CELE Strength of Materials — Shear and Moment DiagramsExam Answer Templates

Marks

1

Topic

Internal Shear and Moment

Difficulty

easy

Template Id

T1

Examiner Tip

A definition answer must contain the operation (algebraic sum), the quantity (moments), and the reference (one side, about the section). Missing any one of these three components typically means zero marks.

Model Answer

The bending moment at a cross-section of a beam is the algebraic sum of the moments of all external forces acting on one side of that section about the centroid of the section. It is positive (sagging) when it causes concavity upward (tension at the bottom fiber) and negative (hogging) when it causes concavity downward.

Question Type

very_short_answer

Answer Structure

• Line 1: State that bending moment is the algebraic sum of moments of forces on one side of a section [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'sum of forces' instead of 'sum of moments' — loses the entire mark.
• Omitting 'one side of the section' makes the definition incomplete.
• Confusing bending moment with torque or shear force.

Key Phrases To Include

• algebraic sum of moments
• one side of the section
• about the centroid / about the section
• positive (sagging) / negative (hogging)

Marks

1

Topic

Sign Convention

Difficulty

easy

Template Id

T2

Examiner Tip

The single most tested 1-mark recall item in SFD/BMD. Memorize: LEFT side, UPWARD = positive shear. Everything else follows.

Model Answer

Shear force is positive when the resultant of forces to the LEFT of the section acts UPWARD (or equivalently, when the left portion tends to slide upward relative to the right portion). This produces a clockwise couple on the element — the positive sense.

Question Type

very_short_answer

Answer Structure

• Line 1: Direction of force on the left portion (upward = positive) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the convention backward (upward on the RIGHT = positive) — common confusion, zero marks.
• Describing the moment convention instead of the shear convention.
• Being vague: writing only 'upward is positive' without specifying which side.

Key Phrases To Include

• resultant to the left acts upward
• positive shear
• clockwise couple
• left portion slides upward relative to right

Marks

1

Topic

Load–Shear–Moment Relationships

Difficulty

easy

Template Id

T3

Examiner Tip

This concept is the pivot of every M_max question. Examiners reward the explicit statement of dM/dx = V as it shows conceptual understanding, not just pattern memorization.

Model Answer

When the shear force diagram crosses zero (changes sign) at a point between supports, the bending moment is at a local maximum or minimum at that point. This is because dM/dx = V; when V = 0, the rate of change of M is zero, indicating an extremum of the moment diagram.

Question Type

very_short_answer

Answer Structure

• Line 1: State that bending moment has a local maximum/minimum where V = 0, citing dM/dx = V [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'shear is maximum' at the zero crossing — this is the opposite of what is true.
• Failing to state it is a LOCAL maximum — for overhanging beams the overall maximum may be elsewhere.
• Omitting the differential relationship when asked 'what does it mean' (examiner expects the reasoning, not just the conclusion).

Key Phrases To Include

• maximum or minimum bending moment
• V = 0
• dM/dx = V
• extremum of the moment diagram

Marks

2

Topic

Simply Supported Beam — UDL

Difficulty

easy

Template Id

T4

Examiner Tip

For a 2-mark numerical, the examiner expects ONE reaction calculation and ONE result per mark. Write compactly but show the formula before substituting numbers.

Model Answer

Given: L = 8 m, w = 10 kN/m Reactions (by symmetry): R_A = R_B = wL/2 = 10(8)/2 = 40 kN Maximum Shear Force (at supports): V_max = R_A = 40 kN Maximum Bending Moment (at midspan, where V = 0): M_max = wL²/8 = 10(8)²/8 = 80 kN·m

Question Type

numerical

Answer Structure

• Step 1: Compute reactions — R_A = R_B = wL/2 [1 mark]
• Step 2: Apply standard formulas for V_max and M_max with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using M_max = wL²/4 (the formula for a point load at midspan) — wrong formula, loses 1 mark.
• Omitting units (kN, kN·m) — typically –0.5 mark per missing unit in board-style rubrics.
• Stating V_max at midspan instead of at the support.

Key Phrases To Include

• R_A = R_B = wL/2
• V_max = 40 kN (at supports)
• M_max = wL²/8 = 80 kN·m
• at midspan (x = L/2)

Marks

2

Topic

Load–Shear–Moment Relationships

Difficulty

medium

Template Id

T5

Examiner Tip

A 2-mark theory question expects BOTH equations AND a one-line interpretation of each. Simply writing the equations with no explanation earns only partial credit.

