A circular hole is to be punched in a plate that has a shear strength of 40
ksi. The working compressive stress for the punch is 50 ksi.
Compute the maximum thickness of a plate in which a hole 2.5 in. in diameter can be
punched.
0.781in
0.8in
0.734in
0.825in
If the plate is 0.25 in. thick, determine the diameter of the smallest hole
that can be punched.
0.8in
0.781in
0.734in
0.825in
### Punching capacity and minimum hole size
The punching force needed to shear a circular hole is
$$P_{\text{shear}}=\tau(\pi dt).$$
The maximum force the punch can carry in compression is
$$P_{\text{punch}}=\sigma_cleft(\frac{\pi d^2}{4}\right).$$
At the limiting condition, set these equal:
$$\taupi dt=\sigma_c\frac{\pi d^2}{4}.$$
After cancelling $\pi d$,
$$t=\frac{\sigma_cd}{4\tau}.$$
For $d=2.5 \text{in.}$,
$$t=\frac{50(2.5)}{4(40)}=\boxed{0.781 \text{in.}}.$$
If $t=0.25 \text{in.}$, solve instead for $d$:
$$d=\frac{4\tau t}{\sigma_c}=\frac{4(40)(0.25)}{50}=\boxed{0.8 \text{in.}}.$$
Find the smallest diameter bolt that can be used in the clevis if P=400kN and the working shear stress for the bolt is 300MPa.
It
Answer:
29.13mm
41.20mm
58.27mm
32.63mm
### Clevis pin in double shear
The clevis pin has two shear planes, so its resisting shear area is
$$A_s=2\left(\frac{\pi d^2}{4}\right).$$
Set the applied load equal to the allowable shear resistance:
$$400,000=300\left[2\left(\frac{\pi d^2}{4}\right)\right].$$
Thus
$$d=\sqrt{\frac{2(400,000)}{\pi(300)}}=\boxed{29.13 \text{mm}}.$$
A practical design would select the next available standard diameter not smaller than this value.
What minimum force (in kN) is required to punch a 25-mm-diameter hole in a plate that is 20mm thick? The shear
strength of the plate is 330MPa. Factor of safety = 1.5.
Answer:
519
345
518
346
### Punching force
The sheared area is the hole perimeter times the plate thickness:
$$A_s=\pi dt=\pi(25)(20)=1570.80 \text{mm}^2.$$
The force required to shear the plate is
$$P=\tau A_s=330(1570.80)=518,364 \text{N}.$$
Therefore the minimum punching force is
$$Papprox\boxed{519 \text{kN}}.$$
The stated factor of safety is used when sizing or checking the punch; the force required to shear the plate is based on the plate shear strength and the punching-shear area.
When solving the minimum safe diameter of the cross-section given various stress conditions, the following
values have been computed: 12.4mm, 13.6mm, 14.2mm, and 15.3mm. What should you use as the minimum safe
diameter?
Answer:
16mm
15mm
12mm
14mm
### Select a safe standard diameter
Every listed stress condition imposes a minimum diameter. To satisfy all conditions at once, use the largest required value:
$$d_{\text{required}}=\max(12.4,13.6,14.2,15.3)=15.3 \text{mm}.$$
The selected diameter must not be less than the required value. The next available standard diameter is therefore
$$\boxed{16 \text{mm}}.$$
