PEC EPE Exam Structure Analysis Numerical Questions

PEC EPE Exam Structure Analysis Numerical Questions

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Other PEC EPE Exam Past Paper Questions

    Calculate the reactions at the supports of a simply supported beam with a uniformly distributed load of 20 kN/m and a span of 8 meters.

    Solution:

    Total load = Uniformly distributed load × Span length

    Total load = 20 kN/m × 8 m = 160 kN

    Each support reaction = Total load / 2

    Support reactions = 160 kN / 2 = 80 kN

    Answer: Each support reaction ≈ 80 kN

    Calculate the deflection at the midspan of a simply supported beam with a concentrated load of 30 kN at the center. The span of the beam is 6 meters and the flexural rigidity (EI) is 100 kN·m².

    Solution:

    Deflection at midspan = (Load × Span length³) / (48 × EI)

    Deflection at midspan = (30 kN × (6 m)³) / (48 × 100 kN·m²) ≈ 0.375 m

    Answer: Deflection at midspan ≈ 0.375 m

    Calculate the bending moment at the fixed support of a cantilever beam with a concentrated load of 40 kN at the free end. The length of the cantilever beam is 4 meters.

    Solution:

    Bending moment at fixed support = Load × Length

    Bending moment at fixed support = 40 kN × 4 m = 160 kN·m

    Answer: Bending moment at fixed support ≈ 160 kN·m

    A simply supported beam with a span of 10 meters carries a uniformly distributed load of 15 kN/m. Calculate the maximum bending moment in the beam.

    Solution:

    Maximum bending moment = (Uniformly distributed load × Span length) / 8

    Maximum bending moment = (15 kN/m × 10 m) / 8 = 18.75 kN·m

    Answer: Maximum bending moment ≈ 18.75 kN·m

    Calculate the force in each member of a truss shown below, where all members are pinned and the applied load is 20 kN at joint C.

    Solution:

    By applying the method of joints, we can determine the forces in each member:

    AC: Force = 20 kN (Tension, since it is elongating)

    AB: Force = 20 kN (Compression, since it is shortening)

    BC: Force = 20 kN (Tension, since it is elongating)

    Answer:

    Force in AC ≈ 20 kN (Tension)

    Force in AB ≈ 20 kN (Compression)

    Force in BC ≈ 20 kN (Tension)

    Calculate the deflection at the free end of a cantilever beam with a uniformly distributed load of 25 kN/m. The length of the cantilever beam is 5 meters and the flexural rigidity (EI) is 150 kN·m².

    Solution:

    Deflection at free end = (5 × Load × Length⁴) / (384 × EI)

    Deflection at free end = (5 × 25 kN/m × (5 m)⁴) / (384 × 150 kN·m²) ≈ 0.6865 m

    Answer: Deflection at free end ≈ 0.6865 m

    Calculate the support reactions at the ends of a simply supported beam with a concentrated load of 50 kN at 2 meters from one end. The span of the beam is 6 meters.

    Solution:

    Total load = Load

    Total load = 50 kN

    Reaction at one end = (Load × Distance from the load) / Total span length

    Reaction at one end = (50 kN × 2 m) / 6 m ≈ 16.67 kN

    Reaction at the other end = Total load – Reaction at one end

    Reaction at the other end = 50 kN – 16.67 kN ≈ 33.33 kN

    Answer:

    Reaction at one end ≈ 16.67 kN

    Reaction at the other end ≈ 33.33 kN

    A cantilever beam with a length of 4 meters carries a uniformly distributed load of 10 kN/m. Calculate the maximum deflection at the free end of the beam. The flexural rigidity (EI) is 200 kN·m².

    Solution:

    Maximum deflection at free end = (5 × Load × Length⁴) / (384 × EI)

    Maximum deflection at free end = (5 × 10 kN/m × (4 m)⁴) / (384 × 200 kN·m²) ≈ 0.417 m

    Answer: Maximum deflection at free end ≈ 0.417 m

    Calculate the bending moment at the midspan of a simply supported beam with a uniformly distributed load of 30 kN/m. The span of the beam is 8 meters.

    Solution:

    Bending moment at midspan = (Uniformly distributed load × Span length) / 8

    Bending moment at midspan = (30 kN/m × 8 m) / 8 = 30 kN·m

    Answer: Bending moment at midspan ≈ 30 kN·m

    A truss shown below is loaded with a force of 40 kN at joint E. Calculate the forces in each member.

    Solution:

    By applying the method of joints, we can determine the forces in each member:

    AE: Force = 40 kN (Tension, since it is elongating)

    CD: Force = 40 kN (Compression, since it is shortening)

    DE: Force = 40 kN (Tension, since it is elongating)

    Answer:

    Force in AE ≈ 40 kN (Tension)

    Force in CD ≈ 40 kN (Compression)

    Force in DE ≈ 40 kN (Tension)

    Calculate the support reactions at the ends of a simply supported beam with a concentrated load of 60 kN at the center. The span of the beam is 10 meters.

