\text { Given For beam } S_{y t}=250 N / mm ^{2} \quad(f s)=5 .
h/b = 2.
\text { For vessel, } \quad S_{u t}=200 N / mm ^{2} \quad(f s)=5 .
p=0.25 N / mm ^{2} \quad D=500 mm .
Step I Calculation of permissible stresses
(a) Steel parts
\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{250}{5}=50 N / mm ^{2} .
Assuming
S_{y c}=S_{y t} .
\sigma_{c}=\frac{S_{y c}}{(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} .
(b) Cast iron parts
\sigma_{t}=\frac{S_{u t}}{(f s)}=\frac{200}{5}=40 N / mm ^{2} .
Step II Free body diagram
The free body diagram of forces acting on various parts of the pressure vessel is shown in Fig. 4.55. This diagram is constructed starting with the forces acting on the cover and then proceeding to screw, beam, pin, link L_2 and the extension of vessel to support the pin. The direction of forces acting on various parts is decided by using the following two principles:
(i) The sum of vertical forces acting on any part must be zero; and
(ii) Action and reaction are equal and opposite.
Step III Diameter of screw
The force acting on the cover, as shown in Fig. 4.55 (a), is given by,
F=\frac{\pi}{4} D^{2} p=\frac{\pi}{4}(500)^{2}(0.25)=49087.39 N .
As shown in Fig. 4.55 (b), the portion of the screw between the beam and the cover is subjected to compressive stress. If a is the stressed area of the screw then the compressive force is given by,
F=a \sigma_{c} \quad \text { or } \quad 49087.39=a(50) .
∴ a=981.75 mm ^{2} .
From the given data, a screw of M42 size (stressed area = 1120 mm²) is suitable. The nominal diameter of the screw is 42 mm and the pitch is 4.5 mm.
Step IV Cross-section of beam
As shown in Fig. 4.56(a), the beam is simply supported with a single concentrated load F at the centre of the span length. Due to symmetry of loading, the reaction at each of the two pins, C and D, is equal to (F/2). The bending moment is maximum at the midpoint of the beam. It is given by,
M_{b}=325 \times\left(\frac{F}{2}\right)=325 \times\left(\frac{49087.39}{2}\right) .
or M_{b}=7976700.88 N – mm .
Since,
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 .
\sigma_{b}=\frac{M_{b} y}{I} .
Substituting,
50=\frac{(7976700.88)(b)}{\left[\frac{2}{3}\left(b^{4}\right)\right]} .
\therefore \quad b=62.08 \text { or } 65 mm \quad h=2 b=130 mm .
As shown in Fig. 4.56(a), the axis of the tapped hole is parallel to h dimension of the section. As shown in Fig 4.56 (c), the solid rectangular section of the beam of thickness b can be split into two halves, each having a width (b/2) at the hole, so that the metallic area in a section through the hole is equal to the area of the solid section (h × b). In this case, the factor of safety will remain unchanged. The diameter of the hole \left(d_{1}\right) is the nominal diameter of the screw. Therefore,
d_{1}=42 mm .
d_{0}=d_{1}+\frac{b}{2}+\frac{b}{2}=42+\frac{65}{2}+\frac{65}{2}=107 mm .
Step V Diameter of pins
As shown in Fig. 4.55(d), the pin is subjected to double shear. The force acting on all four pins at A, B, C and D is same and equal to (F/2). The shear stress in the pin is given by,
\tau=\frac{\left(\frac{F}{2}\right)}{2\left[\frac{\pi}{4} d^{2}\right]} \quad \text { or } \quad 25=\frac{\left(\frac{49087.39}{2}\right)}{2\left[\frac{\pi}{4} d^{2}\right]} .
∴ d = 25 mm.
\text { Step } V \text { I Diameter of links } L_{1} \text { and } L_{2} .
As shown in Fig. 4.55(e), the links are subjected to tensile stresses.
\frac{\pi}{4} d_{2}^{2} \sigma_{t}=\frac{F}{2} \quad \text { or } \quad \frac{\pi}{4} d_{2}^{2}(50)=\frac{49087.39}{2} .
\therefore \quad d_{2}=25 mm .
Step VII Dimensions of support
The extensions or brackets on the vessel are part of the casting of the cylinder. They act as cantilevers. As shown in Fig. 4.54, the maximum length of the cantilever can be taken as (325 – 250) or 75 mm, neglecting the thickness of the cylinder.
Therefore,
M_{b}=75\left(\frac{F}{2}\right)=75\left(\frac{49087.39}{2}\right) .
=1840777 \cdot 13 N – mm .
\text { Assuming } h_{1}=2 b_{1} .
y=\frac{h_{1}}{2}=b_{1} \quad I=\frac{b_{1} h_{1}^{3}}{12}=\frac{b_{1}\left(2 b_{1}\right)^{3}}{12}=\frac{2 b_{1}^{4}}{3} .
\sigma_{b}=\frac{M_{b} y}{I} .
Substituting,
40=\frac{(1840777.13)\left(b_{1}\right)}{\left[\frac{2}{3}\left(b_{1}^{4}\right)\right]} .
\therefore \quad b_{1}=41.02 \text { or } 45 mm \quad h_{1}=2 b_{1}=2 \times 45=90 mm .