Model Answer

The three diagrams are linked by two differential relationships derived from equilibrium of an infinitesimal beam element: (1) dV/dx = −w(x) The slope of the shear diagram at any point equals the negative of the load intensity at that point. A downward UDL (w > 0) makes the shear diagram slope downward. (2) dM/dx = V(x) The slope of the moment diagram at any point equals the shear force at that point. Where V is positive, M increases; where V is negative, M decreases; where V = 0, M has a local extremum. In integrated (area) form: • ΔV = −(area under load diagram between two points) • ΔM = area under shear diagram between two points These relationships allow the SFD and BMD to be constructed graphically without writing individual section equations.

Question Type

short_answer

Answer Structure

• Part A: State and explain dV/dx = −w(x) with physical meaning [1 mark]
• Part B: State and explain dM/dx = V(x) with the consequence at V = 0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing dV/dx = +w(x) (wrong sign) — loses the first mark.
• Stating the relationships without explaining their physical meaning — partial credit only.
• Confusing which relationship applies to shear vs. moment.

Key Phrases To Include

• dV/dx = −w(x)
• dM/dx = V(x)
• slope of the shear diagram
• slope of the moment diagram
• extremum where V = 0
• area under the diagram

Marks

3

Topic

Simply Supported Beam — Point Load

Difficulty

medium

Template Id

T6

Examiner Tip

Show ΣM explicitly before substituting. For the 3-mark structure: 1 mark = reactions, 1 mark = V = 0 location, 1 mark = M_max with location. All three must be explicit to earn all three marks.

Model Answer

Given: L = 6 m, P = 30 kN, a = 2 m from A, b = L − a = 4 m Step 1 — Reactions: ΣM_A = 0: R_B(6) = 30(2) → R_B = 10 kN ↑ ΣF_y = 0: R_A + 10 = 30 → R_A = 20 kN ↑ Step 2 — Location of M_max: Shear to the left of the load = +R_A = +20 kN Shear to the right of the load = 20 − 30 = −10 kN V changes sign AT the load point (x = 2 m from A) → M_max is under the load. Step 3 — Maximum Bending Moment: M_max = R_A × a = 20 × 2 = 40 kN·m Check: M_max = Pab/L = 30(2)(4)/6 = 40 kN·m ✓ Answer: R_A = 20 kN, R_B = 10 kN, M_max = 40 kN·m at x = 2 m from A.

Question Type

numerical

Answer Structure

• Step 1: Equilibrium equations for R_A and R_B with ΣM and ΣF_y [1 mark]
• Step 2: Identify zero-shear location (under the point load) [1 mark]
• Step 3: Compute M_max = R_A × a with units and verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Taking moments about the wrong point without showing the equation — partial method marks only.
• Not stating the location of M_max — loses the third mark.
• Using M_max = PL/4 (midspan formula) — wrong for an off-center load, loses 1 mark.

Key Phrases To Include

• ΣM_A = 0
• ΣF_y = 0
• R_A = 20 kN, R_B = 10 kN
• V changes sign at x = 2 m
• M_max = Pab/L = 40 kN·m

Marks

3

Topic

Load–Shear–Moment Relationships

Difficulty

medium

Template Id

T7

Examiner Tip

A 3-mark theory question is typically marked 1 mark per load type. You earn each mark only if you correctly identify BOTH the SFD shape AND the BMD shape for that load type.

Model Answer

The shape of the SFD and BMD is governed by successive integration of the load diagram (from dV/dx = −w and dM/dx = V). The degree increases by one with each integration: 1. Point Load (concentrated force, degree 0 as a load): • SFD: constant between loads (degree 0 — horizontal line); a sudden step (jump) equal to the load magnitude occurs at the load point. • BMD: linear (degree 1 — straight sloping line) between load points, with a kink directly under the point load. 2. Uniformly Distributed Load — UDL (degree 1 linear load): • SFD: linear (degree 1 — straight sloped line). • BMD: parabolic (degree 2 — curved); opens downward for a downward UDL on a simply supported beam. 3. Uniformly Varying Load — UVL / Triangular Load (degree 1 to degree 2 linearly varying load): • SFD: parabolic (degree 2 — curved). • BMD: cubic (degree 3 — S-shaped or single-peaked curve). Memory rule: each load type produces diagrams one degree higher than the load itself.

