CELE Strength of Materials — Shear and Moment DiagramsDetailed Explanation

Examples

Applications

• Determining support reactions is Step 1 for EVERY structural analysis problem in the board exam.
• In NSCP 2015 Section 405 (ACI 318 equivalent), beam design for flexure and shear requires Mu and Vu — both come from the BMD and SFD.
• Identifying load type (UDL vs. point load) determines the shape of the shear and moment diagrams (linear vs. parabolic).
• Fixed supports in cantilever retaining walls, canopy beams, and balcony slabs generate moment reactions that must be included in the free body diagram.

Misconceptions

• WRONG: 'The UDL resultant acts at the midpoint always.' — TRUE only for a full-span UDL. For a partial UDL, the resultant acts at the midlength of the LOADED portion.
• WRONG: 'A roller prevents both vertical and horizontal movement.' — A roller prevents only vertical translation; it is free to move horizontally.
• WRONG: 'An applied couple creates a shear reaction.' — An applied couple creates only a moment reaction at the fixed support; shear (vertical) reactions come only from transverse forces.
• WRONG: 'A propped cantilever can be solved by taking moments only.' — It has 4 reactions and 3 equations, so one compatibility condition is needed.

Related Concepts

• Static equilibrium (ΣFx = 0, ΣFy = 0, ΣM = 0)
• Free body diagrams
• Centroid of load distributions (for UVL)
• Degree of static indeterminacy
• Beam deflection and boundary conditions

Common Exam Questions

Key Points To Remember

• A roller gives 1 reaction, a pin gives 2, and a fixed support gives 3 — count these first before solving.
• A simply supported beam (pin + roller) is the most common board-exam beam type — 3 reactions, statically determinate.
• UDL resultant W = wL acts at the midlength of the loaded span.
• UVL (triangular) resultant W = ½w₀L acts at L/3 from the LARGER intensity end (NOT the midpoint).
• An applied couple does NOT change the shear diagram — it only creates a sudden jump in the moment diagram.
• Propped cantilevers and fixed beams are statically indeterminate — do NOT try to solve them with statics alone.

Examples

Applications

• Sign convention directly determines whether a beam needs top bars (hogging/negative moment) or bottom bars (sagging/positive moment) in RC design — critical for NSCP 2015 Chapter 4 compliance.
• Identifying hogging vs. sagging zones in an overhanging beam determines bar placement and bar cutoff locations.
• The point where M = 0 (point of inflection / contraflexure) is important for determining bar development lengths per NSCP 2015 Section 425.
• In prestressed concrete beams, the sign of the moment determines whether the prestress force is applied eccentrically above or below the centroidal axis.

Misconceptions

• WRONG: 'Positive moment means the top fiber is in tension.' — OPPOSITE. Positive moment (sagging) puts the BOTTOM fiber in tension and the top fiber in compression.
• WRONG: 'V and M have the same sign at all sections.' — V and M are independent quantities. V can be positive while M is negative (e.g., near the support of a hogging cantilever region).
• WRONG: 'The moment at a pin support is non-zero.' — A pin (and roller) offers no moment resistance, so M = 0 at every pin or roller. This is a critical boundary condition.
• WRONG: 'Cut the beam anywhere — the two sides will give the same V and M.' — TRUE for V and M values, but the signs may appear different depending on which side you use if you forget to apply the consistent sign convention.

Related Concepts

• Normal stress in beams: σ = Mc/I (flexure formula) — requires M
• Shear stress in beams: τ = VQ/Ib — requires V
• Beam deflection equations — derived by double integrating M(x)
• Moment of inertia and section modulus
• Neutral axis location

Common Exam Questions

Key Points To Remember

• ALWAYS cut the beam and isolate ONE portion — use the simpler side (fewer forces) to save time.
• Positive shear = left portion tends to slide UP = clockwise shear pair on the element.
• Positive moment = SAGGING (concave up, smile, tension at bottom fiber).
• Negative moment = HOGGING (concave down, frown, tension at top fiber).
• The sign convention is consistent throughout the analysis — mixing conventions leads to wrong diagrams.
• At a free end, both V and M must equal ZERO — this is a boundary condition check.
• At a pin or roller, the moment M = 0 (no moment resistance). Shear is generally non-zero.
• At a fixed support, V and M are generally both non-zero; their magnitudes come from equilibrium.

