each segment boundary with the value of V and M at that boundary — these become the starting points for the next segment.

Key Points

• The method of sections is the universal exact method: cut, isolate, and apply equilibrium to find V and M at any location.
• Step 1: Compute all support reactions from the complete FBD using ΣFx=0, ΣFy=0, ΣM=0.
• Step 2: Identify all segments where loading is continuous (no jumps) — segment boundaries occur at each support, point load, applied couple, and at the start/end of every distributed load.
• Step 3: For each segment, introduce a cut at distance x from the left, isolate the left portion, draw V and M on the cut face in the positive direction, and write the two equations ΣFy=0 (gives V) and ΣM_cut=0 (gives M).
• Step 4: Plot V(x) and M(x) from these equations to form the SFD and BMD.
• For complex beams, the load–shear–moment relationships (Section 4) are faster and should replace the equation-by-equation approach for the exam.
• The method of sections is still essential for computing V and M at a specific point (e.g., 'find V and M at x = 2 m').

Definitions

Section Title

3. Method of Sections – Step-by-Step Procedure

Common Mistakes

• Cutting through a point load or support instead of between them — always cut within a segment, not at its boundary.
• Forgetting to include the distributed load between the left support and the cut when writing ΣFy — the omitted load = w × x.
• Taking moments about the wrong point (e.g., about A instead of the cut centroid), introducing extra moment-arm errors.
• Working with too many segments — only cut between loading changes, not at arbitrary locations.

Formulas

Exam Tips

• Memorize the area formulas: rectangle = bh; triangle = ½bh; parabola under a triangle load = ⅓bh (specific orientation matters — verify by integration if uncertain).
• At each jump point (point load or reaction), write the value of V just before and just after the jump to define the two endpoints of the next segment.
• The moment area method is fastest: compute M at every key point as M_left + ∫V dx. Start from M = 0 at a simple support or free end.
• For UVL (triangular) load: V is parabolic and M is cubic — sketch the curve shape but compute exact values only at the key points (boundaries and zero-shear point).

Key Points

• These four relationships come from equilibrium of a differential beam element (dx long). They let you sketch the SFD and BMD WITHOUT writing explicit equations for each segment.
• RELATIONSHIP 1 — Slope of shear = negative load intensity: dV/dx = −w(x). A downward UDL makes shear DECREASE linearly; a zero load keeps shear CONSTANT.
• RELATIONSHIP 2 — Slope of moment = shear: dM/dx = V(x). Where V > 0, moment is increasing. Where V < 0, moment is decreasing. WHERE V = 0 (OR CHANGES SIGN), MOMENT IS AT A LOCAL MAXIMUM OR MINIMUM — this is the key to locating Mmax.
• RELATIONSHIP 3 (integrated form) — Change in shear = −(area under load diagram): ΔV = V_B − V_A = −∫w dx. Use this to compute shear at any boundary from a previous known value.
• RELATIONSHIP 4 (integrated form) — Change in moment = area under shear diagram: ΔM = M_B − M_A = ∫V dx. Use this to compute moment at any boundary from a previous known value.
• DEGREE RULE: Each integration raises the polynomial degree by 1. Point load → V is degree 0 (constant), M is degree 1 (linear). UDL → V is degree 1 (linear), M is degree 2 (parabolic). UVL → V is degree 2 (parabolic), M is degree 3 (cubic).
• JUMPS: A downward point load P causes a sudden DROP in V of magnitude P. An applied couple M₀ causes a sudden JUMP in M (clockwise couple causes upward jump if standard convention) but NO change in V.
• The area method requires knowing the VALUE of V or M at one boundary and the SHAPE of the curve in each segment — then you compute the change using area formulas.

