Question 9.28: A cantilever beam carrying a load of 500 N is shown in Fig. ...

\text { Given } \delta_{\max }=0.05 mm \quad E=207000 N / mm ^{2} .

P = 500 N
Step I B-M diagram

\left(M_{b}\right) \text { at } B=500 \times 100=50000 N – mm .

\left(M_{b}\right) \text { at } A=500 \times 200=100000 N – mm .

The bending moment diagram is shown in Fig. 9.51(b).
Step II (M/EI) diagram
Let us denote,

I=\frac{\pi d^{4}}{64} .

For the portion AB,

I^{\prime}=\frac{\pi(1.2 d)^{4}}{64}=2.074 I .

Considering the portion BC,

(M / E I) \text { at } B=50000 / E I mm ^{-1} .

Considering the portion AB,

(M / E I) \text { at } B=50000 / 2.074 E I)=24113 / E I mm ^{-1} .

(M/EI) at A = 100 000/(2.074EI).

=48225 / E I mm ^{-1} .

The (M/EI) diagram shown in Fig. 9.51(c)
Step III Slope and deflection Diagrams
The (M/EI) diagram shown in Fig. 9.51(c) is drawn to the following scale:

S_{x}=(5 / 3) mm \text { per } mm .

S_{y 2}=(1000 / E I) \text { per } mm .

The diagram is divided into six equal parts and their mid-points are denoted as 1, 2, …, 6. The heights of these points are projected on the y_{2} \text {-axis. }
The pole distance H is taken as 50 mm. Join lines

\overline{O_{1} 1}, \overline{O_{1} 2}, \ldots \ldots \ldots \ldots \ldots, \overline{O_{1} 6} .

The slope diagram shown in Fig. 9.51(d) is constructed by the graphical integration method.
From the origin of the slope diagram, draw a line parallel to \overline{O_{1} 1} of the (M/EI) diagram. This is the slope of part 1. Similar procedure is used for other parts. As discussed in the previous section, the scale of the slope diagram is given by,

S_{y 3}=H S_{x} S_{y 2}=50(5 / 3)(1000 / E I) .

= (250 000/3EI) per mm
The deflection diagram shown in Fig. 9.51(e), is constructed by the graphical integration of the slope diagram. Similar procedure is used in this case. The scale of the deflection diagram is given by,

S_{y 4}=H S_{x} S_{y 3}=50(5 / 3)(250000 / 3 E I) .

= (6944.44 × 10³/EI) per mm
As shown in the figure, the actual dimension of the maximum deflection is 105 mm.
Therefore,

\delta_{\max }=(105)\left(6944.44 \times 10^{3} / E I\right) .

=\left(729.166 \times 10^{6} / E I\right) mm .

Step VI Diameter of beam
The permissible deflection is 0.05 mm. Therefore,

0.05=\frac{729.166 \times 10^{6}}{207000 \times\left(\pi d^{4} / 64\right)} .

d = 34.61 mm.