Examples

Applications

• The Method of Sections is the verification step when using faster graphical methods — if you have time in the exam, check your diagram at one or two key points using section cuts.
• For overhanging beams, set up separate segments for the span portion and the overhang portion.
• In RC beam design (NSCP 2015), the critical sections for moment are at the face of support and at midspan; these are specific section cuts.
• In steel beam design (AISC 360), the plastic moment Mp = FyZ is compared to the maximum moment from the BMD.

Misconceptions

• WRONG: 'I can skip computing reactions if I work from the free end of a cantilever.' — For a cantilever, working from the free end is actually valid AND avoids needing the wall reaction. But for any other beam type, you MUST find reactions first.
• WRONG: 'V is always positive in the first segment.' — V starts at +RA which is positive if RA is upward. But if RA were downward (possible in some unusual loading), V would start negative.
• WRONG: 'M can be non-zero at a free end.' — A free end has zero shear and zero moment — always. Violation of this means arithmetic error.

Related Concepts

• Load-Shear-Moment relationships (area method)
• Beam deflection by double integration (EI·d²y/dx² = M)
• Macaulay bracket notation for variable loading
• Conjugate beam method

Common Exam Questions

Key Points To Remember

• Always compute reactions FIRST — this is non-negotiable.
• Segment the beam at EVERY loading change point (supports, point loads, start/end of distributed loads, applied couples).
• Within a no-load segment, V is constant and M is linear.
• Within a UDL segment, V is linear and M is parabolic (opens downward for downward UDL).
• CHECK BOUNDARY CONDITIONS: M = 0 at pins, rollers, and free ends; V = 0 at free ends.
• For speed in the board exam, use the Area Method (next concept) instead of writing V(x) expressions for each segment.

Examples

Applications

• The Area Method is the standard tool in the board exam for quickly drawing SFD and BMD without algebra.
• In RC beam design (NSCP 2015 Section 406), the factored moment Mu and factored shear Vu are read directly from the BMD and SFD to size the beam and determine stirrup spacing.
• In steel design (AISC 360 Chapter F), the required moment strength Mu is the maximum value on the BMD, compared to φbMnx.
• Locating V = 0 (for M_max) determines where to place the maximum stirrup concentration in RC beams.
• The point of inflection (M = 0 on BMD) determines bar cutoff points in RC beams per NSCP 2015.

Misconceptions

• WRONG: 'M_max always occurs at midspan.' — TRUE only for symmetric loading (e.g., full-span UDL on SSB, or center point load). For unsymmetric loading, M_max is where V = 0, which must be computed.
• WRONG: 'The shear area formula ΔM = ∫V dx applies only for UDL.' — It applies for ANY loading — UDL, UVL, point loads (where shear is constant between loads). The shape of the shear diagram determines the type of integration.
• WRONG: 'A downward UDL makes the shear diagram slope upward.' — Downward UDL → dV/dx = −w < 0 → shear slopes DOWNWARD left to right.
• WRONG: 'An applied couple acts like a concentrated shear force.' — An applied couple ONLY affects the moment diagram (sudden jump). It has zero effect on the shear diagram.
• WRONG: 'The parabola in the BMD for a UDL opens upward.' — For a downward UDL on a simply supported beam, the BMD parabola opens DOWNWARD (maximum positive at center, zero at ends) — it is a sagging arch shape.

Related Concepts

• Macaulay's method for beam deflection
• Principle of superposition for combined loading
• Elastic curve equation EI(d²y/dx²) = M(x)
• Influence lines (related concept in structural analysis)
• Three-moment equation for continuous beams

Common Exam Questions

Key Points To Remember

• dV/dx = −w: slope of SFD = negative of load intensity. Downward UDL → SFD slopes DOWNWARD left to right.
• dM/dx = V: slope of BMD = shear. Positive shear → BMD slopes UPWARD. Negative shear → BMD slopes DOWNWARD.
• WHERE V = 0 (shear crosses zero), M is MAXIMUM or MINIMUM — this is the critical design location.
• Change in V = −(area under load diagram): for a UDL of w over length L, ΔV = −wL.
• Change in M = (area under shear diagram): for a triangular shear block of base b and height h, ΔM = ½bh.
• Point load → STEP in SFD. Applied couple → STEP in BMD (not SFD).
• UDL makes the BMD parabolic — the parabola OPENS DOWNWARD for a downward UDL on a simply supported beam (concave down at the midspan hump).
• At free ends and at pin/roller supports: M = 0 (use this as a check).