Question Type

short_answer

Answer Structure

• Part 1: Point load → SFD constant with step, BMD linear [1 mark]
• Part 2: UDL → SFD linear, BMD parabolic [1 mark]
• Part 3: UVL → SFD parabolic, BMD cubic [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating UDL gives a constant SFD — incorrect; UDL gives a LINEAR (sloped) SFD.
• Mixing up which diagram is linear and which is parabolic for a UDL.
• Calling the UVL result a 'quadratic moment diagram' instead of cubic.

Key Phrases To Include

• degree increases by one with each integration
• point load → step in SFD, linear BMD
• UDL → linear SFD, parabolic BMD
• UVL → parabolic SFD, cubic BMD
• dV/dx = −w, dM/dx = V

Marks

3

Topic

Cantilever Beam — Combined Loading

Difficulty

medium

Template Id

T8

Examiner Tip

For cantilever problems, isolate the free end (no unknown reactions there) and sum toward the fixed end. The UDL resultant acts at L/2 from the free end; the point load acts at L. Do NOT confuse these moment arms.

Model Answer

Given: L = 3 m, P = 15 kN (at free end), w = 6 kN/m (full span) Sign convention: shear positive upward-left; moment positive sagging. Cantilever loads cause hogging (negative) moment at the fixed end. Step 1 — Shear at the fixed support (section at x = 0, isolating the right-hand free body): V = P + w·L = 15 + 6(3) = 15 + 18 = 33 kN (The fixed support must provide 33 kN upward.) Step 2 — Bending moment at the fixed support (taking moments of all loads about the fixed end, hogging = negative): M = −[P·L + w·L·(L/2)] M = −[15(3) + 6(3)(1.5)] M = −[45 + 27] M = −72 kN·m Answer: V at fixed support = 33 kN ↑ M at fixed support = 72 kN·m (hogging / negative) Note: The fixed support moment of 72 kN·m is the maximum magnitude bending moment in the beam.

Question Type

numerical

Answer Structure

• Step 1: Draw/describe sign convention and beam configuration [0.5 mark implicit]
• Step 2: Sum vertical forces for shear at fixed end: V = P + wL [1 mark]
• Step 3: Take moments about fixed end for M, applying both load contributions with correct moment arms [2 marks]

Scoring Breakdown

Common Mark Deductions

• Forgetting to include the UDL contribution to shear — loses 1 mark.
• Using L/3 instead of L/2 for the moment arm of the UDL resultant (L/3 is for a triangular load).
• Giving moment as positive when it is clearly hogging — sign error costs 1 mark.
• Treating the moment as wL² instead of wL·(L/2) = wL²/2.

Key Phrases To Include

• V = P + wL = 33 kN
• M = −[P·L + wL·(L/2)]
• M = −72 kN·m (hogging)
• moment arm of UDL resultant = L/2
• fixed end moment is maximum

Marks

2

Topic

Discontinuities — Concentrated Couple

Difficulty

medium

Template Id

T9

Examiner Tip

This is a classic board-exam trap. The examiner is testing whether you know couples affect MOMENT diagrams only, not shear diagrams. State both effects — the non-effect on SFD is worth the same mark as the effect on BMD.

Model Answer

When a concentrated couple (applied moment) M_o acts at a point along a beam: Effect on SFD: None. A couple has no net force resultant, so it does NOT cause any jump or change in the shear force diagram. The shear force value is continuous through the point of the couple. Effect on BMD: The bending moment diagram undergoes a SUDDEN JUMP equal to the magnitude of the couple at that point. The jump is upward (positive step) if the couple is clockwise when viewed from the standard orientation, and downward (negative step) if counterclockwise — consistent with the sign convention for positive moments (sagging = positive). In summary: A concentrated couple → step in BMD only; zero change in SFD.

Question Type

short_answer

Answer Structure

• Part A: SFD is unaffected — no jump, no change [1 mark]
• Part B: BMD has a sudden jump equal to M_o at the point of application [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the couple causes a jump in the SFD — this is the most common error on board exams; loses 1 mark.
• Saying the jump in BMD equals the moment of the couple about the end (incorrect; it equals M_o directly).
• Omitting the direction of the jump (sign of the step).