Definitions

Section Title

4. Load–Shear–Moment Relationships (The Area Method)

Common Mistakes

• Forgetting to subtract the area of the load diagram from V — students add instead of subtract (note the negative sign in dV/dx = −w).
• Using the wrong geometric formula for shear diagram area — triangles (½bh), rectangles (bh), or parabolic segments (⅔bh for concave, ⅓bh for convex) are all needed.
• Applying a concentrated couple to the shear diagram instead of only the moment diagram.
• Not recognizing that the parabolic BMD in a UDL region opens DOWNWARD for positive loading (sagging), not upward.
• Forgetting that upward reactions (supports) cause an UPWARD JUMP in the shear diagram.

Formulas

Exam Tips

• For a simply supported beam with symmetric loading, Mmax is always at midspan and R_A = R_B = half total load. Use this to verify your reaction calculations instantly.
• When the board exam gives an overhanging beam, ALWAYS compute both the span maximum (sagging) and the hogging moment at the interior support — the design moment is the larger of the two absolute values.
• The formula Mmax = Pab/L is faster than section-by-section analysis for an off-center single point load — memorize it.

Key Points

• These standard results appear directly or as sub-problems in board exam questions. Memorize them for speed-checking.
• Simply supported, central point load P (span L): R = P/2 each side; Mmax = PL/4 at midspan; Vmax = P/2 at supports; SFD is rectangular (constant) on each half; BMD is triangular peaking at midspan.
• Simply supported, UDL w over full span L: R = wL/2 each side; Mmax = wL²/8 at midspan; Vmax = wL/2 at supports; SFD is linear (triangular shape); BMD is parabolic (open downward), peaking at midspan.
• Simply supported, point load P at distance a from left (b = L − a): R_A = Pb/L; R_B = Pa/L; Mmax = Pab/L under the load.
• Cantilever, point load P at free end: M_fixed = −PL (hogging, maximum); V = P (constant throughout); SFD is rectangular; BMD is linear.
• Cantilever, UDL w over full length L: M_fixed = −wL²/2 (hogging, maximum); Vmax = wL at fixed end; SFD is linear (triangular); BMD is parabolic.
• Overhanging beam — the maximum moment can occur at the interior support (hogging), NOT within the span. Always check both the span sagging maximum and the support hogging value and use the LARGER for design.

Definitions

Section Title

5. Standard Cases and Key Formulas

Common Mistakes

• Using Mmax = wL²/8 for beams where the UDL does NOT cover the full span — the formula is only valid for a simply supported beam with UDL over the ENTIRE span.
• Assuming Mmax is always at midspan for overhanging beams — it can be at the interior support or at the zero-shear point within the span, whichever is larger.
• Forgetting the negative sign in cantilever moment formulas — the moment at the fixed end is always HOGGING (negative) for downward loads.

Formulas

Exam Tips

• For UVL problems, immediately identify: (a) total resultant = ½w₀L, (b) position = L/3 from the LARGER end (= 2L/3 from the smaller end). This is the fastest way to get reactions.
• For overhanging beams, compute the moment at the interior support FIRST using the overhang segment — it is usually the simpler calculation and immediately gives the hogging moment.
• Always verify your SFD by confirming that the net area under the shear diagram equals the total moment change from one free end to the other (which must be zero for a simply supported beam).