Examples

Applications

• Standard case formulas (PL/4, wL²/8) are used as QUICK CHECKS in the board exam and as reference values in beam design tables.
• In NSCP 2015 (ACI 318-equivalent), Table 406.5 provides approximate moment and shear values for continuous beams — these are based on modifications of the standard cases.
• Overhanging beam behavior (hogging at interior support) is critical for slab cantilever design in residential and commercial buildings.
• Cantilever beam analysis applies to retaining wall design (the wall stem acts as a cantilever), where M_max = wH²/2 governs the stem thickness.

Misconceptions

• WRONG: 'For a cantilever with UDL, M_max = wL²/8 (same as simply supported).' — WRONG. Cantilever with UDL: M_max = wL²/2, which is FOUR TIMES the simply supported value.
• WRONG: 'In an overhanging beam, the maximum moment is always in the sagging zone.' — The hogging moment at the interior support can easily be larger. Always compute both and compare.
• WRONG: 'For a simply supported beam with a UDL over half the span, M_max is still at midspan.' — When the UDL covers only part of the span, M_max shifts and must be found by setting V = 0.

Related Concepts

• Beam design formulas (NSCP 2015 / ACI 318)
• Approximate analysis for continuous beams (NSCP Table 406.5)
• Slab design (one-way and two-way slabs)
• Retaining wall stem design
• Tributary area and load transfer to beams

Common Exam Questions

Key Points To Remember

• Simply supported + central P: M_max = PL/4 at center.
• Simply supported + full UDL w: M_max = wL²/8, V_max = wL/2 — MEMORIZE THESE.
• Simply supported + off-center P: M_max = Pab/L, occurs UNDER the load.
• Cantilever + tip load P: M_max = PL (hogging) at the wall.
• Cantilever + full UDL: M_max = wL²/2 (hogging) at the wall — 4× more than SSB.
• Overhanging beams: CHECK BOTH the positive peak (in span) AND the negative peak (at support).
• Applied couple: JUMP in BMD only. No change in SFD.

Problem

Problem 1 (Simply Supported + Two Point Loads): A simply supported beam of span 10 m carries point loads of 20 kN at 3 m from the left (point C) and 30 kN at 7 m from the left (point D). (a) Find all reactions. (b) Draw the SFD and BMD. (c) Find M_max and its location.

Solution

Step 1: Reactions. ΣM_A = 0: RB(10) = 20(3) + 30(7) = 60 + 210 = 270 → RB = 27 kN RA = 20 + 30 − 27 = 23 kN Check ΣM_B = 0: RA(10) = 20(7) + 30(3) → 23(10) = 140 + 90 = 230 ✔ Step 2: SFD (left to right): • V at A = 0, then jumps to +23 kN (reaction RA) • Segment AC (0 to 3 m): no load, V = +23 kN (constant) • At C (x = 3m): 20 kN load → V drops by 20: V = 23 − 20 = +3 kN • Segment CD (3 to 7 m): no load, V = +3 kN (constant) • At D (x = 7m): 30 kN load → V drops by 30: V = 3 − 30 = −27 kN • Segment DB (7 to 10 m): no load, V = −27 kN (constant) • At B: RB = +27 kN → V = −27 + 27 = 0 ✔ Step 3: BMD (starting M_A = 0, using area of SFD): • ΔM(A to C): Area = +23(3) = +69 kN·m → M at C = +69 kN·m • ΔM(C to D): Area = +3(4) = +12 kN·m → M at D = 69 + 12 = +81 kN·m • ΔM(D to B): Area = −27(3) = −81 kN·m → M at B = 81 − 81 = 0 ✔ BMD is bilinear with three segments — peaks at D. Step 4: M_max location. V is positive throughout the span EXCEPT after D where it becomes −27 kN. Shear crosses zero at D (x = 7 m) — but wait: between C and D, V = +3 kN (positive). At D, V jumps from +3 to −27. The shear changes sign AT x = 7 m (at load D). M_max = 81 kN·m at x = 7 m (under the 30 kN load at D). Note: The moment is also 69 kN·m at x = 3 m (under the 20 kN load) — but 81 > 69, so the critical moment is at D.

