# Design of RC Structures Questions and Answers – Prestressed Concretes Basics – Set 3

This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Prestressed Concretes Basics – Set 3”.

1. In a rectangular section of dimension 25 cm × 30 cm eccentricity of prestress is 7.5 cm. If the net losses are 15% and the final prestress force is 500 kN. What is the value of final stresses due to prestress alone?
a) – 4.0 MPa and 16.70 MPa
b) – 3.40 MPa and 19.60 MPa
c) – 3.40 MPa and 16.70 MPa
d) – 4.0 MPa and 19.60 MPa

Explanation: A = 750 cm2, I = 25 × (303 / 12) = 56250 cm2
r2 = I / A = 75 cm2
yt = yb = 15 cm
Final stress
pσtt = (P / A) [1 – (eyt / r2)]
= (500 / 750) [1 – (7.5 × 15 / 75)]
= -3.40 MPa (tensile)
pσbc = (P / A) [1 + (eyb / r2)]
= (500 / 750) [1 + (7.5 × 15 / 75)]
= 16.70 MPa (compression)
Initial stress
pσtt = -3.40 / 0.85 = -4.0 MPa.
pσbc = 16.70 / 0.85 = 19.60 MPa.

2. If the permissible stresses are σtt’ = -5 N/mm2, σbc’ = 20 N/mm2. What will be the value of eccentricity and initial force of prestress?
a) 8.40 cm and 561.40 kN
b) 8.40 cm and 560.00 kN
c) 8.00 cm and 561.40 kN
d) 9.40 cm and 560.00 kN

Explanation: Tensile stress at top fibre is 5 N/mm2 or 0.5 kN/cm2. Similarly, compressive stress at bottom fibre is 20 N/mm2 or 2 kN/cm2.
pσtt = (P / A) [1 – (eyt / r2)]     (eq.1)
-0.5 = (P / 750) [1 – (15e / 75)]
pσbc = (P / A) [1 + (eyb / r2)]     (eq.2)
2 = (P / 750) [1 + (15e / 75)]
Dividing (eq.1) and (eq.2) gives,
(-0.5 / 2) = [1 – (e / 5)] / [1 + (e / 5)]
e = 8.40 cm and P = 561.40 kN
These are only preliminary values.

3. The locus of the points of application of combined effect of forces on a prestressed section is called pressure line.
a) True
b) False

Explanation: At any section of a beam combined effect of the prestressing force and the externally applied load will result in a distribution of concrete stresses that can be resolved into a single force. The locus of points of application of this force in any beam is called the pressure line. Pressure lines are very useful in understanding the load carrying mechanism in prestressed concrete.

4. A prestressed concrete beam of 150 mm × 300 mm cross – section supports a live load of 6 kN/m over a simple span of 8 m. It has parabolic cable having an eccentricity of 80 mm at the midspan and zero at the ends. If the net resultant stress at the bottom fibre at midspan is zero under the action of dead load, live load and prestress force. What is the value of prestressing force?
a) 395.57 kN
b) 438.57 kN
c) 495.57 kN
d) 492.57 kN

Explanation: Area of section = 150 × 300 = 45000 mm2
Moment of inertia = 150 × (3003 / 12) = 3.375 × 108 mm4
Self weight of beam = 25 × 0.15 × 0.30 = 1.125 kN/m
Live load on beam = 6 kN/m
Maximum bending moment at mid span = (wl2 / 8) = 7.125 × (82 / 8) = 57 kNm
Bending stress at bottom fibre σbt = (M / I)y
σbt = (57 × 106 × 150) / (3.375 × 108) = 25.34 N/mm2
Stress due to prestress at bottom fibre is
pσbc = P / A + Pey / I)
= (P / 45000) + (P × 80 × 150) / (3.375 × 108) = 25.34
For net stress to be zero
P = 438576.92N = 438.57kN

5. A rectangular beam 200 mm × 400 mm spans over 10 m and carries an imposed load of 5 kN/m. If the force of prestress is 500 kN and tendons are centred at 100 mm from the bottom. Assume prestress tendons are parabolically draped. What will be the stresses in the beam at transfer?
a) 10.00 N/mm2 and 1.57 N/mm2
b) 10.93 N/mm2 and 1.80 N/mm2
c) -3.13 N/mm2 and 13.75 N/mm2
d) 10.93 N/mm2 and 1.57 N/mm2

