CELE Strength of Materials — Shear and Moment DiagramsStudy Notes

A beam is a structural member carrying transverse (perpendicular) loads and resisting them primarily through bending and shear. Understanding support types, reaction components, and load classifications is the prerequisite for drawing accurate diagrams. **Support Types and Reactions:** Each support type provides a specific number and type of reaction force or moment: - **Roller Support:** Provides one vertical reaction; restrains vertical translation only. The beam may rotate and translate horizontally freely. - **Pinned (Hinge) Support:** Provides two reactions (vertical and horizontal); restrains both translations but permits rotation. - **Fixed (Cantilever) Support:** Provides three reactions (vertical force, horizontal force, and moment); fully restrains the beam end against translation and rotation. **Load Classifications:** 1. **Concentrated (Point) Load, P:** A single force applied at a discrete point. Units: N, kN. Creates an immediate discontinuity (jump) in the shear diagram. 2. **Uniformly Distributed Load (UDL), w:** Constant load intensity over a length. Units: N/m, kN/m. The resultant force is W = wL, acting at the midpoint of the loaded region. Produces linear variation in shear and parabolic variation in moment. 3. **Uniformly Varying Load (UVL) / Triangular Load:** Load intensity varies linearly from zero at one end to w₀ at the other. Resultant force is (1/2)w₀L, positioned at the centroid of the triangle — critically, at L/3 from the larger-intensity end, not L/2. Results in parabolic shear and cubic moment. 4. **Applied Moment (Couple), M:** A concentrated rotational effect applied without translation. Creates a jump in the moment diagram but no change in shear. **Beam Classification:** - **Statically determinate beams** (simply supported, cantilever, overhanging): solvable using three equilibrium equations alone. - **Statically indeterminate beams** (fixed, propped cantilever, continuous): require additional methods beyond statics (discussed in Structural Theory). For the Civil Engineer Licensure Examination, emphasis is placed on determinate cases, particularly simply supported and cantilever beams.

When a beam is cut at any section and one side is isolated as a free body, internal forces and moments emerge at the cut surface to maintain equilibrium. These internal effects are the shear force V and bending moment M. **Definition by the Method of Sections:** Make an imaginary cut perpendicular to the beam axis at distance x from a reference point. Isolate one side and apply equilibrium: - **Shear Force V(x):** The sum of all transverse (vertical) forces on one side of the cut, or equivalently, the internal vertical force pair at the section that prevents relative sliding. - **Bending Moment M(x):** The sum of all moments (about the section) from external loads on one side, or equivalently, the internal couple that prevents rotation of one part relative to the other. Mathematically: V(x) = Σ(transverse forces on isolated side) M(x) = Σ(moments about the section on isolated side) **Physical Interpretation:** - Shear represents the beam's resistance to sliding failure; maximum shear often governs design for lateral-torsional instability and web yielding in steel beams (per AISC 360-16). - Bending moment represents the beam's resistance to flexural failure; maximum moment governs flexural capacity, determines reinforcement in concrete beams (ACI 318), and is the primary design criterion for most beam problems. **Internal Stress Relationship:** At a given section, if the material behaves elastically: - Shear stress τ = V / A_shear (varies across the section; used for shear reinforcement design in concrete) - Bending stress σ = M·y / I (varies linearly from neutral axis; governs selection of steel grade and concrete compressive strength) For licensed engineers, the stress-strain relationship underlies all design codes (NSCP 2015 for concrete, AISC 360-16 for steel).

