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Chapter 4

Q. 4.18

The mechanism of a bench-shearing machine is illustrated in Fig. 4.57. It is used to shear mild steel bars up to 6.25 mm diameter. The ultimate shear strength of the material is 350 N/mm². The link, lever and pins at B, C and D are made of steel FeE 250 \left(S_{y t}=250 N / mm ^{2}\right) and the factor of safety is 5. The pins at B, C and D are identical and their length to diameter ratio is 1.25.
The permissible bearing pressure at the pins is 10 N/mm². The link has circular cross-section. The cross-section of the lever is rectangular and the ratio of width to thickness is 2:1. Calculate
(i) Diameter of pins at B, C and D;
(ii) Diameter of the link
(iii) Dimensions of the cross-section of the lever.


Verified Solution

\text { Given } \quad S_{v t}=250 N / mm ^{2} \quad(f s)=5 .

For bars to be sheared D = 6.25 mm

S_{u s}=350 N / mm ^{2} .

\text { For pins, } \quad p=10 N / mm ^{2} \quad l / d=1.25 .

For lever, h/b = 2

Step I Calculation of permissible stresses

\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{250}{5}=50 N / mm ^{2} .

\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(250)}{5}=25 N / mm ^{2} .

Step II Calculation of forces
The maximum force P_s required to shear the bar is given by,

P_{s}=(\text { area of bar }) \times(\text { ultimate shear strength }) .

=\frac{\pi}{4}(6.25)^{2} \times 350=10737.87 N .

The free body diagram of forces acting on various parts of the shearing machine is shown in Fig. 4.58.
This diagram is constructed starting with shear force P_{s} acting on the bar and then proceeding to the block, link and the lever.

Since the three pins are identical, we will design the pin for maximum force of 2684.47 N. The dimensions of the pins are determined on the basis of bearing consideration and checked for shear consideration.

The direction of forces acting on the various parts is decided by using the following two principles:
(i) The action and reaction are equal and opposite.
(ii) The sum of vertical forces acting on any part must be equal to zero.
Taking moment of forces acting on the block, as shown in Fig. 4.58(b) about the fulcrum A,

P_{1} \times 400=P_{s} \times 100 \quad \text { or }

P_{1} \times 400=10737.87 \times 100 .

\therefore \quad P_{1}=2684.47 N .


R_{A}+P_{1}=P_{s} \quad \text { or } \quad R_{A}=P_{s}-P_{1} .

R_{A}=10737.87-2684.47=8053.4 N .

Taking moment of forces acting on lever, as shown in Fig. 4.58(d), about the fulcrum D,

P \times 1000=P_{1} \times 100 \quad \text { or } .

P \times 1000=2684.47 \times 100 .

\therefore \quad P=268.45 N .


R_{D}+P=P_{1} \quad \text { or } .

R_{D}=P_{1}-P=2684.47-268.45=2416.02 N .

Step III Diameter of pins
The forces acting on the pins at B, C and D are 2684.47, 2684.47 and 2416.02 N respectively.

Considering bearing pressure on the pin,

R=p \text { (projected area of the pin) }=p\left(d_{1} \times l_{1}\right) .

or    R=p\left(d_{1} \times l_{1}\right)             (a).


d_{1}=\text { diameter of the pin }( mm ) .

l_{1}=\text { length of the pin }( mm ) .

p = permissible bearing pressure (N/mm²).

Substituting values in Eq. (a),

2684.47=10\left(d_{1} \times 1.25 d_{1}\right) .

\therefore \quad d_{1}=14.65 \text { or } 15 mm .

l_{1}=1.25 d_{1}=1.25(15)=18.75 \text { or } 20 mm .

The pin is subjected to double shear stress, which is given by,

\tau=\frac{R}{2\left[\frac{\pi}{4} d_{1}^{2}\right]}=\frac{2684.47}{2\left[\frac{\pi}{4}(15)^{2}\right]}=7.60 N / mm ^{2} .

\therefore \quad \tau<25 N / mm ^{2} .

Step IV Diameter of link
The link is subjected to tensile stress as shown in Fig. 4.58(c). Therefore

\sigma_{t}=\frac{P_{1}}{\left(\frac{\pi}{4} d^{2}\right)} \quad \text { or } \quad 50=\frac{2684.47}{\left(\frac{\pi}{4} d^{2}\right)} .

∴      d = 8.27 or 10 mm.

Step V Dimensions of lever
As shown in Fig. 4.59, the lever is subjected to bending moment. The maximum bending moment occurs at C, which is given by,

M_{b}=P \times 900=268.45 \times 900=241605 N – mm .


h=2 b \quad I=\frac{b h^{3}}{12}=\frac{b(2 b)^{3}}{12}=\left(\frac{2 b^{4}}{3}\right) mm ^{4} .

y=\frac{h}{2}=b mm \quad \sigma_{b}=\frac{M_{b} y}{I} .


50=\frac{(241605)(b)}{\left(\frac{2 b^{4}}{3}\right)} .

\therefore \quad b=19.35 \text { or } 20 mm \quad h=2 b=2(20)=40 mm .

The lever becomes weak due to the pinhole at C and it is necessary to check bending stresses at this critical cross-section, which is shown in Fig. 4.60.

The diameter of the pin is 15 mm, while the length is 20 mm. It is assumed that a gunmetal bush of 2.5 mm thickness is fitted in the hole to reduce friction and wear. Therefore, inner diameter of the boss will be (15+2 \times 2.5) or 20 mm. The outside diameter of the boss is kept twice of the inner diameter, i.e., 40 mm. Therefore,

I=\frac{1}{12}\left[20(40)^{3}-20(20)^{3}\right]=93333.33 mm ^{4}

y = 20 mm.

\sigma_{b}=\frac{M_{b} y}{I}=\frac{(241605)(20)}{(93333.33)}=51.77 N / mm ^{2} .

There is slight difference between bending stress (51.77 N/mm² ) and permissible stress (50 N/mm²).
Since the difference is small, it is neglected and the dimensions of the boss of the lever are kept unchanged.