Unit 1 / Practice Drills
Unit 1 Practice Drills
Force Systems
Resultants, centroids, and moment of inertia. This is the most stable math-first scorer in the subject.
Unit 1 Practice Drills
Drill 1 - Simply Supported Beam Reactions: UDL and Point Loads
5 questions. Same procedure every time - own this and you do not drop marks on beam reactions.
Beam-reaction questions are almost never about algebra difficulty. They are about clean load conversion, correct moment arms, and not losing signs on couples.
Figure 0A
Reference FBD: simply supported beam with reactions and UDL
Key reference
| Load type | Convert to | Acts at |
|---|---|---|
| UDL: w kN/m over length L | W = w × L | Midpoint of loaded length |
| UVL: 0 to w kN/m over length L | W = ½ × w × L | L/3 from the larger end |
| Point load | Already a point load | Where it acts |
| Couple M | No conversion needed | Same moment about every point |
Procedure: draw FBD → convert distributed loads → take moments about A → apply ΣFy = 0 → check.
Question 1 - UDL over full span with one point load
Problem
A simply supported beam AB has a span of 8 m. It carries a UDL of 10 kN/m over the entire span and a point load of 20 kN at 3 m from A. Find RA and RB.
Figure 1A
Beam with full-span UDL and one point load
Step 1 - Convert UDL.
W = 10 × 8 = 80 kN at 4 m from A
Step 2 - ΣMA = 0.
RB × 8 − 80 × 4 − 20 × 3 = 0
RB = (320 + 60) / 8 = 47.5 kN ↑
Step 3 - ΣFy = 0.
RA = 80 + 20 − 47.5 = 52.5 kN ↑
Step 4 - Check.
52.5 + 47.5 = 100 kN = 80 + 20 ✓
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 1B
Converted FBD for the moment equation
Where students lose marks
The UDL resultant acts at the midpoint of the loaded length. Here the UDL covers the full 8 m so midpoint = 4 m from A. If the UDL covered only part of the beam, the midpoint would shift with the loaded length, not stay at beam midspan.
Question 2 - UDL over partial span with two point loads
Problem
A simply supported beam AB has a span of 10 m. It carries: a UDL of 6 kN/m from A to C where C is 4 m from A, a point load of 25 kN at 6 m from A, and a point load of 15 kN at 9 m from A. Find RA and RB.
Figure 2A
Partial UDL over AC with two point loads
Step 1 - Convert UDL.
W = 6 × 4 = 24 kN acting at midpoint of AC = 2 m from A
Step 2 - ΣMA = 0.
RB × 10 − 24 × 2 − 25 × 6 − 15 × 9 = 0
RB × 10 = 48 + 150 + 135 = 333
RB = 33.3 kN ↑
Step 3 - ΣFy = 0.
RA = 24 + 25 + 15 − 33.3 = 30.7 kN ↑
Step 4 - Check.
30.7 + 33.3 = 64 kN = 24 + 25 + 15 ✓
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 2B
Converted FBD with UDL replaced by 24 kN at 2 m
Where students lose marks
The UDL only covers A to C. Its resultant acts at 2 m from A, which is the midpoint of the loaded 4 m, not the midpoint of the 10 m beam. Putting it at 5 m ruins the moment equation.
Question 3 - UDL with an applied clockwise couple
Problem
A simply supported beam AB of span 12 m carries a UDL of 5 kN/m over the full span and a clockwise couple of 60 kNm at 4 m from A. Find RA and RB.
Figure 3A
Full-span UDL with a clockwise couple
Step 1 - Convert UDL.
W = 5 × 12 = 60 kN at 6 m from A
Step 2 - ΣMA = 0 (anticlockwise positive).
RB × 12 − 60 × 6 − 60 = 0
The clockwise couple is negative regardless of where it acts on the beam.
RB × 12 = 360 + 60 = 420
RB = 35 kN ↑
Step 3 - ΣFy = 0.
