___________________________________________________________________
FE Civil Structural Analysis Practice
Exam 2026–2027 | 100 Questions,
Answers & Detailed Rationales
1. A simply supported beam of span LL carries a uniformly distributed load ww
over its entire span. What is the maximum bending moment?
A. wL2/12wL^2/12
B. wL2/10wL^2/10
C. wL2/8wL^2/8
D. wL2/6wL^2/6
Answer: wL2/8wL^2/8
Rationale: For a simply supported beam with a uniform load over the full span,
the maximum positive moment occurs at midspan and equals wL2/8wL^2/8.
2. A simply supported beam carries a concentrated load PP at midspan.
Where does the maximum bending moment occur?
A. At the left support
B. At quarter span
C. At midspan
D. At the right support
,Answer: At midspan
Rationale: The shear force changes sign at the point of symmetry directly
beneath the midspan load, making the bending moment maximum there.
3. A 6-m simply supported beam carries a 20-kN point load located 2 m from
the left support. What is the reaction at the left support?
A. 6.67 kN
B. 10.0 kN
C. 13.33 kN
D. 20.0 kN
Answer: 13.33 kN
Rationale: Taking moments about the right support gives
RA(6)=20(4)R_A(6)=20(4), so RA=13.33R_A=13.33 kN.
4. For a statically determinate beam, which relationship correctly connects
load intensity w(x)w(x), shear V(x)V(x), and moment M(x)M(x)?
A. dV/dx=MdV/dx=M
B. dM/dx=VdM/dx=V
C. dM/dx=wdM/dx=w
D. dV/dx=M′dV/dx=M'
Answer: dM/dx=VdM/dx=V
Rationale: The slope of the bending-moment diagram equals the shear force.
With the common sign convention, dV/dx=−wdV/dx=-w.
5. A point load is applied to a beam. What occurs in the shear-force diagram
at the load location?
A. A parabolic curve develops
B. A sudden jump occurs
C. The diagram remains continuous and horizontal
D. The moment becomes discontinuous
,Answer: A sudden jump occurs
Rationale: A concentrated force produces an abrupt change in shear equal to
the magnitude of the applied force.
6. What occurs in the bending-moment diagram at the location of an applied
concentrated couple?
A. Shear becomes zero
B. Moment becomes zero
C. Moment diagram has a vertical jump
D. Shear diagram becomes parabolic
Answer: Moment diagram has a vertical jump
Rationale: A concentrated moment causes a discontinuity in the bending-
moment diagram equal to the applied couple.
7. A beam has constant EIEI and a known bending-moment function M(x)M(x).
Which equation describes its elastic curve?
A. EI dy/dx=M(x)EI\,dy/dx=M(x)
B. EI d2y/dx2=M(x)EI\,d^2y/dx^2=M(x)
C. EI d3y/dx3=M(x)EI\,d^3y/dx^3=M(x)
D. EI y=M(x)EI\,y=M(x)
Answer: EI d2y/dx2=M(x)EI\,d^2y/dx^2=M(x)
Rationale: Beam curvature is related to bending moment by
EI v′′(x)=M(x)EI\,v''(x)=M(x), subject to the selected sign convention.
8. Which quantity primarily controls the bending stiffness of a prismatic
beam?
A. AA
B. II
C. PP
D. VV
, Answer: II
Rationale: Flexural rigidity is EIEI. The elastic modulus EE describes material
stiffness, while the second moment of area II describes geometric stiffness.
9. A steel beam has its depth doubled while width remains constant.
Approximately how does its strong-axis moment of inertia change?
A. Doubles
B. Quadruples
C. Increases by a factor of 6
D. Increases by a factor of 8
Answer: Increases by a factor of 8
Rationale: For a rectangular section, I=bh3/12I=bh^3/12. Doubling hh increases
II by 23=82^3=8.
10.A rectangular cross section is 300 mm wide and 500 mm deep. What is its
centroidal moment of inertia about the horizontal axis?
A. 3.125×109 mm43.125\times10^9\text{ mm}^4
B. 6.25×109 mm46.25\times10^9\text{ mm}^4
C. 3.125×1010 mm43.125\times10^{10}\text{ mm}^4
D. 1.25×1010 mm41.25\times10^{10}\text{ mm}^4
Answer: 3.125×109 mm43.125\times10^9\text{ mm}^4
Rationale:
I=bh3/12=(300)(5003)/12=3.125×109 mm4I=bh^3/12=(300)(500^3)/12=3.125\ti
mes10^9\text{ mm}^4.
11.The parallel-axis theorem is expressed as:
A. I=Ic−Ad2I=I_c-Ad^2
B. I=Ic+A/d2I=I_c+A/d^2
C. I=Ic+Ad2I=I_c+Ad^2
D. I=Ic+dA2I=I_c+dA^2
FE Civil Structural Analysis Practice
Exam 2026–2027 | 100 Questions,
Answers & Detailed Rationales
1. A simply supported beam of span LL carries a uniformly distributed load ww
over its entire span. What is the maximum bending moment?