Problem

Problem 2 (Cantilever with Combined Loading): A cantilever of length 5 m is fixed at the left wall. It carries a UDL of 4 kN/m over its full length and a point load of 10 kN at the free right end. (a) Find V and M at the fixed support. (b) Draw the SFD and BMD.

Solution

Approach: Work from the FREE END (right) to avoid needing the wall reactions explicitly for V and M calculation. Step 1: SFD from free end (x measured from the right, i.e., from the free end): • At the free end (x = 0): V = 0 (free end boundary condition). Point load 10 kN acts downward. Just inside the free end: V = +10 kN (the 10 kN load must be carried — using the convention that upward internal shear is positive) Let x = distance from the free end (measuring LEFT from the right). V(x) = 10 + 4x (contributions: tip load 10 kN + UDL accumulated over length x) At x = 5 m (at the fixed wall): V = 10 + 4(5) = 10 + 20 = 30 kN ← maximum shear at the wall. Step 2: BMD from free end: • At free end: M = 0 (boundary condition). M(x) = −[10·x + 4x·(x/2)] = −[10x + 2x²] (Negative because the loads cause hogging/negative moment.) At x = 5 m (wall): M = −[10(5) + 2(25)] = −[50 + 50] = −100 kN·m SFD shape: linear from +10 kN at free end to +30 kN at wall (slope = +4 kN/m, rising toward the wall since UDL accumulates). BMD shape: parabolic (quadratic in x) from 0 at free end to −100 kN·m at the wall. Entirely negative (hogging). Results: V_max = 30 kN at the fixed support. M_max = 100 kN·m (hogging) at the fixed support.

Problem

Problem 3 (Overhanging Beam): A beam is supported at A (x = 0) and B (x = 6 m). It overhangs 2 m beyond B to a free end C (x = 8 m). A UDL of 8 kN/m acts over the entire 8-m length. (a) Find reactions. (b) Find M at support B and the maximum positive moment in the span AB. (c) Compare and state the critical moment.

Solution

Step 1: Total load = 8(8) = 64 kN, acting at x = 4 m from A. ΣM_A = 0: RB(6) = 64(4) → RB = 256/6 = 42.67 kN ↑ RA = 64 − 42.67 = 21.33 kN ↑ Step 2: SFD (x from left): • Just right of A: V = +21.33 kN • V(x) = 21.33 − 8x for 0 ≤ x ≤ 6 (in span AB) • V = 0 at: 21.33 = 8x → x = 2.667 m ← max sagging moment location • Just left of B (x = 6⁻): V = 21.33 − 8(6) = 21.33 − 48 = −26.67 kN • Just right of B (x = 6⁺): V = −26.67 + 42.67 = +16 kN • From B to C: V = 16 − 8(x−6) → at C (x=8): V = 16 − 16 = 0 ✔ Step 3: M at x = 2.667 m (max positive moment): Using formula M(x) = RA·x − 8x²/2 = 21.33(2.667) − 4(2.667)² M = 56.88 − 28.44 = +28.44 kN·m (positive, sagging) M at support B (x = 6 m): M_B = RA(6) − 8(6)(3) = 21.33(6) − 144 = 128 − 144 = −16 kN·m (hogging) Alternatively, from the overhang side (cleaner): M_B = −8(2)(1) = −16 kN·m (2-m overhang carries 8×2=16 kN at centroid 1 m from B → hogging moment = 16 kN·m) ✔ At C: M = 0 ✔ (free end) Step 4: Compare moments: Max positive (sagging) = +28.44 kN·m at x = 2.667 m Max negative (hogging) = −16 kN·m at support B The critical (governing) moment for design = 28.44 kN·m (the larger magnitude). Note: In this case, the sagging moment controls — unlike Example 5 in the chapter where the hogging moment was larger. This depends on the relative length of overhang vs. span.

Problem

Problem 4 (UDL on Half Span): A simply supported beam of span L carries a UDL of w kN/m over the LEFT half only (from A to midspan). Derive M_max in terms of w and L and state its location.