The sign convention for shear and moment diagrams is standardized in structural mechanics and must be memorized exactly — sign errors are among the most common board-exam mistakes. **Shear Force Sign Convention:** Consider an element of the beam with a cut at each end. Internal shear forces act on the cut faces. - **Positive Shear V > 0:** The internal shear on the left cut face acts upward (equivalently, on the right face acts downward). Visualize this: the left portion tends to slide upward relative to the right. The internal couple formed by these opposite forces rotates the element **clockwise** — the positive rotational sense. - **Negative Shear V < 0:** The internal shear on the left face acts downward. The element tends to rotate **counterclockwise** — negative. **Mnemonic:** Positive shear "climbs up and to the left" — like a hand grasping the left face and pulling it upward. **Bending Moment Sign Convention:** - **Positive Moment M > 0 (Sagging):** The beam bends with curvature concave upward (smiles). The top fibers experience compression, the bottom fibers tension. Positive moment acts on the right cut face in the **counterclockwise direction** (top pulls, bottom pushes). Common in the interior of simply supported beams. - **Negative Moment M < 0 (Hogging):** The beam bends with curvature concave downward (frowns). The top fibers experience tension, the bottom fibers compression. Negative moment acts on the right cut face in the **clockwise direction**. Common over interior supports and along cantilevers. **Mnemonic:** Positive moment = a smile (happy beam, compression on top). Negative moment = a frown. **Sign Convention Applied to Free-Body Equilibrium:** When isolating the left portion of a cut beam and applying moment equilibrium about the cut section: - If you sum moments clockwise as positive, the internal moment M on the right face of the isolated left portion appears with a sign that follows the convention above. - Always be consistent: some engineers sum counterclockwise as positive; others use the convention stated here. Choose one and stick to it. **Critical Rule for the Exam:** Draw a small free-body diagram of a beam element at the section where you are computing shear and moment. Show the external loads and reactions on that element. Then show the internal shear and moment using the standard symbols. This discipline prevents sign errors.

The method of sections is the exact, always-works procedure for deriving shear and moment equations V(x) and M(x) along the beam. While more time-consuming than using relationships (Section 5), it is essential for complex loadings and verifying results. **Step-by-Step Procedure:** 1. **Compute all reactions using global equilibrium:** - Apply ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0 to the entire beam. - Verify: ΣFᵧ (up) = ΣFᵧ (down); this check prevents errors that propagate through all subsequent calculations. 2. **Identify segment boundaries:** - Divide the beam into segments at every discontinuity: each support, each point load, each applied couple, and at the start and end of each distributed load or load change. - Example: a beam from 0 to 10 m with supports at 0 and 10 m, a point load at 3 m, and a UDL from 4 to 8 m has segment boundaries at 0, 3, 4, 8, and 10 m → **five segments**. 3. **For each segment, cut at distance x and isolate one side:** - Choose the simpler side (fewer loads). Write equations for V(x) and M(x) from equilibrium. - V(x) = Σ(transverse forces on isolated side) - M(x) = Σ(moments about the section on isolated side) - Retain x as a variable; do not substitute numbers yet. 4. **Check continuity at segment boundaries:** - At boundaries without a point load or applied couple, V and M should match from either segment (continuous). - A point load causes V to jump by the load magnitude. - An applied couple causes M to jump by the couple magnitude (V unchanged). 5. **Plot V(x) and M(x):** - Use the equations to evaluate shear and moment at a few key points (supports, boundaries, load positions). - Sketch curves between points, noting the degree of each curve (constant, linear, parabolic, cubic). - Shade or label positive (above axis) and negative (below axis) regions. **Degree of Polynomial:** - **Point load only:** V is piecewise constant (degree 0, horizontal lines); M is piecewise linear (degree 1, straight lines). - **UDL:** V is piecewise linear (degree 1, straight lines); M is piecewise parabolic (degree 2, curved). - **Triangular (UVL) load:** V is piecewise parabolic (degree 2); M is piecewise cubic (degree 3). Each integration increases the degree by one.

