Unit 2 / Practice Drills
Unit 2 Practice Drills
Equilibrium
Beam reactions, free-body diagrams, and classic contact problems. This is the safest high-ROI unit.
Simply Supported Beam Reactions
5 drills. Start from the top. Attempt each problem before revealing any steps.
These problems are drawn from the SPPU FE Engineering Mechanics paper families. The procedure is identical across all five. Own it here and you will not drop marks on beam reactions in the exam.
How to use these drills
Attempt the problem on paper first. Then reveal Step 1. Check your work. Continue step by step. If your answer matches at each step, move on. If it does not, stop at that step and find your error before continuing.
Unit 2 Practice Drills
Drill 1 - UDL over full span with a point load
Problem
A simply supported beam AB has a span of 8 m. It carries a uniformly distributed load of 10 kN/m over its entire length and a point load of 20 kN acting vertically downward at 3 m from A. Find the support reactions at A and B.
Figure 1A
Beam with full-span UDL and point load
Step 1 - Convert the UDL to an equivalent point load.
Total UDL load = w × L = 10 × 8 = 80 kN.
This acts at the midpoint of the beam = 4 m from A.
Check: you should have two loads now: 80 kN at 4 m from A, and 20 kN at 3 m from A.
Step 2 - Take moments about A to find RB.
ΣMA = 0 (anticlockwise positive):
RB × 8 − 80 × 4 − 20 × 3 = 0
RB × 8 = 320 + 60 = 380
RB = 47.5 kN ↑
Step 3 - Apply ΣFy = 0 to find RA.
RA + RB = total downward load
RA + 47.5 = 80 + 20 = 100
RA = 52.5 kN ↑
Step 4 - Check.
RA + RB = 52.5 + 47.5 = 100 kN, which matches the total downward load.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Converted FBD for the moment equation
The UDL is replaced by an equivalent 80 kN point load acting at the midpoint of the loaded 8 m span.
Where students lose marks on this type
The UDL resultant acts at the midpoint of the loaded length, not automatically the midpoint of the beam. Here the UDL covers the full 8 m so it acts at 4 m from A. If the UDL covered only 6 m from A, the resultant would act at 3 m from A, not 4 m.
Unit 2 Practice Drills
Drill 2 - UVL over full span
Problem
A simply supported beam AB has a span of 6 m. It carries a uniformly varying load that starts at 0 kN/m at A and increases to 18 kN/m at B. Find the support reactions at A and B.
Figure 1A
UVL increasing from A to B
Step 1 - Convert the UVL to an equivalent point load.
Total UVL load = ½ × w × L = ½ × 18 × 6 = 54 kN.
This acts at L/3 from the larger end = L/3 from B = 6/3 = 2 m from B = 4 m from A.
Check: you should have one load: 54 kN acting at 4 m from A.
Step 2 - Take moments about A to find RB.
ΣMA = 0:
RB × 6 − 54 × 4 = 0
RB × 6 = 216
RB = 36 kN ↑
Step 3 - Apply ΣFy = 0 to find RA.
RA + 36 = 54
RA = 18 kN ↑
Step 4 - Check.
RA + RB = 18 + 36 = 54 kN.
Notice RA ≠ RB even though the span is symmetric. That is correct because the load is heavier toward B.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Equivalent point load for the UVL
The UVL resultant is 54 kN and acts at L/3 from the larger end B, which is 4 m from A.
Where students lose marks on this type
The UVL resultant acts at L/3 from the larger end. Students often place it at L/3 from A, the zero end, which gives the wrong moment arm and flips the answer.
Unit 2 Practice Drills
Drill 3 - Point load, UDL over partial span, and an applied couple
Problem
A simply supported beam AB has a span of 10 m. It carries a point load of 30 kN at 2 m from A, a UDL of 8 kN/m over 4 m from C to D where C is 4 m from A and D is 8 m from A, and a clockwise couple of 50 kN·m at 7 m from A. Find the support reactions at A and B.
Figure 1A
Point load, partial UDL, and clockwise couple
Step 1 - Convert the UDL to an equivalent point load.
Total UDL = 8 × 4 = 32 kN.
It acts at the midpoint of CD = 4 + 2 = 6 m from A.
Check: you now have three loads: 30 kN at 2 m, 32 kN at 6 m, and a 50 kN·m couple.
Step 2 - Take moments about A to find RB.
