Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 17)
17.
The maximum diameter of the hole that can be punched from a plate of maximum shear stress 1/4th of its maximum crushing stress of punch, is equal to (where t = Thickness of the plate)
t
2t
4t
8t
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
65 comments Page 2 of 7.

Adarsh said:   1 decade ago
min d = [4t(max ss of plate)/max crushing stress of punch].

MANPREET SINGH said:   1 decade ago
Shear stress = Crushing stress in rivet joint.

Rajeeev said:   10 years ago
D = T should be correct answer.

Shiyas said:   10 years ago
Answer is A.

(pi*(D^2) /4)*Crushing stress = pi*D*t*(Crushing stress/4).

===> D=t;

Ashok kumar Chauhan said:   10 years ago
Shear stress = F/pi*d*t.

Crushing stress = 4F/pi*d^2.

Here pi = 22/7.

According to question Shear stress = Crushing stress/4.

That is why d = t.

AMAL said:   10 years ago
Here a hole is punched then we need to take area of that hole that is (in shear) pi*d^2/4 and in crushing pi*d*t. So the right answer is d=t.

Prashant said:   9 years ago
Shear stress * 4 = crushing stress.

(F/pi * d * t) * 4 = (4F/pi * d * d).

therefore, d = t.

Sarath t r said:   9 years ago
Anybody. please help me. Actually what is the equation of crushing stress and how it came?

Praveen said:   9 years ago
@Sarath.

Crushing stress = f\crushing area.
Crushing area = pi * d * d/4.

Mukesh said:   9 years ago
1/4 s.s = c.s
1/4 * F/pi d t = F/pi d d.
d = 4t.


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