Engineering Questions with Answers - Multiple Choice Questions
MCQs on Shear Stress Distribution in Various Sections
1 - Question
A beam has a triangular cross-section, having altitude ”h” and base “b”. If the section is being subjected to a shear force “F”. Calculate the shear stress at the level of neutral axis in the cross section.
a) 4F/5bh
b) 4F/3bh
c) 8F/3bh
d) 3F/4bh
View Answer
Answer: c
Explanation: For a triangular section subjected to a shear force, the shear stress at neutral axis is
= 4/3 × average shear stress
= 4/3 × F/A/2 ; A = bh
= 8F/3bh.
Explanation: For a triangular section subjected to a shear force, the shear stress at neutral axis is
= 4/3 × average shear stress
= 4/3 × F/A/2 ; A = bh
= 8F/3bh.
2 - Question
The maximum shear stress in the rectangular section is ______________ times the average shear stress.
a) 3/4
b) 3/7
c) 5/3
d) 3/2
View Answer
Answer: d
Explanation: The maximum shear stress occurs at the neutral axis. So, y = 0.
Maximum shear stress = 3/2 × F/bd (• Average shear stress = F/bd ).
= 3/2 × average shear stress.
Explanation: The maximum shear stress occurs at the neutral axis. So, y = 0.
Maximum shear stress = 3/2 × F/bd (• Average shear stress = F/bd ).
= 3/2 × average shear stress.
3 - Question
The modular ratio for M20 grade concrete is _____________
a) 16
b) 13
c) 11
d) 07
View Answer
Answer: b
Explanation: According to Indian Standards 456-2000 The modular ratio (m) = 280/3× cbc, For M20; compressive bearing capacity in concrete = 7 N/mm2 & tensile strength = 330 N/mm2.
Modular ratio (m) = 280/21 = 13.33.
4 - Question
In doubly reinforced beam, the maximum shear stress occurs ______________
a) along the centroid
b) along the neutral axis
c) on the planes between neutral axis and tensile reinforcement
d) on the planes between neutral axis and compressive reinforcement
View Answer
Answer: d
Explanation: In continuous beam the moments developed at the supports are greater than a moment’s developed at the mid span, show the maximum bending moment occurs at the supports.
For continuous beams, the maximum shear stress occurs at the planes intersecting the compressive reinforcement and the neutral axis.
Explanation: In continuous beam the moments developed at the supports are greater than a moment’s developed at the mid span, show the maximum bending moment occurs at the supports.
For continuous beams, the maximum shear stress occurs at the planes intersecting the compressive reinforcement and the neutral axis.
5 - Question
A cylindrical section having no joint is known as _______________
a) Proof section
b) Seamless section
c) Target section
d) Mown section
View Answer
Answer: b
Explanation: A cylindrical section having no joint is known as seamless section. Built up section is not that strong as a seamless section of the same thickness.
6 - Question
The efficiency of cylindrical section is the ratio of the strength of joint to the strength of _______________
a) Solid plate
b) Boilerplate
c) Circumferential plate
d) Longitudinal plate
View Answer
Answer: a
Explanation: The strength of plate or strength of rivet whichever is less is called the strength of joint. The ratio of the strength of joint to the strength of steel plate is called the efficiency of the cylinder.
7 - Question
Calculate the modulus of section for a hollow circular column of external diameter 60 mm and 10 mm thickness.
a) 170 m
b) 190 m
c) 250 m
d) 300 m
View Answer
Answer: a
Explanation: Given data :
D = 60 mm ; t = 10 mm & d = 60 – 2×10 = 40 mm
For hollow circular section, modulus of section( Z ) = 3.14 × D4 – d4 / 32 D.
= 17016.3 mm = 170 m.
8 - Question
Determine the modulus of a section for an I section, given the distance from neutral axis is 50 mm and moment of inertia is 2.8×106 mm4.
a) 59m
b) 51m
c) 58m
d) 63m
View Answer
Answer: c
Explanation: The modulus of section is the ratio of the moment of inertia to the distance of the neutral axis.
Given y = 50 mm
I = 2.8×106 mm4.
& Z = I/y = 2.8×106 / 50
= 57.76 ×103 mm
= 57.7 ~ 58 m.
9 - Question
A circular Beam of 0.25 m diameter is subjected to you shear force of 10 kN. Calculate the value of maximum shear stress. [Take area = 176 m2].
a) 0.75 N/mm2
b) 0.58 N/mm2
c) 0.73 N/mm2
d) 0.65 N/mm2
View Answer
Answer: a
Explanation: Given diameter = 0.25 m
Area (A) = 176 m2
Shear Force (F) = 10 kN ~ 10000 N.
For circular cross section the maximum shear stress is equal to 4/3 times of average shear stress
Maximum shear force = 4/3 × F/A
= 4/3 × 10000/176
= 0.75 N/mm2.
10 - Question
The maximum shear stress distribution is _____________ percentage more than average shear stress in circular section.
a) 54 %
b) 60 %
c) 33 %
d) 50 %
View Answer
Answer: c
Explanation: Maximum shear stress occurs at neutral axis; y =0
Maximum shear stress = 16/3 × average shear stress
But 4F / A is the average shear stress.
So, the maximum shear stress = 4/3 times the average shear stress.
Hence the maximum shear stress is 33% more than the average shear stress in a circular section.