ΣMA = 0 (anticlockwise positive):
RB × 10 − 30 × 2 − 32 × 6 − 50 = 0
The clockwise couple is negative in an anticlockwise-positive convention regardless of where it acts.
RB × 10 = 60 + 192 + 50 = 302
RB = 30.2 kN ↑
Step 3 - Apply ΣFy = 0 to find RA.
RA + 30.2 = 30 + 32 = 62
RA = 31.8 kN ↑
Step 4 - Check.
RA + RB = 31.8 + 30.2 = 62 kN, matching the total downward force.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Converted FBD with the couple retained
The 8 kN/m UDL over 4 m becomes a 32 kN point load acting at the midpoint of C-D, which is 6 m from A. The couple stays as a pure moment term.
Where students lose marks on this type
There are two traps here. First, the couple sign: a clockwise couple is negative when anticlockwise is positive. Second, the UDL moment arm: the load covers C to D, not A to B, so its resultant acts at 6 m from A, not 5 m.
Unit 2 Practice Drills
Drill 4 - The April 2025 repeat
This exact problem appeared in April 2025.
Problem
A simply supported beam AB of span 8 m carries two point loads of 50 kN each at 2 m and 5 m from A, and a clockwise couple of 50 kN·m at 6 m from A. Find the reactions at A and B.
Figure 1A
Two point loads and a clockwise couple
This is the April 2025 repeat. No distributed loads need conversion in this one.
Step 1 - List all loads and their positions.
50 kN downward at 2 m from A.
50 kN downward at 5 m from A.
50 kN·m clockwise couple at 6 m from A.
No distributed loads need conversion.
Step 2 - Take moments about A to find RB.
ΣMA = 0 (anticlockwise positive):
RB × 8 − 50 × 2 − 50 × 5 − 50 = 0
RB × 8 = 100 + 250 + 50 = 400
RB = 50 kN ↑
Step 3 - Apply ΣFy = 0 to find RA.
RA + 50 = 50 + 50 = 100
RA = 50 kN ↑
Step 4 - Check.
RA + RB = 50 + 50 = 100 kN.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked FBD showing that the couple remains a moment
There is nothing to convert here. The solution figure is still useful because it shows that the couple enters ΣM_A = 0 but not ΣF_y = 0.
Where students lose marks on this type
This one looks symmetric after the equations are written, but students still lose marks by treating the couple like a force. It is a pure moment term and does not enter the vertical-force balance.
Unit 2 Practice Drills
Drill 5 - Beam with an inclined point load
Problem
A simply supported beam AB has a span of 6 m. A load of 60 kN acts at 4 m from A at an angle of 30° to the vertical. Find the support reactions at A and B. Hinge at A, roller at B.
Figure 1A
Inclined point load on a simply supported beam
Step 1 - Resolve the inclined load into components.
Vertical component = 60 cos 30° = 51.96 kN downward.
Horizontal component = 60 sin 30° = 30 kN horizontal.
Only the vertical component creates vertical reactions at A and B. The horizontal component is resisted by the hinge at A.
Step 2 - Apply ΣFx = 0.
RAx = 30 kN, opposite to the horizontal load component.
Step 3 - Take moments about A to find RB.
ΣMA = 0:
RB × 6 − 51.96 × 4 = 0
RB = 207.84 / 6 = 34.64 kN ↑
Step 4 - Apply ΣFy = 0 to find RAy.
RAy + 34.64 = 51.96
RAy = 17.32 kN ↑
Step 5 - Find the magnitude of the total reaction at A.
|RA| = √(30² + 17.32²) = √1200 = 34.64 kN.
Direction = tan⁻¹(17.32 / 30) = 30° to the horizontal.
Step 6 - Check.
RB + RAy = 34.64 + 17.32 = 51.96 kN, which matches the vertical component.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Resolved components used for equilibrium
Use only the vertical component in the moment equation. The horizontal component is balanced by the hinge reaction R_Ax.
Where students lose marks on this type
Students often substitute the full 60 kN directly into the moment equation instead of using only the vertical component. Resolve the load first, always.
After these five drills
You have covered:
- UDL over full span
- UVL over full span
- Mixed loading with a couple
- A close repeat-family beam variant
- An inclined load
These are the five problem families that cover beam-reaction questions in SPPU FE Engineering Mechanics papers. If you can solve all five without hesitation, you will not drop marks on Unit 2 beam reactions.