Homework 8

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1 Try pointing out the shortcomings existing in the following workpieces from the point of machining shown as Fig. 1, and propose a way to improve them.

2 Try selecting the rough and finish datum surface when machining the surfaces that have the arc marked ▽ in Fig 2, and explain the reason.

3 When machining the part shown in the following figure 3, the dimension 6±0.1mm must be ensured. As it is inconvenient for inspection, dimension of L is measured to indirectly ensure it. Please decide the dimension L and its tolerance.

Fig.3 For exercise 3 Fig.4 For exercise 4
Solution

First, 6±0.1 is resultant dimension.

$6=L+26-36$

$L=-26+36+6=16$

KaTeX parse error: Double subscript at position 8: 0.1=δ_δ_̲{Lmax}+0.05-(-0…

$δ_{Lmax}=0.1-0.05-0.05=0$

$-0.1=δ_{Lmin}-0.05-0$

$δ_{Lmin}=-0.1+0.05=-0.05$

In a word,$L_0^{δ_L }=16_{-0.05}^0 mm$

4 When machining the spindle shown in Fig.4, the depth of keyway t=4_0^(+0.15)mm must be ensured. Its machining process includes：

（1） turning external cylindrical surface to dimension $φ28_{-0.1}^0$ mm;
（2） milling the keyway to dimension $H_0^δH$;
（3） heat treatment;
（4） grinding external cylindrical surface to dimension $φ28_{+0.008}^{+0.024}$ mm.
Given that the coaxial degree is Φ 0.04mm when grinding and turning, try calculating the operational dimension $H_0^δH$ using the maximum and minimum method.
Φ means coaxial degree, but φ means diameter.
Solution

First, $=4_0^{+0.15}$ is resultant dimension.

$4=28/2+H-28.5/2$

So, H=4.25

$0.15=0.024/2+δ_H-(-0.10)/2-(-0.02)$

So, δ_H=0.15-0.024/2-0.10/2-0.02=0.068

$EI_t=ES_R-ES_e-EI_r-EI_H$

$EI_t=0+0.02-0.004-0=0.016$

In a word, $H_0^δ=4.25_{+0.016}^{+0.068} mm$.

5 The material for one part is 2Cr13, and its machining sequence is,

Solution

6. In Fig 5(a), It is the design requirement of a workpiece, and Fig.(b) is the operational drawing (工序图）. The workpiece is machined through the following operations,

Solution,

I


II