Civil Engineering - Estimating and Costing - Discussion
Discussion Forum : 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
Discussion:
13 comments Page 1 of 2.
Deepanshu said:
1 decade ago
Bond length or development length = d*tensilestress/4*bond stress.
D*1400/4*6 = 58.33.
D*1400/4*6 = 58.33.
Pawan said:
1 decade ago
Ld = fi sigma st / 4 tou bd.
Salem basha said:
10 years ago
Ld = 0.87Fy/4 τ bd.
This will gives answer as 50.75.
So answer will be 50 times the diameter.
This will gives answer as 50.75.
So answer will be 50 times the diameter.
Mohan said:
9 years ago
Ld = 1400/(4 * 6).
= 58.3 =>59.
= 58.3 =>59.
(1)
Pawan said:
9 years ago
Bond length = tensile stress * Diameter of rod/(4 times bond strength).
= 1400 * D/(4 * 6),
= 1400 * D/24,
= 59D.
= 1400 * D/(4 * 6),
= 1400 * D/24,
= 59D.
(7)
PRADEEP PATEL said:
8 years ago
Acc. to wsm Ld = (1400*D)/(4*6) = 58.33.
Acc. To LSM Ld=(.87 *1400 *D)/(4*6) = 50.75.
Acc. To LSM Ld=(.87 *1400 *D)/(4*6) = 50.75.
(4)
Kanhaiya kr. said:
8 years ago
(D*1400)÷(4*6)=58.33,
Say=59D.
Say=59D.
(2)
Kundan kumar said:
7 years ago
Ld = .87fy * D/4 * τbd.
(2)
Anand said:
7 years ago
It is [fy*tensile strength]/4*bond stress.
(1)
Alexrampee said:
7 years ago
Fy is not equal to σ.
(1)
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