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MCQOPTIONS
This section includes 144 Mcqs, each offering curated multiple-choice questions to sharpen your ENGINEERING SERVICES EXAMINATION (ESE) knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The scissors lift consists of two sets of symmetrically placed cross members (one in front that is shown and one behind that is not shown) and two hydraulic cylinders (front [labeled DE] and back [not shown]). The uniform platform has a mass of 60 kg with a center of gravity at G1. The 85 kg load (center of gravity at G2) is centered front to back. Determine the force in each of the hydraulic cylinders necessary to maintain equilibrium. These are rollers at B and D. |
| A. | DE = 1.067 kN C |
| B. | DE = 1.139 kN C |
| C. | DE = 0.606 kN C |
| D. | DE = 1.207 kN C |
| Answer» B. DE = 1.139 kN C | |
| 2. |
Determine the horizontal and vertical components of force at pins A and C of the two-member frame. |
| A. | Ax = -212 N, Ay = 388 N, Cx = 212 N, Cy = 212 N |
| B. | Ax = -300 N, Ay = 300 N, Cx = 300 N, Cy = 300 N |
| C. | Ax = -849 N, Ay = 149 N, Cx = 849 N, Cy = 849 N |
| D. | Ax = -1200 N, Ay = 1200 N, Cx = 1200 N, Cy = -600 N |
| Answer» C. Ax = -849 N, Ay = 149 N, Cx = 849 N, Cy = 849 N | |
| 3. |
A horizontal semi-circular beam of radius ‘R’ is fixed at the ends and carries a uniformly distributed load ‘W’ over the entire length. The bending moment at the fixed support is |
| A. | \(\frac{{{\rm{W}}{{\rm{R}}^2}}}{4}\) |
| B. | \(\frac{{{\rm{W}}{{\rm{R}}^2}}}{3}\) |
| C. | \(\frac{{{\rm{W}}{{\rm{R}}^2}}}{2}\) |
| D. | WR2 |
| Answer» C. \(\frac{{{\rm{W}}{{\rm{R}}^2}}}{2}\) | |
| 4. |
For the frame shown in figure the, the shear equation is |
| A. | \(\frac{{({M_{BA}} + {M_{BC}})}}{4} + \frac{{{M_{CD}}}}{4} + P = 0\) |
| B. | \(\frac{{({M_{BA}} + {M_{AB}})}}{4} + \frac{{{M_{CD}}}}{4} + P = 0\) |
| C. | \({M_{BA}} + {M_{AB}} + {M_{CD}} + {M_{DC}} = 0\) |
| D. | \({M_{BA}} + {M_{AB}} + {M_{CD}} + {M_{DC}} = P\) |
| Answer» C. \({M_{BA}} + {M_{AB}} + {M_{CD}} + {M_{DC}} = 0\) | |
| 5. |
Joints I, J, K, L, Q and M of the frame shown in the figure (not drawn to the scale) are pins. Continuous members IQ and LJ are connected through a pin at N. continuous members JM and KQ are connected through a pin at P. The frame has hinge supports at joints R and S. The loads acting at joints I, J and K are along the negative Y direction and the loads acting at joints L and M are along the positive X direction.The magnitude of the horizontal component of reaction (in kN) at S, is |
| A. | 5 |
| B. | 10 |
| C. | 15 |
| D. | 20 |
| Answer» D. 20 | |
| 6. |
A fixed beam AB is subjected to a triangular load varying from zero at end A to W per unit length at end B. What is the ratio of fixed end moment at A to B? |
| A. | 1/3 |
| B. | 1 |
| C. | 2/3 |
| D. | 3/2 |
| Answer» D. 3/2 | |
| 7. |
A continuous beam with uniform flexural rigidity is shown in the figure.The moment at B is |
| A. | 18 kNm |
| B. | 16 kNm |
| C. | 14 kNm |
| D. | 12 kNm |
| Answer» E. | |
| 8. |
Each span of a two-span continuous beam of uniform flexural rigidity is 6 m. All three supports are simple supports. It carries a uniformly distributed load of 20 kN/m over the left span only. The moment at the middle support is |
| A. | 90 kNm Sagging |
| B. | 45 kNm Hogging |