RA = 60 − 35 = 25 kN ↑
Step 4 - Check.
25 + 35 = 60 kN = total UDL load ✓
The couple does not appear in ΣFy because a couple has no net force, only a net moment.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 3B
Converted FBD with UDL replaced by 60 kN
Figure 3C
Sign convention reminder for couples
Where students lose marks
Do not add the couple to ΣFy. A couple has no net force, only a net moment. Also: the sign is set by the rotation sense, not by its position on the beam. Clockwise is negative in an anticlockwise-positive convention.
Question 4 - UVL over full span
Problem
A simply supported beam AB of span 9 m carries a uniformly varying load that is zero at A and increases to 24 kN/m at B. Find RA and RB.
Figure 4A
UVL increasing from A to B
Step 1 - Convert UVL.
W = ½ × 24 × 9 = 108 kN
Acts at L/3 from the larger end (B) = 3 m from B = 6 m from A
Step 2 - ΣMA = 0.
RB × 9 − 108 × 6 = 0
RB = 648 / 9 = 72 kN ↑
Step 3 - ΣFy = 0.
RA = 108 − 72 = 36 kN ↑
Step 4 - Check.
36 + 72 = 108 kN ✓
RB > RA is correct because the UVL is heavier toward B.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 4B
Converted FBD with UVL resultant at 6 m from A
Figure 4C
Wrong placement versus correct placement for the UVL resultant
Where students lose marks
The UVL resultant is placed at L/3 from the larger end, not the zero end. Because the triangle is heavy at B, its centroid is closer to B. That is why the correct resultant is 6 m from A, not 3 m from A.
Question 5 - Mixed loading: UDL, UVL, point load, and couple combined
Problem
A simply supported beam AB of span 10 m carries: a UDL of 8 kN/m from A to D where D is 5 m from A, a UVL from D to B increasing from 0 at D to 20 kN/m at B, a point load of 30 kN at 3 m from A, and an anticlockwise couple of 40 kNm at 7 m from A. Find RA and RB.
This is the most complex beam question type. If you can solve this, the simpler beam problems become routine.
Figure 5A
Mixed loading: UDL, UVL, point load, and anticlockwise couple
Step 1 - Convert UDL (A to D, 5 m).
W₁ = 8 × 5 = 40 kN at midpoint of AD = 2.5 m from A
Step 2 - Convert UVL (D to B, 5 m, zero at D and 20 kN/m at B).
W₂ = ½ × 20 × 5 = 50 kN
Acts at L/3 from B = 5/3 = 1.667 m from B = 8.333 m from A
Step 3 - ΣMA = 0 (anticlockwise positive).
RB × 10 − 40 × 2.5 − 30 × 3 − 50 × 8.333 + 40 = 0
The anticlockwise couple is positive.
RB × 10 = 100 + 90 + 416.65 − 40 = 566.65
RB = 56.67 kN ↑
Step 4 - ΣFy = 0.
RA = 40 + 30 + 50 − 56.67 = 63.33 kN ↑
Step 5 - Check.
63.33 + 56.67 = 120 kN = 40 + 30 + 50 ✓
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 5B
Converted FBD with all equivalent loads shown
Figure 5C
Load-identification checklist before the moment equation
| Load type | Status |
|---|---|
| UDL (A to D) | converted ✓ |
| UVL (D to B) | converted ✓ |
| Point load at 3 m from A | no conversion needed ✓ |
| Couple at 7 m from A | sign checked: positive ✓ |
Where students lose marks
Two traps dominate this question. First, the UVL is only from D to B, so its length is 5 m and its centroid is 1.667 m from B = 8.333 m from A. Second, the couple sign matters: anticlockwise is positive here, so it reduces RB slightly in the final moment balance.
Unit 1 Practice Drills
Drill 2 - Composite Centroid: L-Plate with Circular Cut-Out
1 worked problem. Medium-hard centroid table with two negative areas.