Key Phrases To Include

• concentrated couple has no net force
• SFD is unaffected / no jump in SFD
• BMD has a sudden jump equal to M_o
• step in the moment diagram
• continuous shear through the couple

Marks

5

Topic

Simply Supported Beam — Triangular (UVL) Load

Difficulty

hard

Template Id

T10

Examiner Tip

The UVL problem is the hardest standard type in the SFD/BMD topic. The two most dangerous traps are (1) using L/3 instead of 2L/3 for the centroid and (2) using x/2 instead of x/3 for the partial-triangle moment arm. Write these explicitly — the examiner awards marks for showing the correct centroid, not just the correct final number.

Model Answer

Given: L = 9 m, w_o = 18 kN/m (at B), load intensity at x from A: w(x) = (w_o/L)x = 2x kN/m --- Part (a): Support Reactions --- Total resultant load: W = ½ w_o L = ½(18)(9) = 81 kN Centroid of triangular load: at 2L/3 from the zero end (A) = 2(9)/3 = 6 m from A ΣM_A = 0: R_B(9) = 81(6) → R_B = 54 kN ↑ ΣF_y = 0: R_A + 54 = 81 → R_A = 27 kN ↑ ✓ Check ΣM_B = 0: R_A(9) = 81(9 − 6) = 81(3) = 243; 27(9) = 243 ✓ --- Part (b): Location of Maximum Bending Moment --- M is maximum where V = 0. Shear at distance x from A: V(x) = R_A − (area of load triangle from 0 to x) Partial triangle resultant from 0 to x: ½ × w(x) × x = ½(2x)(x) = x² Set V(x) = 0: 27 − x² = 0 x² = 27 x = √27 = 3√3 ≈ 5.196 m from A --- Part (c): Maximum Bending Moment --- The partial resultant x² acts at 2x/3 from A (centroid of the sub-triangle): M(x) = R_A · x − x² · (x/3) M_max = 27(5.196) − (5.196)²·(5.196/3) M_max = 27(5.196) − (27)(5.196/3) M_max = 5.196[27 − 27/3] M_max = 5.196[27 − 9] M_max = 5.196(18) M_max = 93.53 kN·m Alternative check using M_max = (2R_A/w_o)·R_A − (1/3)(2R_A/w_o)³·(w_o/L): M_max = 2(27²)/(3·18) = 2(729)/54 = 1458/54 = 27.0... (use direct formula above) Final Answers: (a) R_A = 27 kN ↑, R_B = 54 kN ↑ (b) Maximum moment at x = 3√3 ≈ 5.196 m from A (left support) (c) M_max ≈ 93.5 kN·m

Question Type

numerical

Answer Structure

• Step 1: Compute total UVL resultant W = ½w_o·L and locate centroid at 2L/3 from zero end [1 mark]
• Step 2: Apply ΣM_A = 0 and ΣF_y = 0 for R_B and R_A [1 mark]
• Step 3: Write V(x) = R_A − x² expression and set V = 0 to find x [1 mark]
• Step 4: Write M(x) expression using the sub-triangle centroid [1 mark]
• Step 5: Substitute x = 3√3 and compute M_max with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Placing the UVL centroid at L/3 from the zero-intensity end instead of 2L/3 — wrong reactions and everything downstream fails.
• Using the moment arm x/2 instead of x/3 for the sub-triangle resultant in the moment equation.
• Locating M_max at midspan (x = 4.5 m) by assumption instead of solving V(x) = 0.
• Forgetting to square x in the sub-triangle area (using x instead of x²).

Key Phrases To Include

• W = ½w_o·L = 81 kN
• centroid at 2L/3 = 6 m from zero-intensity end
• R_A = 27 kN, R_B = 54 kN
• V(x) = R_A − x² = 0 → x = √27
• M(x) = R_A·x − x²·(x/3)
• M_max = 93.5 kN·m at x = 5.196 m

Marks

5

Topic

Overhanging Beam — UDL

Difficulty

hard

Template Id

T11

Examiner Tip

Overhanging beam problems ALWAYS require checking both the span maximum (where V = 0 in AB) and the support B moment (computed from the overhang). The board exam is designed so that candidates who check only the span will miss the governing case. Always compute M at every support and every zero-shear point.

