This set of Prestressed Concrete Structures Multiple Choice Questions & Answers (MCQs) focuses on “Shear and Principal Stresses”.
1. The shear stress is a function of:
a) Shear force and Cross section
b) Principle stresses and elevation
c) Strain & Compatibility
d) Axial prestress & tension
Explanation: The shear distribution in un cracked structural concrete members linear deformations are assumed to be developed due to shear distribution and the shear stress in a function of shear force and cross section of the members which is given by the equations:
τv =VS/IB, τv =shear stress, V = shear force, S = first moment of inertia, I = moment of inertia, B = width of the beam section.
2. The strength of concrete subjected to pure shear being nearly twice that in:
Explanation: The effect of this shear stress is to induce principal tensile stresses on diagonal planes and in pure shear, the strength of the concrete is twice that of the strength in the tension local failures first appear in the form of diagonal tension cracks in legions of height shear stress.
3. The effect of maximum shear stress (τ v) produces:
a) Principal tensile stresses
b) Principal compression stresses
c) Principal strain stresses
d) Principal span stresses
Explanation: cracks are observed at the point of the development of maximum shear stresses diagonally. The effect of this maximum shear stress (τ v) also produces principle tensile stresses on diagonal plane, the calculation of principle tensile stress resulting from direct at critical sections with or without bending and shear combined shall be carried out it is also done at the material change in width of section and should be less than 0.126(fc)1/2.
4. In a prestressed concrete member, the shear stress is generally accompanied by:
a) Zone stresses
b) Anchorage stresses
c) Direct stresses
d) Bondage stresses
Explanation: In a prestressed concrete member, the shear stresses is generally accompanied by a direct stress in the axial direction of the member, and if transverse, vertical prestressing is adopted, compressive stresses in perpendicular to the axis of the moment will present in addition to the axial pre stresses.
5. The major principal stresses produced on diagonal plane is expressed as:
a) fx + fy/2
b) fx + fy/2 – 1/2 ((fx – fy )2 +4τ v2 )1/2
c) fx + fy/2 + 1/2 ((fx – fy)2 +4τ v2 )1/2
d) fx – fy/2
Explanation: The major principal stress Fmax = fx + fy/2 + 1/2 ((fx – fy )2 +4τv2 )1/2
Minor principal stress Fmin = fx + fy/2 + 1/2 ((fx – fy )2 +4τ v2 )1/2
Fx, Fy are the direct stresses in horizontal & vertical directions respectively.
6. If the direct stresses are compressive, then the magnitude of principal stresses in prestressed concrete member gets:
Explanation: If the direct stresses are compressive, then the magnitude of principal stresses in prestressed concrete member acts reduced considerably and therefore under working loads, these principal stresses have to be compressive in nature in order to eliminate diagonal cracks in concrete.
7. How many ways are there for improving the shear resistance of structural concrete members by prestressing techniques?
Explanation: In general three ways of improving the shear resistance of structural concrete members by prestressing techniques:
Horizontal or axial prestressing by inclined or sloping cables and vertical or transverse prestressing.
8. A prestressed concrete beam span 10mm of rectangular section, 120mm wide & 300mm deep is axially prestressed on effective force of 180kn, uniformly distributed load of 5kn/m include the self weight of member. Find maximum shear stress at support?
Explanation: A= (120×300) = 36×103 mm2, I = 27×107 mm4, Wd = 5kn/m
Shear force at support V = (5×10/2) =25kn
Maximum shear stress at support, τv = (3v/2bh) = (3/2)x(25×103 /120×300) = 1.05n/mm2.
9. A prestressed concrete beam of span 10m of rectangular section, 120mm wide & 300mm deep a curved cable having an eccentricity of 100mm at the centre of span. Find the slope of cable of support:
a) 0.08 radians
b) 0.01 radians
c) 0.04 radians
d) 0.12 radians
Explanation: I = 10mm, l=100mm
Slope of cable at support = (4e/l) = (4×100/10×100) = 0.04 radians.
10. Which type of tensioning is generally uneconomical for vertical prestressing?
a) Post tensioning
b) Pre tensioning
c) Chemical tension
d) Thermal tension
Explanation: Vertical prestressing is not generally adopted because the length of the cables being short, the loss of prestress due to anchorage slip is excessively large post tensioning is generally uneconomical for vertical prestressing due to losses of prestress encountered.
Sanfoundry Global Education & Learning Series – Prestressed Concrete Structures.
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