Explanation: Area of section = 200 × 400 = 80000 mm2
Moment of inertia = 200 × (4003 / 12) = 1.067 × 109 mm4
At transfer
The maximum hogging moment due to prestress = Pe
= 500 × (200 – 100) = 50000 kNmm = 50 kNm
Dead load on beam = 0.2 × 0.4 × 25 = 2 kN/m
Maximum sagging moment at mid span due to dead load, Md = (wl2 / 8) = 2 × (102 / 8) = 25 kNm
Net bending moment = -50 + 25 = -25 kNm (hogging)
Net stress at the bottom and top due to dead load and prestress
σb,t = (P / A) ± (M / I)y
σb,t = [(500 × 1000) / 80000] ± [(25 × 106 × 200) / (1.067 × 109)] = 6.25 ± 4.68
σb = 10.93 N/mm2 (at bottom)
σt = 1.57 N/mm2 (at top)

6. Which of the following is the correct equation for determining the shear stress at any point in a prestressed concrete section?
a) (VAy / Ib)
b) (Ay / Ib)
c) (VAy / Ib3)
d) (VAy / Ib2)

Explanation: The shear stress distribution in an uncracked concrete section is a function of shear force and properties of the cross – section of the member. The shear stress at any point in a section is given by:
τ = (VAy / Ib)
Here V = effective shear force,
Ay = moment of the area above the point under consideration about the centroid,
I = moment of inertia and
b = breadth of the section.

7. Which of the following is the correct equation for determining the principal stress at any point in a prestressed concrete section?
a) [(σx + σy) / 2] ± (1 / 2)√[(σx – σy)2 + 2τ2]
b) [(σx + σy) / 2] ± (1 / 2)√[(σx – σy)2 + 4τ2] / 2
c) [(σx + σy) / 2] ± (1 / 2)√[(σx – σy)2 + 4τ2]
d) [(σx + σy) / 2] ± (1 / 2)√[(σx – σy) + 4τ]

Explanation: The effect of shear stress is to introduce principal tensile stresses on diagonal planes. In prestressed concrete members the shear stress is generally accompanied by a direct stress in the axial direction of the member. The principal stresses are given by:
σ1,2 = [(σx + σy) / 2] ± (1 / 2)√[(σx – σy)2 + 4τ2]
Here σ1,2 = maximum and minimum principal stresses
σx = direct stress in x – direction
σy = direct stress in y – direction
τ = shear stress

8. The principal stresses in a prestressed concrete section are not checked for dead load shear force with shear force due to pestressing.
a) True
b) False

Explanation: The principal stresses in a prestressed concrete section are checked for two conditions:
i. Dead load shear force with shear force due to prestressing, and
Sometimes shear stress may also be accompanied with vertical compressive stress if vertical prestressing is adopted.

9. In prestressed concrete members principal tensile stresses are sometimes reduced to compressive stresses.
a) True
b) False

Explanation: In a post – tensioned beam, the sectional properties (Ay) and I for the condition, dead load shear force with shear force due to prestressing may be computed by based on concrete section alone. As the shear stresses are caused before the beam is grouted. In prestressed concrete members σx and σy being compressive the magnitude of principal tensile stresses is considerably reduced and in some cases they even become compressive.

10. A prestressed concrete beam of rectangular section 150 mm by 300 mm is axially prestressed by a tendon carrying an effective force of 200 kN. The beam supports a uniformly distributed load of 7.5 kN/m including its self weight. What is the value of principle stresses? (The span is 8 m).
a) – 0.21 MPa (tension) or 4.71 MPa (compression)
b) – 4.21 MPa (tension) or 0.21 MPa (compression)
c) – 0.21 MPa (tension) or 0.21 MPa (compression)
d) – 4.71 MPa (tension) or 4.71 MPa (compression)

Explanation: A = 450 cm2, I = 15 × (302 / 12) = 33750 cm4
Shear force at support = 7.5 × 8 / 2 30 kN
Maximum shear stress at support occurs at the neutral axis,
τ = (VAy / Ib)
τmax = 30 × 15 × 15 × 7.5 / 33750 × 7.5
= 0.1 kN/cm2
≡ 1.0 MPa
Maximum shear stress is also given by τmax = 3 / 2 . V / bd
= 3 / 2 . (30 / 15 × 30)
= 0.1 kN/cm2
Axial stress, σx = 200 / 50 = 0.45 kN/cm2 ≡ 4.5 MPa
Principal stresses, σ1,2 = σx / 2 ± √σx2 / 4 + τ2
= 4.5 / 2 ± √4.52 / 4 + 12
= -0.21 MPa (tension) or 4.71 MPa (compression)
For M35 concrete mix the permissible principal tensile stress using a partial safety factor of 1.5 is
= 0.16 √σck MPa
= 0.95 N/mm2 > 0.21 MPa
The beam is safe in shear.

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