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 5 of 7.

Sridevi said:   9 years ago
Yes, agree that d = t.

Indrajeet said:   9 years ago
Option A is the correct answer.

Crushing stress = 4 * Load(P)/φ * D * D.

Shear stress = Load(P)/φ * D * t.

Since Crushing stress = 4 * Shear stress.

After resolve, We get, D = t.

So option A is right.

Deepak said:   9 years ago
S.S = ( 1/4) * C.S.
F/Pi dt = (1/4) * (4F/pi * d^2).
Thus, d = t.

NITESH BORKAR said:   9 years ago
The equation will be:

(Pi/4 * d^2) * Crushing strength,

= Pi * d * t * Shear strength (or crushing strength * 2).

Which comes out to be t only.

Ajitesh said:   9 years ago
σs = (1/4)σc.
F/(Φ d.t) = F/((Φ/4).d^2.t).
d = t.

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

Praveen said:   9 years ago
@Sarath.

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

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

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

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

therefore, 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.


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