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.
Drills 2 to 5 stay premium
Unit 1 keeps the full beam-reaction drill open as the sampler. The composite centroid, area moment of inertia, and resultant drills unlock with the premium pass.