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 3 of 7.
AMAL said:
10 years ago
Here a hole is punched then we need to take area of that hole that is (in shear) pi*d^2/4 and in crushing pi*d*t. So the right answer is d=t.
Pratik said:
7 years ago
Here, it is for a maximum shear stress 1/4th of its maximum crushing stress i.e. (1/4 * maximum shear stress=maximum crushing stress).
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.
Mehul Patel said:
6 years ago
Here given that,
Crushing stress = 1/4 * Shear stress
P/A. = 1/4 * F/As
P/πd2/4 = 1/4 * F/πdt.
d = t. (Here Let P=F).
Crushing stress = 1/4 * Shear stress
P/A. = 1/4 * F/As
P/πd2/4 = 1/4 * F/πdt.
d = t. (Here Let P=F).
Ravikumar said:
7 years ago
The Answer is d = t.
Because shear stress = 1/4 crashing stress.
Now,
p/(πdr) = p/(πdd/4)4.
d=t.
Because shear stress = 1/4 crashing stress.
Now,
p/(πdr) = p/(πdd/4)4.
d=t.
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.
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).
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.
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)
Srikanth said:
1 decade ago
How could it be @Rajesh there is nothing explained in your answer. Can you please explain it briefly?
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