| C. | 90 kNm Hogging |
| D. | 45 kNm Sagging |
| Answer» C. 90 kNm Hogging | |
| 9. |
Bending moment at any section in a conjugate beam gives the actual beams |
| A. | Curvature |
| B. | Bending moment |
| C. | Deflection |
| D. | Slope |
| Answer» D. Slope | |
| 10. |
A propped cantilever beam of length L is subjected to a moment M at the propped end. The support moment at the fixed end will be |
| A. | M |
| B. | M/2 |
| C. | M/3 |
| D. | 2M |
| Answer» C. M/3 | |
| 11. |
For a member fixed at the end, the rotation factor is |
| A. | – 1/2 |
| B. | zero |
| C. | 4EIθ/L |
| D. | 3EIθ/L |
| Answer» B. zero | |
| 12. |
In the conjugate beam method, the fixed support in actual beam is considered as _________ support in the conjugate beam. |
| A. | Free |
| B. | Hinge |
| C. | Fixed |
| D. | Roller |
| Answer» B. Hinge | |
| 13. |
For the structure shown in the figure, the fixed end moment at support ‘A’ is |
| A. | 14·14 kN-m |
| B. | 10 kN-m |
| C. | Zero |
| D. | 3 kN-m |
| Answer» B. 10 kN-m | |
| 14. |
A rigid cantilever frame ABC is loaded and supported as shown in the figure below. The horizontal displacement of point C is |
| A. | \(\frac{{2P{h^3}}}{{3EI}}\) |
| B. | \(\frac{{P{h^2}\left( {2h + L} \right)}}{{2EI}}\) |
| C. | \(\frac{{P{h^3}}}{{3EI}}\) |
| D. | \(\frac{{P{h^2}\left( {h + L} \right)}}{{3EI}}\) |
| Answer» D. \(\frac{{P{h^2}\left( {h + L} \right)}}{{3EI}}\) | |
| 15. |
In the moment distribution method, the carry over moment is equal to |
| A. | double of its corresponding distributed moment and has same sign |
| B. | one-half of its corresponding distributed moment and has same sign |
| C. | one-half of its corresponding distribution moment and has opposite sign |
| D. | None of the above |
| Answer» C. one-half of its corresponding distribution moment and has opposite sign | |
| 16. |
If W is the load on a circular slab of radius R, the maximum circumferential moment at the centre of the slab isA. \(\frac{{W{R^2}}}{{16}}\)B. \(\frac{{2W{R^2}}}{{16}}\)C. \(\frac{{3W{R^2}}}{{16}}\) |
| A. | A Only |
| B. | B Only |
| C. | C Only |
| D. | Zero |
| Answer» D. Zero | |
| 17. |
Fixed end moments at A and B for the fixed beam shown in the figure, subjected to the indicated uniformly varying load, are respectively |
| A. | \(\frac{{W{l^2}}}{{30}}\;and\frac{{W{l^2}}}{{20}}\) |
| B. | \(\frac{{W{l^2}}}{{20}}\;and\frac{{W{l^2}}}{{30}}\) |
| C. | \(\frac{{W{l^2}}}{{12}}\;and\frac{{W{l^2}}}{8}\) |
| D. | \(\frac{{W{l^2}}}{8}\;and\frac{{W{l^2}}}{{12}}\) |
| Answer» B. \(\frac{{W{l^2}}}{{20}}\;and\frac{{W{l^2}}}{{30}}\) | |
| 18. |
A cantilever beam ‘A’ with a rectangular cross-section is subjected to a concentrated load at its free end. If the width and depth of another cantilever beam ‘B’ are twice those of beam ‘A’ and subjected to the same load, then the deflection at the free end of beam ‘B’ as compared to that of ‘A’ will be |
| A. | 6.25% |
| B. | 14.0% |
| C. | 23.6% |
| D. | 28.0% |
| Answer» B. 14.0% | |
| 19. |
A beam of length l is fixed at its both ends and carries two concentrated loads of W each at a distance of l/3 from both ends. The fixed end moment at A will be |
| A. | \(\frac{{ - Wl}}{3}\) |
| B. | \(\frac{{ - 2\;Wl}}{9}\) |
| C. | \(\frac{{ - 6\;Wl}}{{15}}\) |
| D. | \(\frac{{ - 4\;Wl}}{{27}}\) |
| Answer» C. \(\frac{{ - 6\;Wl}}{{15}}\) | |
