Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 4 (Q.No. 18)
18.
An R.C.C. beam of 6 m span is 30 cm wide and has a lever arm of 55 cm. If it carries a U.D.L. of 12 t per m and allowable shear stress is 5 kg/cm2, the beam
Discussion:
11 comments Page 1 of 2.
RIZWAN SHOWKET said:
4 years ago
Well explained, Thanks @Pranay,
According to me;
As we know Share Stress = Share Force/lever arm * width of the beam.
Share Stress= 12*10^3/55*30 = 7.27kg/sqcm.
But allowable share stress is 5 kg/cm2 so, 7.27 exceeds 5, therefore we need to redesign it.
According to me;
As we know Share Stress = Share Force/lever arm * width of the beam.
Share Stress= 12*10^3/55*30 = 7.27kg/sqcm.
But allowable share stress is 5 kg/cm2 so, 7.27 exceeds 5, therefore we need to redesign it.
(6)
Pranay said:
7 years ago
The shear stress = shear force/(lever arm * width of the beam) = (12 * 10^3 *6/2)/(55 *30)= 21.8 kg/cm^2.
Now section needs to be redesigned if shear stress exceeds 20 Kg/cm2.
Now section needs to be redesigned if shear stress exceeds 20 Kg/cm2.
(2)
Sangamesh said:
5 years ago
@Pranay.
What is that 20 kg/cm2? Explain it in detail.
What is that 20 kg/cm2? Explain it in detail.
SRIPATHI REDDYSEKHAR said:
6 years ago
Good explanation, Thanks @Pranay.
(1)
Zahid baloch said:
2 months ago
How? Please explain the answer.
Anop said:
5 years ago
I think it exceeds value.
Sumanta said:
5 years ago
Option D is correct.
(1)
Chajiya said:
7 years ago
It is 21.8 kg/cm^2.
Badhon said:
7 years ago
Explain it please.
Jay said:
7 years ago
Explain please.
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