Model Answer

Given: A at x=0, B at x=5 m, C at x=7 m; w = 8 kN/m over entire 7 m --- Part (a): Support Reactions --- Total load: W = 8(7) = 56 kN acting at midlength x = 3.5 m from A ΣM_A = 0: R_B(5) = 56(3.5) → R_B = 196/5 = 39.2 kN ↑ ΣF_y = 0: R_A + 39.2 = 56 → R_A = 16.8 kN ↑ ✓ ΣM_B = 0: R_A(5) = 8(7)(3.5−0) − R_B(0)... [verify by right segment] Right segment check (B to C): M_B from right = −8(2)(1) = −16 kN·m (hogging) --- Part (b): Zero Shear in Span AB --- V(x) in span AB (0 ≤ x ≤ 5): V(x) = R_A − w·x = 16.8 − 8x Set V = 0: 16.8 − 8x = 0 → x = 16.8/8 = 2.1 m from A --- Part (c): Maximum Positive and Negative Moments --- Maximum POSITIVE moment (sagging) at x = 2.1 m: M(2.1) = R_A(2.1) − w(2.1)²/2 = 16.8(2.1) − 8(2.1)²/2 M = 35.28 − 17.64 = 17.64 kN·m (sagging) ← maximum positive Maximum NEGATIVE moment (hogging) at support B (check from right segment): M_B = −[w × overhang × (overhang/2)] = −[8(2)(1)] = −16 kN·m (hogging) Alternative check M_B from left: M_B = R_A(5) − 8(5)(2.5) = 16.8(5) − 100 = 84 − 100 = −16 kN·m ✓ Final Answers: (a) R_A = 16.8 kN ↑, R_B = 39.2 kN ↑ (b) V = 0 at x = 2.1 m from A (c) M_max(+) = 17.64 kN·m (sagging) at x = 2.1 m; M_max(−) = 16 kN·m (hogging) at B

Question Type

numerical

Answer Structure

• Step 1: Compute total load and apply ΣM_A = 0 for R_B [1 mark]
• Step 2: Apply ΣF_y = 0 for R_A and verify with right-segment moment check [0.5 mark]
• Step 3: Write V(x) for span AB and solve V = 0 for zero-shear location [1 mark]
• Step 4: Compute M at zero-shear point for maximum positive moment [1 mark]
• Step 5: Compute M at support B from the overhang (right segment) for maximum negative moment [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not computing M_B from the overhang segment — the most common omission in overhanging beam problems.
• Taking moments of the UDL only over the span AB (ignoring the overhang contribution to reactions).
• Using the total 7 m length centroid incorrectly in the moment equation for ΣM_A.
• Failing to state which moment governs design.

Key Phrases To Include

• R_B = 39.2 kN, R_A = 16.8 kN
• V(x) = 16.8 − 8x = 0 → x = 2.1 m
• M_max(sagging) = 17.64 kN·m at x = 2.1 m
• M_B = −16 kN·m (hogging at support B)
• check from the overhang (right segment)

Marks

2

Topic

Sign Convention

Difficulty

easy

Template Id

T12

Examiner Tip

The stress-state description (which fiber is in compression, which in tension) is what earns full marks — the smile/frown analogy alone is only a memory aid. For RC design context, mentioning rebar placement location adds valuable depth.

Model Answer

Sagging Moment (Positive Moment): A bending moment that causes the beam to bend concave UPWARD, creating compression at the top fiber and tension at the bottom fiber. Visually, the deflected shape resembles a smile. It is the dominant moment type in the MIDSPAN region of simply supported beams and in the positive-moment regions of continuous beams. Hogging Moment (Negative Moment): A bending moment that causes the beam to bend concave DOWNWARD, creating tension at the top fiber and compression at the bottom fiber. Visually, the deflected shape resembles a frown. It is dominant in CANTILEVERS (throughout their length), at interior supports of continuous beams, and in the overhang region of overhanging beams. Design significance: In reinforced concrete (NSCP 2015 / ACI 318), tensile steel is placed at the BOTTOM for sagging regions and at the TOP for hogging regions.

Question Type

short_answer

Answer Structure

• Part A: Define sagging (positive) — concave up, compression top, tension bottom, smile shape [1 mark]
• Part B: Define hogging (negative) — concave down, tension top, compression bottom, frown shape; typical locations [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the fiber stress distribution for either type.
• Failing to mention the typical structural configuration where each type dominates.
• Using only the smile/frown analogy without describing the fiber stress state — incomplete for full marks.

Key Phrases To Include

• sagging / positive moment
• concave upward
• compression at top, tension at bottom
• hogging / negative moment
• concave downward
• tension at top, compression at bottom
• smile vs frown analogy
• cantilever, interior support

Marks

5

Topic

Simply Supported Beam — Multiple Point Loads

Difficulty

hard

Template Id

T13

Examiner Tip

For diagram-based problems, label ALL key values on the SFD and BMD. An unlabeled sketch earns no diagram marks. Show computations for at least two moment ordinates — the examiner expects verification, not just the peak value.

