Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 24 (Q.No. 34)
34.
A bar of square section is subjected to a pull of 10000 kg. If the maximum allowable shear stress on any section is 500 kg/cm2, then the side of the square section will be
5 cm
10 cm
15 cm
20 cm
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Dhaval Vaghasiya said:   6 years ago
The correct answer is B.

Because...
Max shear stress= Normal stress/2.

Here,
Normal stress= Load /Area=10000/A

So,
Max shear stress= 10000/2A.

So,
A=10000/2*500,

So,
A=10.
If one side is B,
b*b=10.
That's why B=√10.
(2)

Preetam salunke said:   5 years ago
Tau max = 3/2 * τ(avg).
Tau avg= normal load /area.
Normal load = mac pull/2.
Hence, τ avg =10000/2 * area.
Hence,
500 = 3/2 *(10000/2*area).
Hence area =15
Side=√15.
So, the Answer is C.
(11)

Deepak kumar sharma said:   8 years ago
Max. Shear stress= p/2.
Given p/2= 500 =>p = 1000.
p = P/A =>A=P/p.
A = 10000/1000=>10,
A = 10.
Side of square section "a"= √10.
(1)

Sampat tak said:   1 year ago
The answer should be B.

Because the given force is pull not the transverse loading we can apply 1.5tau.

We will apply (tau) max. = 0.5 (normal stress).

Ashish said:   4 years ago
Direct stress = 10000/x2.
Max shear stress= (10000/x2-0)/2.
500=5000/x2,
x2=10,
x= √10.
(5)

Pradip said:   7 years ago
@ ROY.

It has not a shear force or transverse force on beam.

Vakeem Ahmed said:   7 years ago
Yes, right @Umesh.

The answer should be D.

Roy said:   8 years ago
But max shear stress 1.5tav.

Rahul said:   10 years ago
Please give description.

Umesh JIT said:   7 years ago
The Answer should be D.


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