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.
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)
Vinay said:
5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(15)
Prudvi Anemoni said:
5 years ago
D=T should be the correct answer. Analyse the area where crushing force is acting.
(7)
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)
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)
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)
Navneet said:
5 years ago
D = t is the correct answer.
(1)
Nilesh said:
9 years ago
d = t is the right answer.
Akahai said:
9 years ago
A steel punch can be worked to a compressive stress of 800N/mm^2.
Find the least diameter of hole which can be punched through a steel plate of 12mm thick, if its ultimate shear strength is 315N/mm^2.
Can anyone answer this?
Find the least diameter of hole which can be punched through a steel plate of 12mm thick, if its ultimate shear strength is 315N/mm^2.
Can anyone answer this?
Neeraj said:
9 years ago
@Akahai.
7.62 apply same formula and this answer of questions is D=tCrushing force = shearing force.
==> σ [(φ/4)* D^2] = τ[φD t].
7.62 apply same formula and this answer of questions is D=tCrushing force = shearing force.
==> σ [(φ/4)* D^2] = τ[φD t].
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