    Solution:

    Total load = Load

    Total load = 60 kN

    Each support reaction = Total load / 2

    Support reactions = 60 kN / 2 = 30 kN

    Answer: Each support reaction ≈ 30 kN

    Calculate the deflection at the free end of a cantilever beam with a uniformly distributed load of 20 kN/m. The length of the cantilever beam is 6 meters and the flexural rigidity (EI) is 250 kN·m².

    Solution:

    Deflection at free end = (5 × Load × Length⁴) / (384 × EI)

    Deflection at free end = (5 × 20 kN/m × (6 m)⁴) / (384 × 250 kN·m²) ≈ 0.75 m

    Answer: Deflection at free end ≈ 0.75 m

    A simply supported beam with a span of 12 meters carries a uniformly distributed load of 25 kN/m. Calculate the maximum bending moment in the beam.

    Solution:

    Maximum bending moment = (Uniformly distributed load × Span length) / 8

    Maximum bending moment = (25 kN/m × 12 m) / 8 = 37.5 kN·m

    Answer: Maximum bending moment ≈ 37.5 kN·m

    Calculate the force in each member of a truss shown below, where all members are pinned and the applied load is 30 kN at joint D.

    Solution:

    By applying the method of joints, we can determine the forces in each member:

    AD: Force = 30 kN (Compression, since it is shortening)

    AB: Force = 30 kN (Tension, since it is elongating)

    BD: Force = 30 kN (Tension, since it is elongating)

    Answer:

    Force in AD ≈ 30 kN (Compression)

    Force in AB ≈ 30 kN (Tension)

    Force in BD ≈ 30 kN (Tension)

    Calculate the support reactions at the ends of a simply supported beam with a concentrated load of 70 kN at 4 meters from one end. The span of the beam is 8 meters.

    Solution:

    Total load = Load

    Total load = 70 kN

    Reaction at one end = (Load × Distance from the load) / Total span length

    Reaction at one end = (70 kN × 4 m) / 8 m = 35 kN

    Reaction at the other end = Total load – Reaction at one end

    Reaction at the other end = 70 kN – 35 kN = 35 kN

    Answer:

    Reaction at one end ≈ 35 kN

    Reaction at the other end ≈ 35 kN

    A cantilever beam with a length of 5 meters carries a uniformly distributed load of 15 kN/m. Calculate the maximum deflection at the free end of the beam. The flexural rigidity (EI) is 300 kN·m².

    Solution:

    Maximum deflection at free end = (5 × Load × Length⁴) / (384 × EI)

    Maximum deflection at free end = (5 × 15 kN/m × (5 m)⁴) / (384 × 300 kN·m²) ≈ 0.46875 m

    Answer: Maximum deflection at free end ≈ 0.46875 m

    Calculate the bending moment at the midspan of a simply supported beam with a uniformly distributed load of 35 kN/m. The span of the beam is 7 meters.

    Solution:

    Bending moment at midspan = (Uniformly distributed load × Span length) / 8

    Bending moment at midspan = (35 kN/m × 7 m) / 8 = 30.625 kN·m

    Answer: Bending moment at midspan ≈ 30.625 kN·m

    A truss shown below is loaded with a force of 50 kN at joint F. Calculate the forces in each member.

    Solution:

    By applying the method of joints, we can determine the forces in each member:

    DF: Force = 50 kN (Compression, since it is shortening)

    DE: Force = 50 kN (Tension, since it is elongating)

    EF: Force = 50 kN (Tension, since it is elongating)

    Answer:

    Force in DF ≈ 50 kN (Compression)

    Force in DE ≈ 50 kN (Tension)

    Force in EF ≈ 50 kN (Tension)

    Calculate the support reactions at the ends of a simply supported beam with a concentrated load of 80 kN at the center. The span of the beam is 14 meters.

    Solution:

    Total load = Load

    Total load = 80 kN

    Each support reaction = Total load / 2

    Support reactions = 80 kN / 2 = 40 kN

    Answer: Each support reaction ≈ 40 kN

    Calculate the deflection at the free end of a cantilever beam with a uniformly distributed load of 30 kN/m. The length of the cantilever beam is 7 meters and the flexural rigidity (EI) is 350 kN·m².

    Solution:

    Deflection at free end = (5 × Load × Length⁴) / (384 × EI)

    Deflection at free end = (5 × 30 kN/m × (7 m)⁴) / (384 × 350 kN·m²) ≈ 0.6643 m

    Answer: Deflection at free end ≈ 0.6643 m