Key Points

• EXAMPLE 1 — Simply supported, UDL: L = 6 m, w = 12 kN/m. R_A = R_B = 36 kN. Vmax = 36 kN at supports. V = 0 at x = 3 m (midspan). Mmax = ½(3)(36) = 54 kN·m by shear area = wL²/8 = 12(36)/8 = 54 kN·m ✔
• EXAMPLE 2 — Simply supported, off-center point load: L = 8 m, P = 40 kN at a = 3 m. R_A = 40(5)/8 = 25 kN; R_B = 40(3)/8 = 15 kN. SFD: +25 kN from A to load point; drops by 40 to −15 kN; zero at load point. Mmax = 25(3) = 75 kN·m = Pab/L = 40(3)(5)/8 = 75 kN·m ✔
• EXAMPLE 3 — Cantilever, combined loading: L = 4 m, w = 5 kN/m, P = 10 kN at free end. V at fixed end = 5(4) + 10 = 30 kN. M at fixed end = −[5(4)(2) + 10(4)] = −[40 + 40] = −80 kN·m (hogging).
• EXAMPLE 4 — Triangular load (board-hard): L = 6 m, w(x) = 2x kN/m (0 at left, 12 kN/m at right). R_A = 12 kN, R_B = 24 kN. V(x) = 12 − x². Set V = 0: x = √12 = 3.46 m. M(x) = 12x − x³/3. Mmax = 12(3.46) − (3.46)³/3 = 41.6 − 13.8 = 27.7 kN·m at x = 3.46 m.
• EXAMPLE 5 — Overhanging beam: Span AB = 4 m, overhang BC = 2 m, UDL w = 6 kN/m over all 6 m. R_B = 27 kN, R_A = 9 kN. Zero shear in span at x = 9/6 = 1.5 m from A. M⁺ = 9(1.5) − 6(1.5²)/2 = 13.5 − 6.75 = 6.75 kN·m. M at B (from right) = −6(2)(1) = −12 kN·m. Design moment = 12 kN·m (hogging at B governs).
• Key insight from Example 5: The overhang creates hogging at B that often exceeds the span sagging moment — never assume span sagging governs for overhanging beams.

Section Title

6. Worked Board-Style Examples

Common Mistakes

• For the UVL example (Example 4): placing the resultant at midspan (3 m) instead of 4 m from the zero end (L×2/3 = 4 m) — leads to wrong reactions.
• For the overhanging beam (Example 5): computing only the span maximum and missing the larger hogging moment at the interior support — a common exam trap.
• In Example 3 (cantilever): computing the UDL moment as w×L×L (wrong) instead of w×L×(L/2) — forgetting that the UDL resultant acts at the midpoint of the loaded length.

Exam Tips

• Draw a vertical dashed line from the zero-shear point on the SFD down to the BMD to clearly show the Mmax location — this makes your solution easier for examiners to follow.
• For the exam, work through Steps 1–9 in order for every beam problem, even if you feel confident. The structured approach prevents the most common errors.
• Always write the numerical value of V and M at every segment boundary on your diagram — partial credit on the board exam is awarded for correct segment values even if the final Mmax is wrong.

Key Points

• STEP 1: Compute all support reactions. Verify equilibrium.
• STEP 2: Identify all segment boundaries: supports, point loads, starts and ends of distributed loads, and applied couples.
• STEP 3: Starting from the LEFT end, track V and M using the area method segment by segment.
• STEP 4: At each point load or reaction, apply the jump: V_after = V_before + reaction_up (or − load_down).
• STEP 5: Within each segment, compute ΔV = −(area of load diagram) and ΔM = (area of shear diagram). Determine the shape (constant/linear/parabolic/cubic) from the degree rule.
• STEP 6: Find the zero-shear location within any segment where V changes sign. This requires setting V(x) = 0 and solving for x. Then compute Mmax using ΔM = area under shear up to that point.
• STEP 7: At an applied couple M₀, step the moment diagram by ±M₀ (no shear change). Direction of step depends on the couple's orientation.
• STEP 8: VERIFY: M must return to zero at simple supports and free ends. The final shear at the rightmost end (just before the reaction, from the right) must equal the right reaction.
• STEP 9: Label all critical values on the SFD and BMD: maximum shear, maximum moment, zero-shear location, and boundary values.

Definitions

Section Title

7. Constructing the SFD and BMD – Systematic Procedure

Common Mistakes

• Skipping the verification step (Step 8) — a BMD that does not close back to zero at a simple support contains an error that will propagate to beam design calculations.
• Confusing the direction of the moment step for an applied couple — always establish the sign convention for couples before plotting.
• Not labeling the zero-shear location and Mmax on the BMD — board exam solutions require these values to be identified explicitly.