Civil Engineering (CE) Exam  >  Civil Engineering (CE) Questions  >  It is given that:Modulus of elasticity of cem... Start Learning for Free
It is given that:
Modulus of elasticity of cement concrete = 210 × 103 kg/cm2
Poisson’s ratio for concrete = 0.20
Modulus of subgrade reaction, K = 5 kg/m3
Radius of wheel load distribution = 15 cm.
Slab thickness = 20 cm
The distance from the apex of slab corner to section of maximum stress along the corner bisector is,
  • a)
    74 cm
  • b)
    99 cm
  • c)
    86 cm
  • d)
    79 cm
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
It is given that:Modulus of elasticity of cement concrete = 210 ×...
Concept:
Wastergaard defines this term as the radius of relative stiffness as
Where l = radius of relative stiffness, cm
E = modulus of elasticity of cement concrete, kg/cm2
μ = Poisson’s ratio
h = slab thickness
k = subgrade modulus or modulus of subgrade reaction, kg/cm3.
The radius of relative stiffness is used to calculate the distance from the apex of slab corner to the section of maximum stress along the corner bisector (X).
Where a = radius of wheel load distribution
L = radius of relative stiffness.
Calculation:
= 73.5 cm
Now, the distance, X
X = 85.66 cm
X = 85.66 cm
View all questions of this test
Explore Courses for Civil Engineering (CE) exam

Top Courses for Civil Engineering (CE)

It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer?
Question Description
It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer?.
Solutions for It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE). Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.
Here you can find the meaning of It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice It is given that:Modulus of elasticity of cement concrete = 210 × 103kg/cm2Poisson’s ratio for concrete = 0.20Modulus of subgrade reaction, K = 5 kg/m3Radius of wheel load distribution = 15 cm.Slab thickness = 20 cmThe distance from the apex of slab corner to section of maximum stress along the corner bisector is,a)74 cmb)99 cmc)86 cmd)79 cmCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
Explore Courses for Civil Engineering (CE) exam

Top Courses for Civil Engineering (CE)

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev