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 1 of 7.
Sachin Sharma said:
8 years ago
I also agree with d=t
Beacuse, shearing stress is given 1/4th of crushing stress hence and shearing force is calculated by shearing stress into the shearing area and hence shearing area is (π/4 d^2) x t Hence shearing force will be,
(πd) x t x (shearing stress)
and crushing stress of punch will be equal to π/4 d^2) x σc
now (πd) x t x (σc/4) =π/4 d^2) x σc
Hence after solving d=t.
Beacuse, shearing stress is given 1/4th of crushing stress hence and shearing force is calculated by shearing stress into the shearing area and hence shearing area is (π/4 d^2) x t Hence shearing force will be,
(πd) x t x (shearing stress)
and crushing stress of punch will be equal to π/4 d^2) x σc
now (πd) x t x (σc/4) =π/4 d^2) x σc
Hence after solving 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.
Koribilli hemanth said:
9 years ago
Answer was wrong.
The right answer is T = D.
Shear force = pi * D * T * shear stress.
Crushing force = (pi/4) * D^2 * crushing stress.
The condition is that "crushing stress = 4 * shear stress " at equilibrium condition " shear force = crushing force"
pi * D * T * shear stress = (pi/4) * D^2 * crushing stress.
By canceling the terms finally we will get D = T.
The right answer is T = D.
Shear force = pi * D * T * shear stress.
Crushing force = (pi/4) * D^2 * crushing stress.
The condition is that "crushing stress = 4 * shear stress " at equilibrium condition " shear force = crushing force"
pi * D * T * shear stress = (pi/4) * D^2 * crushing stress.
By canceling the terms finally we will get D = T.
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)
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.
Shubham Soni said:
9 years ago
Shear stress = P/π/4 * D^2 = 4P/π.D^2.
Crushing Stress = P/D.t.
So, Shear stress = 1/4 of Crushing Stress.
Therefore, 4P/pi.D^2 = P/D.t.
Hence, 4/pi. D = 1/t.
D = 4t/π.
i.e. D is directly proportional to the 4t.
Hence the answer is 4t.
Crushing Stress = P/D.t.
So, Shear stress = 1/4 of Crushing Stress.
Therefore, 4P/pi.D^2 = P/D.t.
Hence, 4/pi. D = 1/t.
D = 4t/π.
i.e. D is directly proportional to the 4t.
Hence the answer is 4t.
Bharadwaz said:
8 years ago
Almost all are saying using shear in rivets formula,
But in question it is mentioned punching so shear is load(p)/area(pi*d*t).
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D,
Shear=crushing/4,
P/pi*D*t.=(4*P/pi*D*D)*(1/4),
P/pi*D*t=P/pi*D*D,
d=t.
But in question it is mentioned punching so shear is load(p)/area(pi*d*t).
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D,
Shear=crushing/4,
P/pi*D*t.=(4*P/pi*D*D)*(1/4),
P/pi*D*t=P/pi*D*D,
d=t.
Rajesh said:
1 decade ago
Max.shear stress of plate = 1/4 of max crushing stress.
And plate thickness = t.
We know, shear stress = L*t and crushing stress = d*L.
Therefore,L*t = 1/4 of max. crushing stress.
or, 4*t*L = crushing stress=d*L.
or, 4t = dia of hole.
And plate thickness = t.
We know, shear stress = L*t and crushing stress = d*L.
Therefore,L*t = 1/4 of max. crushing stress.
or, 4*t*L = crushing stress=d*L.
or, 4t = dia of hole.
Abhilasha B R Gowda said:
7 years ago
The max. shear stress = (1/4) * max. crushing stress.
(P/A) = (1/4) * (P/dt).
A= area of hole= π*d^2.
substitute A and cancel P & d,
(P/π*d^2) = (1/4) * (P/dt).
4t = π*d.
'd' is directly proportional to 4t.
The answer is d=4t.
(P/A) = (1/4) * (P/dt).
A= area of hole= π*d^2.
substitute A and cancel P & d,
(P/π*d^2) = (1/4) * (P/dt).
4t = π*d.
'd' is directly proportional to 4t.
The answer is d=4t.
UPPU HARISH said:
1 decade ago
I couldn't understand this.
If we take a punch punching a hole in a plate then,
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D.
When these two are equated with the relation given then we get d=t.
Please tell me where I am wrong?
If we take a punch punching a hole in a plate then,
Shear stress = P/pi*D*t.
Crushing stress = 4*P/pi*D*D.
When these two are equated with the relation given then we get d=t.
Please tell me where I am wrong?
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