# 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

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.