Model Answer

Given: L = 10 m, P₁ = 20 kN at x = 3 m, P₂ = 30 kN at x = 7 m --- Step 1: Support Reactions --- ΣM_A = 0: R_B(10) = 20(3) + 30(7) = 60 + 210 = 270 → R_B = 27 kN ↑ ΣF_y = 0: R_A + 27 = 20 + 30 = 50 → R_A = 23 kN ↑ --- Step 2: Shear Force Values at Key Points --- At A (x = 0⁺): V = +R_A = +23 kN Just left of P₁ (x = 3⁻): V = +23 kN Just right of P₁ (x = 3⁺): V = 23 − 20 = +3 kN [drop of 20 kN] Just left of P₂ (x = 7⁻): V = +3 kN Just right of P₂ (x = 7⁺): V = 3 − 30 = −27 kN [drop of 30 kN] At B (x = 10⁻): V = −27 kN At B: V rises by R_B = +27 → V = 0 ✓ Shear crosses zero at x = 7 m (under P₂) — wait, check: V changes from +3 to −27 at x = 7 m. Zero crossing IS at x = 7 m. --- Step 3: Bending Moment Values at Key Points --- M_A = 0 (pin support) M at x = 3 m: M = R_A(3) = 23(3) = 69 kN·m M at x = 7 m: M = R_A(7) − P₁(7−3) = 23(7) − 20(4) = 161 − 80 = 81 kN·m M_B = 0 (pin support) Check from right at x = 7: M = R_B(10−7) = 27(3) = 81 kN·m ✓ --- Step 4: SFD and BMD Description --- SFD: Starts at +23 kN at A; constant to x=3 m; drops to +3 kN; constant to x=7 m; drops to −27 kN; constant to B (where +27 kN reaction closes it to zero). Shape: horizontal steps with jumps at each load. BMD: Starts at 0 at A; rises linearly to 69 kN·m at x=3 m; continues rising (but at a lower rate) to 81 kN·m at x=7 m; falls linearly to 0 at B. Shape: triangular/piecewise linear with peak under P₂. --- Final Answer --- R_A = 23 kN, R_B = 27 kN M_max = 81 kN·m at x = 7 m from A (under the 30 kN load, where V = 0 / changes sign)

Question Type

diagram_based

Answer Structure

• Step 1: Compute R_A and R_B from ΣM_A = 0 and ΣF_y = 0 [1 mark]
• Step 2: Compute shear values at each key point (A, just left/right of P₁, just left/right of P₂, B) [1 mark]
• Step 3: Describe/draw SFD shape with all values labeled [1 mark]
• Step 4: Compute moment values at key points (M at 3 m and 7 m) [1 mark]
• Step 5: Identify M_max = 81 kN·m at x = 7 m and describe BMD shape [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assuming M_max is at midspan (x = 5 m) without checking — for two asymmetric loads, M_max is under the larger load, not at midspan.
• Not labeling the SFD and BMD axes and key values — loses the diagram mark.
• Arithmetic error in moment at x = 7 m (forgetting to subtract P₁'s contribution).
• Forgetting to verify equilibrium by checking R_A + R_B = total load.

Key Phrases To Include

• ΣM_A = 0 → R_B = 27 kN
• R_A = 23 kN
• V drops by 20 kN at x = 3 m
• V drops by 30 kN at x = 7 m
• M at x = 3 m = 69 kN·m
• M_max = 81 kN·m at x = 7 m
• V changes sign at x = 7 m

Marks

3

Topic

Standard Formulas — Simply Supported and Cantilever

Difficulty

easy

Template Id

T14

Examiner Tip

These three formulas are the most frequently tested recall items in board-exam MCQ and identification-type questions. Memorize them with their locations AND signs. An answer that gives the formula but the wrong location earns only half marks.

