Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 1 (Q.No. 30)
30.
According to I.S. : 456 specifications, the safe diagonal tensile stress for M 150 grade concrete, is
Discussion:
31 comments Page 1 of 4.
SATISH SUMAN said:
7 years ago
Focus on the term, Safe Diagonal Tensile Stress.
1) SAFE :- F.O.S. of Concrete= Ch. Strength/1.5 = 150/1.5=100 (Its Compressive strength).
2)TENSILE STRESS :- Tensile Strength=0.7*Fck^0.5= 0.7*100^0.5=0.7*10=7 (Its Tensile stress along the neutral Axis).
3)DIAGONAL:- Diagonal Shear failure exists at 45-degree angle to the neutral axis line.
So, Corresponding Diagonal strength=Longitudinal stress*Cos45=7 cos45= 4.95 i.e 5 Ans.
1) SAFE :- F.O.S. of Concrete= Ch. Strength/1.5 = 150/1.5=100 (Its Compressive strength).
2)TENSILE STRESS :- Tensile Strength=0.7*Fck^0.5= 0.7*100^0.5=0.7*10=7 (Its Tensile stress along the neutral Axis).
3)DIAGONAL:- Diagonal Shear failure exists at 45-degree angle to the neutral axis line.
So, Corresponding Diagonal strength=Longitudinal stress*Cos45=7 cos45= 4.95 i.e 5 Ans.
(2)
Susmita De said:
7 years ago
As per thumb full of WSM permissible stress in;
Bending compression =Fck/3.
Direct compression = Fck /3.75.
Bending tension =Fck/7.
Direct tension = 0.5 x Flexural strength.
And because it has been stated that diagonal tension so I think it is referring to bending tension which is Fck/7= 21 which is nearer to 20.
Bending compression =Fck/3.
Direct compression = Fck /3.75.
Bending tension =Fck/7.
Direct tension = 0.5 x Flexural strength.
And because it has been stated that diagonal tension so I think it is referring to bending tension which is Fck/7= 21 which is nearer to 20.
(1)
Lucky Nagar said:
5 years ago
Direct Tensile Stress = (0.5 to 0.625) * fck.
The factor of safety for stress (FOS) = 1.5.
We are moving forward by considering that tensile stress of concrete is minimum therefore use
Direct Tensile Stress = 0.5fck/FOS =0.5 * 15/1.5 = 5 N/mm^2.
The factor of safety for stress (FOS) = 1.5.
We are moving forward by considering that tensile stress of concrete is minimum therefore use
Direct Tensile Stress = 0.5fck/FOS =0.5 * 15/1.5 = 5 N/mm^2.
(1)
Shubham Chaturvedi said:
7 years ago
The tensile strength is 0.7 root over fck, it's bending tensile strength not tensile strength. Also known as flexure strength and in ques it is M15 not M150.
Safe diagonal tension = fck/3 = 15/3 = 5.
Safe diagonal tension = fck/3 = 15/3 = 5.
AKHIL said:
5 years ago
From Clause 6.2.2 of IS 456 : 2000.
Diagonal Tensile Strength = 1/2 * Flexural strength (fcr).
Flexural strength ( fcr ) = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck,
= 0.35 * 150 = 5.
Diagonal Tensile Strength = 1/2 * Flexural strength (fcr).
Flexural strength ( fcr ) = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck,
= 0.35 * 150 = 5.
(2)
Anujit said:
3 years ago
Diagonal Tensile Strength = 1/2 * Flexural strength.
Flexural strength = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck, = 0.35 * 150 = 4.28~5.
Flexural strength = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck, = 0.35 * 150 = 4.28~5.
(5)
Saugat Oli said:
7 years ago
Diagonal Tensile Strength = 1/2 * Flexural strength.
Flexural strength = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck,
= 0.35 * 150 = 5.
Flexural strength = 0.7 * fck,
So, Diagonal Tensile strength = 0.35* fck,
= 0.35 * 150 = 5.
Sunil g said:
6 years ago
Safe diagonal tensile stress is nothing but safe shear stress.
So, the answer should be 25kgf/cm2 for M15 as max shear stress is 2.5N/mm2.
So, the answer should be 25kgf/cm2 for M15 as max shear stress is 2.5N/mm2.
(2)
Sandeep Singh said:
3 years ago
M150 means 15Mpa.
Design strength 15/FOS=15/3=5Mpa=50kg/cm^2.
Tensile strength=0.7fck=0.7*50kg/cm^2.
= 4.95kg/cm^2.
Approx 5kg/cm^2.
Design strength 15/FOS=15/3=5Mpa=50kg/cm^2.
Tensile strength=0.7fck=0.7*50kg/cm^2.
= 4.95kg/cm^2.
Approx 5kg/cm^2.
(7)
Prashant singh said:
9 years ago
The tensile strength of concrete is (0.7) multiply (root of fck).
Means (0.7) multiply (root of 15) = 2.7,
Therefore 3.
Means (0.7) multiply (root of 15) = 2.7,
Therefore 3.
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