This is the repeated centroid-table pattern: choose one reference corner, mark holes as negative areas, and do not substitute until the whole table is built.
Key reference
Use the same origin for every x and y coordinate.
Question 1 - Composite centroid with a cut-out and a hole
Problem
A composite lamina consists of an outer rectangle 180 mm × 140 mm, a rectangular cut-out 100 mm × 80 mm removed from the top-right corner, and a circular hole of diameter 40 mm whose centre is 45 mm from the left edge and 40 mm from the bottom edge. Find the centroid of the remaining lamina from the bottom-left corner O.
Figure 1A
Composite lamina with a rectangular cut-out and circular hole
Step 1 - Split the section and assign signs.
A₁ = 180 × 140 = 25200 mm² at (x₁, y₁) = (90, 70)
A₂ = -(100 × 80) = -8000 mm² at (x₂, y₂) = (130, 100)
A₃ = -π × 20² = -1256.64 mm² at (x₃, y₃) = (45, 40)
Step 2 - Find the total area.
ΣA = 25200 − 8000 − 1256.64 = 15943.36 mm²
Step 3 - Find x̄.
ΣAx = 25200 × 90 − 8000 × 130 − 1256.64 × 45 = 1171451.2
x̄ = ΣAx / ΣA = 1171451.2 / 15943.36 = 73.48 mm
Step 4 - Find ȳ.
ΣAy = 25200 × 70 − 8000 × 100 − 1256.64 × 40 = 913734.4
ȳ = ΣAy / ΣA = 913734.4 / 15943.36 = 57.31 mm
Step 5 - Sense-check the location.
The large rectangular cut-out at the top-right pulls the centroid left and downward from the outer rectangle centre. The circular hole near the bottom-left shifts it slightly back toward the right and upward.
Centroid G = (73.48 mm, 57.31 mm) from O
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 1B
Solved centroid location for the composite lamina
Where students lose marks
Students often subtract the hole area in ΣA but forget to subtract its moment terms in ΣAx and ΣAy. The sign must stay negative in every column.
Unit 1 Practice Drills
Drill 3 - Area Moment of Inertia: T-Section About x-x
1 worked problem. Classic centroid plus parallel-axis theorem workflow.
Area moment of inertia questions go wrong when students jump straight to I = bh³ / 12 for the whole section. Split it first, find the centroidal axis, then move each rectangle with Ad².
Key reference
Use the dimension perpendicular to the axis as h.
Question 1 - T-section centroid and area MOI
Problem
A T-section has a flange 120 mm × 20 mm and a web 20 mm × 100 mm attached centrally below the flange. Find the area moment of inertia of the section about its centroidal horizontal axis x-x.
Figure 1A
T-section for centroid and area moment of inertia
Step 1 - Split the section into flange and web.
Flange: A₁ = 120 × 20 = 2400 mm², y₁ = 110 mm from base
Web: A₂ = 20 × 100 = 2000 mm², y₂ = 50 mm from base
Step 2 - Find the centroidal x-x axis first.
ȳ = (A₁y₁ + A₂y₂) / (A₁ + A₂)
ȳ = (2400 × 110 + 2000 × 50) / 4400 = 82.73 mm from base
Step 3 - Write the local rectangle moments of inertia.
I₁ = bh³/12 = 120 × 20³ / 12 = 80000 mm⁴
I₂ = bh³/12 = 20 × 100³ / 12 = 1666666.67 mm⁴
Step 4 - Apply the parallel-axis theorem.
d₁ = 110 − 82.73 = 27.27 mm, d₂ = 82.73 − 50 = 32.73 mm
Iₓ = I₁ + A₁d₁² + I₂ + A₂d₂²
Iₓ = 80000 + 2400(27.27)² + 1666666.67 + 2000(32.73)²
Step 5 - Add the contributions.