The relationships between load, shear, and moment are derived from differential equilibrium of an infinitesimal beam element. These are the key to rapid, accurate diagram sketching without writing equations — essential for time-pressured board exams. **Differential Relationships:** Consider an element of beam of length dx with load w(x) acting downward. Applying equilibrium: dV/dx = −w(x) [Slope of shear diagram = negative of load intensity] dM/dx = V(x) [Slope of moment diagram = shear] **Integrated Form (Areas and Changes):** Integrating over a distance from point A to point B: ΔV = V_B − V_A = −∫[A to B] w(x) dx = −(area under load diagram from A to B) ΔM = M_B − M_A = ∫[A to B] V(x) dx = (area under shear diagram from A to B) **Practical Interpretation for Board Exams:** 1. **Slope of SFD = negative load intensity at that point:** - No load (w = 0): shear is horizontal (flat). - Downward UDL (w > 0): shear has negative slope (line slopes downward to the right). - Upward load (w < 0): shear has positive slope. 2. **Slope of BMD = shear at that point:** - When V > 0 (positive shear): moment diagram has positive slope (rising to the right). - When V < 0: moment diagram has negative slope (falling). - When V = 0: moment diagram is horizontal (stationary) — **this is where M reaches a local extremum (maximum or minimum).** 3. **Change in shear = negative area under load diagram:** - A downward UDL from A to B reduces shear by the area w·(B−A). - A point load P at position C causes shear to drop by P. 4. **Change in moment = area under shear diagram:** - A positive (above-axis) shear region adds area to moment. - A negative shear region subtracts area. - The magnitude of moment change equals the area magnitude. **Key Insight for Finding M_max:** Locate the point where V = 0 or changes sign. At that point, the slope of M becomes zero (horizontal tangent) → **M is a local extremum there.** For simply supported beams with downward loads, where shear crosses from positive to negative, that crossing point is M_max (sagging). Over interior supports of continuous beams, where shear crosses from negative to positive, that crossing is M_max (hogging in magnitude). **Degree Rule (from Calculus):** Each integration raises the polynomial degree by 1: - If load is piecewise constant (degree 0), then V is piecewise linear (degree 1) and M is piecewise parabolic (degree 2). - If load is piecewise linear (degree 1, triangular), then V is piecewise parabolic (degree 2) and M is piecewise cubic (degree 3). **Discontinuities (Jumps):** - **Point load P at position c:** Shear diagram jumps by −P (downward load causes downward jump). - **Applied couple M_c at position c:** Moment diagram jumps by M_c (no effect on shear). **Sketching Without Equations (the "fast method"):** 1. Draw reaction arrows and label magnitudes (use method of sections if needed). 2. Start at the left support with V = R_A. Follow the load diagram: as you move right, reduce V by the area under the load. Mark where V crosses zero (M_max is here). 3. Start moment at M = 0 (at a support, usually). Follow the shear: as you move right, add area under shear to moment. Where shear crosses zero, mark the moment value — this is M_max. 4. Use the degree rule to sketch curves: straight lines between point loads; parabolas under UDL; no curves where w = 0. 5. Check: at the far end, V and M should match the known reaction and support moment (if fixed) — this verifies correctness.

For the board exam, memorizing these standard results allows instant verification and rapid problem-solving. Use them as a first check; if your answer differs significantly, re-examine your calculation. **Simply Supported Beam, Central Point Load P (symmetric):** - Span: L - Reactions: R_A = R_B = P/2 - Maximum shear: V_max = P/2 (at supports) - Maximum moment: M_max = PL/4 (at midspan, x = L/2) - Shear diagram: two horizontal lines, one above (P/2) and one below (−P/2), stepping at midspan - Moment diagram: two triangles meeting at the center, symmetric, with peak pointing up **Simply Supported Beam, Uniformly Distributed Load w over Entire Span:** - Total load: W = wL - Reactions: R_A = R_B = wL/2 - Maximum shear: V_max = wL/2 (at supports) - Maximum moment: M_max = wL²/8 (at midspan) - Shear diagram: straight line from +wL/2 at A to −wL/2 at B, crossing zero at L/2 - Moment diagram: parabola opening downward (concave down in shape) with peak at L/2. The curvature matches the load diagram (also parabolic by the degree rule). **Simply Supported Beam, Point Load P at Distance a from Left (b = L − a):** - Reactions: R_A = Pb/L (downward load closer to B → smaller R_A); R_B = Pa/L - Maximum moment: M_max = Pab/L (at x = a, directly under the load, provided a < b) - Shear diagram: positive constant (+Pb/L) from A to the load, then negative constant (−Pa/L) from the load to B - Moment diagram: linear rise from 0 to M_max over distance a, then linear fall to 0 over distance b - **Key:** M_max always occurs **under the point load**, regardless of symmetry **Cantilever Beam, Point Load P at the Free End:** - Maximum shear: V = P (constant along the beam, in the direction of the load) - Maximum moment: M_max = −PL (at the fixed end, negative = hogging). Magnitude: |M| = PL - Shear diagram: horizontal line at P (if P is downward, show as negative by convention, V = −P; but magnitude is P) - Moment diagram: straight line from 0 at the free end to −PL at the fixed end **Cantilever Beam, Uniformly Distributed Load w over Entire Length:** - Maximum shear: V_max = wL (at the fixed end) - Maximum moment: M_max = −wL²/2 (at the fixed end, hogging, magnitude wL²/2) - Shear diagram: straight line from 0 at the free end to −wL at the fixed end (negative because w is downward) - Moment diagram: parabola, concave down, from 0 at the free end to −wL²/2 at the fixed end **Overhanging Beam: Two Supports at A and B, Overhang Beyond B:** - Critical consideration: **M_max can occur at the interior support (hogging), not within the span.** Always check both interior and exterior regions. - Use the method of sections or load–shear–moment relationships to find V = 0 points (candidates for M_max). - Typical issue on board exams: students find sagging moment in the span but miss the larger hogging moment at the support. **Propped Cantilever (Fixed at A, Roller Support at B):** - Indeterminate structure: requires superposition or other advanced methods (beyond scope of simple diagrams). - However, SFD and BMD can be sketched once reactions are known (from Structural Theory course or provided).

