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 4 of 7.
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.
Prashant said:
9 years ago
Shear stress * 4 = crushing stress.
(F/pi * d * t) * 4 = (4F/pi * d * d).
therefore, d = t.
(F/pi * d * t) * 4 = (4F/pi * d * d).
therefore, d = t.
Jammy khan said:
1 decade ago
Shear stress = F/A.
F = 1/4, A = L*thickness.
Put the value in given equation,
Shear stress = (4/1)*(L*thickness).
F = 1/4, A = L*thickness.
Put the value in given equation,
Shear stress = (4/1)*(L*thickness).
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.
Akshay said:
1 decade ago
How is shear stress equal to l*t?
Kuruu said:
1 decade ago
@Akshay I want to explain you about the question.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Actually @Rajesh had done exactly no need again just check once if at all u didnt get that will see.
Satya said:
1 decade ago
Shear stress = Pi/4*d2*load.
Crushing stress = load*d*t.
Given that s.s = 1/4*c.s.
Solving this we get d = t/pi.
Crushing stress = load*d*t.
Given that s.s = 1/4*c.s.
Solving this we get d = t/pi.
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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