| 20. |
Considering the symmetry of a rigid frame as shown below, the magnitude of the bending moment (in kNm) at P (preferably using the moment distribution method) is |
| A. | 170 |
| B. | 172 |
| C. | 176 |
| D. | 178 |
| Answer» D. 178 | |
| 21. |
Clapeyron's theorem is also know as the theory of - |
| A. | 2 - Moments |
| B. | 1 - Moment |
| C. | None of these |
| D. | 3 - Moment |
| Answer» E. | |
| 22. |
A beam of length L carries a uniformly distributed load throughout its length. In which of the following condition will the strain energy be maximum? |
| A. | Cantilever beam |
| B. | Simply supported beam |
| C. | Propped cantilever |
| D. | Fixed beam |
| Answer» B. Simply supported beam | |
| 23. |
A single-bay portal frame of height h, fixed at the base, is subjected to a horizontal displacement Δ at the top. The base moments developed are each proportional to |
| A. | \(\frac{1}{h}\) |
| B. | \(\frac{1}{{{h^2}}}\) |
| C. | \(\frac{1}{{{h^3}}}\) |
| D. | \(\frac{1}{{{h^4}}}\) |
| Answer» C. \(\frac{1}{{{h^3}}}\) | |
| 24. |
Members AB and BC shown in the figure below are identical. Due to a moment, M applied at B, what is the value of axial force in the member AB? |
| A. | M/L (Compression) |
| B. | M/L (Tension) |
| C. | 0.75M/L (Compression) |
| D. | 0.75M/L (Tension) |
| Answer» E. | |
| 25. |
A fixed beam AB of span L is subjected to a clockwise moment at a distance 'a' from end A. Fixed end moment at end A will be- |
| A. | \(\frac{M}{{{L^2}}}\left( {L - a} \right)\) |
| B. | \(\frac{M}{{2{L^2}}}\left( {L - a} \right)\) |
| C. | \(\frac{M}{{{3L^2}}}\left( {L - 3a} \right)\) |
| D. | \(\frac{M}{{{L^2}}}\left( {L - a} \right)\left( {L - 3a} \right)\) |
| Answer» E. | |
| 26. |
Force method in structural analysis always ensures |
| A. | compatibility of deformation |
| B. | equilibrium of forces |
| C. | kinematically admissible strains |
| D. | overall safety |
| Answer» C. kinematically admissible strains | |
| 27. |
All members of the frame shown below have the same flexural rigidity EI and length L. If a moment M is applied at joint B, the rotation of the joints is |
| A. | \(\frac{{ML}}{{12EI}}\) |
| B. | \(\frac{{ML}}{{11EI}}\) |
| C. | \(\frac{{ML}}{{8EI}}\) |
| D. | \(\frac{{ML}}{{7EI}}\) |
| Answer» C. \(\frac{{ML}}{{8EI}}\) | |
| 28. |
A fixed beam is loaded as in figure. The fixed end moment at support A is |
| A. | \(\frac{{w{L^2}}}{{30}}\) |
| B. | \(\frac{{w{L^2}}}{{20}}\) |
| C. | \(\frac{{w{L^2}}}{{10}}\) |
| D. | \(\frac{{w{L^2}}}{8}\) |
| Answer» C. \(\frac{{w{L^2}}}{{10}}\) | |
| 29. |
All members of the frame shown below has equal flexural rigidity EI. Calculate the rotation of joint O if moment M is applied? |
| A. | \(\frac{{ML}}{{13EI}}\) |
| B. | \(\frac{{2ML}}{{13EI}}\) |
| C. | \(\frac{{3ML}}{{13EI}}\) |
| D. | \(\frac{{ML}}{{12EI}}\) |
| Answer» C. \(\frac{{3ML}}{{13EI}}\) | |
| 30. |
A beam of span 5 m, fixed at A and B, carries a point load of 50 kN at 2 m from 'A'. The fixed end moments at the supports 'A' and 'B', respectively, are; |
| A. | 24 kNm clockwise and 36 kNm clockwise |
| B. | 24 kNm anticlockwise and 36 kNm anticlockwise |
| C. | 36 kNm clockwise and 24 kNm anticlockwise |
| D. | 36 kNm anticlockwise and 24 kNm clockwise |
| Answer» E. | |
| 31. |