Solution

Step 1: Reactions. Total load W = w(L/2) at centroid x = L/4 from A. ΣM_A = 0: RB·L = w(L/2)(L/4) → RB = wL/8 ↑ RA = wL/2 − wL/8 = 3wL/8 ↑ Step 2: SFD. For 0 ≤ x ≤ L/2 (loaded half): V(x) = RA − wx = 3wL/8 − wx V = 0 when: 3wL/8 = wx → x = 3L/8 Since 3L/8 < L/2, the zero-shear point is WITHIN the loaded zone. ✔ For L/2 ≤ x ≤ L (unloaded half): V(x) = RA − w(L/2) = 3wL/8 − wL/2 = −wL/8 (constant) SFD check at B: V just left of B = −wL/8, then + RB = +wL/8 jump → V = 0 ✔ Step 3: M_max at x = 3L/8. M(x) = RA·x − wx²/2 = (3wL/8)x − wx²/2 At x = 3L/8: M_max = (3wL/8)(3L/8) − w(3L/8)²/2 M_max = 9wL²/64 − 9wL²/128 M_max = 18wL²/128 − 9wL²/128 = 9wL²/128 Result: M_max = 9wL²/128 at x = 3L/8 from the left support. Note: This is LESS than wL²/8 (the full-span UDL maximum) — specifically it is 9/128 ÷ 1/8 = 72/128 = 56.25% of the full-span value, which makes physical sense since only half the load is applied.

Problem

Problem 5 (Applied Couple Effect): A simply supported beam of span L = 6 m has a clockwise couple M₀ = 30 kN·m applied at x = 2 m from the left support A. (a) Find the reactions. (b) Draw the SFD and BMD.

Solution

Step 1: Reactions. The couple provides no net force, so only moments are needed. ΣM_A = 0: RB(6) = 30 → RB = 5 kN ↓ (downward at B to balance the clockwise couple) ΣF_y = 0: RA + RB = 0 (no transverse loads) → RA = −RB = +5 kN ↑ Step 2: SFD. The applied couple does NOT appear in the shear equation (it has no transverse force component). • Just right of A: V = +5 kN • No transverse loads anywhere → V = +5 kN throughout (constant) • At B: −RB = +5 − 5 = 0 ✔ (Note RB is downward, so just before B: V = +5; after B: V = +5 − 5 = 0 if we treat the downward RB as −5.) Actually: RA = +5 kN upward, RB = 5 kN downward. SFD: V = +5 kN from A to B (constant, since no transverse loads). Jump at B by −5 to zero. ✔ Step 3: BMD. The couple causes a JUMP at x = 2 m. • M at A = 0 • Area of SFD from A to x = 2 m: +5(2) = +10 kN·m → M just BEFORE the couple = +10 kN·m • Clockwise couple M₀ = 30 kN·m applied at x = 2 m → BMD JUMPS UP by +30 kN·m • M just AFTER the couple = 10 + 30 = +40 kN·m • Area of SFD from x = 2 m to B: +5(4) = +20 kN·m → this area is ADDED but shear is positive... Wait: M at B = 40 + 5(4) = 40 + 20 = 60 kN·m ≠ 0 — check sign of RB! Revise: RB is DOWNWARD → V between A and B: V = RA = +5 kN upward → slope of BMD = +5 kN throughout. But area must reduce M back to 0 at B. Recheck reactions: ΣM_B = 0: RA(6) = M₀ = 30 → RA = 5 kN ↑ ✔ ΣM_A = 0: RB(6) = M₀ = 30 → but direction of RB? For clockwise M₀ at x=2m: ΣM_A = M₀ − RB(6) = 0 → RB = +5 kN ↑ ΣFy: RA + RB = 0 → RA = −5 kN ↓ Actually, for a simply supported beam with ONLY a clockwise couple M₀: To resist a clockwise couple, the beam needs a counterclockwise couple of reactions: RA = M₀/L downward, RB = M₀/L upward (or vice versa depending on direction). For M₀ clockwise: RB = M₀/L = 30/6 = 5 kN ↑; RA = 5 kN ↓ SFD: V = −5 kN throughout (RA is downward). Jump at B by +5 → 0 ✔ BMD: • M at A = 0. • SFD area A to x=2m: (−5)(2) = −10 kN·m → M just before couple = −10 kN·m • Clockwise couple M₀ = +30 kN·m → M jumps UP: M just after couple = −10 + 30 = +20 kN·m • SFD area x=2m to B: (−5)(4) = −20 kN·m → M at B = 20 − 20 = 0 ✔