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)
Discussion:
65 comments Page 7 of 7.
Navneet said:
5 years ago
D = t is the correct answer.
(1)
Amanullah said:
5 years ago
d=4t is the correct answer
We know that Shearing pull Ps = pie/4* d^2*τ.
Therefore tao(shear stress) = 4Ps/pie d^2.
Same, crushing pull Pc = d*t*σ.
Therefore σ (crushing stress) = Pc/dt.
Now, as per question τ = 1/4 σ.
Solving this, we found d = 16/π *t.
So, dmax = 5.06t its coming, taking nearer value, its 4t.
We know that Shearing pull Ps = pie/4* d^2*τ.
Therefore tao(shear stress) = 4Ps/pie d^2.
Same, crushing pull Pc = d*t*σ.
Therefore σ (crushing stress) = Pc/dt.
Now, as per question τ = 1/4 σ.
Solving this, we found d = 16/π *t.
So, dmax = 5.06t its coming, taking nearer value, its 4t.
(5)
Prudvi Anemoni said:
5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(7)
Vinay said:
5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(15)
Kalyan gali said:
4 years ago
Shear area = πDT.
Crushing area = πD^2/4,
shear stress= 1/4crushing stress
p/πDT=P/πD^2/4,
D = 4T.
Crushing area = πD^2/4,
shear stress= 1/4crushing stress
p/πDT=P/πD^2/4,
D = 4T.
(22)
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