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 6 of 7.
Bhavik said:
6 years ago
The correct answer is d=t.
Surya said:
6 years ago
The correct answer is option A) d=t.
Juhi said:
6 years ago
While punching a hole in a plate.
Shear stress = F/c.s area.
F/(π/4)d^2.
Crushing stress= F/2π*r*h.
= F/π*d*t (since h is the thickness of the plate).
Now,
Shear stress = 4* crushing stress.
4F/(π*d*d)= 4*F/(π*d*t).
d= t.
Hence the dia of the smallest hole that can be punched will be equal to the thickness of the plate.
Shear stress = F/c.s area.
F/(π/4)d^2.
Crushing stress= F/2π*r*h.
= F/π*d*t (since h is the thickness of the plate).
Now,
Shear stress = 4* crushing stress.
4F/(π*d*d)= 4*F/(π*d*t).
d= t.
Hence the dia of the smallest hole that can be punched will be equal to the thickness of the plate.
Pratipalsinh said:
6 years ago
D = t is not correct answer, in this condition D = t is the smallest diameter of the hole that can be punched in a plate, not maximum.
Jitesh said:
6 years ago
Yes, the answer will be d=t as the shearing area will be π*d*t and normal area for the plate will be π*d^2.
JAYADEB DASH said:
6 years ago
The answer is d = t.
1. Max shear stress in plate = 1/4 max crushing stress in the punch.
2. Shear stress in plate = p/ dt.
3. Crushing stress in punch = 4p/ d^2.
4. Now p/' dt = 4p/4 d^2.
And d = t.
1. Max shear stress in plate = 1/4 max crushing stress in the punch.
2. Shear stress in plate = p/ dt.
3. Crushing stress in punch = 4p/ d^2.
4. Now p/' dt = 4p/4 d^2.
And d = t.
(1)
Subhendu said:
6 years ago
We know that Shearing pull Ps = pie/4* d^2*tao.
Therefore tao(shear stress) = 4Ps/pie d^2.
Same, crushing pull Pc = d*t*sigma.
Therefore sigma (crushing stress) = Pc/dt.
Now, as per question tao = 1/4 sigma.
Solving this, we found d = 16/pie *t.
So, dmax = 5.06t its coming, taking nearer value, its 4t.
I think this may clear your doubt.
Therefore tao(shear stress) = 4Ps/pie d^2.
Same, crushing pull Pc = d*t*sigma.
Therefore sigma (crushing stress) = Pc/dt.
Now, as per question tao = 1/4 sigma.
Solving this, we found d = 16/pie *t.
So, dmax = 5.06t its coming, taking nearer value, its 4t.
I think this may clear your doubt.
C.Felices said:
6 years ago
The answer is d=t.
Tarik said:
5 years ago
The answer is d=t.
Phani said:
5 years ago
Crushing=4.πd^2.
Shearing=πd.t.
From shearing=1/4.crushing.
d=t.
Shearing=πd.t.
From shearing=1/4.crushing.
d=t.
(2)
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