Model Answer

(a) Simply supported beam, central point load P, span L: M_max = PL/4, occurring at MIDSPAN (x = L/2) (b) Simply supported beam, full-span UDL w, span L: M_max = wL²/8, occurring at MIDSPAN (x = L/2) (c) Cantilever, point load P at free end, length L: M_max = PL (hogging, negative), occurring at the FIXED SUPPORT

Question Type

very_short_answer

Answer Structure

• Part a: M_max = PL/4 at midspan [1 mark]
• Part b: M_max = wL²/8 at midspan [1 mark]
• Part c: M_max = PL (hogging) at fixed support [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing wL²/4 for the UDL case — mixing with the point load formula, loses 1 mark.
• Not specifying 'at the fixed support' for the cantilever case — location is part of the required answer.
• Omitting the sign (hogging) for the cantilever moment.

Key Phrases To Include

• PL/4 at midspan
• wL²/8 at midspan
• PL at the fixed support
• hogging / negative (for cantilever)
• x = L/2

Marks

5

Topic

Simply Supported Beam — Partial UDL

Difficulty

hard

Template Id

T15

Examiner Tip

Partial-UDL problems require TWO separate segment analyses. The maximum moment is NOT at midspan and NOT at the end of the distributed load — it is where V = 0 within the loaded region. Show both segments, check M at the boundaries, and explicitly argue which segment contains the maximum.

Model Answer

Given: L = 12 m, UDL w (kN/m) from x = 0 to x = 6 m only --- Step 1: Support Reactions --- Resultant of UDL: W = w(6) = 6w kN, acting at centroid x = 3 m from A ΣM_A = 0: R_B(12) = 6w(3) → R_B = 18w/12 = 1.5w kN ↑ ΣF_y = 0: R_A + 1.5w = 6w → R_A = 4.5w kN ↑ --- Step 2: Segment AB (0 ≤ x ≤ 6 m) — under the UDL --- V(x) = R_A − wx = 4.5w − wx = w(4.5 − x) M(x) = R_A·x − wx²/2 = 4.5wx − wx²/2 --- Step 3: Segment BC (6 m ≤ x ≤ 12 m) — no load --- V(x) = R_A − w(6) = 4.5w − 6w = −1.5w (constant) [Alternatively: V = −R_B = −1.5w from the right] M(x) = R_A·x − 6w(x − 3) = 4.5wx − 6wx + 18w = 18w − 1.5wx [Check: M(12) = 18w − 18w = 0 ✓; M(6) = 18w − 9w = 9w ✓] --- Step 4: Location of Maximum Moment --- In Segment AB: set V(x) = 0: 4.5w − wx = 0 → x = 4.5 m (this is within 0–6 m ✓) In Segment BC: V = −1.5w (constant negative, no zero crossing) → M is decreasing, no local max. Therefore, M is maximum at x = 4.5 m. --- Step 5: Maximum Bending Moment --- M_max = M(4.5) = 4.5w(4.5) − w(4.5)²/2 M_max = 20.25w − 10.125w M_max = 10.125w kN·m Alternatively: M_max = R_A²/(2w) = (4.5w)²/(2w) = 20.25w²/2w = 10.125w ✓ Final Answers: R_A = 4.5w kN, R_B = 1.5w kN V(x) in AB: w(4.5 − x); in BC: −1.5w (constant) M(x) in AB: w(4.5x − x²/2); in BC: w(18 − 1.5x) M_max = 10.125w kN·m (= 81w/8) at x = 4.5 m from A

Question Type

numerical

Answer Structure

• Step 1: Find W = 6w, centroid at 3 m, compute R_A = 4.5w and R_B = 1.5w [1 mark]
• Step 2: Derive V(x) and M(x) for Segment AB (0 to 6 m) [1 mark]
• Step 3: Derive V(x) and M(x) for Segment BC (6 to 12 m) [1 mark]
• Step 4: Set V = 0 in AB → x = 4.5 m; confirm no zero in BC [1 mark]
• Step 5: Substitute x = 4.5 into M(x) for AB → M_max = 10.125w kN·m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Treating the UDL as acting over the full span — wrong resultant, wrong reactions, cascading error.
• Using x = 6 m as the maximum moment point (it is only a kink point, not a maximum).
• Not deriving separate expressions for each segment — a single equation fails to capture the step at x = 6 m.
• Incorrect moment arm for the UDL resultant (e.g., using 6 m instead of 3 m from A).

Key Phrases To Include

• R_A = 4.5w kN, R_B = 1.5w kN
• UDL resultant 6w at centroid x = 3 m
• V(x) = w(4.5 − x) in segment AB
• V = −1.5w in segment BC (constant)
• V = 0 at x = 4.5 m
• M_max = 10.125w = 81w/8 kN·m