A. wL2/12wL^2/12
B. wL2/10wL^2/10
C. wL2/8wL^2/8
D. wL2/6wL^2/6
Answer: wL2/8wL^2/8
Rationale: For a simply supported beam with a uniform load over the full span,
the maximum positive moment occurs at midspan and equals wL2/8wL^2/8.
2. A simply supported beam carries a concentrated load PP at midspan.
Where does the maximum bending moment occur?
A. At the left support
B. At quarter span
C. At midspan
D. At the right support
,Answer: At midspan
Rationale: The shear force changes sign at the point of symmetry directly
beneath the midspan load, making the bending moment maximum there.
3. A 6-m simply supported beam carries a 20-kN point load located 2 m from
the left support. What is the reaction at the left support?
A. 6.67 kN
B. 10.0 kN
C. 13.33 kN
D. 20.0 kN
Answer: 13.33 kN
Rationale: Taking moments about the right support gives
RA(6)=20(4)R_A(6)=20(4), so RA=13.33R_A=13.33 kN.
4. For a statically determinate beam, which relationship correctly connects
load intensity w(x)w(x), shear V(x)V(x), and moment M(x)M(x)?
A. dV/dx=MdV/dx=M
B. dM/dx=VdM/dx=V
C. dM/dx=wdM/dx=w
D. dV/dx=M′dV/dx=M'
Answer: dM/dx=VdM/dx=V
Rationale: The slope of the bending-moment diagram equals the shear force.
With the common sign convention, dV/dx=−wdV/dx=-w.
5. A point load is applied to a beam. What occurs in the shear-force diagram
at the load location?
A. A parabolic curve develops
B. A sudden jump occurs
C. The diagram remains continuous and horizontal
D. The moment becomes discontinuous
,Answer: A sudden jump occurs
Rationale: A concentrated force produces an abrupt change in shear equal to
the magnitude of the applied force.
6. What occurs in the bending-moment diagram at the location of an applied
concentrated couple?
A. Shear becomes zero
B. Moment becomes zero
C. Moment diagram has a vertical jump
D. Shear diagram becomes parabolic
Answer: Moment diagram has a vertical jump
Rationale: A concentrated moment causes a discontinuity in the bending-
moment diagram equal to the applied couple.
7. A beam has constant EIEI and a known bending-moment function M(x)M(x).
Which equation describes its elastic curve?
A. EI dy/dx=M(x)EI\,dy/dx=M(x)
B. EI d2y/dx2=M(x)EI\,d^2y/dx^2=M(x)
C. EI d3y/dx3=M(x)EI\,d^3y/dx^3=M(x)
D. EI y=M(x)EI\,y=M(x)
Answer: EI d2y/dx2=M(x)EI\,d^2y/dx^2=M(x)
Rationale: Beam curvature is related to bending moment by
EI v′′(x)=M(x)EI\,v''(x)=M(x), subject to the selected sign convention.
8. Which quantity primarily controls the bending stiffness of a prismatic
beam?
A. AA
B. II
C. PP
D. VV
, Answer: II
Rationale: Flexural rigidity is EIEI. The elastic modulus EE describes material
stiffness, while the second moment of area II describes geometric stiffness.
9. A steel beam has its depth doubled while width remains constant.
Approximately how does its strong-axis moment of inertia change?
A. Doubles
B. Quadruples
C. Increases by a factor of 6
D. Increases by a factor of 8
Answer: Increases by a factor of 8
Rationale: For a rectangular section, I=bh3/12I=bh^3/12. Doubling hh increases
II by 23=82^3=8.
10.A rectangular cross section is 300 mm wide and 500 mm deep. What is its
centroidal moment of inertia about the horizontal axis?
A. 3.125×109 mm43.125\times10^9\text{ mm}^4
B. 6.25×109 mm46.25\times10^9\text{ mm}^4
C. 3.125×1010 mm43.125\times10^{10}\text{ mm}^4
D. 1.25×1010 mm41.25\times10^{10}\text{ mm}^4
Answer: 3.125×109 mm43.125\times10^9\text{ mm}^4
Rationale:
I=bh3/12=(300)(5003)/12=3.125×109 mm4I=bh^3/12=(300)(500^3)/12=3.125\ti
mes10^9\text{ mm}^4.
11.The parallel-axis theorem is expressed as:
A. I=Ic−Ad2I=I_c-Ad^2
B. I=Ic+A/d2I=I_c+A/d^2
C. I=Ic+Ad2I=I_c+Ad^2
D. I=Ic+dA2I=I_c+dA^2