A 3 m long simply supported beam of uniform cross section is subjected to a uniformly distributed load of w = 20 kN/m in the central 1 m as shown in the figure.If the flexural rigidity (EI) of the beam is 30 x 106 N-m2, the maximum slope (expressed in radians) of the deformed beam is |
| A. | 0.681 × 10-3 |
| B. | 0.943 × 10-3 |
| C. | 0.3611 × 10-3 |
| D. | 5.910 × 10-3 |
| Answer» D. 5.910 × 10-3 | |
| 32. |
For a linear elastic structural analysis system, minimization of potential energy yields: |
| A. | compatibility conditions |
| B. | constitutive relations |
| C. | equilibrium equations |
| D. | strain-displacement relations |
| Answer» B. constitutive relations | |
| 33. |
By what other name is the virtual work load method used to find the displacement and rotation at a point of a structure known? |
| A. | Unit load method |
| B. | Castigliano’s method |
| C. | Actual work method |
| D. | Euler's method |
| Answer» B. Castigliano’s method | |
| 34. |
A two span continuous beam ABC is as shown in figure below. The distribution factors at joint B are |
| A. | 0.4, 0.6 |
| B. | 0.6, 0.4 |
| C. | 0.5, 0.5 |
| D. | 0.55, 0.45 |
| Answer» D. 0.55, 0.45 | |
| 35. |
A continuous loaded beam ABC rests on simple supports at A,B and C without yielding (AB = L1, and BC = L2). Which one of the following represents correct equation of claperyron's three moment's theorem? |
| A. | \({M_A}{L_1} + {M_B}\left( {{L_1} + {L_2}} \right) + {M_c}{L_2} + \frac{{6{a_1}{x_1}}}{{{L_1}}} + \frac{{6{A_2}{x_2}}}{{{L_2}}} = 0\) |
| B. | \({M_B}\left( {{L_1} + {L_2}} \right) + \frac{{3{A_1}{x_!}}}{{{L_1}}} + \frac{{3{A_2}{x_2}}}{{{L_2}}} = 0\) |
| C. | \({M_A}{L_1} + 2{M_B}\left( {{L_1} + {L_2}} \right) + {M_c}{L_2}\frac{{12{A_1}{x_!}}}{{{L_1}}} + \frac{{12A + 2\;{x_2}}}{{{L_2}}} = 0\) |
| D. | \({M_A}{L_1} + 2{M_B}\left( {{L_1} + {L_2}} \right) + {M_c}{L_2}\frac{{4{A_1}{x_1}}}{{{L_1}}} + \frac{{4{A_2}{x_2}}}{{{L_2}}} = 0\) |
| Answer» C. \({M_A}{L_1} + 2{M_B}\left( {{L_1} + {L_2}} \right) + {M_c}{L_2}\frac{{12{A_1}{x_!}}}{{{L_1}}} + \frac{{12A + 2\;{x_2}}}{{{L_2}}} = 0\) | |
| 36. |
In column analogy method, what is the area of an analogous column for a fixed beam of span 2L and flexural rigidity 2EI? |
| A. | L/EI |
| B. | L/2EI |
| C. | 2L/EI |
| D. | L/4EI |
| Answer» B. L/2EI | |
| 37. |
A continuous beam ABC is shown in the figure. End supports are simple (i.e., A and C) and span AB = span BC = L. There is a concentrated load ‘W’ at the centre of the span AB while no load over the span BC. EI is the same for both spans. What is the moment at the continuous support B? |
| A. | \(-\frac{WL}{16}\) |
| B. | \(-\frac{5WL}{32}\) |
| C. | \(-\frac{3WL}{32}\) |
| D. | \(-\frac{3WL}{16}\) |
| Answer» D. \(-\frac{3WL}{16}\) | |
| 38. |
Assertion (A): The slope deflection method is a stiffness method in which the joint displacements are found by applying the equilibrium conditions at each joint.Reason (R): The displacement at a joint of a member is independent of the displacement the member at the far end of the joint. |
| A. | Both assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A). |
| B. | Both Assertion (A) and Reason (R) are individually true but Reason (R) is NOT the correct explanation of Assertion (A). |
| C. | Assertion (A) is true but Reason (R) is false. |
| D. | Assertion (A) is the false but Reason (R) is true. |
