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.
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)
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)
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)
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)
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)
Bairagi said:
4 years ago
Thanks @Satish Suman.
(1)
Sindhu said:
4 years ago
Diagonal tension stress:large shear force & And less bending moment.
(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)
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)
Noblie said:
7 years ago
Thnak you @Satish.
(1)
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