Iₓ = 5673939.39 mm⁴
Iₓ ≈ 5.674 × 10⁶ mm⁴ about the centroidal x-x axis
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 1B
Centroidal axis x-x used for the parallel-axis theorem
Where students lose marks
The most common mistake is using the total section depth in every rectangle formula. Each component keeps its own b and h, and the shift to the common axis happens only through the Ad² term.
Unit 1 Practice Drills
Drill 4 - Resultant of Concurrent Forces
1 worked problem. Medium-hard force-resolution drill with a quadrant check.
This is the cleanest force-systems scorer in the unit. The exam marks go missing when signs are guessed instead of read from the geometry.
Key reference
Question 1 - Three-force concurrent system at O
Problem
Three concurrent forces act at a point O: 12 kN at 25° above the positive x-axis, 7 kN at 120° from the positive x-axis, and 9 kN vertically downward. Find the magnitude and direction of the single resultant.
Figure 1A
Three concurrent forces acting at O
Step 1 - Resolve each force into x and y components.
F₁ = 12 kN at 25°: (12 cos 25°, 12 sin 25°) = (10.876, 5.071)
F₂ = 7 kN at 120°: (7 cos 120°, 7 sin 120°) = (-3.500, 6.062)
F₃ = 9 kN downward: (0, -9)
Step 2 - Sum the horizontal components.
ΣFₓ = 10.876 − 3.500 + 0 = 7.376 kN
Step 3 - Sum the vertical components.
ΣFᵧ = 5.071 + 6.062 − 9 = 2.134 kN
Step 4 - Find the resultant magnitude.
R = √[(ΣFₓ)² + (ΣFᵧ)²] = √[(7.376)² + (2.134)²] = 7.678 kN
Step 5 - Find the direction and quadrant.
θ = tan⁻¹(ΣFᵧ / ΣFₓ) = tan⁻¹(2.134 / 7.376) = 16.1°
Both ΣFₓ and ΣFᵧ are positive, so the resultant lies in the first quadrant.
Resultant = 7.678 kN at 16.1° above the +x-axis
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 1B
Resultant direction after summing x and y components
Where students lose marks
The 120° force does not have both components positive. In the second quadrant its x-component is negative and its y-component is positive. One wrong sign changes the whole resultant direction.
Unit 1 Practice Drills
Drill 5 - Resultant of Parallel Forces and Line of Action
1 worked problem. Replace a parallel force system by one equivalent force.
This is where magnitude alone is not enough. The examiner usually wants both the single resultant and where it acts, so moment balance has to survive right to the final line.
Key reference
Upward forces reduce both the net force and the net moment if their sense opposes the chosen convention.
Question 1 - Equivalent resultant and its position
Problem
Along a straight line from a reference point O, three vertical forces act: 15 kN downward at 1.5 m, 25 kN downward at 4 m, and 10 kN upward at 6 m. Replace the system by a single resultant and give its position from O.
Figure 1A
Parallel vertical forces on a straight line from O
Step 1 - Find the single equivalent vertical force.
R = 15 + 25 − 10 = 30 kN downward
Step 2 - Take moments about O.
ΣMₒ = 15(1.5) + 25(4) − 10(6)
ΣMₒ = 22.5 + 100 − 60 = 62.5 kNm
Step 3 - Equate the moment of the resultant.
R × x = 62.5 → 30x = 62.5
Step 4 - Solve for the line of action.
x = 62.5 / 30 = 2.083 m from O
Step 5 - State the final equivalent system.
The original three-force system is equivalent to one downward force located between the two larger downward loads, which is physically sensible.
Equivalent resultant = 30 kN downward acting 2.08 m from O
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal solution figure
Open this only after you have attempted the setup yourself.
Figure 1B
Single equivalent resultant replacing the original force system
Where students lose marks
Students often subtract the upward force in ΣF but then forget to subtract its moment term. The sign of the force and the sign of its moment must stay consistent with the same convention.