These examples demonstrate the full process from reactions to final diagrams, at the level expected in the PRC Civil Engineer Licensure Examination.

Board exams frequently test understanding through trick questions and common misconceptions. Awareness of these pitfalls improves accuracy dramatically. **Pitfall 1: Forgetting the Shear Jump at a Point Load** A point load creates a discontinuity (jump) in the shear diagram equal to the load magnitude. Many students draw a smooth curve through the load point, forgetting the step. This error cascades: if the shear diagram is wrong, the moment diagram location of zero shear (and thus M_max) is also wrong. *Prevention:* Always mark point load locations on the shear diagram. Draw a vertical line at the load; shear jumps by −P (for downward P). **Pitfall 2: Misunderstanding the Sign Convention** Confusing which direction is "positive" shear or moment causes diagrams to be flipped. A simply supported beam with downward loads should show positive moment in the interior (sagging, smile). If your BMD is inverted (frown in the interior), your sign convention is backwards. *Prevention:* Draw a small free-body element with forces and moments labeled. Enforce the convention consistently throughout. **Pitfall 3: Wrong Centroid of Triangular Load** A triangular load with intensity 0 at the left and w_o at the right has its resultant (1/2)w_o·L acting at x = L/3 from the left, NOT at L/2. Students often place the resultant at the geometric midpoint. *Prevention:* Memorize: "centroid of triangle is at 1/3 from the **larger** end." Verify by integration: centroid = ∫x·w(x)dx / ∫w(x)dx. **Pitfall 4: Ignoring Applied Couples** An applied couple (external moment) creates a jump in the moment diagram but **no change in shear**. Students sometimes mistakenly add it to shear. *Prevention:* Remember: couples affect moment only. Check the SFD for continuity across an applied-couple location. **Pitfall 5: Maximum Moment in Overhanging Beams** In an overhanging beam, the maximum moment can occur at the interior support (hogging), not within the main span (sagging). Many students find the sagging moment within the span and miss the larger hogging moment at the support. *Prevention:* Check all regions: interior span (sagging), at supports (hogging), and overhang. Compute M at critical points: where V = 0, at supports, at load positions. Compare magnitudes. **Pitfall 6: Not Verifying Reactions** Reactions set the starting point for all diagrams. A wrong reaction propagates to every subsequent calculation. Yet, many students skip equilibrium checks. *Prevention:* Always verify: ΣF_y (up) = ΣF_y (down), and take moments about two different points to double-check R_A and R_B. **Pitfall 7: Confusing Cantilever Fixed Support Moment** A cantilever fixed at one end has a reaction moment at the fixed end (usually hogging). This reaction moment is not an externally applied couple; it's the internal moment required to balance the loads. Students sometimes forget to include it when drawing the BMD. *Prevention:* For a cantilever, always compute the reaction moment: M_fixed = −Σ(moment of external loads about fixed end). Include this as the BMD value at x = 0. **Pitfall 8: Forgetting Boundary Conditions in SFD and BMD** At a simply supported end, V ≠ 0 (there's a reaction) but M = 0. At a cantilever fixed end, V and M are both non-zero. At a free end, both V and M are typically zero unless there's a load or couple right at the end. *Prevention:* Sketch the support symbols and reactions before drawing diagrams. Use them as anchoring points. **Pitfall 9: Polynomial Degree Mistakes** A constant load (no UDL, only point loads) should produce a linear moment diagram, not parabolic. Students sometimes confuse the degree rule. *Prevention:* Point load → V constant, M linear. UDL → V linear, M parabolic. UVL → V parabolic, M cubic. Sketch only straight lines or the appropriate curve.