| Answer» C. Assertion (A) is true but Reason (R) is false. | |
| 39. |
A fixed beam AB is subjected to a triangular load varying from zero at end A to W per unit length at end B. the ratio of fix end moments at A to B will be |
| A. | 1/2 |
| B. | 2/3 |
| C. | 4/9 |
| D. | 1/4 |
| Answer» C. 4/9 | |
| 40. |
If the deflection at the free end of a uniformly loaded cantilever beam of length 1 m is equal to 7.5 mm, then the slope at the free end is: |
| A. | 0.01 radians |
| B. | 0.015 radians |
| C. | 0.02 radians |
| D. | 0.03 radians |
| Answer» B. 0.015 radians | |
| 41. |
For a fixed beam AB, the support B settles by δ downward, then what is the direction of rotation of point A and B? |
| A. | –ve, -ve |
| B. | +ve, +ve |
| C. | +ve, -ve |
| D. | –ve, +ve |
| Answer» C. +ve, -ve | |
| 42. |
Fixed end moments developed at both the ends in a fixed beam of span L and flexural rigidity El, when its right-side support settles down by Δ, is |
| A. | \(\frac{{6EI{\rm{\Delta }}}}{{{L^2}}}\) (sagging) |
| B. | \(\frac{{12EI{\rm{\Delta }}}}{{{L^2}}}\) (sagging) |
| C. | \(\frac{{6EI{\rm{\Delta }}}}{{{L^2}}}\) (hogging) |
| D. | \(\frac{{12EI{\rm{\Delta }}}}{{{L^2}}}\) (hogging) |
| Answer» D. \(\frac{{12EI{\rm{\Delta }}}}{{{L^2}}}\) (hogging) | |
| 43. |
A uniform simply supported beam is subjected to a clockwise moment M at the left end. The moment required at the right end of the beam so that the rotation of right end is zero is equal to |
| A. | 2 M |
| B. | M |
| C. | \(\frac{{\rm{M}}}{2}\) |
| D. | \(\frac{{\rm{M}}}{3}\) |
| Answer» D. \(\frac{{\rm{M}}}{3}\) | |
| 44. |
Horizontal stiffness coefficient ( K11) of a bar ‘AB’ is given by –(Where A is cross section area and E is Young’s Modulus) |
| A. | AE/L |
| B. | AE/√2L |
| C. | 2AE/L |
| D. | AE/2L |
| Answer» E. | |
| 45. |
In the moment distribution method, the distribution factor for the end span, on the fixed end side is: |
| A. | 0 |
| B. | 1/4 |
| C. | 1/2 |
| D. | 1 |
| Answer» B. 1/4 | |
| 46. |
In column analogy method, the area of an analogous column for a fixed beam of span L and flexural rigidity EI is taken as |
| A. | L/EI |
| B. | L/2EI |
| C. | L/4EI |
| D. | L/8EI |
| Answer» B. L/2EI | |
| 47. |
If in a pin-jointed plane frame (m + r) > 2j, then the frame is (Where ‘m’ is number of members, ‘r’ is reaction components and ‘j’ is number of joints) |
| A. | table and statically determinate |
| B. | table and statically indeterminate |
| C. | nstable |
| D. | one of the above |
| Answer» C. nstable | |
| 48. |
The Castigliano's second theorem can be used to compute deflections |
| A. | n statically determinate structures only |
| B. | or any type of structure |
| C. | t the point under the load only |
| D. | or beams and frames only |
| Answer» C. t the point under the load only | |
| 49. |
The deflection at any point of a perfect frame can be obtained by applying a unit load at the joint in |
| A. | ertical direction |
| B. | orizontal direction |
| C. | nclined direction |
| D. | he direction in which the deflection is required |
| Answer» E. | |
| 50. |
Effects of shear force and axial force on plastic moment capacity of a structure are respectively to |
| A. | ncrease and decrease |
| B. | ncrease and increase |
| C. | ecrease and increase |
| D. | ecrease and decrease |
| Answer» E. | |