机械制造作业考研题目答案分享——工艺尺寸链

Homework 8

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文章目录

• 版权声明
• 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.
• 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：
• 5 The material for one part is 2Cr13, and its machining sequence is,
• 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,

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…

$${\delta }_{Lmax}=0.1-0.05-0.05=0$$

$$-0.1={\delta }_{Lmin}-0.05-0$$

$${\delta }_{Lmin}=-0.1+0.05=-0.05$$

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

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 $\phi 28_{-0.1}^0$ mm;
（2） milling the keyway to dimension $H_0^{\delta }H$;
（3） heat treatment;
（4） grinding external cylindrical surface to dimension $\phi 28_{+0.008}^{+0.024}$ mm.
Given that the coaxial degree is Φ 0.04mm when grinding and turning, try calculating the operational dimension $H_0^{\delta }H$ 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

$$0.15=0.024/2+{\delta }_H-(-0.10)/2-(-0.02)$$

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

$$E{I}_t=E{S}_R-E{S}_e-E{I}_r-E{I}_H$$

$$E{I}_t=0+0.02-0.004-0=0.016$$

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

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