## PEC EPE Exam Structure Analysis Numerical Questions

Click here for PEC EPE Exam Structure Analysis MCQs

### 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 ร Span length

Total load = 20 kN/m ร 8 m = 160 kN

Each support reaction = Total load / 2

Support reactions = 160 kN / 2 = 80 kN

Answer: Each support reaction โ 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 kNยทmยฒ.**

Solution:

Deflection at midspan = (Load ร Span lengthยณ) / (48 ร EI)

Deflection at midspan = (30 kN ร (6 m)ยณ) / (48 ร 100 kNยทmยฒ) โ 0.375 m

Answer: Deflection at midspan โ 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 ร Length

Bending moment at fixed support = 40 kN ร 4 m = 160 kNยทm

Answer: Bending moment at fixed support โ 160 kNยทm

**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 ร Span length) / 8

Maximum bending moment = (15 kN/m ร 10 m) / 8 = 18.75 kNยทm

Answer: Maximum bending moment โ 18.75 kNยทm

**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 โ 20 kN (Tension)

Force in AB โ 20 kN (Compression)

Force in BC โ 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 kNยทmยฒ.**

Solution:

Deflection at free end = (5 ร Load ร Lengthโด) / (384 ร EI)

Deflection at free end = (5 ร 25 kN/m ร (5 m)โด) / (384 ร 150 kNยทmยฒ) โ 0.6865 m

Answer: Deflection at free end โ 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 ร Distance from the load) / Total span length

Reaction at one end = (50 kN ร 2 m) / 6 m โ 16.67 kN

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

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

Answer:

Reaction at one end โ 16.67 kN

Reaction at the other end โ 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 kNยทmยฒ.**

Solution:

Maximum deflection at free end = (5 ร Load ร Lengthโด) / (384 ร EI)

Maximum deflection at free end = (5 ร 10 kN/m ร (4 m)โด) / (384 ร 200 kNยทmยฒ) โ 0.417 m

Answer: Maximum deflection at free end โ 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 ร Span length) / 8

Bending moment at midspan = (30 kN/m ร 8 m) / 8 = 30 kNยทm

Answer: Bending moment at midspan โ 30 kNยทm

**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 โ 40 kN (Tension)

Force in CD โ 40 kN (Compression)

Force in DE โ 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 โ 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 kNยทmยฒ.**

Solution:

Deflection at free end = (5 ร Load ร Lengthโด) / (384 ร EI)

Deflection at free end = (5 ร 20 kN/m ร (6 m)โด) / (384 ร 250 kNยทmยฒ) โ 0.75 m

Answer: Deflection at free end โ 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 ร Span length) / 8

Maximum bending moment = (25 kN/m ร 12 m) / 8 = 37.5 kNยทm

Answer: Maximum bending moment โ 37.5 kNยทm

**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 โ 30 kN (Compression)

Force in AB โ 30 kN (Tension)

Force in BD โ 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 ร Distance from the load) / Total span length

Reaction at one end = (70 kN ร 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 โ 35 kN

Reaction at the other end โ 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 kNยทmยฒ.**

Solution:

Maximum deflection at free end = (5 ร Load ร Lengthโด) / (384 ร EI)

Maximum deflection at free end = (5 ร 15 kN/m ร (5 m)โด) / (384 ร 300 kNยทmยฒ) โ 0.46875 m

Answer: Maximum deflection at free end โ 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 ร Span length) / 8

Bending moment at midspan = (35 kN/m ร 7 m) / 8 = 30.625 kNยทm

Answer: Bending moment at midspan โ 30.625 kNยทm

**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 โ 50 kN (Compression)

Force in DE โ 50 kN (Tension)

Force in EF โ 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 โ 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 kNยทmยฒ.**

Solution:

Deflection at free end = (5 ร Load ร Lengthโด) / (384 ร EI)

Deflection at free end = (5 ร 30 kN/m ร (7 m)โด) / (384 ร 350 kNยทmยฒ) โ 0.6643 m

Answer: Deflection at free end โ 0.6643 m