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 4 of 7.

Sharan said:   1 decade ago
Shear stress = force/area.

Area = length*thickness.

So, 1/4 shear stress = l*t.

Stress = 4*t*l.

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?

Ananth said:   9 years ago
Shear stress = Load(P)/φ * D * t.
φ * D * t = ?

(area= L*T or (φ * D2)/4).

Shiyas said:   10 years ago
Answer is A.

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

===> D=t;

Vinay said:   5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(15)

Prudvi Anemoni said:   5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(7)

Praveen said:   9 years ago
@Sarath.

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

Phani said:   5 years ago
Crushing=4.πd^2.
Shearing=πd.t.
From shearing=1/4.crushing.
d=t.
(2)

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


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