Mastery of shear and moment diagrams is foundational for Civil Engineer licensure success. These diagrams appear in nearly every structural design problem: reinforced concrete beams (ACI 318), steel beams (AISC 360), and Filipino design standards (NSCP 2015). The following summary and strategy guide your preparation. **Core Competencies You Must Demonstrate:** 1. **Reaction Calculation:** Using ΣF = 0 and ΣM = 0, find all support reactions. Verify your answer by checking equilibrium about a second point or summing forces in the opposite direction. 2. **Segment Identification:** Identify all boundaries (supports, point loads, load changes) and classify each segment (uniform load, point load only, etc.). 3. **Shear Diagram Sketching:** Use either the method of sections (exact) or the load–shear–moment relationships (fast). Mark shear at key points, show jumps at point loads, and use the degree rule (constant → V const, UDL → V linear, etc.). 4. **Moment Diagram Sketching:** Plot moment at segment boundaries. Use the load–shear–moment relationship: area under SFD = change in moment. Identify where V = 0 (extremum of M). 5. **Locating M_max:** Search for V = 0 or V crossing zero. At such points, M reaches a local extremum. Compare M at all critical points: supports, load positions, V = 0 locations. Report the largest magnitude as M_max. 6. **Verification:** Check boundary values (M = 0 at simply supported ends; M and V non-zero at fixed ends; both zero at free ends unless loaded there). Verify continuity (except at point loads and applied couples). **Time Management During the Exam:** - **First 1 minute:** Read the problem, sketch the beam, mark supports and loads. - **Next 2 minutes:** Compute reactions using global equilibrium. Double-check. - **Next 3 minutes:** Draw the SFD using the load–shear–moment relationships (faster than method of sections). - **Next 2 minutes:** Mark V = 0 locations on the SFD. - **Next 2 minutes:** Draw the BMD, using the SFD to find slopes and areas. - **Final 1 minute:** Verify M_max location and magnitude. Check boundary conditions. - **Total: ~11 minutes per beam problem** (typical exam problem). **What the PRC Examiners Test:** Examiners expect you to: - Find M_max quickly (it's the design driver). - Sketch diagrams that are qualitatively correct (shapes, signs, continuity). - Identify critical sections for design (where M is maximum, where V is maximum, where combined stress is worst). - Apply the diagrams to downstream design (e.g., "find the required beam section given M_max"). **Recommended Resources for Practice:** - Work 10–15 problems of varying complexity: simply supported with UDL, cantilever with point load, overhanging with triangular load, etc. - For each, draw both diagrams; find V_max and M_max; verify reactions and boundary conditions. - Compare your results with a solutions manual or instructor feedback. - Memorize the standard formulas (Section 6) and use them as sanity checks. **Red Flags — If You Can't Do These, Review the Chapter:** 1. Can you identify the sign of shear at a point without hesitation? 2. Can you sketch the SFD for a cantilever with both distributed and point loads in under 2 minutes? 3. Can you locate M_max without writing equations, using only the load–shear–moment relationships? 4. Can you explain why the moment diagram under a UDL is parabolic (degree rule)? 5. Can you identify the centroid of a triangular load correctly? If you struggle with any of these, revisit the corresponding section before moving to the next chapter.