RCC Sheet 43

1) Question.

After prestressing process is completed, a loss of stress is due to
Answer.
- 1
  
  shrinkage of concrete
- 2
  
  elastic shortening of concrete
- 3
  
  creep of concrete
- 4
  
  all the above
Answer: 4

Explanation:
2) Question.

If k is wobble correction factor, $μ$ is coefficient of friction between the duet surface and the curve of tendon of radius R, the tension ratio at a distance x from either end, is
Answer.
- 1
  
  $1 + k x - \frac{μ x}{R}$
- 2
  
  $1 + k x + \frac{μ x}{R}$
- 3
  
  $1 - k x - \frac{μ x}{R}$
- 4
  
  $1 + k x + \frac{μ x}{R}$
Answer: 3

Explanation:
3) Question.

The system in which high tensile alloy steel bars (silica managanese steel) are used as prestressing tendons, is konwn as
Answer.
- 1
  
  Freyssinet system
- 2
  
  Magnel-Blaton system
- 3
  
  C.C.L. standard system
- 4
  
  Lee-McCall system
Answer: 4

Explanation:
4) Question.

Post tensioning system
Answer.
- 1
  
  was widely used in earlier days
- 2
  
  is not economical and hence not generally used
- 3
  
  is economical for large spans and is adopted now a days
- 4
  
  none of these
Answer: 4

Explanation:
5) Question.

A prestressed rectangular beam which carries two concentrated W at L/3 from either end, is provided with a bent tendon with tension P such that central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip h is
Answer.
- 1
  
  $\frac{W L}{P}$
- 2
  
  $\frac{W L}{2 P}$
- 3
  
  $\frac{W L}{3 P}$
- 4
  
  $\frac{3 W L}{5 P}$
Answer: 4

Explanation:
6) Question.

If a bent tendon is required to balance a concentrated load W at the centre of the span L, the central dip h must be at least
Answer.
- 1
  
  $\frac{W L}{P}$
- 2
  
  $\frac{W L}{2 P}$
- 3
  
  $\frac{W L}{3 P}$
- 4
  
  $\frac{W L}{2 P}$
Answer: 4

Explanation:
7) Question.

If a rectangular prestressed beam of an effective span of 5 meters and carrying a total load 3840 kg/m, is designed by the load balancing method, the central dip of the parabolic tendon should be
Answer.
- 1
  
  5 cm
- 2
  
  10 cm
- 3
  
  15 cm
- 4
  
  20 cm
Answer: 2

Explanation:
8) Question.

If the maximum dip of a parabolic tendon carrying tension P is h and the effective length of the prestressed beam is L, the upward uniform pressure will be
Answer.
- 1
  
  $\frac{8 h P}{l}$
- 2
  
  $\frac{8 h P}{l^{2}}$
- 3
  
  $\frac{8 h l}{P}$
- 4
  
  $\frac{8 h l}{P^{2}}$
Answer: 2

Explanation:
9) Question.

If the loading on a prestressed rectangular beam, is uniformly distributed, the tendon to be provided should be
Answer.
- 1
  
  straight below centroidal axis
- 2
  
  parabolic with convexity downward
- 3
  
  parabolic with convexity upward
- 4
  
  straight above centroidal axis
Answer: 2

Explanation:
10) Question.

In a prestressed beam carrying an external load W with a bent tendon is having angle of inclination $θ$ and prestressed load P. The net downward load at the centre is
Answer.
- 1
  
  $W - 2 P \cos θ$
- 2
  
  $W - P \cos θ$
- 3
  
  $W - P \sin θ$
- 4
  
  $W - 2 P \sin θ$
Answer: 4

Explanation:
11) Question.

If the tendon is placed at an eccentricity e below the centroidal axis of the longitudial axis of a rectangular beam(sectional modulus Z and stressed load P in tendon) the stress at the extreme top edge
Answer.
- 1
  
  $i s i n c r e a s e d b y \frac{P Z}{e}$
- 2
  
  $i s i n c r e a s e d b y \frac{P e}{Z}$
- 3
  
  $i s d e c r e a s e d b y \frac{P e}{Z}$
- 4
  
  remains unchanged
Answer: 3

Explanation:
12) Question.

P is the prestressed force applied to endton of a rectangular prestressed beam whose area of cross section is A and sectional modulus is Z. The minumum stress f on the beam subjected to a maximum bending moment M is
Answer.
- 1
  
  $f = \frac{P}{A} - \frac{Z}{M}$
- 2
  
  $f = \frac{A}{P} - \frac{M}{Z}$
- 3
  
  $f = \frac{P}{A} - \frac{M}{Z}$
- 4
  
  $f = \frac{P}{A} - \frac{M}{6 Z}$
Answer: 3

Explanation:
13) Question.

P is the prestressed force applied to the tendon of a rectangular prestressed beam whose area of cross section is A and sectional modulus is Z. The maximum stress f in the beam, subjected to a maximum bending moment M, is
Answer.
- 1
  
  $f = \frac{P}{A} + \frac{Z}{M}$
- 2
  
  $f = \frac{A}{P} + \frac{M}{Z}$
- 3
  
  $f = \frac{P}{A} + \frac{M}{Z}$
- 4
  
  $f = \frac{P}{A} + \frac{M}{6 Z}$
Answer: 3

Explanation:
14) Question.

If A is the sectional area of a prestressed rectangular beam provided with a tendon prestressed by a force P through its centroidal longitudinal axis, the compressive stress in concrete, is
Answer.
- 1
  
  P/A
- 2
  
  A/P
- 3
  
  P/2A
- 4
  
  2A/P
Answer: 1

Explanation:
15) Question.

Pick up the assumption for the design of a prestressed concrete member from the
Answer.
- 1
  
  A transverse plane section remains a plane after bending
- 2
  
  During deformation limits, Hook's law is equally applicable to concrete as well as to steel
- 3
  
  Variation of stress in reinforcement due to changes in external loading is negligible
- 4
  
  All the above
Answer: 4

Explanation:
16) Question.

High strength concrete is used in prestressed member
Answer.
- 1
  
  to overcome high bearing stresses developed at the ends
- 2
  
  to ovecome bursting stresses at the ends
- 3
  
  to provide high bond stresses
- 4
  
  all the above
Answer: 4

Explanation:
17) Question.

The minimum cube stength of concrete used for a prestressed member, is
Answer.
- 1
  
  $50 k g / c m^{2}$
- 2
  
  $150 k g / c m^{2}$
- 3
  
  $150 k g / c m^{2}$
- 4
  
  $350 k g / c m^{2}$
Answer: 4

Explanation:
18) Question.

In a prestressed member it is advisable to use
Answer.
- 1
  
  low strength concrete only
- 2
  
  high strength concrete only
- 3
  
  low strength concrete but high tensile steel
- 4
  
  high strength concrete and high tensile steel
Answer: 4

Explanation:
19) Question.

A pre-stressed concrete member is perferred because
Answer.
- 1
  
  its diamensions are not decided from the diagonal tensile stress
- 2
  
  large size of long beams carrying large shear force need not be adopted
- 3
  
  removal of cracks in the members due to shrinkage
- 4
  
  all the above
Answer: 4

Explanation:
20) Question.

In favourable circumstrances a 15 cm concrete cube after 28 days, attains a maximum crushing strenth
Answer.
- 1
  
  $100 k g / c m^{2}$
- 2
  
  $200 k g / c m^{2}$
- 3
  
  $300 k g / c m^{2}$
- 4
  
  $400 k g / c m^{2}$
Answer: 4

Explanation:

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