Civil Engineering - Estimating and Costing - Discussion

Discussion :: Estimating and Costing - Section 1 (Q.No.45)

45. 

If tensile stress of a steel rod of diameter D is 1400 kg/cm2 and bond stress is 6 kg/cm2, the required bond length of the rod is

[A]. 30 D
[B]. 40 D
[C]. 50 D
[D]. 53 D
[E]. 59 D

Answer: Option E

Explanation:

No answer description available for this question.

Deepanshu said: (Feb 11, 2014)  
Bond length or development length = d*tensilestress/4*bond stress.

D*1400/4*6 = 58.33.

Pawan said: (Sep 18, 2014)  
Ld = fi sigma st / 4 tou bd.

Salem Basha said: (Jan 18, 2016)  
Ld = 0.87Fy/4 τ bd.

This will gives answer as 50.75.

So answer will be 50 times the diameter.

Mohan said: (Feb 24, 2017)  
Ld = 1400/(4 * 6).
= 58.3 =>59.

Pawan said: (Feb 24, 2017)  
Bond length = tensile stress * Diameter of rod/(4 times bond strength).
= 1400 * D/(4 * 6),
= 1400 * D/24,
= 59D.

Pradeep Patel said: (Jun 13, 2017)  
Acc. to wsm Ld = (1400*D)/(4*6) = 58.33.
Acc. To LSM Ld=(.87 *1400 *D)/(4*6) = 50.75.

Kanhaiya Kr. said: (Dec 10, 2017)  
(D*1400)÷(4*6)=58.33,
Say=59D.

Kundan Kumar said: (Apr 19, 2018)  
Ld = .87fy * D/4 * τbd.

Anand said: (Jul 2, 2018)  
It is [fy*tensile strength]/4*bond stress.

Alexrampee said: (Sep 24, 2018)  
Fy is not equal to σ.

Jinish Patel said: (Nov 6, 2019)  
The increase of length due to bending stress is called bond length.

Bond length = (diameter/4)X stress/strength.
= 1400x D/ 4x6.
= 59 D.

Ranu Kumar said: (Jan 5, 2020)  
Answer will be 50D.

Sabina said: (Mar 9, 2020)  
Tensile stress = 1400.
Bond stress = 6.
Bond length = 1400*D/4*6 = 58.33.
Hence, the answer is 59D.

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