MCQOPTIONS
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MCQOPTIONS
This section includes 19 Mcqs, each offering curated multiple-choice questions to sharpen your Prestressed Concrete Structures knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The deflection due to self weight and imposed loads are __________ |
| A. | 5(g+q)L4/384EI |
| B. | 5(g+q)L4/384EI |
| C. | 5(g+q)L4/384EI |
| D. | 5(g+q)L4/384EI |
| Answer» B. 5(g+q)L4/384EI | |
| 2. |
The deflection is computed in a way similar to sloping tendon is given as __________ |
| A. | 2PL2/24EI |
| B. | 4PL2/24EI |
| C. | PL3/24EI (-2e1+e2) |
| D. | PL2/24EI (e1+e2) |
| Answer» D. PL2/24EI (e1+e2) | |
| 3. |
The deflection of a beam with parabolic tendon is given as __________ |
| A. | –5PeL2/48EI |
| B. | –10PeL2/48EI |
| C. | –15PeL2/48EI |
| D. | –3PeL2/48EI |
| Answer» B. –10PeL2/48EI | |
| 4. |
A tendon with a trapezoidal profile considering the bending moment and deflection at the centre of the beam is obtained by __________ |
| A. | –Pe/6EI(2l12+6l1l2+3l22) |
| B. | –Pe/6EI(2l12+6l1l2+3l22) |
| C. | –Pe/6EI(2l12+6l1l2+3l22) |
| D. | –Pe/6EI(2l12+6l1l2+3l22) |
| Answer» C. –Pe/6EI(2l12+6l1l2+3l22) | |
| 5. |
A straight tendon at a uniform eccentricity below the centroidal axis is given as __________ |
| A. | –PeL2/4EI |
| B. | –PeL2/8EI |
| C. | –PeL2/14EI |
| D. | –PeL2/16EI |
| Answer» C. –PeL2/14EI | |
| 6. |
The problems involving unsymmetrical loading can be solved by __________ |
| A. | Mohr’s theorem |
| B. | Kennedy’s theorem |
| C. | Row’s theorem |
| D. | Casagrande’s theorem |
| Answer» B. Kennedy’s theorem | |
| 7. |
Which of the following deflections are directly obtained by Mohr’s second area theorem? |
| A. | Simply supported beam |
| B. | Uniformly distributed load |
| C. | Point beams |
| D. | Fixed beams |
| Answer» B. Uniformly distributed load | |
| 8. |
Which of the following is the equation given by Mohr’s second theorem? |
| A. | Mid span/flexural rigidity |
| B. | Moment of area of bending moment diagram/flexural rigidity |
| C. | End span/flexural rigidity |
| D. | Thickness/flexural rigidity |
| Answer» C. End span/flexural rigidity | |
| 9. |
Which of the following is the equation given Mohr’s first theorem? |
| A. | Area of bending moment deflection/flexural rigidity |
| B. | Moment/flexural rigidity |
| C. | Deflection/flexural rigidity |
| D. | Loads/flexural rigidity |
| Answer» B. Moment/flexural rigidity | |
| 10. |
The short term deflections are also known as __________ |
| A. | Cracked |
| B. | Un cracked |
| C. | Instantaneous |
| D. | Non instantaneous |
| Answer» D. Non instantaneous | |
| 11. |
THE_DEFLECTION_IS_COMPUTED_IN_A_WAY_SIMILAR_TO_SLOPING_TENDON_IS_GIVEN_AS:?$ |
| A. | 2PL<sup>2</sup>/24EI |
| B. | 4PL<sup>2</sup>/24EI |
| C. | PL<sup>3</sup>/24EI (-2e<sub>1</sub>+e<sub>2</sub>) |
| D. | PL<sup>2</sup>/24EI (e<sub>1</sub>+e<sub>2</sub>) |
| Answer» D. PL<sup>2</sup>/24EI (e<sub>1</sub>+e<sub>2</sub>) | |
| 12. |
The_deflection_due_to_self_weight_and_imposed_loads_are:$ |
| A. | 5(g+q)L<sup>4</sup>/384EI |
| B. | 5(g+q)L<sup>4</sup>/384EI |
| C. | 5(g+q)L<sup>4</sup>/384EI |
| D. | 5(g+q)L<sup>4</sup>/384EI |
| Answer» B. 5(g+q)L<sup>4</sup>/384EI | |
| 13. |
The deflection of a beam with parabolic tendon is given as? |
| A. | –5PeL<sup>2</sup>/48EI |
| B. | –10PeL<sup>2</sup>/48EI |
| C. | –15PeL<sup>2</sup>/48EI |
| D. | –3PeL<sup>2</sup>/48EI |
| Answer» B. ‚Äö√Ñ√∂‚àö√ë‚àö¬®10PeL<sup>2</sup>/48EI | |
| 14. |
A tendon with a trapezoidal profile considering the bending moment and deflection at the centre of the beam is obtained by: |
| A. | –Pe/6EI(2l<sub>1</sub><sup>2</sup>+6l<sub>1</sub>l<sup>2</sup>+3l<sub>2</sub><sup>2</sup>) |
| B. | –Pe/6EI(2l<sub>1</sub><sup>2</sup>+6l<sub>1</sub>l<sup>2</sup>+3l<sub>2</sub><sup>2</sup>) |
| C. | –Pe/6EI(2l<sub>1</sub><sup>2</sup>+6l<sub>1</sub>l<sup>2</sup>+3l<sub>2</sub><sup>2</sup>) |
| D. | –Pe/6EI(2l<sub>1</sub><sup>2</sup>+6l<sub>1</sub>l<sup>2</sup>+3l<sub>2</sub><sup>2</sup>) |
| Answer» C. ‚Äö√Ñ√∂‚àö√ë‚àö¬®Pe/6EI(2l<sub>1</sub><sup>2</sup>+6l<sub>1</sub>l<sup>2</sup>+3l<sub>2</sub><sup>2</sup>) | |
| 15. |
A straight tendon at a uniform eccentricity below the centroidal axis is given as: |
| A. | –PeL<sup>2</sup>/4EI |
| B. | –PeL<sup>2</sup>/8EI |
| C. | –PeL<sup>2</sup>/14EI |
| D. | –PeL<sup>2</sup>/16EI |
| Answer» C. ‚Äö√Ñ√∂‚àö√ë‚àö¬®PeL<sup>2</sup>/14EI | |
| 16. |
The problems involving unsymmetrical loading can be solved by: |
| A. | Mohr’s theorem |
| B. | Kennedy’s theorem |
| C. | Row’s theorem |
| D. | Casagrande’s theorem |
| Answer» B. Kennedy‚Äö√Ñ√∂‚àö√ë‚àö¬•s theorem | |
| 17. |
Which of the following deflections are directly obtained by Mohr’s second area theorem?$ |
| A. | Simply supported beam |
| B. | Uniformly distributed load |
| C. | Point beams |
| D. | Fixed beams |
| Answer» B. Uniformly distributed load | |
| 18. |
Which of the following is the equation given Mohr’s first theorem?$ |
| A. | Area of bending moment deflection/flexural rigidity |
| B. | Moment/flexural rigidity |
| C. | Deflection/flexural rigidity |
| D. | Loads/flexural rigidity |
| Answer» B. Moment/flexural rigidity | |
| 19. |
The short term deflections are also known as: |
| A. | Cracked |
| B. | Un cracked |
| C. | Instantaneous |
| D. | Non instantaneous |
| Answer» D. Non instantaneous | |