# Civil Engineering - RCC Structures Design - Discussion

Discussion Forum : RCC Structures Design - Section 2 (Q.No. 14)
14.
An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 metres and carries a U.D.L. of 3000 kg/m inclusive of its self weight. If the lever arm constant for the section is 0.865, the maximum intensity of shear stress, is
8.3 kg/cm2
7.6 kg/cm2
21.5 kg/cm2
11.4 kg/cm2
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Khattak said:   2 years ago
Shear stress = F/(b x J x d).
= 0.5WL/(b x J x d).
= 0.5 x 30 x 600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].
= 9000/1081.25.
= 8.32.

Mark said:   2 years ago
Max Shear force = (wl/2).
Max.intensity of shear force = (wl/2)Ã·(b*d*j).

Ashwini said:   3 years ago
Maximum shear force=wl/2=3000*6/2 = 9000kg.
Maximum intensity of shear force = V/Bjd.
= 9000/25*0.865 * 50 = 8.32 kg/cm2.

Abhishek said:   5 years ago
But according to IS 456,
shear stress = V/bd
Which comes out to be 7.2 kg/cm^2.

And if we use old method, then
Shear stress = V/b jd.
Which comes out to be 8.32 kg/cm^2.

Shilendra singh said:   6 years ago
Maximum intensity of shear stress
= shear force/(leaver arm * width of beam).

So
Tc,max=V/job.
V=wl/2.
=3000 * 6/2 = 9000kg.

Leaver arm= 0.865times of depth of beam.
= 0.865 * 50= 43.25.
T.c max=9000/(43.25 * 25) = 8.323kg/cm^2.

Sahil said:   6 years ago

Mahesh said:   7 years ago
I think answer D is right because stress P/A and maximum stress = 1.5 x stress.

P = 30 * 600/2 = 9000 kg.

A = 25 * 50 = 1250.

Stress = 9000/1250 = 7.20 kg/m^2.

Maximum stress = 1.5 * 7.20 = 10.80 kg/ m^2.

10.80 is near to 11 so, d is the answer.

Er. Udit said:   7 years ago
Shear stress = F/(b x J x d).

= 0.5WL/(b x J x d).

= 0.5x30x600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].

= 9000/1081.25.

= 8.32 , Therefore the answer (A) is correct.

Jass said:   8 years ago
What is the formula?