1. Which of the following influence the deflections of prestressed concrete members?
a) Cable profile
b) Wall profile
c) Type of aggregates
d) Type of cement
Explanation: The deflections of prestressed concrete members are influenced by the following salient features: imposed load and self weight, magnitude of the prestressing force, cable profile, and second moment of area of cross section, modulus of elasticity of concrete, shrinkage, creep, relaxation, span, and fixity conditions.
2. Which type of deflections is solved by Mohr’s theorem?
a) Instantaneous
b) Long
c) Middle span
d) End span
Explanation: The computations of short term or instantaneous deflections, which occur immediately after the transfer of prestress and on application of loads is conveniently done by Mohr’s theorem from certain surveys it is concluded that the maximum deflection should be noted including with limiting deflection span ratio.
3. In the pre cracking stage, the deflections are computed by ____________
a) Prestressing force
b) Sectional area
c) Diameter
d) Second moment of area
Explanation: In the pre cracking stage the whole cross section is effective and the deflections in this stage are computed by using the second moment of area of the gross concrete section, as the length of the structural member increases the deflections can be controlled to a maximum extent possible.
4. The computations in post cracking stage are by considering ____________
a) Moment of inertia
b) Moment curvature
c) Moment design
d) Moment area
Explanation: The computations of deflections in this stage are made by considering moment curvature relationships which involve the section, properties of the cracked beam, based on prestressing forces and live loads the deflections of prestressed concrete members can be calculated if the longitudinal distribution of curvatures and magnitude of beam can be known at that particular time.
5. The short term deflections are also known as __________
a) Cracked
b) Un cracked
c) Instantaneous
d) Non instantaneous
Explanation: Short term deflections of prestressed members are also known as instantaneous deflections governed by distribution of bending moment throughout the span and flexural rigidity of member, these theorems are applied for determining the deflections due to prestressing force, imposed loads and self weight.
6. 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
Explanation: When the beam AB is subjected to a bending moment distribution due to prestressing force or self weight or imposed loads, ACB is the centre line of the deformed structure under the system of given loads, According to Mohr’s first theorem
Slope = area of bending moment deflection/flexural rigidity, θ = A/EI.
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
Explanation: The deflections of symmetrically loaded and simply supported beam at the mid span point are directly obtained from the second moment area theorem since the tangent is horizontal at this span, In most of cases of prestressed beams tendons are located with eccentricities towards the soffit of the beam to counteract the sagging bending moments due to transverse loads.
8. 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
Explanation: More complicated problems involving unsymmetrical loading may be solved by combining both the moment area theorems Mohr’s first theorem and second theorem, since the bending moment at every section is the product of prestressing force and eccentricity the tendon profile itself will represent the shape of the bending moment diagram.
9. The deflection of a beam with parabolic tendon is given as __________
a) –5PeL2/48EI
b) –10PeL2/48EI
c) –15PeL2/48EI
d) –3PeL2/48EI
Explanation: The deflection of the beam with parabolic tendons having an eccentricity e at the center and zero at the supports is given by a = –5PeL2/48EI, a beam with a parabolic tendon having an eccentricity e1 at the centre of span and e2 at the support sections and the resultant deflection at the centre is obtained as the sum of the upward deflection of a beam with a parabolic tendon of eccentricity e1+e2 at the centre and zero at the supports and the downward deflection of a beam subjected to a uniform sagging bending moment of intensity pe2 throughout the length, the resultant stress becomes a = PL2/48EI(-5e1+e2).
10. 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)
Explanation: The deflection in sloping tendon is computed in a way similar to:
A = (-PL2/12EI (e1+e2)) + (Pe2L2/8EI)
A = (PL3/24EI (-2e1+e2)).