Question

Find the maximum diameter of hole that can be punch in a M.S. plate 10mm thick
having ultimate shear stress of 220 N/mm2. The allowable crushing stress in punch material is 350 N/mm2.

Answer 1

Question #81952. Find the maximum diameter of hole that can be punch in a M.S. plate $10\mathrm{mm}$ thick having ultimate shear stress of $220~\mathrm{N / mm2}$ . The allowable crushing stress in punch material is $350~\mathrm{N / mm2}$ .

Answer.

Interaction of punch with plate is shown on the pic. 1.

Picture 1 - The punch with plate

It's necessary to provide the condition for the diameter $d$ of punch/plate to crash the plate (to make hole), but not to damage the punch.

Allowable stresses for punch are

$$\left[ \sigma_ {p u} \right] = 3 5 0 \cdot 1 0 ^ {6} P a.$$

Ultimate shear stresses for the plate are

$$\left[ \tau_ {p l} \right] = 2 2 0 \cdot 1 0 ^ {6} P a.$$

Normal stresses acting in punch are

$$\sigma_ {p u} = \frac {P}{A _ {p u}}; \quad A _ {p u} = 0. 2 5 \pi d ^ {2}; \quad \sigma_ {p u} = \frac {P}{0 . 2 5 \pi d ^ {2}},$$

where $A_{pu}$ - punch area.

Shear stresses acting on the shear surface of the plate are

$$\tau_ {p l} = \frac {P}{A _ {p l}}; \quad A _ {p l} = \pi d t; \quad \tau_ {p l} = \frac {P}{\pi d t},$$

where $A_{pl}$ - shear surface plate area.

Equating stresses (1), (2) accordingly with stresses defined in (3), (4) we shall have next

$$\left[ \sigma_{pu} \right] = \sigma_{pu} \Rightarrow \left[ \sigma_{pu} \right] = \frac{P}{0.25 \pi d^{2}} \Rightarrow P = 0.25 \pi d^{2} \left[ \sigma_{pu} \right].$$ $$\left[ \tau_{pl} \right] = \tau_{pl} \Rightarrow \left[ \tau_{pl} \right] = \frac{P}{\pi dt} \Rightarrow P = \pi dt \left[ \tau_{pl} \right].$$

Equating the force $P$, defined in (5), (6) we shall have the nonlinear equation relatively $d$, presented below

$$0.25 \pi d^{2} \left[ \sigma_{pu} \right] = \pi dt \left[ \tau_{pl} \right]$$

Solving (7) (see MathCAD file “Question #81952”) we have that the necessary diameter to provide punching of plate without damaging of the punch (but making hole in plate) is $d = 0.025142857142857142857 \mathrm{~m} \approx 25.14 \mathrm{~mm}$

Answer: $d = 25.14 \mathrm{~mm}$.